Essential Mathematics for the Australian Curriculum, 10 & 10A [2 ed.] 9781107568907, 1107568900

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10& 10A

ESSENTIAL MATHEMATICS FOR THE AUSTRALIAN CURRICULUM SECOND EDITION

DAVID GREENWOOD | SARA WOOLLEY JENNY GOODMAN | JENNIFER VAUGHAN STUART PALMER Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

10& 10A

ESSENTIAL MATHEMATICS FOR THE AUSTRALIAN CURRICULUM SECOND EDITION

DAVID GREENWOOD | SARA WOOLLEY JENNY GOODMAN | JENNIFER VAUGHAN STUART PALMER Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

477 Williamstown Road, Port Melbourne, VIC 3207, Australia Cambridge University Press is part of the University of Cambridge. It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence. www.cambridge.edu.au Information on this title: www.cambridge.org/9781107568907 © David Greenwood, Sara Woolley, Jenny Goodman, Jennifer Vaughan, 2015 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2011 Reprinted 2012, 2013, 2014, 2015 Second edition published 2015 Reprinted 2016 Cover designed by Sardine Design Typeset by Diacritech Printed in China by C & C Offset Printing Co. Ltd. A Cataloguing-in-Publication entry is available from the catalogue of the National Library of Australia at www.nla.gov.au ISBN 978-1-107-56890-7 Paperback Additional resources for this publication at www.cambridge.edu.au/hotmaths Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this publication, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL) under the Act. For details of the CAL licence for educational institutions contact: Copyright Agency Limited Level 15, 233 Castlereagh Street Sydney NSW 2000 Telephone: (02) 9394 7600 Facsimile: (02) 9394 7601 Email: [email protected] Reproduction and communication for other purposes Except as permitted under the Act (for example a fair dealing for the purposes of study, research, criticism or review) no part of this publication may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher at the address above. Cambridge University Press has no responsibility for the persistence or accuracy of URLS for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Information regarding prices, travel timetables and other factual information given in this work is correct at the time of first printing but Cambridge University Press does not guarantee the accuracy of such information thereafter. Disclaimer All material identified by is material subject to copyright under the Copyright Act 1968 (Cth) and is owned by the Australia Curriculum, Assessment and Reporting Authority 2015. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

able of ontents About the authors Introduction and overview of this series Working with unfamiliar problems

1

inear relations 1 1 1 1 1 1 1

1

2

2

Reviewing algebra CONSOLIDATING Algebraic fractions Solving linear equations Inequalities Graphing straight lines CONSOLIDATING Finding an equation of a line Length and midpoint of a line segment Progress quiz Perpendicular and parallel lines Simultaneous equations – substitution Simultaneous equations – elimination Further applications of simultaneous equations Half planes EXTENDING Investigation Problems and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

1 1 1 1

eometr 2 2 2 2

ix x xiv

Geometry review CONSOLIDATING Congruent triangles Investigating parallelograms using congruence Similar figures CONSOLIDATING

4 10 15 21 25 34 41 46 47 53 58

Number and Algebra Patterns and algebra Linear and non-linear relationships

62 67 75 78 79 80 82 85

86

Measurement and Geometry

88 97

Geometric reasoning

103 108

iii Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

2 2 2 2 2 2

3

114 122 124 131 139 144 150 155 157 158 159 160 162

ndices and surds

164

Number and Algebra

3 3 3 3 3 3 3 3

166 172 176 181 186 190 195 200 204 205 210 217 222 226 230 232 233 234 235 236

Patterns and algebra

238

Measurement and Geometry

240 248 253

Pythagoras and trigonometry

3 3 3 3 3

4

Proving similar triangles Progress quiz Circles and chord properties 10A Angle properties of circles – theorems 1 and 2 10A Angle properties of circles – theorems 3 and 4 10A Tangents EXTENDING Intersecting chords, secants and tangents EXTENDING Investigation Problems and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

Irrational numbers and surds 10A Adding and subtracting surds 10A Multiplying and dividing surds 10A Binomial products 10A Rationalising the denominator 10A Review of index laws CONSOLIDATING Negative indices Scientific notation CONSOLIDATING Progress quiz Rational indices 10A Exponential growth and decay Compound interest Comparing interest using technology Exponential equations 10A Investigation Problems and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

rigonometr 4 4 4

Trigonometric ratios Finding angles Applications in two dimensions

Money and financial mathematics Real numbers Linear and non-linear relationships

iv Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

4 4 4 4 4 4 4 4

5

Bearings Applications in three dimensions 10A Obtuse angles and exact values 10A Progress quiz The sine rule 10A The cosine rule 10A Area of a triangle 10A The four quadrants 10A Graphs of trigonometric functions 10A Investigation Problems and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

uadratic e uations 5 5 5 5 5 5 5 5 5

Expanding expressions CONSOLIDATING Factorising expressions Factorising trinomials of the form x 2 + bx + c Factorising quadratic trinomials of the form ax 2 + bx + c 10A Factorising by completing the square Solving quadratic equations Progress quiz Applications of quadratics Solving quadratic equations by completing the square The quadratic formula Investigation Problems and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

emester review 1

259 266 272 278 280 286 291 296 303 313 315 316 317 318 321

322

Number and Algebra

324 330

Patterns and algebra Linear and non-linear relationships

336 341 345 350 356 357 361 365 370 372 373 374 375 376

378

v Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

6

easurement 6 6 6 6 6

Using units of measurement

Parabolas and other graphs

462

Number and Algebra

7 7 7 7 7 7

464 472 479 485 491 496 503 505 514 521 527 534 540 543 544 545 547 549

Patterns and algebra

7 7 7 7 7

Review of length CONSOLIDATING Pythagoras’ theorem Area CONSOLIDATING Surface area – prisms and cylinders Surface area – pyramids and cones 10A Progress quiz Volume – prisms and cylinders Volume – pyramids and cones 10A Spheres 10A Limits of accuracy EXTENDING Investigation Problems and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

Measurement and Geometry

390 397 404 412 419 425 427 434 440 447 451 454 455 456 457 460

6 6 6 6

7

388

Exploring parabolas Sketching parabolas with transformations Sketching parabolas using factorisation Sketching by completing the square Sketching using the quadratic formula 10A Applications involving parabolas Progress quiz Intersection of lines and parabolas 10A Graphs of circles Graphs of exponentials Graphs of hyperbolas Further transformations of graphs 10A Investigation Problems and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

Pythagoras and trigonometry

Linear and non-linear relationships

vi Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

8

Probabilit

550

Statistics and Probability

8 8 8 8

552 559 568 573 579 580 587 596 601 604 605 606 607 609

Chance

610

Statistics and Probability

612 617 626 632 638 644 645 651 659 666 671 673 674 675 676 678

Data representation and interpretation

8 8 8

9

Review of probability CONSOLIDATING Unions and intersections The addition rule Conditional probability Progress quiz Two-step experiments using tables Using tree diagrams Independent events Investigation Problems and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

tatistics 9 9 9 9 9 9 9 9 9

Collecting and using data Review of statistical graphs CONSOLIDATING Summary statistics Box plots Standard deviation 10A Progress quiz Time-series data Bivariate data and scatter plots Line of best fit by eye Linear regression with technology 10A Investigation Problems and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

vii Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

10

ogarithms and pol nomials 10 10 10 10 10 10 10 10 10

Logarithms 10A Logarithm laws 10A Exponential equations using logarithms 10A Polynomials 10A Expanding and simplifying polynomials 10A Division of polynomials 10A Progress quiz The remainder and factor theorems 10A Solving polynomial equations 10A Graphs of polynomials 10A Investigation Problems and challenges Review: Chapter summary Multiple-choice questions Short-answer questions Extended-response questions

emester review 2 Answers

680

Number and Algebra

682 686 690 694 699 702 705 706 710 715 721 723 724 725 725 727

Patterns and algebra Real numbers Linear and non-linear relationships

728 738

viii Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

bout the authors avid reenwood is the Head of

athematics at Trinity Grammar School in elbourne and has 21 years’ experience teaching mathematics from Years 7 to 12. He has run numerous workshops within Australia and overseas regarding the implementation of the Australian Curriculum and the use of technology for the teaching of mathematics. He has written more than 20 mathematics titles and has a particular interest in the sequencing of curriculum content and working with the Australian Curriculum proficiency strands.

ara Woolle was born and educated in Tasmania. She completed an Honours degree in athematics at the University of Tasmania before completing her education training at the University of elbourne. She has taught mathematics in ictoria from Years 7 to 12 since 2006, has written more than 10 mathematics titles and specialises in lesson design and differentiation. ennifer aughan has taught secondary mathematics for over 30 years in New South Wales, Western Australia, ueensland and New ealand and has tutored and lectured in mathematics at ueensland University of Technology. She is passionate about providing students of all ability levels with opportunities to understand and to have success in using mathematics. She has taught special needs students and has had extensive experience in developing resources that make mathematical concepts more accessible hence, facilitating student confidence, achievement and an enjoyment of maths. enn oodman has worked for 20 years in comprehensive state and selective high schools in New South Wales and has a keen interest in teaching students of differing ability levels. She was awarded the ones edal for education at Sydney University and the Bourke prize for athematics. She has written for Cambridge NSW and was involved in the Spectrum and Spectrum Gold series.

tuart Palmer has been a head of department in two schools and is now an educational consultant who conducts professional development workshops for teachers all over New South Wales and beyond. He is a Project fficer with the athematical Association of New South Wales, and also works with pre-service teachers at The University of Sydney and The University of Western Sydney.

ix Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Introduction This second edition of Essential Mathematics for the Australian Curriculum has been developed into a complete resources pack comprising a revised and updated print textbook, a new interactive textbook with a host of cutting-edge features, and an online teaching suite. The successful elements of the first edition have been retained and improved upon. These include: • logical sequencing of chapters and development of topics • careful structuring of exercises according to the four Australian Curriculum proficiency strands plus enrichment • graduated difficulty of exercise questions within the overall exercise and within proficiency groups • Let’s Start and Key Ideas to help introduce concepts and key skills. Additions and revisions to the text include: • New topics reflecting updates to the Australian Curriculum and state syllabuses • Revision and extension topics are marked as ‘Consolidating’ or ‘Extending’ to help customise the course to each classroom’s needs • The working programs have been subtly embedded in each exercise to differentiate three student pathways: Foundation, Standard and Advanced • A ‘Progress quiz’ has been added about two-thirds of the way into each chapter, allowing students to check and consolidate their learning – in time to address misunderstandings or weaknesses prior to completing the chapter • Pre-tests have been revised and moved to the interactive textbook. Features of the all-new interactive textbook: • Seamlessly blended with Cambridge HOTmaths, allowing enhanced learning opportunities in blended classrooms, revision of previous years’ work, and access to Scorcher • Every worked example in the book is linked to a high-quality video demonstration, supporting both inclass learning and the ‘flipped classroom’ • A searchable dictionary of mathematical terms and pop-up definitions in the text • Hundreds of interactive widgets, walkthroughs and games • Automatically-marked quizzes and assessment tests, with saved scores • Printable worksheets (HOTsheets) suitable for homework or class group work. Features of the Online Teaching Suite, also powered by Cambridge HOTmaths: • A test generator, with ready-made tests • Printable worked solutions for all questions • A powerful learning management system with task-setting, progress-tracking and reporting functions. The chart on the next pages shows how the components of this resource are integrated.

x Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Guide to the working programs in the exercises

FLUENCY

UNDERSTANDING

16

PROBLEM-SOLVING

—

ENRICHMEMENT

—

Gradients within exercises and proficiency strands

REASONING

The working programs that were previously available in separate supporting documents have been updated, refined and embedded in the exercises for this second edition of Essential Mathematics for the Australian Curriculum. The suggested worki ng programs provide three pathways through the course to allow differentiation for Foundation, Standard and Advanced students. As with the first edition, each exercise is structured in subsections that match the four Australian Curriculum proficiency strands (Understanding, Fluency, Problem-solving and Reasoning) as well as Enrichment (Challenge). The questions suggested for each pathway are listed in three columns at the top Foundation Standard Advanced of each subsection: • The left column (lightest shaded colour) is 1–2 2 — the Foundation pathway 3–5(½) 3–6(½) 3–6(½) • The middle column (medium shaded 7–8 8–10 8–10 colour) is the Standard pathway • The right column (darkest shaded colour) 11 11–13 13–15 is the Advanced pathway.

The working programs make use of the gradients that have been seamlessly integrated into the exercises. A gradient runs through the overall structure of each exercise – where there is an increasing level of mathematical sophistication required from Understanding through to Reasoning and Enrichment – but also within each proficiency strand; the first few questions in Fluency, for example, are easier than the last few, and the last Problem-solving question is more challenging than the first Problem-solving question.

The right mix of questions Questions in the working programs are selected to give the most appropriate mix of types of questions for each learning pathway. Students going through the Foundation pathway will likely need more practice at Understanding and Fluency, but should also attempt the easier Problem-solving and Reasoning questions. An Advanced student will likely be able to skip the Understanding questions, proceed through the Fluency questions (often half of each question), focus on the Problem-solving and Reasoning questions, and attempt the Enrichment question. A Standard student would do a mix of everything.

Choosing a pathway There are a variety of ways of determining the appropriate pathway for students through the course. Schools and individual teachers should follow the method that works for them. If required, the chapter pre-tests (now found online) can also be used as a diagnostic tool. The following are recommended guidelines: • A student who gets 40% or lower in the pre-test should complete the Foundation questions • A student who gets between 40% and 85% in the pre-test should complete the Standard questions • A student who gets 85% or higher in the pre-test should complete the Advanced questions. For schools that have classes grouped according to ability, teachers may wish to set one of the Foundation, Standard or Advanced pathways as their default setting for their entire class and then make individual alterations depending on student need. For schools that have mixed-ability classes, teachers may wish to set a number of pathways within the one class, depending on previous performance and other factors. * The nomenclature used to list questions is as follows: • 3, 4: complete all parts of questions 3 and 4 • 1-4: complete all parts of questions 1, 2, 3 and 4 • 10(½): complete half of the parts from question 10 (a, c, e, ... or b, d, f, ...)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

• 2-4(½): complete half of the parts of questions 2, 3 and 4 • 4(½), 5: complete half of the parts of question 4 and all parts of question 5 • — : do not complete any of the questions in this section.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

An overview of the Essential Mathematics for the Australian Curriculum complete learning suite All non-core topics are marked Consolidating or Extending to assist with course planning

Examples with fully worked solutions and explanations show the thinking behind each step





3

 

hapter 1 Whole numbers





Number and Algebra



37

Investigations

1H Estimating and rounding

Example 13 Using leading digit approximation

ONSOLIDATING

Estimate the answers to these problems by rounding each number to the leading digit. a 42 × 7 b 95 × 326

any theoretical and practical problems do not need precise or exact answers. n such situations reasonable estimations can provide enough information to solve the problem.

SO L UT IO N

The total revenue from the ustralian pen tennis tournament depends on crowd numbers. Estimates would be used before the tournament begins to predict these numbers. n estimate for the total revenue might be 8 million.

42 × 7 ≈ 40 × 7

b

95 × 326 ≈ 100 × 300 = 30 000

ere is a photo of a crowd at a tennis match. escribe how you might estimate the number of people in the photo. hat is your answer ow does your answer differ from those of others in your class

Problems and challenges

EXPL A NA T IO N

a

= 280

Let s start The tennis crowd

The leading digit in 42 is the 4 in the ‘tens’ column. The nearest ‘ten’ to 95 is 100, and the leading digit in 326 is in the ‘hundreds’ column.

Chapter summaries

Example 14 Estimating with operations

PRINT TEXTBOOK

Estimate the answers to these problems by rounding both numbers as indicated. a 115 × 92 (to the nearest 100) b 2266 ÷ 9 (to the nearest 10) How can you estimate the number of spectators?

SO L UT IO N

e ideas

Estimates or approximations to the answers of problems can be found by rounding numbers to the nearest 10, 100, 1000 etc. f the next digit is 0, 1, 2, 3 or 4, then round down. f the next digit is 5, 6, 7, 8 or 9, then round up. Leading digit approximation rounds the rst digit to the nearest 10 or 100 or 1000 etc. e.g. For 932 use 900 For 968 use 1000 The symbol ≈ means ‘approximately equal to’. The symbol  can also be used.

a

b



2266 ÷ 9 ≈ 2270 ÷ 10

Exercise 1H

ound these numbers as indicated. 86 (to the nearest 10)

1

b 4142 (to the nearest 100)

SO LUT IO N

EX P LANAT IO N

a 86 ≈ 90

The digit after the 8 is greater than or e ual to 5, so round up.

b 4142 ≈ 4100

The digit after the 1 is less than or e ual to 4, so round down.

Key ideas summarise key knowledge and skills for each lesson

2266 rounds to 2270 and 9 rounds to 10.

Answers 1, 2

State whether these numbers have been rounded up or down. a 59 ≈ 60 b 14 ≈ 10 d 255 ≈ 260 e 924 ≈ 900

2

—

c 137 ≈ 140 f 1413 ≈ 1000

2 For the given estimates, decide if the approximate answer is going to give a larger or smaller result compared to the true answer. a 58 + 97 ≈ 60 + 100 b 24 × 31 ≈ 20 × 30 c 130 – 79 ≈ 130 – 80 d 267 – 110 ≈ 270 – 110





115 rounds to 100 and 92 rounds to 100.

= 227

Example 12 Rounding a

Chapter reviews

EXPL A NA T IO N

115 × 92 ≈ 100 × 100 = 10 000

 



Let’s start activities get students thinking critically and creatively about the topic

‘Working with unfamiliar problems’ poster





Working programs subtly embedded in each Proficiency Strand to provide three learning pathways through the book

Downloadable

PDF TEXTBOOK

Each topic in the print book comes with interactive HOTmaths widgets, walkthroughs and HOTsheets in the interactive textbook

37

UNDERSTANDING

3



For more detail, see the guide in the online Interactive Textbook

Included with print textbook and interactive textbook

Note-taking Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Search functions

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Easy navigation within sections without scrolling

Walkthroughs, Scorcher and levelled question sets in every section

Interactive navigation and searching

Tasks sent by teacher

INTERACTIVE TEXTBOOK POWERED BY HOTmaths

Access to interactive resources at any time without leaving the page

Hundreds of interactive widgets

Online tests sent by teacher

Student reporting

Class reporting

Video demonstration for every worked example Access to all HOTmaths Australian Curriculum courses

ONLINE TEACHING SUITE POWERED BY HOTmaths

Teacher’s copy of interactive textbook

Test generator and readymade tests

Student results

Printable chapter tests and worksheets

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Plus: • worked solutions • teaching program • curriculum grids and more

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

xiv

Working with unfamiliar problems: Part 1

Working with unfamiliar problems: Part 1 The questions on the next four pages are designed to provide practice in solving unfamiliar problems. Use the ‘Working with unfamiliar problems’ poster at the back of this book to help you if you get stuck. In Part 1, apply the suggested strategy to solve these problems, which are in no particular order. Clearly communicate your solution and final answer. 1 Discover the link between Pascal’s triangle and expanded binomial products and use this pattern to help you expand (x + y)6 . Pascal’s triangle (x + y)0

1

(x + y)1

1

2

1

(x + y)

(x + y)3

For Question 1, try looking for number patterns and algebraic patterns.

1

1 2

3

1 3

1

2 How many palindromic numbers are there between 101 and 103 ? 3 Find the smallest positive integer values for x so that 60x is: i a perfect square ii a perfect cube iii divisible by both 8 and 9.

For Questions 2 and 3, try making a list or table.

4 A Year 10 class raises money at a fete by charging players $1 to flip their dollar coin onto a red and white checked tablecloth with 50 mm squares. If the dollar coin lands fully inside a red square the player keeps their $1. What is the probability of keeping the $1? How much cash is likely to be raised from 64 players? 25 mm

5 The shortest side of a 60◦ set square is 12 cm. What is the length of the longest side of this set square?

For Questions 4–8, try drawing a diagram to help you visualise the problem.

6 A Ferris wheel with diameter 24 metres rotates at a constant rate of 60 seconds per revolution. a Calculate the time taken for a rider to travel: i from the bottom of the wheel to 8 m vertically above the bottom ii from 8 m to 16 m vertically above the bottom of the wheel. b What fraction of the diameter is the vertical height increase after each one-third of the ride from the bottom to the top of the Ferris wheel? 7 ABCD is a rectangle with AB = 16 cm and AD = 12 cm. X and Y are points on BD such that AX and CY are each perpendicular to the diagonal BD. Find the length of the interval XY. 8 How many diagonal lines can be drawn inside a decagon (i.e. a 10-sided polygon)? Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Working with unfamiliar problems: Part 1

9

The symbol ! means factorial. e.g. 4! = 4 × 3 × 2 × 1 = 24. Simplify 9! ÷ 7! without the use of a calculator.

For Question 9, try to break up the numbers to help simplify.

For Question 10, try to set up an equation

10 In 2017 Charlie’s age is the sum of the digits of his birth year 19xy and Bob’s age is one less than triple the sum of the digits of his birth year 19yx. Find Charlie’s age and Bob’s age on their birthdays in 2017.

11

Let D be the difference between the squares of two consecutive positive integers. Find an expression for the average of the two integers in terms of D.

12 For what value of b is the expression 15ab + 6b – 20a – 8 equal to zero for all values of a?

For Questions 11–13, try using algebra as a tool to work out the unknowns.

13 Find the value of k given that the area enclosed by the lines y = x + 3, x + y + 5 = 0, x = k and the y-axis is 209 units2 . √ 14 The diagonal of a cube is 27 cm. Calculate the volume and surface area of this cube.

For Questions 14 and 15, try using concrete, everyday materials to help you understand the problem.

15 Two sides of a triangle have lengths 8 cm and 12 cm, respectively. Determine between which two values the length of the third side would fall. Give reasons for your answer. 16 When 1089 – 89 is expressed as a single number, what is the sum of its digits? 17

Determine the reciprocal of this product: 1 1 1 1 ). (1 – )(1 – )(1 – ) . . . (1 – 2 3 4 n+1

18 Find the value of

For Questions 16–19, try using a mathematical procedure to find a shortcut to the answer.

10022 – 9982 , without using a calculator. 1022 – 982

19 In the diagram at right, AP = 9 cm, PC = 15 cm, BQ = 8.4 cm and QC = 14 cm. Also, CD || QP || BA. Determine the ratio of the sides AB to DC.

A

D P

B

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Q

C

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xv

Working with unfamiliar problems: Part 2

Working with unfamiliar problems: Part 2 For the questions in Part 2, again use the ‘Working with unfamiliar problems’ poster at the back of this book, but this time choose your own strategy (or strategies) to solve each problem. Clearly communicate your solution and final answer.

nit

nit

xu

s

1 The Koch snowflake design starts with an equilateral triangle. A smaller equilateral triangle is built onto the middle third of each side and its base is erased. This procedure can be repeated indefinitely. B B B

A

s

xu

xvi

x units

A

C

C

A

C

a For a Koch snowflake with initial triangle side length x units, determine expressions for the exact value of: i the perimeter after 5 procedure repeats and after n procedure repeats ii the sum of areas after 3 procedure repeats and the change in area after n procedure repeats. b Comment on perimeter and area values as n → ∞. Give reasons for your answers.

2 Two sides of a triangle have lengths in the ratio 3 : 5 and the third side has length 37 cm. If each side length has an integer value, find the smallest and largest possible perimeters, in cm. 3 The midpoints of each side of a regular hexagon are joined to form a smaller regular hexagon with side length k cm. Determine a simplified expression in terms of k for the exact difference in the perimeters of the two hexagons. 4 Angle COD is 66◦ . Find the size of angle CAD.

C B

66°

A

D

O

1 5 The graph of y = ax2 + 2x + 3 has an axis of symmetry at x = . Determine the maximum 4 possible value of y. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Working with unfamiliar problems: Part 2

6 Find the value of x and y given that 5x = 125y – 3 and 81x+1 = 9y × 3. 7 A rectangular prism has a surface area of 96 cm2 and the sum of the lengths of all its edges is 64 cm. Determine the exact sum of the lengths of all its internal diagonals (i.e. diagonals not on a face). 8 In a Year 10 maths test, six students gained 100%, all students scored at least 75% and the mean mark was 82.85%. If the results were all whole numbers, what is the smallest possible number of students in this class? List the set of results for this class size. 9 Determine the exact minimum vertical distance between the line y = 2x and the parabola y = 2x2 – 5x – 3. 10 A + B = 6 and AB = 4. Without solving for A and B, determine the values of: 1 1 b A2 + B2 c (A – B)2 d a (A + 1)(B + 1) + A B 11 If f (1) = 5 and f(x + 1) = 2f (x), determine the value of f(8). 12 Four rogaining markers, PQRS, are in an area of bushland with level ground. Q is 1.4 km east of P, S is 1 km from P on a true bearing of 168◦ and R is 1.4 km from Q on a true bearing of 200◦ . To avoid swamps, Lucas runs the route PRSQP. Calculate the distances (in metres) and the true bearings from P to R, from R to S, from S to Q and from Q to P. Round your answers to the nearest whole number. 13 Consider all points (x, y) that are equidistant from the point (4, 1) and the line y = –3. Find the rule relating x and y and then sketch its graph, labelling all significant features. (Note: Use the distance formula) 14 A ‘rule of thumb’ useful for 4WD beach driving is that the proportion of total tide height change 1 2 3 3 after either high or low tide is in the first hour, in the second hour, in the third hour, 12 12 12 12 2 1 in the fifth hour and in the sixth hour. 12 12 a Determine the accuracy of this ‘rule of thumb’ using the following equation for tide height: h = 0.7 cos(30t) + 1, where h is in metres and t is time in hours after high tide. b Using h = A cos(30t) + D, show that the proportion of total tide height change between any two given times t1 and t2 is independent of the values of A and D. in the fourth hour,

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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1

Linear relations

Chapter

What you will learn

Australian curriculum

1A 1B 1C 1D 1E

NUMBER AND ALGEBRA

Reviewing algebra (Consolidating) Algebraic fractions Solving linear equations Inequalities Graphing straight lines (Consolidating) 1F Finding an equation of a line 1G Length and midpoint of a line segment 1H Perpendicular and parallel lines 1I Simultaneous equations – substitution 1J Simultaneous equations – elimination 1K Further applications of simultaneous equations 1L Half planes (Extending) Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Patterns and algebra Factorise algebraic expressions by taking out a common algebraic factor. Apply the four operations to simple algebraic fractions with numerical denominators. Substitute values into formulas to determine an unknown. Linear and non-linear relationships Solve problems involving linear equations, including those derived from formulas. Solve linear inequalities and graph their solutions on a number line. Solve linear simultaneous equations, using algebraic and graphical techniques, including using digital technology. Solve problems involving parallel and perpendicular lines. Solve linear equations involving simple algebraic fractions.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Online resources • Chapter pre-test • Videos of all worked examples • Interactive widgets • Interactive walkthroughs • Downloadable HOTsheets • Access to HOTmaths Australian Curriculum courses

Iron ore mining The revenue and costs of mining iron ore for Australian mining companies depends on many factors, including the quality of iron ore, the number of qualified staff, the amount of equipment, the price of iron and the depth of the iron ore in the ground. Maximising profit therefore requires the need to balance all the cost factors against the revenue.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

This is achieved by using a mathematical process called linear programming, which involves graphing all the various constraints as straight lines to create a feasible region on an x–y plane. The profit equation is then graphed over this feasible region to determine how maximum profit can be achieved. Such simple mathematical analysis can save mining companies millions of dollars.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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/VNCFS BOE "MHFCSB

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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/VNCFS BOE "MHFCSB

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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$IBQUFS  -JOFBS SFMBUJPOT

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Up for a challenge? If you get stuck on a question, check out the 'Working with unfamiliar problems' poster at the end of the book to help you.

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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



Midpoint and length of a line segment

Parallel and perpendicular lines

Line segment joining (x1, y1) and (x2, y2) has: Midpoint M =

Parallel × 10-3 lines have the same gradient; e.g. y = 2x + 7 and y = 2x − 4 For perpendicular lines with gradients m1 and m2: m1m2 = −1 or m2 = − m1 1 e.g. y = 3x + 4 and y = − 13 x + 2

x1 + x2 y1 + y2 , 2 2

Length/distance d = (x2 − x1)2 + (y2 − y1)2

Finding a rule for a graph For points (x1, y1) and (x2, y2)

To find line equation y = mx + c: - find gradient, m - substitute a point if c is unknown using y = mx + c or y − y1 = m (x − x1) Graphs of linear relations In the form y = mx + c, m is the gradient, c is the y-intercept. m = rise run

Algebra review Add/subtract like terms only. e.g. 2x + 4x + y = 6x + y Multiply: 3x × 2y = 6xy 2 Divide : 4ab ÷ (2b) = 4ab

y

Linear relations

12b

= 2a

y-intercept (x = 0)

Algebraic fractions

5a

5a

5a

Multiply: cancel common factors first. e.g.

x

x-intercept (y = 0)

Add/subtract: find LCD. e.g. 2 + 3 = 10 + 3a = 10 + 3a 5

rise run

O

Expand: a(b + c) = ab + ac a(b − c) = ab − ac

a

y2 − y1 x2 − x1

m=

Solving linear equations

1 2 1 3x + 6 × = 1 × 3(x + 2) = 3 84 4 x+2 4 x+2 1

Some steps include: - using inverse operations - collecting like terms on one side - multiplying by the LCD.

Divide: multiply by the reciprocal of the fraction following the ÷.

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To graph: a use gradient and y-intercept or b locate x-intercept (y = 0) and y-intercept (x = 0)

Linear inequalities Equation contains ‘=’ sign. Inequalities have >, ≥, < or ≤. On a number line: x >3 x ≤1

x 0 1 2 3 4 −2 −1 0 1 2

x

Solve as per linear equations except when multiplying or dividing by a negative, reverse the inequality sign.

Simultaneous equations Solve two equations in two pronumerals using: - substitution when one pronumeral is the subject; e.g. y = x + 4. - elimination when adding or subtracting multiples of equations eliminates one variable.

Applications of simultaneous equations - Define unknowns using pronumerals. - Form two equations from the information in the problem. - Solve simultaneously. - Answer the question in words.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Half planes (Ext) Inequalities with two variables represent the region above or below the line. For y ≥ x − 3, shade above the line. For y > x − 3, use a dashed line. For 2x + 3y < 6, test (0, 0) to decide which region to shade. 2(0) + 3(0) < 6 (true) Include (0, 0); i.e. shade below dashed line.

y

O

x 3

−3

y

2 O

x 3

Two or more half planes form an intersecting region.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter

2

Geometry

What you will learn

Australian curriculum

2A Geometry review (Consolidating) 2B Congruent triangles 2C Investigating parallelograms using congruence 2D Similar figures (Consolidating) 2E Proving similar triangles 2F Circles and chord properties (10A) 2G Angle properties of circles – theorems 1 and 2 (10A) 2H Angle properties of circles – theorems 3 and 4 (10A) 2I Tangents (Extending) 2J Intersecting chords, secants and tangents (Extending)

MEASUREMENT AND GEOMETRY

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Geometric reasoning Formulate proofs involving congruent triangles and angle properties. Apply logical reasoning, including the use of congruence and similarity, to proofs and numerical exercises involving plane 16x16 shapes. (10A) Prove and apply angle and chord properties of circles.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Online resources • Chapter pre-test • Videos of all worked examples • Interactive widgets • Interactive walkthroughs • Downloadable HOTsheets • Access to HOTmaths Australian Curriculum courses

Burj Khalifa building, Dubai Every building that people have walked into throughout the history of the civilised world was built with the help of geometry. The subject of geometry is intertwined with the structural engineering methods applied to the construction of all buildings. The ancient Greek mathematician and inventor Archimedes (born 287 bc) used simple geometry to design and build structures and machines to help advance society.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Today, structural engineers design and build structures using complex mathematics and computer software based on the same geometrical principles. Very tall buildings like the Burj Khalifa in Dubai, which is 828 m tall, need to be designed to take into account all sorts of engineering problems – earthquakes, wind, sustainability and aesthetics, to name just a few.

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88

Chapter 2 Geometry

2A Geometry review

CONSOLIDATING

Based on just five simple axioms (i.e. known or self-evident facts) the famous Greek mathematician Euclid (about 300 ) was able to deduce many hundreds of propositions (theorems and constructions) systematically presented in the 13-volume book collection called the Elements. All the basic components of school geometry are outlined in these books, including the topics angle sums of polygons and angles in parallel lines, which are studied in this section.

Let’s start: Three unknown angles This hexagon contains a pair of parallel sides and three unknown angles, a, b and c. • Find the value of a using the given angles in the hexagon. Hint: Add a construction line parallel to the two parallel sides so 35° that the angle of size a◦ is divided into two smaller angles. Give reasons throughout your solution. a° • Find the value of b, giving a reason. • What is the angle sum of a hexagon? Use this to find the value of c. •

Key ideas

160° b°

c°

75°

Can you find a different method to find the value of c, using parallel lines?

Angles at a point • Complementary (sum to 90◦ ) • Supplementary (sum to 180◦ ) • Revolution (360◦ ) • Vertically opposite angles (equal)

° °

Angles in parallel lines

°

a° b°

° Alternate angles are equal.

•

Corresponding angles are equal.

Cointerior angles are supplementary. a + b = 180

If two lines, AB and CD, are parallel, we write AB || CD.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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89

Measurement and Geometry

Triangles • Angle sum is 180◦ . To prove this, draw a line parallel to a base and then mark the alternate angles in parallel lines. Note that angles on a straight line are supplementary.

a°

Key ideas

c° b° c°

a° a + b + c = 180

• Triangles classified by angles. Acute: all angles acute

Obtuse: one angle obtuse

Right: one right angle

• Triangles classified by side lengths. Scalene Isosceles (3 different sides)

(2 equal sides)

Equilateral (3 equal sides)

60° 60°

60°

Quadrilaterals (Refer to section 2C for more details on quadrilaterals.) • Parallelograms are quadrilaterals with two pairs of parallel sides. • Rectangles are parallelograms with all angles 90◦ . • Rhombuses are parallelograms with sides of equal length. • Squares are parallelograms that are both rectangles and rhombuses. • Kites are quadrilaterals with two pairs of equal adjacent sides. • Trapeziums are quadrilaterals with at least one pair of parallel sides. Polygons have an angle sum given by S = 180(n – 2), where n is the number of sides. • Regular polygons have equal sides and equal angles. 180(n – 2) A single interior angle = n An exterior angle is supplementary to an interior angle. • For a triangle, the exterior angle theorem states that the exterior angle is equal to the sum of the two opposite interior angles. c=a+b a° c° b°

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

n 3 4 5 6 7 8 9 10

Name triangle quadrilateral pentagon hexagon heptagon octagon nonagon decagon

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90

Chapter 2 Geometry

Example 1 Using exterior angles and angle sums Find the value of x in the following, giving reasons. a b 150°

95°

85°

170°

x°

55°

100° 2x° SOL UTI ON

EX P L A NA TI ON

a

Use the exterior angle theorem for a triangle.

x + 85 = 150 (exterior angle theorem) x = 65

b

Use the rule for the angle sum of a polygon (5 sides, so n = 5).

S = 180(n – 2) = 180 × (5 – 2) = 540 2x + 100 + 170 + 95 + 55 = 540

The sum of all the angles is 540◦ .

2x + 420 = 540 2x = 120 ∴ x = 60

Example 2 Working with regular polygons and parallel lines Find the value of the pronumeral, giving reasons. a a regular hexagon b

A

B

25° C a°

a° 140° D

E

SOL UTI ON

EX P L A NA TI ON

a

Use the angle sum rule for a polygon with n = 6.

S = 180(n – 2) = 180 × (6 – 2) = 720 a = 720 ÷ 6

In a regular hexagon there are 6 equal angles.

= 120

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Measurement and Geometry

A

b Construct a third parallel line, CF. ∠BCF = 25◦ (alternate angles in || lines)

B C

∠FCD = 180◦ – 140◦ ◦

= 40 (cointerior angles in || lines)

F 140°

E D ∠ABC and ∠FCB are alternate angles in parallel lines. ∠FCD and ∠EDC are cointerior angles in parallel lines.

∴ a = 25 + 40 = 65

Exercise 2A

1–3

3

—

UNDERSTANDING

1

25° 25° 40°

List the names of the polygons with 3 to 10 sides, inclusive.

2 Decide whether each of the following is true or false. a The angle sum of a quadrilateral is 300◦ . b A square has 4 lines of symmetry. c An isosceles triangle has two equal sides. d An exterior angle on an equilateral triangle is 120◦ . e A kite has two pairs of equal opposite angles. f A parallelogram is a rhombus. g A square is a rectangle. h Vertically opposite angles are supplementary. i Cointerior angles in parallel lines are supplementary. 3 Find the values of the pronumerals, giving reasons. a b 70°

a°

c

71° a°

130°

b° a° d

e

81°

c° b°

b°

a°

a°

f 85° b° 78°

a°

67°

g

a°

b°

74°

h

b° 73° a°

i

b° a° 30°

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91

92

Chapter 2 Geometry

4, 5–6(½), 7

Example 1a

4, 5–6(½), 7

4 Using the exterior angle theorem, find the value of the pronumeral. a b c x° 40° 152° x° 80° 100°

5 Find the value of the pronumeral, giving reasons. a b

4, 5–6(½), 7

FLUENCY

2A

38° x° 18°

c

150°

x°

x°

x°

145° d

e

80° x°

f

5x°

20°

95°

g

100°

45°

x° h

2x° i

67°

x°

x°

62°

71°

j

k

l

x°

x°

x° 59°

Example 1b

38°

6 Find the value of x in the following, giving reasons. a b x° x° 120° 130° 70° 75°

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

c

95° x°

120°

110°

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93

d

e

95°

f

100° 140° 120°

125°

(2x + 55)° 120°

2A

(x + 100)°

(3x + 15)°

5x° 80°

FLUENCY

Measurement and Geometry

120°

7 Find the size of an interior angle of these polygons if they are regular. a pentagon b octagon c decagon 8(½), 9, 10

Example 2b

8 Find the value of the pronumeral a, giving reasons. a A b B 35°

a° 120°

C D

A 51°

E

C a°

8(½), 9–11

E

c

A

E

C 65°

a°

164°

160°

C

A a°

D 143°

B

D E

f

D

B

a°

B

e

290°

A

62° D

B d

8(½), 10–12

PROBLEM-SOLVING

Example 2a

C

a°

31°

D 143°

E B

C

E

A 9 a Find the size of an interior angle of a regular polygon with 100 sides. b What is the size of an exterior angle of a 100-sided regular polygon? 10 Find the number of sides of a polygon that has the following interior angles. a 150◦ b 162◦ c 172.5◦ 11 In this diagram, y = 4x. Find the values of x and y.

x° y°

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94

Chapter 2 Geometry

12 Find the value of x in this diagram, giving reasons. (Hint: Form isosceles and/or equilateral triangles.)

PROBLEM-SOLVING

2A 60° 130° x°

13–15

14–17

REASONING

13, 14

13 The rule for the sum of the interior angles of a polygon, S◦ , is given by S = 180(n – 2). a Show that S = 180n – 360. b Find a rule for the number of sides n of a polygon with an angle sum S; i.e. write n in terms of S. c Write the rule for the size of an interior angle I◦ of a regular polygon with n sides. d Write the rule for the size of an exterior angle E◦ of a regular polygon with n sides.

B

14 Prove that the exterior angle of a triangle is equal to the sum of the two opposite interior angles by following these steps.

b°

a Write ∠BCA in terms of a and b and give a reason. b Find c in terms of a and b using ∠BCA and give a reason.

A

c°

a°

D

C

15 a Explain why in this diagram ∠ABD is equal to b◦ . b Using ∠ABC and ∠BCD, what can be said about a, b and c? c What does your answer to part b show?

D

A b°

B

c°

a° C

16 Give reasons why AB and DE in this diagram are parallel; i.e. AB || DE.

A

B C

D

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

E

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95

REASONING

Measurement and Geometry

17 Each point on the Earth’s surface can be described by a line of longitude (degrees east or west from Greenwich, England) and a line of latitude (degrees north or south from the equator). Investigate and write a brief report (providing examples) describing how places on the Earth can be located with the use of longitude and latitude.

Multilayered reasoning

—

18–20

ENRICHMENT

—

2A

18 Find the value of the pronumerals, giving reasons. a b 45°

60° 290° a°

70° a° 19 Give reasons why ∠ABC = 90◦ .

A

C

B

20 In this diagram ∠AOB = ∠BOC and ∠COD = ∠DOE. Give reasons why ∠BOD = 90◦ .

D

C

E

B O A

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96

Chapter 2 Geometry

Exploring triangles with dynamic geometry 1

Construct a triangle and illustrate the angle sum and the exterior angle theorem.

Dynamic geometry instructions and screens a

Construct a line AD and triangle ABC, as shown.

b Measure all the interior angles and the exterior angle ∠BCD. c

Use the calculator to check that the angle sum is 180◦ and that ∠BCD = ∠BAC + ∠ABC.

d Drag one of the points to check that these properties are retained for all triangles.

B 90.29° 142.26° 37.74° 51.97°

D

C

A

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97

Measurement and Geometry

2B Congruent triangles In geometry it is important to know whether or not two objects are in fact identical in shape and size. If two objects are identical, then we say they are congruent. Two shapes that are congruent will have corresponding (i.e. matching) sides equal in length and corresponding angles also equal. For two triangles it is not necessary to know every side and angle to determine if they are congruent. Instead, a set of minimum conditions is enough. There are four sets of minimum conditions for triangles and these are known as the tests for congruence of triangles.

Are the Deira Twin Towers in Dubai congruent?

Let’s start: Which are congruent? Consider these four triangles. 1 ABC with ∠A = 37◦ , ∠B = 112◦ and AC = 5 cm.

2 DEF with ∠D = 37◦ , DF = 5 cm and ∠E = 112◦ .

3 GHI with ∠G = 45◦ , GH = 7 cm and HI = 5 cm. 4 JKL with ∠J = 45◦ , JK = 7 cm and KL = 5 cm.

Sarah says that only ABC and DEF are congruent. George says that only GHI and JKL are congruent and Tobias says that both pairs (ABC, DEF and GHI, JKL) are congruent. • • •

Discuss which pairs of triangles might be congruent, giving reasons. What drawings can be made to support your argument? Who is correct: Sarah, George or Tobias? Explain why. Two objects are said to be congruent when they are exactly the same size and shape. For two congruent triangles ABC and DEF, we write ABC ≡ DEF. • When comparing two triangles, corresponding sides are equal in length and corresponding angles are equal. • When we prove congruence in triangles, we usually write vertices in matching order.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Key ideas

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98

Chapter 2 Geometry

Key ideas

Two triangles can be tested for congruence using the following conditions. • Corresponding sides are equal (SSS).

• Two corresponding sides and the included angle are equal (SAS).

• Two corresponding angles and a side are equal (AAS).

×

×

• A right angle, the hypotenuse and one other pair of corresponding sides are equal (RHS).

AB || CD means AB is parallel to CD. AB ⊥ CD means AB is perpendicular to CD.

Example 3 Proving congruence in triangles Prove that these pairs of triangles are congruent. a b E B

A

80°

80°

C D

F

D

B

×

F

A E

× C

SOL UTI ON a

EX P LA NA TI ON List all pairs of corresponding sides and angles.

AB = DE (given) S ◦

∠BAC = ∠EDF = 80 (given) A AC = DF (given) S So, ABC ≡ DEF (SAS) b

The two triangles are therefore congruent, with two pairs of corresponding sides and the included angle equal.

∠ABC = ∠DEF (given) A

List all pairs of corresponding sides and angles.

∠ACB = ∠DFE (given) A

The two triangles are therefore congruent with two pairs of corresponding angles and a corresponding side equal.

AB = DE (given) S So, ABC ≡ DEF (AAS) Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

Example 4 Using congruence in proof In this diagram, ∠A = ∠C = 90◦ and AB = CB. a Prove ABD ≡ CBD.

3 cm

c

A

D

b Prove AD = CD.

State the length of CD.

C B

SOL UTI ON

EX P L A NA TI ON

a ∠A = ∠C = 90◦ (given) R BD is common H AB = CB (given) S ∴ ABD ≡ CBD (RHS)

Systematically list corresponding pairs of equal angles and lengths.

b ABD ≡ CBD so AD = CD (corresponding sides in congruent triangles)

Since ABD and CBD are congruent, the matching sides AD and CD are equal.

c CD = 3 cm

AD = CD from part b above.

1

1, 2

1, 2

—

Which of the tests (SSS, SAS, AAS or RHS) would be used to decide whether the following pairs of triangles are congruent? a b c 5 6 4 85° 3 × 2

7

5 85°

2

×

UNDERSTANDING

Exercise 2B

4

6

3 7

d

e

10

8

7 3

8

3

10

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

7

f

6

15

15 6

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99

100

Chapter 2 Geometry

UNDERSTANDING

2B 2 Assume these pairs of figures are congruent and find the value of the pronumeral in each case. a b x x 2 4

2

5

5

5

6

d

× 25

2x 8

7

e

3x

6

8

Example 3

×

5x

4x + 1

f

7 9

3 Prove that these pairs of triangles are congruent. a A b D

3, 4

3, 4–5(½)

F

A

3–5(½)

FLUENCY

c

B

E B

E

C

D

C

F C

c

D

d

E

C

E D

B A

F

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

×

A

B × F

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FLUENCY

Measurement and Geometry

4 Find the value of the pronumerals in these diagrams, which include congruent triangles. a b x y 12 11 5.2 y 7.3 x

a

c

101

2B

16

d

1.3 9

y x

b

2.4

5 Prove that each pair of triangles is congruent, giving reasons. Write the vertices in matching order. a b A B D C

×

A

B

× C

D c

B

A

D

d

C

e

B

A

B

A

E

D

C

D

f

A

C

O D

B

C

Example 4

6 In this diagram, O is the centre of the circle and ∠AOB = ∠COB. a Prove AOB ≡ COB. b Prove AB = BC. c State the length of AB.

6, 7

6–8

A ´ O ´

C

PROBLEM-SOLVING

6

B 10 mm

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102

Chapter 2 Geometry

7 In this diagram, BC = DC and AC = EC. a Prove ABC ≡ EDC. b Prove AB = DE. c Prove AB || DE. d State the length of DE.

5 cm

A

PROBLEM-SOLVING

2B B

C E

D A

D

C

B

9a,b

9 Prove the following for the given diagrams. Give reasons. a AB || DE b OB bisects AC; i.e. AB = BC E B

9

c AD || BC

D

C

O

C D

A

A

C

B

e ∠OAD = ∠OBC

d AD = AC

E

9(½)

REASONING

8 In this diagram, AB = CD and AD = CB. a Prove ABD ≡ CDB. b Prove ∠DBC = ∠BDA. c Prove AD || BC.

C

D

B

f

B

A

A AC ⊥ BD; i.e. ∠ACD = ∠ACB

B

D O

A

C

A

Draw your own diagram

C

B

—

—

10

10 a A circle with centre O has a chord AB. M is the midpoint of the chord AB. Prove OM ⊥ AB. b Two overlapping circles with centres O and C intersect at A and B. Prove ∠AOC = ∠BOC. c ABC is isosceles with AC = BC, D is a point on AC such that ∠ABD = ∠CBD, and E is a point on BC such that ∠BAE = ∠CAE. AE and BD intersect at F. Prove AF = BF.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

ENRICHMENT

D

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Measurement and Geometry

103

2C Investigating parallelograms using congruence Recall that parallelograms are quadrilaterals with two pairs of parallel sides. We therefore classify rectangles, squares and rhombuses as special types of parallelograms, the properties of which can be explored using congruence. By dividing a parallelogram into ‘smaller’ triangles, we can prove congruence for pairs of these triangles and use this to deduce the properties of the shape.

Let’s start: Aren’t they the same proof? Here are two statements involving the properties of a parallelogram. 1

A parallelogram (with parallel opposite sides) has opposite sides of equal length.

2

A quadrilateral with opposite sides of equal length is a parallelogram. • Are the two statements saying the same thing? • Discuss how congruence can be used to help prove each statement. • Formulate a proof for each statement.

Some vocabulary and symbols: • If AB is parallel to BC, then we write AB || CD. • If AB is perpendicular to CD, then we write AB ⊥ CD. • To bisect means to cut in half. Parallelogram properties and tests: • Parallelogram – a quadrilateral with opposite sides parallel. Properties of a parallelogram - opposite sides are equal in length - opposite angles are equal - diagonals bisect each other

Key ideas

Tests for a parallelogram - if opposite sides are equal in length - if opposite angles are equal - if one pair of opposite sides are equal and parallel - if diagonals bisect each other

• Rhombus – a parallelogram with all sides of equal length. Properties of a rhombus - all sides are equal length - opposite angles are equal - diagonals bisect each other at right angles - diagonals bisect the interior angles

Tests for a rhombus - if all sides are equal length - if diagonals bisect each other at right angles

• Rectangle – a parallelogram with all angles 90◦ . Properties of a rectangle - opposite sides are of equal length - all angles equal 90° - diagonals are equal in length and bisect each other

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Tests for a rectangle - if all angles are 90° - if diagonals are equal in length and bisect each other

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104

Chapter 2 Geometry

Key ideas

• Square – a parallelogram that is a rectangle and a rhombus. Properties of a square - all sides are equal in length - all angles equal 90° - diagonals are equal in length and bisect each other at right angles - diagonals bisect the interior angles

Tests for a square - if all sides are equal in length and at least one angle is 90° - if diagonals are equal in length and bisect each other at right angles

Example 5 Proving properties of quadrilaterals a

Prove that a parallelogram (with opposite sides parallel) has equal opposite angles.

b Use the property that opposite sides of a parallelogram are equal to prove that a rectangle (with all angles 90◦ ) has diagonals of equal length. SOL UTI ON

C

D

a

A

b

EX P L A NA TI ON Draw a parallelogram with parallel sides and the segment BD.

B

∠ABD = ∠CDB (alternate angles in || lines) A ∠ADB = ∠CBD (alternate angles in || lines) A

Prove congruence of ABD and CDB, noting alternate angles in parallel lines.

BD is common S ∴ ABD ≡ CDB (AAS) ∴ ∠DAB = ∠BCD, so opposite angles are equal.

Note also that BD is common to both triangles. Corresponding angles in congruent triangles.

D

C

A

B

First, draw a rectangle with the given properties.

Consider ABC and BAD. AB is common S ∠ABC = ∠BAD = 90◦ A BC = AD (opposite sides of a parallelogram are equal in length) S ∴ ABC ≡ BAD (SAS) ∴ AC = BD, so diagonals are of equal length.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Choose ABC and BAD, which each contain one diagonal. Prove congruent triangles using SAS.

Corresponding sides in congruent triangles.

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Measurement and Geometry

105

Example 6 Testing for a type of quadrilateral a

Prove that if opposite sides of a quadrilateral are equal in length, then it is a parallelogram.

b Prove that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

SOL UTI ON a

EX P L A NA TI ON

C

D

B A AB = CD (given) S BC = DA (given) S AC is common S ∴ ABC ≡ CDA (SSS)

∴ ∠BAC = ∠DCA So AB || DC (since alternate angles are equal). Also ∠ACB = ∠CAD. ∴ AD || BC (since alternate angles are equal). ∴ ABCD is a parallelogram. b D

Prove that they are congruent using SSS.

Choose corresponding angles in the congruent triangles to show that opposite sides are parallel. If alternate angles between lines are equal then the lines must be parallel. All angles at the point E are 90◦ , so it is easy to prove that of all four smaller triangles are congruent using SAS.

C E A

First, label your quadrilateral and choose two triangles, ABC and CDA.

B

ABE ≡ CBE ≡ ADE ≡ CDE by SAS ∴ AB = CB = CD = DA and since ∠ABE = ∠CDE, then AB || DC. Also, as ∠CBE = ∠ADE, then AD || BC.

Corresponding sides in congruent triangles. All sides are therefore equal. Prove opposite sides are parallel using pairs of equal alternate angles.

∴ ABCD is a rhombus.

All sides are equal and opposite sides parallel.

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106

Chapter 2 Geometry

Exercise 2C

2, 3

—

UNDERSTANDING

1

1–3

Name the special quadrilateral given by these descriptions. a a parallelogram with all angles 90◦ b a quadrilateral with opposite sides parallel c a parallelogram that is a rhombus and a rectangle d a parallelogram with all sides of equal length

2 List all the special quadrilaterals that have these properties. a All angles 90◦ . b Diagonals are equal in length. c Diagonals bisect each other. d Diagonals bisect each other at 90◦ . e Diagonals bisect the interior angles. b A kite is not a parallelogram.

Example 5

4–6

4–6

C

D

4 Complete these steps to prove that a parallelogram (with opposite parallel sides) has equal opposite sides. a Prove ABC ≡ CDA. b Hence, prove opposite sides are equal.

B

A 5 Complete these steps to prove that a parallelogram (with opposite equal parallel sides) has diagonals that bisect each other. a Prove ABE ≡ CDE. b Hence, prove AE = CE and BE = DE.

C

D E

B

A

D

A

6 Complete these steps to prove that a rhombus (with sides of equal length) has diagonals that bisect the interior angles. a Prove ABD ≡ CDB. b Hence, prove BD bisects both ∠ABC and ∠CDA.

B

7 Example 6

7, 8

C 8, 9

C

D

7 Complete these steps to prove that if the diagonals in a quadrilateral bisect each other, then it is a parallelogram. a Prove ABE ≡ CDE. b Hence, prove AB || DC and AD || BC.

E

A

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

FLUENCY

4, 5

B

PROBLEM-SOLVING

3 Give a reason why: a A trapezium is not a parallelogram.

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8 Complete these steps to prove that if one pair of opposite sides is equal and parallel in a quadrilateral, then it is a parallelogram. a Prove ABC ≡ CDA. b Hence, prove AB || DC.

D

C

107

2C

B

A

9 Complete these steps to prove that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. a Give a brief reason why ABE ≡ CBE ≡ ADE ≡ CDE. b Hence, prove ABCD is a rhombus.

C

D E

B

A

10, 11

11, 12

REASONING

10

PROBLEM-SOLVING

Measurement and Geometry

10 Prove that the diagonals of a rhombus (i.e. a parallelogram with sides of equal length): a intersect at right angles b bisect the interior angles. 11 Prove that a parallelogram with one right angle is a rectangle. 12 Prove that if the diagonals of a quadrilateral bisect each other and are of equal length, then it is a rectangle. —

—

13 In this diagram, E, F, G and H are the midpoints of AB, BC, CD and DA, respectively, and ABCD is a rectangle. Prove that EFGH is a rhombus.

13

D

G

H

A

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

C

ENRICHMENT

Rhombus in a rectangle

F

E

B

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108

Chapter 2 Geometry

2D Similar figures

CONSOLIDATING

You will recall that the transformation called enlargement involves reducing or enlarging the size of an object. The final image will be of the same shape but of different size. This means that matching pairs of angles will be equal and matching sides will be in the same ratio, just as in an accurate scale model.

Let’s start: The Q1 tower The Q1 tower, pictured here, is located on the Gold Coast and was the world’s tallest residential tower up until 2011. It is 245 m tall. • • •

Key ideas

Measure the height and width of the Q1 tower in this photograph. Can a scale factor for the photograph and the actual Q1 tower be calculated? How? How can you calculate the actual width of the Q1 tower using this photograph? Discuss.

Similar figures have the same shape but are of different size. • Corresponding angles are equal. • Corresponding sides are in the same proportion (or ratio). Scale factor =

image length original length

The symbols ||| or ∼ are used to describe similarity and to write similarity statements. For example, ABC ||| DEF or ABC ∼ DEF. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

109

Example 7 Finding and using scale factors These two shapes are similar. a Write a similarity statement for the two shapes. EH FG b Complete the following: = . ---- ---c Find the scale factor. d

Find the value of x.

e

Find the value of y.

D 3 cm

F

y cm

5 cm

E

C x cm

A 2 cm B

G 7 cm H

SOL UTI ON

EX P L A NA TI ON

a ABCD ||| EFGH or ABCD ∼ EFGH

Use the symbol ||| or ∼ in similarity statements.

EH FG = AD BC

c

EF 5 = or 2.5 AB 2

d

x = 3 × 2.5

Ensure you match corresponding vertices. EF and AB are matching sides and both lengths are given. EH corresponds to AD, which is 3 cm in length. Multiply by the scale factor.

= 7.5 e

y = 7 ÷ 2.5

DC corresponds to HG, which is 7 cm in length.

= 2.8

Exercise 2D 1

1–3

3

—

These two figures are squares.

D

A

C

1 cm

H

G

B

UNDERSTANDING

b

E

F 3 cm a Would you say that the two squares are similar? Why? b What is the scale factor when the smaller square is enlarged to the size of the larger square? c If the larger square is enlarged by a factor of 5, what would be the side length of the image?

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110

Chapter 2 Geometry

UNDERSTANDING

2D 2 Find the scale factor for each shape and its larger similar image. a b

5 4

8 8

c

12

5

d

7 9

8

14 21

7 3 The two triangles shown opposite are similar. a In ABC, which vertex corresponds to (matches) vertex E? b In ABC, which angle corresponds to ∠D? c In DEF, which side corresponds to BC? d Write a similarity statement for the two triangles. Write matching vertices in the same order.

C

×

5 These two shapes are similar. a Write a similarity statement for the two shapes. EF ---b Complete the following: . = ---- CD c Find the scale factor. d Find the value of x. e Find the value of y.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

D

FLUENCY

5, 6

J 2 cm I

A

H

2 cm 1 cm

x cm

B

F

H 9m

12 m B ym

G

3 cm

D

A

B

A

4–6

4 These two shapes are similar. a Write a similarity statement for the two shapes. AB DE b Complete the following: = . ---- ---E c Find the scale factor. d Find the value of x. y cm D e Find the value of y.

C

E ×

4, 5 Example 7

F

xm C

16 m F

G

14 m

E

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111

6

FLUENCY

Measurement and Geometry

Find the value of the pronumeral in each pair of similar figures. Round to one decimal place where necessary. a b 0.4 0.6 10

0.8

x

2D

x

5 25

c

d

x

12

20

4

1.5 10

x

8 18

e

x

11

f

4.2 5.7

× x

7

10.7 ×

7, 8

8, 9

7 A 50 m tall structure casts a shadow 30 m in length. At the same time, a person casts a shadow of 1.02 m. Estimate the height of the person. (Hint: Draw a diagram of two triangles.) 8

A BMX ramp has two vertical supports, as shown. a Find the scale factor for the two triangles in the diagram. b Find the length of the inner support.

1m

PROBLEM-SOLVING

7, 8

1 m 60 cm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter 2 Geometry

2D Find the value of the pronumeral if the pairs of triangles are similar. Round to one decimal place in part d. a b 2 1 1.2 x 2 0.4

x

0.5 c

d

3 8

x

3.6

x 5

1.3

1.1

10

10, 11

11, 12

REASONING

9

PROBLEM-SOLVING

112

10 In this diagram, the two triangles are similar and AB || DE.

6

A

B

y

3 1.5

C x D

2

E

a Which side in ABC corresponds to DC? Give a reason. b Write a similarity statement by matching the vertices. c Find the value of x. d Find the value of y. 11 Decide whether each statement is true or false. a All circles are similar. c All rectangles are similar. e All parallelograms are similar. g All kites are similar. i All equilateral triangles are similar.

b d f h j

All squares are similar. All rhombuses are similar. All trapeziums are similar. All isosceles triangles are similar. All regular hexagons are similar.

12 These two triangles each have two given angles. Decide whether they are similar and give reasons.

85° 75°

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

20° 85°

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Measurement and Geometry

—

—

13, 14

13 Shown here is a cube of side length 2 and its image after enlargement.

ENRICHMENT

Length, area, volume and self-similarity

113

2D

2

3 a Write down the scale factor for the side lengths as an improper fraction. b Find the area of one face of: i the smaller cube ii the larger cube. c

Find the volume of: i the smaller cube

ii the larger cube.

d Complete this table. Cube Small Large Scale factor (fraction)

e f

Length 2 3

Area

Volume

How do the scale factors for Area and Volume relate to the scale factor for side length? b If the side length scale factor was , write down expressions for: a i the area scale factor ii the volume scale factor.

14 An object that is similar to a part of itself is said to be self-similar. These are objects that, when magnified, reveal the same structural shape. A fern is a good example, as shown in this picture.

Self-similarity is an important area of mathematics, having applications in geography, econometrics and biology. Use the internet to explore the topic of self-similarity and write a few dot points to summarise your findings.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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114

Chapter 2 Geometry

2E Proving similar triangles As with congruent triangles, there are four tests for proving that two triangles are similar. When solving problems involving similar triangles, it is important to recognise and be able to prove that they are in fact similar.

Let’s start: How far did the chicken travel?

Similar triangles are used in this bridge structure.

A chicken is considering how far it is across a road, so it places four pebbles in certain positions on one side of the road. Two of the pebbles are directly opposite a rock on the other side of the road. The number of chicken paces between three pairs of the pebbles is shown in the diagram. •

rock road 5

Has the chicken constructed any similar triangles? If so, discuss why they are similar. • What scale factor is associated with the two triangles? • Is it possible to find how many paces the chicken must take to get across the road? If so, show a solution. • Why did the chicken cross the road? Answer: To explore similar triangles.

Key ideas

7 10

Two objects are said to be similar if they are of the same shape but of different size. • For two similar triangles ABC and DEF, we write ABC ||| DEF or ABC ∼ DEF. • When comparing two triangles, corresponding sides are opposite equal angles. Two triangles can be tested for similarity by considering the following conditions. • All pairs of corresponding sides are in the same ratio (or proportion) (SSS). 6 5 2 = = =2 2.5 1 2 3 2.5 1 5

3 •

6 Two pairs of corresponding sides are in the same ratio and the included corresponding angles are equal (SAS). 4 2 = = 2 and ∠A = ∠D 2 D 4 A 2 1 B E 1 2 C F

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

• Three corresponding angles are equal (AAA). (Remember that two pairs of corresponding equal angles implies that all three pairs of corresponding angles are equal.)

115

Key ideas

D

A ×

∠A = ∠D

×

∠B = ∠E

C

B

∠C = ∠F

E

F

• The hypotenuses of two right-angled triangles and another pair of corresponding sides are in the same ratio (RHS). A D

3

5

B

6

C

10

∠B = ∠E = 90◦ 10 6 = =2 5 3

F

E

Note: If the test AAA is not used, then at least two pairs of corresponding sides in the same ratio are required for all the other three tests.

Example 8 Using similarity tests to prove similar triangles Prove that the following pairs of triangles are similar. a b A E F B C

1.5

2 20°

3

4

A

E 40°

40°

65°

D

D

SOL UTI ON

F

EX P L A NA TI ON

DF 4 = = 2 (ratio of corresponding sides) S AC 2 DE 3 = = 2 (ratio of corresponding sides) S AB 1.5 ∠BAC = ∠EDF = 20◦ (given corresponding angles) A ∴ ABC ||| DEF (SAS)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

B

C

20°

a

65°

DF and AC are corresponding sides and DE and AB are corresponding sides, and both pairs are in the same ratio. The corresponding angle between the pair of corresponding sides in the same ratio is also equal. The two triangles are therefore similar.

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116

Chapter 2 Geometry

b ∠ABC = ∠FDE = 65◦ (given corresponding angles) A ◦ ∠ACB = ∠FED = 40 (given corresponding angles) A ∴ ABC ||| FDE (AAA)

There are two pairs of given corresponding angles. If two pairs of corresponding angles are equal, then the third pair must also be equal (due to angle sum). The two triangles are therefore similar.

Example 9 Establishing and using similarity A cone has radius 2 cm and height 4 cm. The top of the cone is cut horizontally through D. a Prove ADE ||| ABC.

b

If AD = 1 cm, find the radius DE.

A

D B

E

2 cm

4 cm C

SOL UTI ON

EX P L A NA TI ON

a ∠BAC is common A ∠ABC = ∠ADE (corresponding angles in parallel lines) A ∴ ADE ||| ABC (AAA)

All three pairs of corresponding angles are equal.

b

DE AD = BC AB DE 1 = 2 4 ∴ DE =

2 4

Therefore, the two triangles are similar. Given the triangles are similar, the ratio of corresponding sides must be equal.

Solve for DE.

= 0.5 cm

Architecture and engineering are just two fields in which the skills covered in this chapter can be used.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

1

1–3

3

—

UNDERSTANDING

Exercise 2E

C

This diagram includes two similar triangles. a Which vertex in DEF corresponds to vertex B? b Which angle in ABC corresponds to ∠F? c Which side in ABC corresponds to DE? d Write a similarity statement for the two triangles.

E

B

A

2 This diagram includes two similar triangles. a Which angle in CDE corresponds to ∠B in ABC and why? b Which angle in ABC corresponds to ∠E in CDE and why? c Which angle is vertically opposite ∠ACB? d Which side on ABC corresponds to side CE on CDE? e Write a similarity statement, being sure to write matching vertices in the same order.

117

D

F

B

A

C D

E

3 Which similarity test (SSS, SAS, AAA or RHS) would be used to prove that these pairs of triangles are similar? a b 1 ×

1 2 c

×

0.5 7

14 13 26

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

1.2

d

5.4

3.6

0.7

1.8

2.1

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Chapter 2 Geometry

4, 5(½)

2E Example 8

4 Prove that the following pairs of triangles are similar. a b B E

65° 70°

A

B

65° D

1

70°

A

4–6(½)

4–6(½)

C

F

3 120°

FLUENCY

118

6

C E 120° 2

F

D c

A

8

D

E

d

28

A

5 B

B 16 F

32

C

10

C

4

4

E 7 8

D

F 5 Find the value of the pronumerals in these pairs of similar triangles. a b

13.5

2

9

3 1 x

13

c

x 24

d

b

12

3.6

a

5

1.2 x

8

6.6 e

f

0.4

0.15 0.13

a y

x 0.375

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

18 16

38.4 b

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FLUENCY

Measurement and Geometry

6 For the following proofs, give reasons at each step. a Prove ABC ||| EDC. b Prove ABE ||| ACD. A A B

C

d Prove AEB ||| CDB.

7.5

D

E

E A

2 3

B

7, 8 Example 9

C

5

D

8, 9

9, 10

7 A right cone with radius 4 cm has a total height of 9 cm. It contains an amount of water, as shown. a Prove EDC ||| ADB. b If the depth of water in the cone is 3 cm, find the radius of the water surface in the cone.

A

4 cm

B

9 cm E

C

PROBLEM-SOLVING

c Prove BCD ||| ECA. A B C ×

×

D

C

E

2E

E

B

D

119

D 8 A ramp is supported by a vertical stud AB, where A is the midpoint of CD. It is known that CD = 4 m and that the ramp is 2.5 m high; i.e. DE = 2.5 m. a Prove BAC ||| EDC. b Find the length of the stud AB.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

E B C

A

D

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Chapter 2 Geometry

PROBLEM-SOLVING

2E 9 At a particular time in the day, Felix casts a shadow 1.3 m long and Curtis, who is 1.75 m tall, casts a shadow 1.2 m long. Find Felix’s height, correct to two decimal places.

10 To determine the width of a chasm, a marker (A) is placed directly opposite a rock (R) on the other side. Point B is placed 3 m away from point A, as shown. Marker C is placed 3 m along the edge of the chasm, and marker D is placed so that BD is parallel to AC. Markers C and D and the rock are collinear (i.e. lie in a straight line). If BD measures 5 m, find the width of the chasm (AR).

R chasm C A 3m

3m D B

11

11, 12

5m

11–13

11 Aiden and May come to a river and notice a tree on the opposite bank. Separately they decide to place rocks (indicated with dots) on their side of the river to try to calculate the river’s width. They then measure the distances between some pairs of rocks, as shown.

May’s rock placement

Aiden’s rock placement

tree

tree width

REASONING

120

river

river

width

10 m 5m a b c d

12 m

8m

25 m

15 m Have both Aiden and May constructed a pair of similar triangles? Give reasons. Use May’s triangles to calculate the width of the river. Use Aiden’s triangles to calculate the width of the river. Which pair of triangles did you prefer to use? Give reasons.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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REASONING

Measurement and Geometry

12 There are two triangles in this diagram, each showing two given angles. Explain why they are similar.

42° 91°

121

2E

47°

13 Prove the following, giving reasons. 7 b AE = AC 5

B

B

A

A O

1

D

2 2

Proving Pythagoras’ theorem

D

C

—

14 In this figure ABD, CBA and CAD are right angled.

a Prove ABD ||| CBA. Hence, prove AB2 = CB × BD. b Prove ABD ||| CAD. Hence, prove AD2 = CD × BD. c Hence, prove Pythagoras’ theorem AB2 + AD2 = BD2 .

C

5 E

—

14

D C

A

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

ENRICHMENT

a OB = 3OA

B

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122

Chapter 2 Geometry

Progress quiz 38pt 2A

1

Find the size of each pronumeral in the following polygons, giving reasons. a b c x° x° 58° 92° 136° x° x° x° w° x° x° 89° d

e

f

2w° x°

72°

x°

107°

43°

83° 38pt 2B

2

a

Prove that ABC is congruent to QBP.

A

P

B

C b 38pt 2C

3

Q

Prove that B is the midpoint of CP.

A

Prove that a rhombus has its diagonals perpendicular to each other.

B

D

C

38pt 2C

4

Prove that if the opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram and the opposite sides are equal in length.

38pt 2D

5

For the two similar triangles shown: a Write a similarity statement. b

Find the scale factor.

c

Find the value of x.

A 6 E 9 D

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

5 B x C

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Measurement and Geometry

38pt 2E

6

Prove that the following pairs of triangles are similar. a b A D

4

5

12

A

123

C

O

15 B

C

B F

E D

38pt 2E

7

A cone with radius 6 cm and height 10 cm is filled with water to a height of 5 cm. a Prove that EDC is similar to ADB. b Find the radius of the water’s surface (EC).

A

B

E

C

D 38pt 2E

8

Prove that if the midpoints, Q and P, of two sides of a triangle ABC are joined as shown, then 1 QP is that of CB. (First prove similarity.) 2

A

Q

C

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

P

B

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124

Chapter 2 Geometry

2F Circles and chord properties

10A

Although a circle appears to be a very simple object, it has many interesting geometrical properties. In this section we look at radii and chords in circles, and then explore and apply the properties of these objects. We use congruence to prove many of these properties.

Let’s start: Dynamic chords This activity would be enhanced with the use of computer dynamic geometry. Chord AB sits on a circle with centre O. M is the midpoint of chord AB. Explore with dynamic geometry software or discuss the following.

O

• Is OAB isosceles and if so why? • Is OAM ≡ OBM and if so why? • Is AB ⊥ OM and if so why? • Is ∠AOM = ∠BOM and if so why?

minor sector

centre m aj

or

ar

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

M

B

mi n

major sector

arc or

ter diame r adi us

Circle language

chord

Key ideas

A

major segment

minor segment

c

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Measurement and Geometry

125

Key ideas

An angle is subtended by an arc or chord if the arms of the angle meet the endpoints of the arc or chord.

A

A

C O

B ∠AOB is subtended at the centre by the minor arc AB.

B ∠ABC is subtended at the circumference by the chord AC.

Chord theorem 1: Chords of equal length subtend equal angles at the centre of the circle. • If AB = CD, then ∠AOB = ∠COD. • Conversely, if chords subtend equal angles at the centre of the circle, then the chords are of equal length.

D

C θ O

θ

B

A Chord theorem 2: Chords of equal length are equidistant (i.e. of equal distance) from the centre of the circle. • If AB = CD, then OE = OF. • Conversely, if chords are equidistant from the centre of the circle, then the chords are of equal length.

D F C O B E A

Chord theorem 3: The perpendicular from the centre of the circle to the chord bisects the chord and the angle at the centre subtended by the chord. • If OM ⊥ AB, then AM = BM and ∠AOM = ∠BOM. • Conversely, if a radius bisects the chord (or angle at the centre subtended by the chord), then the radius is perpendicular to the chord.

A O

θ θ

M

B

Chord theorem 4: The perpendicular bisectors of every chord of a circle intersect at the centre of the circle. • Constructing perpendicular bisectors of two chords will therefore locate the centre of a circle.

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126

Chapter 2 Geometry

Example 10 Using chord theorems For each part, use the given information and state which chord theorem is used. a Given AB = CD and OE = 3 cm, find OF.

A 3 cm

E B

O C F

D b Given OM ⊥ AB, AB = 10 cm and ∠AOB = 92◦ , find AM and ∠AOM.

O A M

B

SOL UTI ON

EX P L A NA TI ON

a OF = 3 cm (using chord theorem 2)

Chords of equal length are equidistant from the centre.

b Using chord theorem 3: AM = 5 cm ∠AOM = 46◦

The perpendicular from the centre to the chord bisects the chord and the angle at the centre subtended by the chord. 10 ÷ 2 = 5 and 92 ÷ 2 = 46.

Example 11 Proving chord theorems Prove chord theorem 3 in that the perpendicular from the centre of the circle to the chord bisects the chord and the angle at the centre subtended by the chord. SOL UTI ON

EX P L A NA TI ON

A O

First, draw a diagram to represent the situation. The perpendicular forms a pair of congruent triangles.

M B

∠OMA = ∠OMB = 90◦ (given) R OA = OB (both radii) H OM is common S ∴ OMA ≡ OMB (RHS)

∴ AM = BM and ∠AOM = ∠BOM Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Corresponding sides and angles in congruent triangles are equal.

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Measurement and Geometry

1

1–4

3, 4

—

Construct a large circle, then draw and label these features. a a chord b radius c centre d minor sector

UNDERSTANDING

Exercise 2F

e major sector

2 Give the size of the angle in this circle subtended by the following. a chord AB b minor arc BC c minor arc AD d chord DC

127

D C

75° 55°

A

B 3 In this circle chords AB and CD are of equal length. a Measure ∠AOB and ∠COD. b What do you notice about your answers from part a? Which chord theorem does this relate to? c Measure OE and OF. d What do you notice about your answers from part c? Which chord theorem does this relate to?

A C

O F D

4 In this circle OM ⊥ AB. a Measure AM and BM. b Measure ∠AOM and ∠BOM. c What do you notice about your answers from parts a and b? Which chord theorem does this relate to?

O A

B

M

5, 7(½)

5, 6, 7(½)

5 For each part, use the information given and state which chord theorem is used. a Given AB = CD and ∠AOB = 70◦ , b Given AB = CD and OF = 7.2 cm, find OE. find ∠DOC. A C

B 70° O

F

O B

D D

C

FLUENCY

5, 6 Example 10

B

E

E A

c Given OZ ⊥ XY, XY = 8 cm and ∠XOY = 102◦ , find XZ and ∠XOZ.

Y

Z X O

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter 2 Geometry

2F 6 The perpendicular bisectors of two different chords of a circle are constructed. Describe where they intersect.

FLUENCY

128

7 Use the information given to complete the following. a Given ∠AOB = ∠COD and CD = 3.5 m, b Given OE = OF and AB = 9 m, find CD. find AB. B D C E C 3.5 m A

110° 110° O

O F D

A

B c Given M is the midpoint of AB, find ∠OMB. d Given ∠AOM = ∠BOM, find ∠OMB. B

A

M

A O

O

M B 8(½), 9

8 Find the size of each unknown angle a◦ . a b

c

a° a°

9, 10

71°

a°

260°

PROBLEM-SOLVING

8

20° d

e

a°

f

a°

18°

a°

9 Find the length OM. (Hint: Use Pythagoras’ theorem.)

O A

M

10 m B

16 m

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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10 In this diagram, radius AD = 5 mm, radius BD = 12 mm and chord CD = 8 mm. Find the exact length of AB, in surd form.

PROBLEM-SOLVING

Measurement and Geometry

C B

A

129

2F

D

Example 11

11, 12

12, 13

11 a

Prove chord theorem 1 in that chords of equal length subtend equal angles at the centre of the circle. b Prove the converse of chord theorem 1 in that if chords subtend equal angles at the centre of the circle then the chords are of equal length.

REASONING

11

12 a Prove that if a radius bisects a chord of a circle then the radius is perpendicular to the chord. b Prove that if a radius bisects the angle at the centre subtended by the chord, then the radius is perpendicular to the chord. 13 In this circle ∠BAO = ∠CAO. Prove AB = AC. (Hint: Construct two triangles.)

A

O

Common chord proof

—

14 For this diagram, prove CD ⊥ AB by following these steps. a Prove ACD ≡ BCD. b Hence, prove ACE ≡ BCE. c Hence, prove CD ⊥ AB.

C

—

14

B C

E

D

ENRICHMENT

B

A

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130

Chapter 2 Geometry

Exploring chord theorems with dynamic geometry 1

Construct a circle with centre O and any given radius.

2 Construct chords of equal length by rotating a chord about the centre by a given angle. 3 Illustrate the four chord properties by constructing line segments, as shown below. Measure corresponding angles and lengths to illustrate the chord properties. • Chord theorem 1: Chords of equal length subtend equal angles at the centre of the circle. • Chord theorem 2: Chords of equal length are equidistant from the centre of the circle. • Chord theorem 3: The perpendicular from the centre of the circle to the chord bisects the chord and the angle at the centre subtended by the chord. • Chord theorem 4: The perpendiculars of every chord of a circle intersect at the centre of the circle. 4 Drag the circle or one of the points on the circle to check that the properties are retained. Chord theorem illustrations Chord theorem 1

Chord theorem 2

3.90 cm

3.68 cm

2.73 cm

67.2°

3.68 cm 67.2°

2.73 cm

O

O

Chord theorem 3

3.90 cm

Chord theorem 4

2.61 cm 51.4° O

51.4°

O 2.61 cm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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131

Measurement and Geometry

2G Angle properties of circles – theorems 1 and 2

10A

The special properties of circles extend to the pairs of angles formed by radii and chords intersecting at the circumference. In this section we explore the relationship between angles at the centre and at the circumference subtended by the same arc.

Let’s start: Discover angle properties – theorems 1 and 2 This activity can be completed with the use of a protractor and pair of compasses, but would be enhanced by using computer dynamic geometry software. •

A

First, construct a circle and include two radii and two chords, as shown. The size of the circle and position of points A, B and C on the circumference can vary. • Measure ∠ACB and ∠AOB. What do you notice? • Now construct a new circle with points A, B and C at different points on the circumference. (If dynamic software is used simply drag the points.) Measure ∠ACB and ∠AOB once again. What do you notice? • Construct a new circle with ∠AOB = 180◦ so AB is a diameter. What do you notice about ∠ACB? Circle theorem 1: Angles at the centre and circumference • The angle at the centre of a circle is twice the angle at a point on the circle subtended by the same arc.

B

O C

Key ideas A 2θ

θ

B For example: 1 A

2

70°

140°

3

A

210° 48° 105°

B

B Circle theorem 2: Angle in a semicircle • The angle in a semicircle is 90◦ . This is a specific case of theorem 1, where ∠ACB is known as the angle in a semicircle.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

96°

A B B

A

C

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132

Chapter 2 Geometry

Example 12 Applying circle theorems 1 and 2 Find the value of the pronumerals in these circles. a b A

B

a°

70°

O

B

C

a°

126°

A C

SOL UTI ON

EX P L A NA TI ON

a

From circle theorem 1, ∠BOC = 2∠BAC.

2a = 126 ∴ a = 63

b ∠ABC is 90◦ . ∴ a + 90 + 70 = 180

AC is a diameter, and from circle theorem 2 ∠ABC = 90◦ .

∴ a = 20

Example 13 Combining circle theorems with other circle properties Find the size of ∠ACB.

C

O 25°

A

SOL UTI ON

B

EX P L A NA TI ON

∠OAB = 25◦ ◦

∠AOB = 180 – 2 × 25

◦

AOB is isosceles.

Angle sum of a triangle is 180◦ .

= 130◦ ∴ ∠ACB = 65◦

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

The angle at the circumference is half the angle at the centre subtended by the same arc.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Measurement and Geometry

1

1–3

2, 3

—

Name another angle that is subtended by the same arc as ∠ABC in these circles. a b c B D C D D B A C

A

UNDERSTANDING

Exercise 2G

133

A C d

A

e

A

f

F

D

B D

F C

B

A

C

C

E E

B D

2 For this circle, O is the centre. a b c d

A

Name the angle at the centre of the circle. Name the angle at the circumference of the circle. If ∠ACB = 40◦ , find ∠AOB using circle theorem 1. If ∠AOB = 122◦ , find ∠ACB using circle theorem 1.

3 For this circle AB is a diameter. a b c d

E

B

What is the size of ∠AOB? What is the size of ∠ACB using circle theorem 2? If ∠CAB = 30◦ , find ∠ABC. If ∠ABC = 83◦ , find ∠CAB.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

B O C

A C O B

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter 2 Geometry

4–6(½)

2G Example 12a

4 Find the value of x in these circles. a b x°

4–7(½)

c

x° x°

20°

100°

4–7(½)

FLUENCY

134

40° d

e

f

125° 120°

x°

225°

x° x°

g

h

x°

i

x°

x°

76° 36°

240°

Example 12b

5 Find the value of x in these circles. a b

c

80°

65° 20°

x°

x°

x°

6 Find the size of both ∠ABC and ∠ABD. a b A 18° A C 20° 68°

D

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

18° 25°

B

D

B

A

c

C

D 37° C

B

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b Find ∠ABC and ∠ADC.

A

D

A

B

A

D

B C

8

8 Find ∠ABC. a B

160°

A

A

b

C

8–9(½)

B

D

A

e

C Example 13

9 Find ∠OAB. a A

B

b

12°

C

c

25°

18° B

A

21°

A

B

e

O 26°

O

O

32°

A

19°

C

C

C d

24° O

20° B

O

B

O

A

O

D

A

f

B

61°

O 80° C

C d

8–9(½)

A

c

171° B

O

O

D

85°

C

115°

C

B

2G

O

O

150°

O

c Find ∠AOD and ∠ABD.

135

PROBLEM-SOLVING

7 a Find ∠ADC and ∠ABC.

FLUENCY

Measurement and Geometry

f

O

B A

65°

57° A

B

48°

B

O

36°

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter 2 Geometry

10

2G 10 a

10, 11

11–13

For the first circle shown, use circle theorem 1 to find ∠ACB.

A

140°

B

O

REASONING

136

C b For the second circle shown, use circle theorem 1 to find ∠ACB. c For the second circle what does circle theorem 2 say about ∠ACB? d Explain why circle theorem 2 can be thought of as a special case of circle theorem 1.

A

180° O

B

C 11 The two circles shown illustrate circle theorem 1 for both a minor arc and minor arc AB a major arc. B a When a minor arc is used, answer true or false. A 2a° i ∠AOB is always acute. ii ∠AOB can be acute or obtuse. iii ∠ACB is always acute. iv ∠ACB can be acute or obtuse. b When a major arc is used, answer true or false. i ∠ACB can be acute. ii ∠ACB is always obtuse. iii The angle at the centre (2a◦ ) is a reflex angle. iv The angle at the centre (2a◦ ) can be obtuse.

O

a°

C major arc AB 2a° O a°

A

B

C B

12 Consider this circle. a Write reflex ∠AOC in terms of x. b Write y in terms of x.

x° y°

A

C

O

13 Prove circle theorem 1 for the case illustrated in this circle by following these steps and letting ∠OCA = x◦ and ∠OCB = y◦ . a Find ∠AOC in terms of x, giving reasons. b Find ∠BOC in terms of y, giving reasons. c Find ∠AOB in terms of x and y. d Explain why ∠AOB = 2∠ACB.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

A O x° y°

B

C

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Measurement and Geometry

—

—

14, 15

14 Question 13 sets out a proof for circle theorem 1 using a given illustration. Now use a similar technique for these cases. a Prove ∠AOB = 2∠ACB; b Prove reflex ∠AOB = 2∠ACB; i.e. prove ∠AOB = 2x◦ . i.e. prove reflex ∠AOB = 2(x + y)◦ .

ENRICHMENT

Proving all cases

137

2G

C x°

O

O

A A

B

x° y°

B

C

c Prove ∠AOB = 2∠ACB; i.e. prove ∠AOB = 2y◦ .

A O x° y° B C

A

15 Prove circle theorem 2 by showing that x + y = 90.

O

C

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

y°

x° B

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138

Chapter 2 Geometry

Exploring circle theorems with dynamic geometry software 1

Construct a circle with centre O and any given radius.

2 Illustrate the four angle properties by constructing line segments, as shown. Measure corresponding angles to illustrate the angle properties. • Circle theorem 1: The angle at the centre of a circle is twice the angle at a point on the circle subtended by the same arc. • Circle theorem 2: The angle in a semicircle is 90◦ . • Circle theorem 3: Angles at the circumference of a circle subtended by the same arc are equal. • Circle theorem 4: Opposite angles in a cyclic quadrilateral are supplementary. 3 Drag the circle or one of the points on the circle to check that the properties are retained. Circle theorem illustrations Circle theorem 1

Circle theorem 2

90.0° 114.8° O

O

57.4°

Circle theorem 3

Circle theorem 4

77.9°

104.8° 32.1°

O

O 32.1°

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

102.1°

75.2°

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Measurement and Geometry

2H Angle properties of circles – theorems 3 and 4

139

10A

A

When both angles are at the circumference, there are two more important properties of pairs of angles in a circle to consider.

D

You will recall from circle theorem 1 that in this circle ∠AOD = 2∠ABD and also ∠AOD = 2∠ACD. This implies that ∠ABD = ∠ACD, which is an illustration of circle theorem 3 – angles at the circumference subtended by the same arc are equal.

O B

The fourth theorem relates to cyclic quadrilaterals, which have all four vertices sitting on the same circle. This also will be explored in this section.

C

Let’s start: Discover angle properties – theorems 3 and 4 Once again, use a protractor and a pair of compasses for this exercise or use computer dynamic geometry software. • Construct a circle with four points at the circumference, as shown. A • Measure ∠ABD and ∠ACD. What do you notice? Drag A, B, C or D and compare the angles.

B • •

Now construct this cyclic quadrilateral (or drag point C if using dynamic geometry software). Measure ∠ABC, ∠BCD, ∠CDA and ∠DAB. What do you notice? Drag A, B, C or D and compare angles.

A

D

C D

B C

The study of angle properties of circles by ancient Greek mathematicians laid the foundations of trigonometry. Much of the work was done by Ptolemy, who lived in the city of Alexandria, Egypt. The photo shows ancient monuments that existed in Ptolemy’s time and are still standing in Alexandria today. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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140

Chapter 2 Geometry

Key ideas

D

Circle theorem 3: Angles at the circumference • Angles at the circumference of a circle subtended by the same arc are equal. – As shown in the diagram, ∠C = ∠D but note also that ∠A = ∠B.

θ

C θ

A B

Circle theorem 4: Opposite angles in cyclic quadrilaterals • Opposite angles in a cyclic quadrilateral are supplementary (sum to 180◦ ). A cyclic quadrilateral has all four vertices sitting on the same circle. a + c = 180

b°

c°

a° d°

b + d = 180

Example 14 Applying circle theorems 3 and 4 Find the value of the pronumerals in these circles. a b a°

125°

30°

128° b°

a°

SOL UTI ON

EX P L A NA TI ON

a a = 30

The a◦ and 30◦ angles are subtended by the same arc. This is an illustration of circle theorem 3.

b

The quadrilateral is cyclic, so opposite angles sum to 180◦ .

a + 128 = 180 ∴ a = 52

This is an illustration of circle theorem 4.

b + 125 = 180 ∴ b = 55

1

1–3

2, 3

Name another angle that is subtended by the same arc as ∠ABD. a A b c B A D

—

UNDERSTANDING

Exercise 2H

E

A

C B

C

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

B E

D

D C

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

A

2 For this circle, answer the following. a Name two angles subtended by arc AD. b Using circle theorem 3, state the size of ∠ACD if ∠ABD = 85◦ . c Name two angles subtended by arc BC. d Using circle theorem 3, state the size of ∠BAC if ∠BDC = 17◦ .

D B

4 Find the value of x in these circles. a b

A

C

71°

D

4–5(½)

4–5(½)

c

x° 110°

20° x°

d

e

x°

f

x°

35°

40° 22.5°

Example 14b

C

63°

x°

37°

2H

FLUENCY

Example 14a

141

B

3 Circle theorem 4 states that opposite angles in a cyclic quadrilateral are supplementary. a What does it mean when we say two angles are supplementary? b Find ∠ABC. c Find ∠BCD. d Check that ∠ABC + ∠BCD + ∠CDA + ∠DAB = 360◦ .

4–5(½)

UNDERSTANDING

Measurement and Geometry

5 Find the value of the pronumerals in these circles. a b 120°

x°

x°

x°

c

x° 150° 70°

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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142

Chapter 2 Geometry

d

e

92° 85°

FLUENCY

2H f

y° x°

57°

108°

x°

95° x°

6 Find the value of x. a

b

6–7(½)

6–7(½)

PROBLEM-SOLVING

6

c

x° 18°

x°

31° 100°

110°

x°

27° d

e

f

53°

x° 33°

20° 50°

x°

58°

x°

74°

85°

g

h

i

147° x°

108° 150°

x°

100°

x°

7 Find the values of the pronumerals in these circles. a b

a°

a°

36°

30°

b°

d

c

b° 54°

50°

75°

a°

c°

e

a° 52° b°

f

41° a° 73°

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b°

b°

62°

a° b°

100° 270° 85°

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Measurement and Geometry

8, 9

8 ∠ABE is an exterior angle to the cyclic quadrilateral BCDE. a If ∠ABE = 80◦ , find ∠CDE. b If ∠ABE = 71◦ , find ∠CDE. c Prove that ∠ABE = ∠CDE using circle theorem 4.

9, 10

A

B C

E

REASONING

8

143

2H

D A

9 Prove that opposite angles in a cyclic quadrilateral are supplementary by following these steps. a Explain why ∠ACD = x◦ and ∠DAC = y◦ . b Prove that ∠ADC = 180◦ – (x + y)◦ . c What does this say about ∠ABC and ∠ADC?

x° y°

B

D

C 10 If ∠BAF = 100◦ , complete the following. a Find: i ∠FEB ii ∠BED iii ∠DCB

B

C

E

D

A 100°

b Explain why AF || CD.

F

—

—

11

11 Consider a triangle ABC inscribed in a circle. The construction line BP is a diameter and PC is a chord. If r is the radius, then BP = 2r.

B c A

ENRICHMENT

The sine rule

a O

b P

C

a What can be said about ∠PCB? Give a reason. b What can be said about ∠A and ∠P? Give a reason. c If BP = 2r, use trigonometry with ∠P to write an equation linking r and a. a d Prove that 2r = , giving reasons. sin A

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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144

Chapter 2 Geometry

2I Tangents

EXTENDING

When a line and a circle are drawn, three possibilities arise: they could intersect 0, 1 or 2 times.

0 points

1 point

2 points

If the line intersects the circle once then it is called a tangent. If it intersects twice it is called a secant. The surface of the road is a tangent to the tyre.

Let’s start: From secant to tangent This activity is best completed using dynamic computer geometry software. • Construct a circle with centre O and a secant line that intersects at A and B. Then measure ∠BAO. • Drag B to alter ∠BAO. Can you place B so that line AB is a tangent? In this case what is ∠BAO?

A O

B

Key ideas

A tangent is a line that touches a circle at a point called the point of contact.

radius

point of contact

tangent • A tangent intersects the circle exactly once. • A tangent is perpendicular to the radius at the point of contact. • Two different tangents drawn from an external point to the circle create line segments of equal length. A secant is a line that cuts a circle twice.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

A P B

PA = PB

secant

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Measurement and Geometry

Key ideas

B

Alternate segment theorem: The angle between a tangent and a chord is equal to the angle in the alternate segment. ∠APY = ∠ABP and ∠BPX = ∠BAP

X

A

b° a° a° b° P

145

Y

Example 15 Finding angles with tangents Find the value of a in these diagrams that include tangents. a b

A

O P

67°

a°

O 220°

A

a°

B

B SOL UTI ON

EX P L A NA TI ON

a ∠BAO = 90◦ a + 90 + 67 = 180

BA is a tangent, so OA ⊥ BA. The sum of the angles in a triangle is 180◦ .

∴ a = 23 b ∠PAO = ∠PBO = 90◦ Obtuse ∠AOB = 360◦ – 220◦ = 140◦ a + 90 + 90 + 140 = 360

PA ⊥ OA and PB ⊥ OB. Angles in a revolution sum to 360◦ . Angles in a quadrilateral sum to 360◦ .

∴ a = 40

Example 16 Using the alternate segment theorem In this diagram XY is a tangent to the circle. a Find ∠BPX if ∠BAP = 38◦ . b

A B

Find ∠ABP if ∠APY = 71◦ .

Y

P

SOL UTI ON

EX P L A NA TI ON

a ∠BPX = 38◦

The angle between a tangent and a chord is equal to the angle in the alternate segment.

X

A 38° 71° 71° b ∠ABP = 71◦

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Y

P

B 38° X

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146

Chapter 2 Geometry

1

1–3

2, 3

—

A

a How many times does a tangent intersect a circle? b At the point of contact, what angle does the tangent make with the radius? c If AP is 5 cm, what is the length BP in this diagram?

P

UNDERSTANDING

Exercise 2I

B 2 For this diagram use the alternate segment theorem and name the angle that is: a equal to ∠BPX b equal to ∠BAP c equal to ∠APY d equal to ∠ABP

A

B

Y

X P

3 What is the interior angle sum for: a a triangle? b a quadrilateral?

Example 15a

4 Find the value of a in these diagrams that include tangents. a b B O

B

Example 15b

4, 5(½), 6–8

a°

A

O

a° O

a°

A

O

B

O

a°

A

a°

P

B B

d

205°

152°

O

B c

20°

a°

A

5 Find the value of a in these diagrams that include two tangents. a b A A 130°

P

B

c

28°

71°

4, 5(½), 6–8

FLUENCY

4, 5(½), 6, 7

243° O A

a°

P

P

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Example 16

A

6 In this diagram, XY is a tangent to the circle. Use the alternate segment theorem to find:

B

a ∠PAB if ∠BPX = 50◦ b ∠APY if ∠ABP = 59◦

Y

7 Find the value of a, b and c in these diagrams involving tangents. a b c

c° 73°

b°

P

65°

b° a° c° 83°

42°

a°

FLUENCY

Measurement and Geometry

147

2I

X

c° a°

69°

b°

26°

8 Find the length AP if AP and BP are both tangents. a b P 5 cm B

A P 11.2 cm

9–10(½)

9 Find the value of a. All diagrams include one tangent line. a b 25° 73° a° O

9–10(½)

10(½), 11

c

PROBLEM-SOLVING

B

A

a°

24°

a° d

e

31°

f

a°

O

a°

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

32° O

O

a°

38°

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter 2 Geometry

2I g

h

18° a°

a°

O

i

122°

a°

O O 30°

PROBLEM-SOLVING

148

10 Find the value of a in these diagrams involving tangents. a b 55°

a°

a° 80°

c

d

a°

O

70°

73°

a° e

f

38° 117°

a° O a° 11 Find the length of CY in this diagram.

A 3 cm

Z

C

B

X

7 cm

Y

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Measurement and Geometry

12, 13

13–15

REASONING

12, 14

12 Prove that AP = BP by following these steps. a Explain why OA = OB. b What is the size of ∠OAP and ∠OBP? c Hence, prove that OAP ≡ OBP. d Explain why AP = BP.

O

A

149

2I

B

P B

13 Prove the alternate angle theorem using these steps. First, let ∠BPX = x◦ , then give reasons at each step.

A

a Write ∠OPB in terms of x. b Write obtuse ∠BOP in terms of x. c Use circle theorem 1 from angle properties of a circle to write ∠BAP in terms of x.

O x°

X

P

Y A

14 These two circles touch with a common tangent XY. Prove that AB || DC. You may use the alternate segment theorem.

B P Y

X D

C

15 PT is a common tangent. Explain why AP = BP.

T A

B

P

—

16 In this diagram, ABC is right angled, AC is a diameter and PM is a tangent at P, where P is the point at which the circle intersects the hypotenuse. a Prove that PM bisects AB; i.e. that AM = MB. b Construct this figure using dynamic geometry software and check the result. Drag A, B or C to check different cases.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

—

16

ENRICHMENT

Bisecting tangent

C P O

A

M

B

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150

Chapter 2 Geometry

2J Intersecting chords, secants and tangents

EXTENDING

In circle geometry, the lengths of the line segments (or intervals) formed by intersecting chords, secants or tangents are connected by special rules. There are three situations in which this occurs: 1

intersecting chords

2

intersecting secant and tangent

3

intersecting secants.

Let’s start: Equal products

Architects use circle and chord geometry to calculate the dimensions of constructions, such as this glass structure.

B

Use dynamic geometry software to construct this figure and then measure AP, BP, CP and DP. • Calculate AP × CP and BP × DP. What do you notice? • Drag A, B, C or D. What can be said about AP × CP and BP × DP for any pair of intersecting chords?

A

P

C

D

Key ideas

A

When two chords intersect as shown, then AP × CP = BP × DP or ac = bd.

B

a P d

D

b c C

When two secants intersect at an external point P as shown, then AP × BP = DP × CP.

A P

D

When a secant intersects a tangent at an external point as shown, then AP × BP = CP2 .

B C

B A P

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

C

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Measurement and Geometry

151

Example 17 Finding lengths using intersecting chords, secants and tangents Find the value of x in each figure. a b 1 x 9m 3 5 8m

c

19 m xm

x cm

5 cm 7 cm

SOL UTI ON

EX P L A NA TI ON

a

Equate the products of each pair of line segments on each chord.

x×3 = 1×5 3x = 5 x=

b

5 3

8 × (x + 8) = 9 × 28

Multiply the entire length of the secant (19 + 9 = 28 and x + 8) by the length from the external point to the first intersection point with the circle. Then equate both products. Expand brackets and solve for x.

8x + 64 = 252 8x = 188 x= = c

188 8 47 2

5 × (x + 5) = 72

Square the length of the tangent and then equate with the product from the other secant.

5x + 25 = 49 5x = 24 24 5

Exercise 2J 1

1–3

3

State these lengths for the given diagram. a AP b DP c AC d BD

B

2 A

2 Solve these equations for x. a x×2 = 7×3 c (x + 3) × 7 = 6 × 9

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

—

C 4 P 3

D 6

UNDERSTANDING

x=

b 4x = 5 × 2 d 7(x + 4) = 5 × 11

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152

Chapter 2 Geometry

3 Complete the rules for each diagram. a b D A

P

A

C

×

A

D C

AP ×

=

×

AP ×

4–6(a, b) Example 17a

Example 17b

4 Find the value of x in each figure. a b 2 4 x 10

4

c

x

10 cm

x 16

11 cm

c

11 cm

9 cm x cm

17 m

6 Find the value of x in each figure. a b

6 cm

7

8

23 m

2

4–7(b)

15

5 Find the value of x in each figure. a b xm 9m 14 m xm 8m

x cm

=

4–6(a, c)

5

18 m

Example 17c

C

P

B AP × CP =

B

c

FLUENCY

B

P

UNDERSTANDING

2J

c

xm 14 mm

5m

x mm

21 mm

3m

√ 7 Find the exact value of x, in surd form. For example, 7. a b x 8 5

4

x 7

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Measurement and Geometry

8(½)

c

8 11 10

d

x

9

e

x+3 7

f

20

8

7 x+5

x

7

19

9 For each diagram, derive the given equations. a x2 + 5x – 56 = 0 b x2 + 11x – 220 = 0

8

2J

5 x+1 x+1 13

x−2

12

x

PROBLEM-SOLVING

b

8(½), 9

x

11

c x2 + 23x – 961 = 0

10

x

31

12 23

10, 11

10–12

12–14

REASONING

8 Find the exact value of x. a 3 5 x 7

8(½)

153

P

10 Explain why AP = BP in this diagram, using your knowledge from this section.

B

A

B

11 In this diagram AP = DP. Explain why AB = DC.

A P

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

D

C

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154

Chapter 2 Geometry

12 Prove that AP × CP = BP × DP by following these steps. a What can be said about the pair of angles ∠A and ∠D and also about the pair of angles ∠B and ∠C? Give a reason. b Prove ABP ||| DCP. c Complete: AP .... = .... CP d Prove AP × CP = BP × DP.

REASONING

2J C

B P A

D

B

13 Prove that AP × BP = DP × CP by following these steps. a Consider PBD and PCA. What can be said about ∠B and ∠C? Give a reason. b Prove PBD ||| PCA. c Prove AP × BP = DP × CP.

A

C D

P 14 Prove that AP × BP = CP2 by following these steps. a Consider BPC and CPA. Is ∠P common to both triangles? b Explain why ∠ACP = ∠ABC. c Prove BPC ||| CPA. d Prove AP × BP = CP2 .

B A C P

—

—

15

15 Two touching circles have radii r1 and r2 . The horizontal distance between their centres is d. Find a rule for d in terms of r1 and r2 .

r1

ENRICHMENT

Horizontal wheel distance

r2 d

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Measurement and Geometry

155

Investigation Some special points of triangles and Euler’s line From Year 9, you may recall finding the circumcentre and the centroid of a triangle. Here we will review the construction of these two points and consider in further detail some of their properties, as well as their relationship with a third point – the orthocentre. This is best done using a dynamic computer geometry software package. The circumcentre of a triangle The perpendicular bisectors of each of the three sides of a triangle meet at a common point called the circumcentre. B A

N

C a

Construct and label a triangle ABC and measure each of its angles.

b

Construct the perpendicular bisector of each side of the triangle.

c

Label the point of intersection of the lines N. This is the circumcentre.

d

By dragging the points of your triangle, observe what happens to the location of the circumcentre. Can you draw any conclusions about the location of the circumcentre for some of the different types of triangles; for example, equilateral, isosceles, right-angled or obtuse?

e

Construct a circle centred at N with radius NA. This is the circumcircle. Drag the vertex A to different locations. What do you notice about vertices B and C in relation to this circle?

The centroid of a triangle The three medians of a triangle intersect at a common point called the centroid. A median is the line drawn from a vertex to the midpoint of the opposite side. a

Construct and label a new triangle ABC and measure each of its angles.

b

Mark the midpoint of each side of the triangle and construct a segment from each vertex to the midpoint of the opposite side.

c

Observe the common point of intersection of these three medians and label this point R – the centroid. Point R is the centre of gravity of the triangle.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

A b

c R B

a

C

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156

Chapter 2 Geometry

d

Label the points of your triangle as shown and complete the table below by measuring the required lengths. Drag the vertices of your triangle to obtain different triangles to use for the table. Lengths Triangle 1 Triangle 2 Triangle 3 Triangle 4

AR

Ra

BR

Rb

CR

Rc

e

From your table, what can you observe about how far along each median the centroid lies?

f

Explain why each median divides the area of triangle ABC in half. What can be said about the area of the six smaller triangles formed by the three medians?

The orthocentre of a triangle The three altitudes of a triangle intersect at a common point called the orthocentre. An altitude of a triangle is a line drawn from a vertex to the opposite side of the triangle, meeting it at right angles. a

Construct a triangle ABC and measure each angle.

b

For each side, construct a line that is perpendicular to it and that passes through its opposite vertex.

c

Label the point where these three lines intersect as O. This is the orthocentre.

A

O C B

d

By dragging the vertices of the triangle, comment on what happens to the location of the orthocentre for different types of triangles.

e

Can you create a triangle for which the orthocentre is outside the triangle? Under what circumstances does this occur?

All in one a

Construct a large triangle ABC and measure the angles. On this one triangle use the previous instructions to locate the circumcentre (N), the centroid (R) and the orthocentre (O).

b

By dragging the vertices, can you make these three points coincide? What type of triangle achieves this?

c

Construct a line joining the points N and R. Drag the vertices of the triangle. What do you notice about the point O in relation to this line?

d

The line in part c is called Euler’s line. Use this line to determine the ratio of the length NR to the length RO.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

Up for a challenge? If you get stuck on a question, check out the 'Working with unfamiliar problems' poster at the end of the book to help you.

Problems and challenges 1

In a triangle ABC, angle C is a right angle, D is the midpoint of AB and DE is perpendicular to AB. The length of AB is 20 and the length of AC is 12. What is the area of triangle ACE?

A

D

12

C 2

20

E

B

A

In this diagram, AB = 15 cm, AC = 25 cm, BC = 30 cm and ∠AED = ∠ABC. If the perimeter of ADE is 28 cm, find the lengths of BD and CE.

E D C

B 3

157

C

Other than straight angles, name all the pairs of equal angles in the diagram shown.

B D F O A E 4

A person stands in front of a cylindrical water tank and has a viewing angle of 27◦ to the sides of the tank. What percentage of the circumference of the tank can they see?

5

An isosceles triangle ABC is such that its vertices lie on the circumference of a circle. AB = AC and the chord from A to the point D on the circle intersects BC at E. Prove that AB2 – AE2 = BE × CE.

6

D, E and F are the midpoints of the three sides of ABC. The straight line formed by joining two midpoints is parallel to the third side and half its length. A a Prove ABC ||| FDE. GHI is drawn in the same way such that G, H and I are the midpoints of the sides of DEF.

b c

Find the ratio of the area of: i ABC to FDE

ii ABC to HGI

Hence, if ABC is the first triangle drawn, what is the ratio of the area of ABC to the area of the nth triangle drawn in this way?

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

H

E I B

D G

F

C

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Chapter summary

158

Chapter 2 Geometry

Similar figures All corresponding angles are equal, corresponding sides are in the same ratio; i.e. same shape but different in size. Tests for similar triangles: SSS, SAS, AAA, RHS. Written as ABC III DEF or ABC ∼ DEF.

Angle properties of circles (10A)

Polygons Angle sum of polygon S = 180(n − 2), where n is the number of sides.

2θ

These triangles are identical, written ABC ≡ DEF. Tests for congruence are SSS, SAS, AAS and RHS.

θ

Geometry

B AB is a chord.

a°

d°

E O C

O B A

D

F

B

Tangents (Ext) O

O

point of contact tangent

Intersecting chords, secants, tangents (Ext)

B

b

C

P

AP × BP = CP × DP

B

D

B

P

A

CP 2

B

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

A

P

A

secant

d

OAB is isosceles given OA and OB are radii.

O

Circle theorem 4 Perpendicular bisectors of every chord of a circle intersect at the centre.

c

c°

A

Circle theorem 3 Perpendicular from centre to chord bisects chord and angle at 0.

a

Opposite angles in cyclic quadrilaterals are supplementary. a + c = 180 b + d = 180

b°

A

Circle theorem 1 Chords of equal length subtend equal angles.

Angles at circumference subtended by the same arc are equal.

θ

Circles and chords (10A)

ab = cd

The angle in a semicircle is 90°.

O

Congruent triangles

Circle theorem 2 If AB = CD, then OE = OF.

Angle at centre is twice angle at the circumference subtended by the same arc.

θ O

C

AP × BP = CP is a tangent.

X

°

P

°

A Y

A tangent touches a circle once and is perpendicular to the radius at point of contact.

Tangents PA and PB have equal length; i.e. PA = PB. Alternate segment theorem: angle between tangent and chord is equal to the angle in the alternate segment.

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Multiple-choice questions 38pt 2A

38pt 2B

1

The value of a in the polygon shown is: A 46 B 64 C 72 D 85 E 102

112°

122°

a°

2a°

2 The test that proves that ABD ≡ CBD is:

A

C D

159

Chapter review

Measurement and Geometry

B A RHS 38pt 2D

B SAS

C SSS

D AAA

E AAS

D 5.5

E 5.2

3 The value of x in the diagram shown is:

8.8 9.6

x 6 A 4.32 38pt 2F 10A

B 4.5

10A

38pt 2G 10A

C 3.6

4 The name given to the dashed line in the circle with centre O is:

O A a diameter D a tangent

38pt 2F

7.2

B a minor arc E a secant

C a chord

5 A circle of radius 5 cm has a chord 4 cm from the centre of the circle. The length of the chord is: A 4.5 cm B 6 cm C 3 cm D 8 cm E 7.2 cm 6 The values of the pronumerals in the diagram are: A a = 55, b = 35 B a = 30, b = 70 C a = 70, b = 35 D a = 55, b = 70 E a = 40, b = 55

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b°

O

70°

a°

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Chapter review

160

Chapter 2 Geometry

38pt 2H 10A

38pt 2I Ext

38pt 2I Ext

7 A cyclic quadrilateral has one angle measuring 63◦ and another angle measuring 108◦ . Another angle in the cyclic quadrilateral is: A 63◦ B 108◦ C 122◦ D 75◦ E 117◦

x cm

8 For the circle shown at right with radius 5 cm, the value of x is: A 13 B 10.9 C 12 D 17 E 15.6

O 12 cm

9 By making use of the alternate segment theorem, the value of ∠APB is: A 50◦ B 45◦ C 10◦ ◦ ◦ D 52 E 25

B

A

58° 70°

X 38pt 2J Ext

P

10 The value of x in the diagram is: A 7.5 B 6 C 3.8 D 4 E 5

Y

3 x 12

Short-answer questions 38pt 2A

1

Determine the value of each pronumeral. a 85° 150° x°

b

y°

y°

y°

y° y°

c

y°

d

325°

118° y°

x°

x° 270° 97° 92°

38pt 2A

2 Find the value of ∠ABC by adding a third parallel line. a b D A B

C

39°

86° 126° D

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

C

B E

A

73°

E

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Measurement and Geometry

3 Prove that each pair of triangles is congruent, giving reasons. a B F

D

A C b

E A

c

A

D B 38pt 2C

38pt 2E

Chapter review

38pt 2B

C

D

161

B C A

4 Complete these steps to prove that if one pair of opposite sides is equal and parallel in a quadrilateral, then it is a parallelogram. a Prove ABC ≡ CDA, giving reasons. b Hence, prove AD || BC.

B

D

C

5 In each of the following, identify pairs of similar triangles by proving similarity, giving reasons, and then use this to find the value of x. a b 9.8 C F A 14.7 B E 100° 100° B 7 E 13 x 10.5 10 3 x A D D C 8 c d D E

x A 4.2

10 55° 7

35°

3 C 6.3

D

A

x

B

C

B 38pt 2F 10A

6 Find the value of each pronumeral and state the chord theorem used. a b 10 cm c a° x cm O O 65° 7 cm

10 cm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

O x 12

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Chapter review

162

Chapter 2 Geometry

38pt 2G/H 10A

7 Use the circle theorems to help find the values of the pronumerals. a b c

65°

50°

a°

a°

O

b°

d

e

f

b°

95° O

230°

36°

38pt 2I

110° 80°

a°

55° a°

a°

b°

8 Find the value of the pronumerals in these diagrams involving tangents and circles. a

Ext

b

O x°

d cm

32°

65°

a° O

z°

c

b°

O

c° 36°

8 cm

t°

y° 38pt 2J Ext

9 Find the value of x in each figure. a b

6

4 2

x 4

3

c

x

x

3

6 4

Extended-response questions 1

The triangular area of land shown is to be divided into two areas such that AC || DE. The land is to be divided so that AC : DE = 3 : 2. a Prove that ABC ||| DBE. C b If AC = 1.8 km, find DE. E c If AD = 1 km and DB = x km: i Show that 2(x + 1) = 3x. ii Solve for x. d For the given ratio, what percentage of the land area does D A DBE occupy? Answer to one decimal place.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

B

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10A

2 The diagram below shows two intersecting circles sharing common tangents AR and AS. The distance between the centres O and O of the two circles is 15 cm. Other measurements are as shown. Given that the two centres O and O and the point A are in a straight line, complete the following. R

Q

53.13° T

O

8 cm

A

O′

4 cm P

a b c d

S Find the values of these angles. i ∠ROS ii ∠RAS Use the rule for intersecting secants and tangents to help find the diameter of the smaller circle. Hence, what is the distance from A to O? By first finding AR, determine the perimeter of AROS. Round to one decimal place.

163

Chapter review

Measurement and Geometry

Two water ripples intersecting.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter

3

Indices and surds

What you will learn

Australian curriculum

3A 3B 3C 3D 3E 3F 3G 3H 3I 3J 3K 3L

NUMBER AND ALGEBRA

Irrational numbers and surds (10A) Adding and subtracting surds (10A) Multiplying and dividing surds (10A) Binomial products (10A) Rationalising the denominator (10A) Review of index laws (Consolidating) Negative indices Scientific notation (Consolidating) Rational indices (10A) Exponential growth and decay Compound interest Comparing interest using technology 3M Exponential equations (10A)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Patterns and algebra Simplify algebraic products and quotients using index laws. Money and financial mathematics Connect the compound interest formula to repeated applications of simple interest using appropriate digital technologies. Real numbers (10A) Define rational and irrational numbers and perform operations with surds and fractional indices. 16x16 Linear and non-linear relationships (10A) Solve simple exponential equations.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Online resources • Chapter pre-test • Videos of all worked examples • Interactive widgets • Interactive walkthroughs • Downloadable HOTsheets • Access to HOTmaths Australian Curriculum courses

Investment returns Does an average investment return of 15% sound good to you? Given this return is compounded annually, a $10 000 investment would grow to over $40 000 after 10 years. This is calculated by multiplying the investment total by 1.15 (to return the original amount plus the 15%) for each year. Using indices, the total investment value after n years would be given by Value = 10 000 × 1.15n . This is an example of an exponential relation that uses indices to connect variables. We can use such a rule to introduce the set of special numbers called surds.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

If, for example, you wanted to calculate the average investment return that delivers $100 000 after 10 years from a $10 000 investment, then you would need to calculate x in the equation 

100 000 = 10 000 ×  1+ This gives x =

(

10

)

x   100 

10

.

10 − 1 × 100 ≈ 25.9,

representing an annual return of 25.9%. Indices and surds are commonplace in the world of numbers, especially where money is involved!

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166

Chapter 3 Indices and surds

3A Irrational numbers and surds

10A

You will recall that when using Pythagoras’ theorem to find unknown lengths in 5 1 right-angled triangles, many answers expressed in exact form are surds. The length of √ the hypotenuse in this triangle, for example, is 5, which is a surd. 2 √  A surd is a number that uses a root sign , sometimes called a radical sign. They are irrational numbers, a meaning that they cannot be expressed as a fraction in the form , where a and b are integers and b π 0. b Surds, together with other irrational numbers such as pi, and all rational numbers (fractions) make up the entire set of real numbers, which can be illustrated as a point on a number line.

−3 (rational)

−3

−2

1

− 2 (rational)

√2 (irrational) π (irrational)

−1

1

0

2

3

4

Let’s start: Constructing surds √ Someone asks you: ‘How do you construct a line that is 10 cm long?’ Use these steps to answer this question.

C

• First, draw a line segment AB that is 3 cm in length. • Construct segment BC so that BC = 1 cm and AB ⊥ BC. You may wish to use a set square or pair of compasses. • Now connect point A and point C and measure the length of the segment. • Use Pythagoras’ theorem to check the length of AC.

1 cm A

3 cm

B

Use this idea to construct line segments with the following lengths. You may need more than one triangle for parts d to f. √ √ √ a 2 b 17 c 20 √ √ √ d 3 e 6 f 22

Key ideas

All real numbers can be located as a point on a number line. Real numbers include: • rational numbers (i.e. numbers that can be expressed as fractions) 3 4 ˙ 0.19 For example: , – , –3, 1.6, 2.7, 7 39 The decimal representation of a rational number is either a terminating or recurring decimal. • irrational numbers (i.e. numbers that cannot be expressed as fractions) √ √ √ For example: 3, –2 7, 12 – 1, p , 2p – 3 The decimal representation of an irrational number is an infinite non-recurring decimal.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Number and Algebra

√  Surds are irrational numbers that use a root sign . √ √ √ √ • For example: 2, 5 11, – 200, 1 + 5 √ √ √ • These numbers are not surds: 4 (= 2), 3 125 (= 5), – 4 16 (= –2). √ The nth root of a number x is written n x. √ √ • If n x = y then yn = x. For example: 5 32 = 2 since 25 = 32.

167

Key ideas

The following rules √ apply to surds. √ • ( x)2 = x and x2 = x when x ≥ 0. √ √ √ • xy = x × y when x ≥ 0 and y ≥ 0. √  x x = √ when x ≥ 0 and y > 0. • y y When a factor of a number is a perfect square we call that factor a square factor. Examples of perfect squares are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, . . . When simplifying surds, look for square factors of the number under the root sign and then use √ √ √ a × b = a × b.

Example 1 Defining and locating surds Express each number as a decimal and decide if they are rational or irrational. Then locate all the numbers on the same number line. √ 3 a – 3 b 137% c 7 SOL UTI ON

EX P L A NA TI ON

√ a – 3 = –1.732050807. . . √ – 3 is irrational.

Use a calculator to express as a decimal. The decimal does not terminate and there is no recurring pattern.

137 = 1.37 100 137% is rational.

b 137% =

c

137% is a fraction and can be expressed using a terminating decimal. 3 is an infinitely recurring decimal. 7

3 = 0.428571 7 3 is rational. 7

Use the decimal equivalents to locate each number on the real number line. 3 7

−√3 −2

−1

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

0

1.37 1

2

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168

Chapter 3 Indices and surds

Example 2 Simplifying surds Simplify the following. √ √ a 32 b 3 200 SO L U T I O N a

√ √ 32 = 16 × 2 √ √ = 16 × 2 √ =4 2

√ √ 3 200 = 3 100 × 2 √ √ = 3 × 100 × 2 √ = 3 × 10 × 2 √ = 30 2 √ √ 5 40 5 4 × 10 c = 6 6 √ √ 5 × 4 × 10 = √6 5 10 10  = 63 √
5 10 = 3 √ √ 75 75 = √ d 9 9 √ √ 25 × 3 5 3 = = √ 3 9

b

c

√ 5 40 6

d

√

75 9

EX P L A N A T I O N When simplifying, choose the highest square factor of 32 (i.e. 16 rather than 4) as there is less work to do to arrive at the same answer. Compare with √ √ √ √ √ √ 32 = 4 × 8 = 2 8 = 2 4 × 2 = 2 × 2 2 = 4 2 Select the appropriate factors of 200 by finding its highest square factor: 100. √ √ √ Use x × y = x × y and simplify.

Select the appropriate factors of 40. The highest square factor is 4.

Cancel and simplify.

Use

√

√ x x =√ . y y

Then select the factors of 75 that include a square number and simplify.

Example 3 Expressing as a single square root of a positive integer Express these surds as a square root of a positive integer. √ √ a 2 5 b 7 2 SO L U T I O N a

b

√ √ √ 2 5 = 4× 5 √ = 20 √ √ √ 7 2 = 49 × 2 √ = 98

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

EX P L A N A T I O N √ Write 2 as 4 and then combine the two surds √ √ √ using x × y = xy. √ Write 7 as 49 and combine.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Number and Algebra

1

1, 2(½)

2(½)

—

Choose the correct word(s) from the words given in red to make the sentence true. a b c d

A number that cannot be expressed as a fraction is a rational/irrational number. A surd is an irrational number that uses a root/square symbol. The decimal representation of a surd is a terminating/recurring/non-recurring decimal. √ 25 is a surd/rational number.

UNDERSTANDING

Exercise 3A

169

2 Write down the highest square factor of these numbers. For example, the highest square factor of 45 is 9. a 20 b 18 c 125 d 24 e 48 f 96 g 72 h 108

Example 1

Example 2a

Example 2b,c

3–8(½)

3–8(½)

3 Express each number as a decimal and decide if it is rational or irrational. Then locate all the numbers on the same number line. √ 2 c d –124% a 5 b 18% 5 √ √ 5 e 1 f – 2 g 2 3 h p 7 4 Decide if these numbers are surds. √ √ a 7 b 2 11 √ √ 3 9 –5 3 e f 2 2 5 Simplify the following surds. √ √ a 12 b 45 √ √ e 75 f 500 √ √ i 128 j 360 6 Simplify the following. √ √ a 2 18 b 3 20 √ 4 125

√ e 3 98

f

√ 24 i 4 √ 3 44 m 2 √ 6 75 q 20

√ 54 j 12 √ 5 200 n 25 √ 4 150 r 5

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

√ c 2 25 √ g 1– 3

√ d –5 144 √ √ h 2 1+ 4

√ c 24 √ g 98 √ k 162

√ d 48 √ h 90 √ l 80

√ c 4 48 √ 45 g 3 √ 80 k 20 √ 2 98 o 7 √ 2 108 s 18

√ d 2 63 √ 28 h 2 √ 99 l 18 √ 3 68 p 21 √ 3 147 t 14

FLUENCY

3–8(½)

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170

Chapter 3 Indices and surds

Example 3

7 Simplify the following.   8 12 a b 9 49   10 21 e f 9 144   15 27 i j 27 4

c g k

 



18 25

d

26 32

h

45 72

l

8 Express these surds as a square root of a positive integer. √ √ √ a 2 3 b 4 2 c 5 2 √ √ √ e 3 5 f 6 3 g 8 2 √ √ √ i 9 10 j 5 5 k 7 5

  

11 25 28 50 56 76

√ d 3 3 √ h 10 7 √ l 11 3

9, 10

9, 10, 12(½)

9 Simplify by searching for the highest square factor. √ √ √ a 675 b 1183 c 1805

d

11, 12(½)

PROBLEM-SOLVING

Example 2d

FLUENCY

3A

√ 2883

10 Determine the exact side length, in simplest form, of a square with the given area. a 32 m2 b 120 cm2 c 240 mm2 11 Determine the exact radius and diameter of a circle, in simplest form, with the given area. a 24p cm2 b 54p m2 c 128p m2 12 Use Pythagoras’ theorem to find the unknown length in these triangles, in simplest form. a b c 6m 12 mm 2 cm 1 mm 3m 4 cm e

√7 m

f

√20 mm 2m

3 mm

13

4 cm

13, 14

10 cm

14, 15

√ √ 13 Ricky uses the following working to simplify 72. Show how Ricky could have simplified 72 using fewer steps. √ √ 72 = 9 × 8 √ =3 8 √ = 3 4×2 √ = 3×2× 2 √ =6 2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

REASONING

d

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REASONING

Number and Algebra

14 a List all the factors of 450 that are perfect squares. √ b Now simplify 450 using the highest of these factors.

171

3A

15 Use Pythagoras’ theorem to construct a line segment with the given lengths. You can use only a ruler and a set square or compasses. Do not use a calculator. √ √ √ √ a 10 cm b 29 cm c 6 cm d 22 cm —

—

16

√ 16 We will prove that 2 is irrational by the method called ‘proof by contradiction’. Your job is to follow and understand the proof, then copy it out and try explaining it to a friend or teacher. a

Before we start, we first need to show that if a perfect square a2 is even then a is even. We do this by showing that if a is even then a2 is even and if a is odd then a2 is odd. If a is even then a = 2k, where k is an integer. So a2 = (2k)2

ENRICHMENT

√ Proving that 2 is irrational

If a is odd then a = 2k + 1, where k is an integer. So a2 = (2k + 1)2

= 4k2

= 4k2 + 4k + 1

= 2 × 2k2 , which must be even.

= 2 × (2k2 + 2k) + 1, which must be odd.

∴ If a2 is even then a is even. √ √ b Now, to prove 2 is irrational let’s suppose that 2 is instead rational and can be written in a the form in simplest form, where a and b are integers (b π 0) and at least one of a or b is odd. b √ a ∴ 2= b a2 (squaring both sides) b2 a2 = 2b2 ∴ a2 is even and, from part a above, a must be even. If a is even then a = 2k, where k is an integer. ∴ If a2 = 2b2 So 2 =

Then (2k)2 = 2b2 4k2 = 2b2 2k2 = b2 ∴ b2 is even and therefore b is even.

a in simplest b √ form will have at least one of a or b being odd.) Therefore, the assumption that 2 can be √ a written in the form must be incorrect and so 2 is irrational. b

This is a contradiction because at least one of a or b must be odd. (Recall that

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172

Chapter 3 Indices and surds

3B Adding and subtracting surds

10A

We can apply our knowledge of like terms in algebra to help simplify expressions involving the addition and subtraction of surds. Recall that 7x and 3x are like terms, so 7x + 3x = 10x. The pronumeral x represents any number. When x = 5 √ then 7 × 5 + 3 × 5 = 10 × 5, and when x = 2 √ √ √ then 7 2 + 3 2 = 10 2. Multiples of the same surd are called ‘like surds’ and can be collected (i.e. counted) in the same way as we collect like terms in algebra. The term ‘surd’ can be traced back to the great Persian mathematician Al-Khwarizmi, who was born here in the ancient city of Khiva in the 9th century.

√ √ Let’s start: Can 3 2 + 8 be simplified? To answer this question, first discuss these points. √ √ • Are 3 2 and 8 like surds? √ • How can 8 be simplified? √ √ √ √ • Now decide whether 3 2 + 8 can be simplified. Discuss why 3 2 – 7 cannot be simplified.

Key ideas

Like surds are multiples of the same surd. √ √ √ √ √ √ For example: 3, –5 3, 12 = 2 3, 2 75 = 10 3 Like surds can be added and subtracted. Simplify all surds before attempting to add or subtract them.

Example 4 Adding and subtracting surds Simplify the following. √ √ a 2 3+4 3

√ √ √ √ b 4 6+3 2–3 6+2 2

SOL UTI ON

EX P L A NA TI ON

√ √ √ a 2 3+4 3=6 3 √ √ √ √ √ √ b 4 6+3 2–3 6+2 2= 6+5 2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Collect the like surds by adding the coefficients: 2 + 4 = 6. √ Collect like surds involving 6: √ √ √ √ 4 6–3 6=1 6= 6 √ Then collect those terms with 2.

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Number and Algebra

173

Example 5 Simplifying surds to add or subtract √ √ √ b 2 5 – 3 20 + 6 45

SO L U T I O N

b

√ √ √ √ 5 2 – 8 = 5 2 – 4×2 √ √ =5 2–2 2 √ =3 2 √ √ √ √ √ √ 2 5 – 3 20 + 6 45 = 2 5 – 3 4 × 5 + 6 9 × 5 √ √ √ = 2 5 – 6 5 + 18 5 √ = 14 5

Exercise 3B 1

√ First, look to simplify surds: 8 has a highest √ square factor of 4 and can be simplified to 2 2. Then subtract like surds. Simplify the surds and then collect like surds. √ √ √ √ Note that 3 4 × 5 = 3 × 4 × 5 = 6 5.

1–2(½), 3, 4

Decide if the following pairs of numbers are like surds. √ √ √ √ a 3, 2 3 b 5, 5 c 2 2, 2 √ √ √ √ √ √ e 2 3, 5 3 f 3 7, 3 5 g –2 5, 3 5

2 Recall your basic skills in algebra to simplify these expressions. a 11x – 5x b 8y – y c 2x – 7x e – 4a + 21a f –8x + 3x g 4t – 5t + 2t √ 3 a Simplify the surd 48. b Hence, simplify the following. √ √ √ √ i 3 + 48 ii 48 – 7 3 √ 4 a Simplify the surd 200. b Hence, simplify the following. √ √ √ √ i 200 – 5 2 ii –2 200 + 4 2

5–7(½) Example 4a

4

5 Simplify the following. √ √ a 2 5+4 5 √ √ c 7 2–3 2 √ √ e 7 5+4 5 √ √ √ g 4 10 + 3 10 – 10 √ √ √ i 21 – 5 21 + 2 21 √ √ √ k –2 13 + 5 13 – 4 13

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b d f h j l

—

√ √ d 4 6, 6 √ √ h – 7, –2 7

UNDERSTANDING

a

EX P L A N A T I O N

d 3b – 5b h 7y + y – 10y

√ √ iii 5 48 – 3 3

√ √ iii 3 200 – 30 2

5–7(½)

√ √ 5 3–2 3 √ √ 8 2–5 2 √ √ 6 3–5 3 √ √ √ 6 2–4 2+3 2 √ √ √ 3 11 – 8 11 – 11 √ √ √ 10 30 – 15 30 – 2 30

5–7(½)

FLUENCY

Simplify these surds. √ √ a 5 2– 8

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Chapter 3 Indices and surds

Example 4b

Example 5a

6 Simplify the following. √ √ √ √ a 2 3+3 2– 3+2 2 √ √ √ √ c 3 5–4 2+ 5–3 2 √ √ √ √ e 2 3+2 7+2 3–2 7 √ √ √ √ g 2 2 – 4 10 – 5 2 + 10

b d f h

7 Simplify the following. √ √ √ √ a 8– 2 b 8+3 2 √ √ √ √ e 4 18 – 5 2 f 2 75 + 2 3 √ √ √ √ i 24 + 54 j 2 125 – 3 45

√ √ c 27 + 3 √ √ g 3 44 + 2 11 √ √ k 3 72 + 2 98

√ √ √ √ 5 6 + 4 11 – 2 6 + 3 11 √ √ √ √ 5 2+2 5–7 2– 5 √ √ √ √ 5 11 + 3 6 – 3 6 – 5 11 √ √ √ √ – 4 5 – 2 15 + 5 15 + 2 5

8(½) Example 5b

8 Simplify the following. √ √ √ a 2 + 50 + 98 √ √ √ c 5 7 + 2 5 – 3 28 √ √ √ √ e 150 – 96 – 162 + 72 √ √ √ √ g 7 3 – 2 8 + 12 + 3 8 √ √ √ √ i 3 49 + 2 288 – 144 – 2 18

b d f h j

FLUENCY

3B

√ √ d 20 – 5 √ √ h 3 8 – 18 √ √ l 3 800 – 4 200 8(½), 10(½)

8–10(½)

√ √ √ 6 – 2 24 + 3 96 √ √ √ 2 80 – 45 + 2 63 √ √ √ √ 12 + 125 – 50 + 180 √ √ √ √ 36 – 108 + 25 – 3 3 √ √ √ √ 2 200 + 3 125 + 32 – 3 242

PROBLEM-SOLVING

174

9 Simplify these surds that involve fractions. Remember to use the LCD (lowest common denominator). √ √ √ √ √ √ 3 3 5 5 2 2 a + b + c – 2 3 4√ 3 5 6 √ √ √ √ √ 7 7 2 3 2 2 3 3 d – e – f + 4√ 12√ 5 2√ 7√ 2 √ √ 7 5 4 5 3 3 8 3 –5 10 3 10 g – h – i + 6 9 10 15 6 8 10 Find the perimeter of these rectangles and triangles, in simplest form. a b √5 cm 3√8 cm √12 cm c

√8 cm

d

√2 cm

e

√2 cm

√20 cm

f

√3 m √27 m

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

√27 cm

√75 cm

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Number and Algebra

11, 12(½)

12(½), 13

REASONING

11

√ √ 11 a Explain why 5 and 20 can be thought of as like surds. √ √ b Explain why 3 72 and 338 can be thought of as like surds.

175

3B

12 Prove that each of the following simplifies to zero by showing all steps. √ √ √ √ √ √ a 5 3 – 108 + 3 b 6 + 24 – 3 6 √ √ √ √ √ √ c 6 2 – 2 32 + 2 2 d 8 – 18 + 2 √ √ √ √ √ √ √ √ e 2 20 – 7 5 + 45 f 3 2 – 2 27 – 50 + 6 3 + 8 13 Prove that the surds in these expressions cannot be added or subtracted. √ √ √ √ √ √ a 3 12 – 18 b 4 8 + 20 c 50 – 2 45 √ √ √ √ √ √ d 5 40 + 2 75 e 2 200 + 3 300 f 80 – 2 54 —

—

14 To simplify the following, you will need to simplify surds and combine using a common denominator. √ √ √ √ √ √ 8 2 12 3 3 5 20 a – b + c – 3 5 4 6 4 3 √ √ √ √ √ √ 2 75 3 3 98 5 2 63 4 7 d – e – f – 4 2 5 2 9 5 √ √ √ √ √ √ 2 18 72 54 24 27 108 g – h + i – 3 2 4 7 5 10 √ √ √ √ √ √ 5 48 2 147 2 96 600 3 125 2 80 j + k – l – 6 3 5 7 14 21

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14

ENRICHMENT

Simplifying both surds and fractions

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176

Chapter 3 Indices and surds

3C Multiplying and dividing surds

10A

√ √ √ √ √ √ When simplifying surds such as 18, we write 18 = 9 × 2 = 9 × 2 = 3 2, where we use the fact √ √ √ that xy = x × y. This can be used in reverse to simplify the product of two surds. A similar process is used for division.

Let’s start: Exploring products and quotients When adding and subtracting surds we can combine like surds only. Do you think this is true for multiplying and dividing surds? √ √ √ • Use a calculator to find a decimal approximation for 5 × 3 and for 15. √ √ √ • Use a calculator to find a decimal approximation for 2 10 ÷ 5 and for 2 2. • What do you notice about the results from above? Try other pairs of surds to see if your observations are consistent.

Key ideas

When multiplying surds, use the following result. √ √ √ √ √ √ • x × y = xy • More generally: a x × b y = ab xy When dividing surds, use the following result. √  x x • √ = • More generally: y y Use the distributive law to expand brackets. •

√  a x a x √ = b y b y

a(b + c) = ab + ac

Example 6 Simplifying a product of two surds Simplify the following. √ √ a 2× 3

√ √ b 2 3 × 3 15

SOL UTI ON

EX P L A NA TI ON

√ √ √ 2× 3 = 2×3 √ = 6 √ √ √ b 2 3 × 3 15 = 2 × 3 × 3 × 15 √ = 6 45 √ = 6 9×5 √ √ = 6× 9× 5 √ = 18 5

√ √ √ Use x × y = xy.

a

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

√ √ √ Use a x × b y = ab xy. √ Then simplify the surd 45, which has a √ highest square factor of 9, using 9 = 3. √ √ √ Alternatively, using 15 = 3 × 5: √ √ √ √ √ 2 3 × 3 15 = 2 × 3 × 3 × 3 × 5 √ = 2×3×3× 5 √ = 18 5

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Number and Algebra

177

Example 7 Simplifying surds using division Simplify these surds. √ √ a – 10 ÷ 2

b

SO L U T I O N √ √ – 10 ÷ 2 = –

EX P L A N A T I O N √

√ √ Use x ÷ y =

10 2 √ =– 5 √ √ 12 18 12 18 √ = 3 3 3 3 √ =4 6

a

b

√ 12 18 √ 3 3

√

x . y

√ √ a x a x Use √ = . b y b y

Example 8 Using the distributive law Use the distributive law to expand the following and then simplify the surds where necessary. √ √ √ √ √ √ a 3(3 5 – 6) b 3 6(2 10 – 4 6) SO L U T I O N

EX P L A N A T I O N

√ √ √ √ √ 3(3 5 – 6) = 3 15 – 18 √ √ = 3 15 – 9 × 2 √ √ = 3 15 – 3 2 √ √ √ √ √ b 3 6(2 10 – 4 6) = 6 60 – 12 36 √ = 6 4 × 15 – 12 × 6 √ = 12 15 – 72

√ √ √ Expand the brackets 3 × 3 5 = 3 15 and √ √ √ 3 × 6 = 18. √ Then simplify 18.

Exercise 3C 1

1–3(½)

Copy and complete. √ √ √ 15 a 15 ÷ 3 = --√ = c

Expand the brackets and simplify the surds. √ √ Recall that 6 × 6 = 6.

--√ √ √ 6× 5 = 6×--=√ ---

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

2–3(½)

√ √ b 42 ÷ 7 =

√

42 ---

=√

—

UNDERSTANDING

a

--√ √ √ d 11 × 2 = 11 × - - =√ ---

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Chapter 3 Indices and surds

Example 6a

2 Simplify the following. √ √ a 3× 5 √ √ d 5× 7 √ √ g – 6 × (– 11)

√ √ b 7× 3 √ √ e 2 × (– 15) √ √ h – 3 × (– 2)

Example 7a

3 Simplify the following. √ √ a 20 ÷ 2

b

√ √ d – 30 ÷ (– 6) g

√ 40 √ 8

√ √ c 2 × 13 √ √ f – 6× 5 √ √ i 10 × 7

√ √ 18 ÷ 3 √ 15 e √ 5 √ 26 h –√ 2

c f i

4–6(½)

Example 6b

Example 7b

Example 8

4 Simplify the following. √ √ a 7× 3 √ √ d 3× 3 √ √ g 14 × 7 √ √ j 10 × 5

√ √ b 2× 5 √ √ e 5× 5 √ √ h 2 × 22 √ √ k 12 × 8

5 Simplify the following. √ √ a 2 5 × 15 √ √ d –5 10 × 30 √ √ g 3 14 × 2 21 √ √ j –2 7 × (–3 14)

b e h k

6 Simplify √ the following. 6 14 √ a 3 7 √ 8 2 d – √ 2 26

√ 15 12 √ b 5 2 √ 3 3 e – √ 9 21

√ √ 3 7 × 14 √ √ 3 6 × (– 18) √ √ – 4 6 × 5 15 √ √ 4 15 × 2 18

UNDERSTANDING

3C

√ 33 ÷ (– 11) √ 30 √ 3 √ 50 –√ 10

√

4–7(½)

4–7(½)

√ √ c 10 × 3 √ √ f 9× 9 √ √ i 3 × 18 √ √ l 5 × 20 c f i l

c f

FLUENCY

178

√ √ 4 6 × 21 √ √ 5 3 × 15 √ √ 2 10 × (–2 25) √ √ 9 12 × 4 21 √ 4 30 √ 8 6 √ 12 70 √ 18 14

7 Use the distributive law to expand the following and then simplify the surds where necessary. √ √ √ √ √ √ a 3( 2 + 5) b 2( 7 – 5) √ √ √ √ √ √ c – 5( 11 + 13) d –2 3( 5 + 7) √ √ √ √ √ √ e 3 2(2 13 – 11) f 4 5( 5 – 10) √ √ √ √ √ √ g 5 3(2 6 + 3 10) h –2 6(3 2 – 2 3) √ √ √ √ √ √ i 3 7(2 7 + 3 14) j 6 5(3 15 – 2 8) √ √ √ √ √ √ k –2 8(2 2 – 3 20) l 2 3(7 6 + 5 3)

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Number and Algebra

8 Simplify the following. √ a (2 7)2 √ √ √ d 2(3 – 3) – 8 √ √ g 44 – 2( 11 – 1)

8(½), 9

√ b (–3 2)2 √ √ √ √ e 8( 6 + 2) – 3 √ √ √ h 24 – 2 2( 3 – 4)

√ c –(5 3)2 √ √ √ f 5( 2 + 1) – 40 √ √ √ √ i 2 3( 6 – 3) – 50

9 Determine the unknown side of the following right-angled triangles. a b c 2√6 2√2 3√6

4

8(½), 9, 10

PROBLEM-SOLVING

9

179

3C

8 2√7

√ 10 a The perimeter of a square is 2 3 cm. Find its area. b Find the length of a diagonal of a square that has an area of 12 cm2 .

√ √ √ 11 Use x × y = xy to prove the following results. √ √ √ √ a 6× 6 = 6 b – 8 × 8 = –8 √ √ 12 8 × 27 could be simplified in two ways, as shown. Method A √ √ √ √ 8 × 27 = 4 × 2 × 9 × 3 √ √ = 2 2×3 3 √ = 2×3× 2×3 √ =6 6

11, 12

12, 13

√ √ c – 5 × (– 5) = 5

√

REASONING

11

Method B √ √ 8 × 27 = 8 × 27 √ = 216 √ = 36 × 6 √ =6 6

a Describe the first step in method A. b Why is it useful to simplify surds before multiplying, as in method A? c Multiply by first simplifying each surd. √ √ √ √ √ √ i 18 × 27 ii 24 × 20 iii 50 × 45 √ √ √ √ √ √ iv 54 × 75 v 2 18 × 48 vi 108 × (–2 125) √ √ √ √ √ √ vii – 4 27 × (– 28) viii 98 × 300 ix 2 72 × 3 80

Pythagoreans are attributed to the discovery of irrational numbers.

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3C

Chapter 3 Indices and surds

REASONING

180

√ 12 13 √ could be simplified in two ways. 3 Method A √ √ 12 12 √ = 3 3 √ = 4

Method B √ √ 12 2 3 √ = √ 3 3 =2

=2 Choose √ of another student. √ a method to simplify these surds. √ Compare your method with that 27 20 162 a √ c – √ b √ 3 5 2 √ √ √ 2 2 2 45 5 27 √ d – √ e f √ 5 8 15 5 75

—

—

14 Look at this example before simplifying the following. √ √ (2 3)3 = 23 ( 3)3 √ √ √ = 2×2×2× 3× 3× 3 √ = 8×3× 3 √ = 24 3 √ √ a (3 2)3 b (5 3)3 √ √ c 2(3 3)3 d ( 5)4 √ √ 4 e (– 3) f (2 2)5 √ √ g –3(2 5)3 h 2(–3 2)3 √ 3 √ (2 7) i 5(2 3)4 j √ √4 (3 2)3 (3 2)4 k l √4 2 √ √ 3 √4 (5 2) (2 3) (2 3)2 (–3 2)4 m × n × 4 3 9 3 √ 2 √ 3 √ 5 √ 3 (2 5) (–2 3) (3 3) (5 2) o × p ÷ 5 24 2 √ √ √ √4 (2 5)4 (2 3)3 (2 2)3 (2 8)2 q ÷ r ÷ √ 50 5 9 ( 27)3

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14

ENRICHMENT

Higher powers

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Number and Algebra

3D Binomial products

181

10A

In the previous section we used the distributive law to expand surds of the form a(b + c). Now we will extend this to expand and simplify the product of two binomial terms, including perfect squares and the difference of perfect squares. We expand binomial products in the same way that we expand the product of binomial expressions in algebra, such as (x + 1)(x – 3).

The distributive law is used in most areas of mathematics.

Let’s start: Show the missing steps You are told that the following three equations are all true. Provide all the missing steps to show how to obtain the right-hand side from each given left-hand side. √ √ √ • (2 – 3)(5 + 3) = 7 – 3 3 √ √ • (3 + 2)2 = 17 + 12 2 √ √ • (5 + 3 5)(5 – 3 5) = –20 Use the distributive law to expand binomial products and simplify by collecting terms where possible. •

Key ideas

(a + b)(c+ d) = ac + ad + bc + bd

Expanding squares • (a + b)2 = (a + b)(a + b) = a2 + ab + ba + b2 •

= a2 + 2ab + b2 (a – b)2 = (a – b)(a – b) = a2 – ab – ba + b2 = a2 – 2ab + b2

Forming a difference of squares • (a + b)(a – b) = a2 – ab + ba – b2 = a2 – b2

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182

Chapter 3 Indices and surds

Example 9 Expanding binomial products Expand and simplify. √ √ a (4 + 3)( 3 – 2)

√ √ b (2 5 – 1)(3 5 + 4)

SO L U T I O N a

b

EX P L A N A T I O N

√ √ √ √ (4 + 3)( 3 – 2) = 4 3 – 8 + 3 – 2 3 √ =2 3–5 √ √ (2 5 – 1)(3 5 + 4) √ √ = 6×5 + 8 5 – 3 5 – 4 √ = 30 + 5 5 – 4 √ = 26 + 5 5

Use (a + b)(c + d) = ac + ad + bc + bd and note √ √ that 3 × 3 = 3. Simplify by collecting like surds. Use the distributive law and collect like surds. √ √ √ √ Recall 2 5 × 3 5 = 2 × 3 × 5 × 5 = 6×5 = 30

Example 10 Expanding perfect squares Expand and simplify. √ a (2 – 7)2

√ √ b (3 2 + 5 3)2

SO L U T I O N a

(2 –

EX P L A N A T I O N

√ 2 √ √ 7) = (2 – 7)(2 – 7) √ √ =4–2 7–2 7+7 √ = 11 – 4 7

Alternatively: √ √ (2 – 7)2 = 4 – 2 × 2 7 + 7 √ = 11 – 4 7 √ √ b (3 2 + 5 3)2 √ √ √ √ = (3 2 + 5 3)(3 2 + 5 3) √ √ = 9 × 2 + 15 × 6 + 15 × 6 + 25 × 3 √ = 18 + 30 6 + 75 √ = 93 + 30 6 Alternatively: √ √ √ (3 2 + 5 3)2 = 9 × 2 + 2 × 15 6 + 25 × 3 √ = 18 + 30 6 + 75 √ = 93 + 30 6

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Note that (a – b)2 = (a – b)(a – b). Expand and simplify using the distributive law. √ √ Recall that 7 × 7 = 7. Use (a – b)2 = a2 – 2ab + b2 .

Use the distributive law to expand and simplify.

Use (a – b)2 = a2 – 2ab + b2 . √ √ √ (3 2)2 = 3 2 × 3 2 = 9 × 2 = 18

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Number and Algebra

183

Example 11 Expanding to form a difference of perfect squares Expand and simplify. √ √ a (2 + 5)(2 – 5)

√ √ √ √ b (7 2 – 3)(7 2 + 3)

SO L U T I O N a

EX P L A N A T I O N

√ √ √ √ (2 + 5)(2 – 5) = 4 – 2 5 + 2 5 – 5 = –1 Alternatively: √ √ √ (2 + 5)(2 – 5) = 22 – ( 5)2

Expand using the distributive law and cancel the two middle terms. Use (a + b)(a – b) = a2 – b2 .

=4–5

b

= –1 √ √ √ √ (7 2 – 3)(7 2 + 3) √ √ = 49 × 2 + 7 6 – 7 6 – 3

Use the distributive law and then cancel the two middle terms.

= 98 – 3 = 95 Alternatively: √ √ √ √ √ √ (7 2 – 3)(7 2 + 3) = (7 2)2 – ( 3)2

Use (a – b)(a + b) = a2 – b2 .

= 49 × 2 – 3 = 95

1

Simplify the following. √ √ a 3× 7 √ d ( 11)2 √ g (5 5)2

2 Simplify the following. √ √ a 5 2–5 2 √ √ d 2 2– 8

1–3(½)

3(½)

√ √ b – 2× 5 √ 2 e ( 13) √ h (7 3)2

√ √ c 2 3×3 2 √ 2 f (2 3) √ i (9 2)2

√ √ b –2 3 + 2 3 √ √ e 2 27 – 4 3

√ √ c 6× 7 – 7×6 √ √ f 5 12 – 48

—

UNDERSTANDING

Exercise 3D

3 Use the distributive law (a + b)(c + d) = ac + ad + bc + bd to expand and simplify these algebraic expressions. a (x + 2)(x + 3) b (x – 5)(x + 1) c (x + 4)(x – 3) d (2x + 1)(x – 5) e (3x + 2)(2x – 5) f (6x + 7)(x – 4) g (x + 4)(x – 4) h (2x – 3)(2x + 3) i (5x – 6)(5x + 6) 2 2 j (x + 2) k (2x – 1) l (3x – 7)2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter 3 Indices and surds

4–7(½)

4–7(½)

FLUENCY

4–7(½)

3D Example 9a

4 Expand and simplify. √ √ a (2 + 2)( 2 – 3) √ √ d (5 – 3)(2 + 3) √ √ g ( 5 – 2)(4 – 5)

√ √ b (4 + 5)( 5 – 2) √ √ e (3 + 7)(4 – 7) √ √ h (4 – 10)(5 – 10)

√ √ c ( 6 + 2)( 6 – 1) √ √ f ( 2 – 5)(3 + 2) √ √ i ( 7 – 4)( 7 – 4)

Example 9b

5 Expand and simplify. √ √ a (5 2 – 1)(3 2 + 3) √ √ d (7 6 + 4)(2 – 3 6) √ √ g (4 3 – 5)(2 – 3 3)

√ √ b (4 3 + 3)(2 3 – 1) √ √ e (2 10 – 3)( 10 – 5) √ √ h (1 – 5 2)(3 – 4 2)

√ √ c (6 5 – 5)(2 5 + 7) √ √ f (2 7 – 3)(3 – 4 7) √ √ i (4 5 – 3)(3 – 4 5)

Example 10a

Example 11a

6 Expand and simplify these perfect squares. √ √ a (3 – 5)2 b (2 – 6)2 √ √ d ( 11 + 2)2 e ( 3 + 5)2 √ √ √ √ 2 g ( 7 + 2) h ( 11 – 5)2 √ √ √ √ j ( 13 + 19)2 k ( 17 + 23)2

c f i l

7 Expand and simplify these differences of perfect squares. √ √ √ √ a (3 – 2)(3 + 2) b (5 – 6)(5 + 6) √ √ √ √ d ( 7 – 1)( 7 + 1) e ( 8 – 2)( 8 + 2) √ √ √ √ √ √ √ √ g ( 5 + 2)( 5 – 2) h ( 11 + 5)( 11 – 5)

√ √ c (4 + 3)(4 – 3) √ √ f ( 10 – 4)( 10 + 4) √ √ √ √ i ( 3 – 7)( 3 + 7)

8(½), 10(½)

√ (4 + 7)2 √ ( 5 – 7)2 √ √ ( 10 – 3)2 √ √ ( 31 – 29)2

8–10(½)

Example 10b

8 Expand and simplify these perfect squares. √ √ √ √ a (2 5 + 7 2)2 b (3 3 + 4 7)2 √ √ √ 2 √ d (8 2 – 4 3) e (6 3 – 3 11)2 √ √ √ √ g (2 3 + 3 6)2 h (3 10 + 5 2)2

Example 11b

9 Expand and simplify these differences of perfect squares. √ √ √ √ √ √ √ a (3 11 – 2)(3 11 + 2) b (2 5 – 3)(2 5 + 3) √ √ √ √ √ √ √ √ c (4 3 + 7)(4 3 – 7) d (5 7 – 2 3)(5 7 + 2 3) √ √ √ √ √ √ √ √ e (4 5 – 3 6)(4 5 + 3 6) f (4 10 – 5 6)(4 10 + 5 6) √ √ √ √ √ √ √ √ g (5 8 – 10 2)(5 8 + 10 2) h (3 7 – 4 6)(3 7 + 4 6) √ √ √ √ √ √ √ √ i (2 10 + 4 5)(2 10 – 4 5) j (2 15 + 5 6)(2 15 – 5 6)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

√ √ c (5 6 + 3 5)2 √ 2 √ f (3 7 – 2 6) √ √ i (5 3 – 2 8)2

8–10(½)

PROBLEM-SOLVING

184

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PROBLEM-SOLVING

Number and Algebra

10 Find the area of these rectangles and triangles, in expanded and simplified form. a b

√3 + 1 m 2 + √3 cm

185

3D

√3 − 1 m

c

d

2√5 m 3√3 − 1 mm

√5 + 6 m

5√2 − 2√3 mm e

f

7√2 − 1 cm √3 − 1 cm

5√6 − 2√3 mm

11, 12(½)

12(½), 13

REASONING

11

11 Show that the following simplify to an integer. √ √ √ √ a ( 2 + 7)( 2 – 7) √ √ √ √ b (2 3 – 5)(2 3 + 5) √ √ √ √ c (5 11 – 7 3)(5 11 + 7 3) 12 Use your knowledge of the simplification of surds to fully simplify the following. √ √ √ √ √ √ a (3 – 2 7)( 21 – 4 3) b (2 6 + 5)( 30 – 2 5) √ √ √ √ √ √ c (3 5 + 1)( 7 + 2 35) d (4 2 + 7)(3 14 – 5) √ √ √ √ √ √ e (3 3 + 4)( 6 – 2 2) f (5 – 3 2)(2 10 + 3 5) 13 Is it possible for (a + b)2 to simplify to an integer if at least one of a or b is a surd? If your answer is yes, give an example.

—

14 Fully expand and simplify these surds. √ √ √ √ a (2 3 – 2)2 + ( 3 + 2)2 √ √ √ √ √ √ c ( 3 – 4 5)( 3 + 4 5) – ( 3 – 5)2 √ √ √ e ( 3 – 2 6)2 + (1 + 2)2 √ √ √ √ √ √ g (2 3 – 3 2)(2 3 + 3 2) – ( 6 – 2)2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b d f h

—

√ √ √ √ ( 5 – 3)2 + ( 5 + 3)2 √ √ –10 3 – (2 3 – 5)2 √ √ (2 7 – 3)2 – (3 – 2 7)2 √ √ √ 2 √ √ 2 2(2 5 – 3 3) + ( 6 + 5)

14

ENRICHMENT

Expansion challenge

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186

Chapter 3 Indices and surds

3E Rationalising the denominator

10A

As you know, it is easier to add or subtract fractions when the fractions are √ expressed with the same 3–1 1 denominator. In a similar way, it is easier to work with surds such as √ and √ when they are expressed 5 2 using a whole number in the denominator. The process that removes a surd from the denominator is called ‘rationalising the denominator’ because the denominator is being converted from an irrational number to a rational number.

Let’s start: What do I multiply by? 1 When trying to rationalise the denominator in a surd like √ , you must multiply the surd by a 2 chosen number so that the denominator is converted to a whole number. •

First, √ is equivalent to. √ decide what each of the following 3 2 i √ ii √ 3 2 √ √ • Recall that x × x = x and simplify the following. √ √ √ √ i 5× 5 ii 2 3 × 3 1 • Now, decide what you can multiply √ by so that: 2 1 – the value of √ does not change, and 2 – the denominator becomes a whole number. • Repeat this for: 3 1 ii √ i √ 2 3 5

Key ideas

√ 21 iii √ 21 √ √ iii 4 7 × 7

Rationalising a denominator involves multiplying by a number equivalent to 1, which changes the denominator to a whole √ number. √ y x y x x √ = √ ×√ = y y y y

Example 12 Rationalising the denominator Rationalise the denominator in√the following. 3 2 2 a √ b √ 3 5 SOL UTI ON

a

√ 2 3 2 √ = √ ×√ 3 3 3 √ 2 3 = 3

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

c

√ 2 6 √ 5 2

d

√ 1– 3 √ 3

EX P L A NA TI ON

Choose the appropriate fraction equivalent to 1 √ 3 to multiply by. In this case, choose √ since 3 √ √ 3 × 3 = 3. Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Number and Algebra

c

d

Choose the √ appropriate fraction. In this √ √ 5 case, use √ since 5 × 5 = 5. Recall √ √ √5 √ 2 × 5 = 2 × 5 = 10. √ 2 Choose the appropriate fraction; i.e. √ . 2 √ √ 5 × 2 × 2 = 5 × 2 = 10 √ Simplify the surd 12 and cancel.

Expand using the distributive law: √ √ √ √ √ √ (1 – 3) × 3 = 1 × 3 – 3 × 3 = 3 – 3

Exercise 3E 1

Simplify. √ 6 a √ 6 √ 8 e –√ 2

1–3(½)

√ 11 b √ 11 √ 3 27 f – √ 3

2 Write the missing number. √ a 3× = 3 √ d 2 5 × = 10 √ 1 3 g √ × = 3 3

3

√ 2 5 √ 4 5 √ 72 g √ 2

√ 7 3 d – √ 14 3 √ 3 45 h – √ 9 5

c

√ × 5=5 √ e 4 3 × = 12 √ 1 7 h √ × = 7 7

b

—

UNDERSTANDING

b

√ √ √ 3 2 3 2 5 √ = √ ×√ 5 5 5 √ 3 10 = 5 √ √ √ 2 6 2 6 2 √ = √ ×√ 5 2 5 2 2 √ 2 12 = 10 √ 1
2 4 × 3 = 1 05 √ 2 3 = 5 √ √ √ 1– 3 1– 3 3 √ = √ ×√ 3 3 3 √ 3–3 = 3

187

c f i

√

√ 10 × 10 = √ × 3 7 = 21 √ 2 2 13 √ × = 13 13

3 Use a calculator to find a decimal approximation to each number in the following pairs of numbers.√What do you notice? √ √ √ 1 7 5 5 3 11 11 11 55 √ , a √ , b √ , c 5 7 7 3 3 5

Example 12a

4 Rationalise the denominators. 1 1 a √ b √ 7 2 5 e √ 3

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

f

8 √ 2

3 c √ 11 √ 5 g √ 3

4–7(½)

4–7(½)

4 d √ 5 √ 2 h √ 7

FLUENCY

4–7(½)

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Chapter 3 Indices and surds

FLUENCY

3E √ a 5 Rewrite each of the following in the form √ and then rationalise the denominators. b     2 5 6 2 a b c d 3 7 11 5     7 6 10 17 e f g h 3 7 3 2 Example 12b

6 Rationalise the denominators. √ 4 2 a √ 7 √ 3 6 d √ 7

Example 12c

7 Rationalise the denominators. √ √ 4 7 2 3 a √ b √ 5 3 3 2 √ √ 5 12 2 7 √ √ f e 3 27 3 35

√ 5 2 b √ 3 √ 7 3 e √ 10

√ 3 5 c √ 2 √ 2 7 f √ 15

c g

√ 5 7 √ 3 5 √ 9 6 √ 2 3 9

Example 12d

8 Rationalise the denominators. √ √ 1+ 2 3+ 5 √ √ b a 7 3 √ √ √ √ 5+ 2 10 – 7 √ √ e f 7 3 √ √ √ √ 6 – 10 4 2–5 3 √ √ j i 6 5

√ 2– 3 √ c 5 √ √ 2+ 7 √ g 6 √ √ 3 5+5 2 √ k 10

d h

√ 4 5 √ 5 10 √ 7 90 √ 2 70

8(½), 9

8(½), 9, 10(½)

√ 3– 5 √ d 2 √ √ 5+ 2 √ h 10 √ √ 3 10 + 5 3 √ l 2 √

PROBLEM-SOLVING

188

9 Determine the exact value of the area of the following shapes. Express your answers using a rational denominator. a b c √2 mm 2 cm 2 1 √3 m mm 3 √5 5 cm

2 m √6

√3 mm

10 Simplify the following by first rationalising denominators and then using a common denominator. 1 1 2 1 3 3 c √ –√ a √ +√ b √ +√ 7 3 3 5 2 2 2 5 2 3 1 5 √ + √ e √ + √ f d √ – √ 2 3 3 2 2 5 5 3 3 2 4 3 √ √ √ √ √ √ 5 2 4 7 7 2 2 7 10 6 4 2 √ + √ √ – √ h i g √ – √ 5 7 3 2 3 5 3 3 3 5 3 6 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

11

12(½), 13

REASONING

11, 12(½)

√ x 11 Explain why multiplying a number by √ does not change its value. x 12 Rationalise the denominators and simplify the following. √ √ 3+a 6+a √ √ a b 7 5 √ √ 3 – 3a 5 – 5a √ √ d e 3 5 √ √ 3a + 3 4a + 5 √ √ g h 10 6 13 To explore how to simplify a number such as a

Simplify: √ √ i (4 – 2)(4 + 2)

ii (3 –

189

3E

√ 2+a √ c 6 √ 7 – 7a √ f 7 √ 2a + 7 √ i 14

3 √ , first answer these questions. 4– 2

√ √ 7)(3 + 7)

√ √ √ √ iii (5 2 – 3)(5 2 + 3)

b What do you notice about each question and answer in part a above? 3 √ by to rationalise the denominator. c Now decide what to multiply 4– 2 d Rationalise the denominator in these expressions. √ √ 3 –3 2 2 6 √ √ i iii √ √ ii √ iv √ 3–1 6–2 5 4– 2 4– 3 —

—

14

14 Rationalise the denominators in the √following by forming a ‘difference of two perfect squares’. 2 2–1 2 =√ ×√ For example: √ 2+1 2+1 2–1 √ 2( 2 – 1) √ = √ ( 2 + 1)( 2 – 1) √ 2 2–2 = √2 – 1 =2 2–2 5 3+1 3 √ 1– 3

a √ e

2 √ 11 – 2 √ 6 m √ 6–1

i

√

q √

a √ a– b

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

4 b √ 3–1 7 √ f 6– 7

3 c √ 5–2 4 √ g 3 – 10

6 √ √ 2+ 5 √ 3 2 n √ 7–2 √ √ a– b r √ √ a+ b

4 k √ √ 3+ 7 √ 2 5 o √ 5+2 √ a s √ √ a+ b

j

ENRICHMENT

Binomial denominators

4 √ 1– 2 7 √ h 2– 5 √ 2 √ l 7+1 d

b √ a+ b √ ab √ √ a– b

p √ t

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190

Chapter 3 Indices and surds

3F Review of index laws

CONSOLIDATING

From your work in Year 9 you will recall that powers (i.e numbers with indices) can be used to represent repeated multiplication of the same factor. For example, 2 × 2 × 2 = 23 and 5 × x × x × x × x = 5x4 . The five basic index laws and the zero power will be revised in this section.

Let’s start: Recall the laws Try to recall how to simplify each expression and use words to describe the index law used.

Key ideas

• 53 × 57

• x4 ÷ x 2

• (a7 )2  4 x • 3

• (2a)3 • (4x2 )0

Recall that a = a1 and 5a = 51 × a1 . The index laws • Law 1: am × an = am + n am

Retain the base and add the indices.

•

Law 2: am ÷ an =

•

Law 3: (am )n = am × n

Retain the base and multiply the indices.

•

Law 4: (a × b)m = am × bm  m am a Law 5: = m b b

Distribute the index number across the bases.

•

an

= am – n

The zero power: a0 = 1

Retain the base and subtract the indices.

Distribute the index number across the bases. Any number (except 0) to the power of zero is equal to 1.

Example 13 Using index law 1 Simplify the following using the first index law. a x5 × x4

b 3a2 b × 4ab3

SOL UTI ON

EX P L A NA TI ON

a x5 × x4 = x9

There is a common base of x, so add the indices.

b 3a2 b × 4ab3 = 12a3 b4

Multiply coefficients and add indices for each base a and b. Recall that a = a1 .

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

191

Example 14 Using index law 2 Simplify the following using the second index law. a m7 ÷ m5 b 4x2 y5 ÷ (8xy2 ) SO L U T I O N

EX P L A N A T I O N

a m7 ÷ m5 = m2

Subtract the indices when dividing terms with the same base.

b

4x2 y5 ÷ (8xy2 ) = =

4x2 y5 8xy2

First, express as a fraction. Divide the coefficients and subtract the indices of x and y (i.e. x2 – 1 y5 – 2 ).

xy3 2

1 = xy3 2

Example 15 Combining index laws Simplify the following using the index laws. a

(a3 )4

b

(2y5 )3

c

(

3x2 5y2 z

)3

d

3(xy2 )3 × 4x4 y2 8x2 y

SO L U T I O N

EX P L A N A T I O N

a (a3 )4 = a12

Use index law 3 and multiply the indices.

b (2y5 )3 = 23 y15 = 8y c

d

(

3x2 5y2 z

)3

Use index law 4 and multiply the indices for each base 2 and y. Note: 2 = 21 .

15

=

33 x6 53 y6 z3

=

27x6 125y6 z3

Raise the coefficients to the power 3 and multiply the indices of each base by 3.

3(xy2 )3 × 4x4 y2 3x3 y6 × 4x4 y2 = 8x2 y 8x2 y =

12x7 y8 8x2 y

=

3x5 y7 2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Remove brackets first by multiplying indices for each base. Simplify the numerator using index law 1. Simplify the fraction using index law 2, subtracting indices of the same base.

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192

Chapter 3 Indices and surds

Example 16 Using the zero power Evaluate, using the zero power. a 4a0

b 2p0 + (3p)0

SO L U T I O N

EX P L A N A T I O N

4a0 = 4 × 1

a

Any number to the power of zero is equal to 1.

=4 2p0 + (3p)0 = 2 × 1 + 1

b

Note: (3p)0 is not the same as 3p0 .

=3

1

1(½), 2, 3

Simplify, using index form. a 3×3×3×3 c 8×8×8 e 3×y×y×5×y×y

3

—

UNDERSTANDING

Exercise 3F

b 7×7×7×7×7×7 d 2×x×x×3×x f 2×b×a×4×b×a×a

2 Copy and complete this table. x 2x

4

3

2

1

0

23 = 8

3 Copy and complete.

c

2 2 × 23 = 2 × 2 × - - - - - = 2---

b

( 2 )3 a = a×a×- - - - - -×- - - - - = a---

x5 x × x × x × - - - - - = x3 -----= x ---

d

(2x)0 × 2x0 = - - - - × - - - =2

4–7(½) Example 13

4 Simplify, using the first index law. a a5 × a4 b 2 3 d 7m × 2m e 1 2 g p ×p h 5 j 2x2 y × 3xy2 k 4 2 4 m 3x × 5xy × 10y n

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

x3 × x2 2s4 × 3s3 1 4 2 3 c × c 4 3 3a2 b × 5ab5 2rs3 × 3r4 s × 2r2 s2

4–7(½)

4–7(½)

c b × b5 f t8 × 2t8 3 3s i s× 5 5 l 3v7 w × 6v2 w o 4m6 n7 × mn × 5mn2

FLUENCY

a

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Example 15a–c

Example 16

5 Simplify, using the second index law. a x5 ÷ x2 b a7 ÷ a6

c q9 ÷ q6

d b5 ÷ b

e

y8 y3

f

d8 d3

g

j7 j6

h

m15 m9

i

2x2 y3 ÷ x

j

3r5 s2 ÷ (r3 s)

k 6p4 q2 ÷ (3q2 p2 )

l

16m7 x5 ÷ (8m3 x4 )

m

5a2 b4 a2 b

n

8st4 2t3

o

2v5 8v3

p

7a2 b 14ab

q

–3x4 y 9x3 y

r

–8x2 y3 16x2 y

6 Simplify using the third, fourth and fifth index laws. a (x5 )2 b (t3 )2 c 2 3 e (4t ) f (2u2 )2 g ( 2 )2 ( 3 )3 a x i j k 3 b y4 ( 2 )2 ( 2 )3 3a b 3f m n o 5g 2pq3 7 Evaluate the following using the zero power. a 8x0 b 3t 0 3 3 0 e 5(g h ) f 8x0 – 5

4(a2 )3 (3r3 )3 ( 2 3 )2 x y z4 ( 3 )3 at 3g4

Example 15d

d (10ab2 )0 h 7x0 – 4(2y)0 8(½), 9

8 Use appropriate index laws to simplify the following. a x6 × x5 ÷ x3 b x2 y ÷ (xy) × xy2 c x4 n7 × x3 n2 ÷ (xn) e g

m2 w × m3 w2 m4 w3 2 9x y3 × 6x7 y5 12xy6

d f h

16a8 b × 4ab7 32a7 b6

j

k –5(a2 b)3 × (3ab)2

l

i

m

4m2 n × 3(m2 n)3 6m2 n

n

o

2(ab)2 × (2a2 b)3 4ab2 × 4a7 b3

p

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

3F

d 5(y5 )3 h (3p4 )4 ( 4 2 )4 u w l v2 ( 2 3 )4 4p q p 3r

c (5z)0 g 4b0 – 9 8(½), 9

193

x2 y3 × x2 y4 x3 y5 r4 s7 × r4 s7 r4 s7 2 4x y3 × 12x2 y2 24x4 y

8–10(½)

PROBLEM-SOLVING

Example 14

FLUENCY

Number and Algebra

(3m2 n4 )3 × mn2 (

( ( ) ) )2 3 4f 2 g × f 2 g4 ÷ 3 fg2

(7y2 z)2 × 3yz2 7(yz)2

(2m3 )2 (6n5 )2 × 4 0 3(mn ) (–2n)3 m4

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194

Chapter 3 Indices and surds

b –(3)3

10 Simplify: (( ) )2 3 a x2

c (–3)4

(( ) )7 3 b a5

d –34

c

11(½)

((

a2 b

)3 )5

11(½), 12

12, 13

REASONING

9 Simplify: a (–3)3

PROBLEM-SOLVING

3F

11 Evaluate without the use of a calculator. a e

133 132 252 54

b f

187 186 362 64

c g

98 96 272 34

d h

410 47 322 27

12 When Billy uses a calculator to raise –2 to the power 4 he gets –16 when the answer is actually 16. What has he done wrong?

13 Find the value of a in these equations in which the index is unknown. a 2a = 8 b 3a = 81 c 2a + 1 = 4 d (–3)a = –27 e (–5)a = 625 f (– 4)a – 1 = 1

—

—

14–16

14 If x4 = 3, find the value of: a x8 b x4 – 1

c 2x16

d 3x4 – 3x8

15 Find the value(s) of x. a x4 = 16

c 22x = 16

d 22x – 3 = 16

b 2x – 1 = 16

16 Find the possible pairs of positive integers for x and y when: a xy = 16 b xy = 64 c xy = 81

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

ENRICHMENT

Indices in equations

d xy = 1

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Number and Algebra

195

3G Negative indices The study of indices can be extended to include negative powers. Using the second index law and the fact that a0 = 1, we can establish rules for negative powers. a0 ÷ an = a0 – n (index law 2)

a0 ÷ an = 1 ÷ an (as a0 = 1) 1 = n a

also

= a–n Therefore: a–n =

1 an

1 = 1 ÷ a–n a–n 1 = 1÷ n a an = 1× 1 n =a

Also:

Therefore:

1 = an . a–n Negative indices can be used to describe very small numbers, such as the mass of a grain of sand.

Let’s start: The disappearing bank balance Due to fees, an initial bank balance of $64 is halved every month.

• • •

Balance ($)

64

32

Positive indices only

26

25

Positive and negative indices

26

16

8

4

2

1

1 2

1 4

1 8

1 22 24

2–1

Copy and complete the table and continue each pattern. Discuss the differences in the way indices are used at the end of the rows. 1 What would be a way of writing using positive indices? 16

• What would be a way of writing

a–m =

1 am

1 = am a–m

1 using negative indices? 16

For example, 2–3 = For example,

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

1 1 = . 23 8

Key ideas

1 = 23 = 8. 2–3

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196

Chapter 3 Indices and surds

Example 17 Writing expressions using positive indices Express each of the following using positive indices. a

b– 4

b 3x– 4 y2

SO L U T I O N a b– 4 =

d

1 b4

3a2 a5

)3

d

x –3

Use a–n = 3y2 x4

(

3a2 a5

)3

=

1 . an

x is the only base with a negative power. 3 1 y2 3y2 × × = 4 1 x4 1 x

5 = 5 × x3 x –3 = 5x3 (

5

EX P L A N A T I O N

b 3x– 4 y2 =

c

c

Use

33 a6 a15

1 5 1 = an and note that –3 = 5 × –3 . a –n x x

Use index laws 3, 4 and 5 to apply the power 3 to each base in the brackets.

= 27a –9 27 = 9 a

Apply index law 2 to subtract the indices of a. 1 Use a–n = n to express with a positive index. a Alternatively, first simplify the power of 3.

3a2 and then raise to a5

Example 18 Simplifying more complex expressions Simplify the following and express your answers using positive indices. ( –2 )–3 ( )3 ( –2 3 )2 (p–2 q)4 p 2m3 5m n a × b ÷ –1 3 3 2 – 4 r 5p q q r n SO L U T I O N

a

(p–2 q)4 × 5p–1 q3

EX P L A N A T I O N (

p–2 q3

)–3

=

p–8 q4 p6 × 5p–1 q3 q–9

=

p–2 q4 5p–1 q–6

p–1 q10 5 10 q = 5p

=

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Deal with brackets first by multiplying the power to each of the indices within the brackets. Use index laws 1 and 2 to combine indices of like bases. Simplify each numerator and denominator first: p–8 + 6 = p–2 and q3 + (–9) = q–6 . Then p–2 – (–1) q4 – (–6) = p–1 q10 . 1 Use a–n = n to express p–1 with a positive index. a

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Number and Algebra

(

b

2m3 r2 n– 4

)3

÷

(

5m–2 n3 r

)2

=

23 m9 52 m– 4 n6 ÷ r2 r6 n–12

Multiply the bracket power to each of the indices within the bracket.

8m9 r2 × r6 n–12 25m– 4 n6 8m13 r– 4 = 25n–6 8m13 n6 = 25r 4

Multiply by the reciprocal of the divisor.

=

Use index laws 1 and 2 to combine indices of like bases. Write the answer with positive powers.

Exercise 3G

1–3

3

—

UNDERSTANDING

1

Write the next three terms in these patterns. a 23 , 22 , 21 , 20 , 2–1 , b x2 , x1 , x0 , , , c 3x, 3x0 , , ,

2 Recall that a

1 9

,

197

,

1 1 = . Similarly, write these denominators using positive indices. 4 22 1 5 2 b c d – 25 16 27

3 Write each rule for negative indices by completing each statement. 1 a a–b = b = a–b

Example 17a,b

Example 17c

4 Express the following using positive indices. a x–5 b a– 4 2 –3 e 3a b f 4m3 n–3 1 –2 3 1 2 –4 5 i p q r j d e f 3 5 5 Express the following using positive indices. 1 2 a b x–2 y–3 2b4 3m2 e f d –3 n– 4

c 2m– 4 g 10x–2 y5 z 3 2 –6 7 k u v w 8

c g

4 m–7 4b4 3a–3

4–8(½)

4–8(½)

d 3y–7 h 3x– 4 y–2 z3 2 3 –5 –2 l b c d 5

d h

FLUENCY

4–7(½)

3 b–5 5h3 2g–3

6 Use index laws 1 and 2 to simplify the following. Write your answers using positive indices. a x3 × x–2 b a7 × a– 4 c 2b5 × b–9 d 3y–6 × y3 – 4 2 –5 4 – 4 –2 e 3a × 2a f 4x × 3x g 5m × (–2m ) h –3a–7 × 6a–3 2x–2 3x–3 5s–2 m 3s i

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

j n

7d –3 10d –5 4f –5 3f –3

k o

6c–5 12c–5 3d –3 6d –1

l p

3b–2 4b– 4 15t – 4 18t –2

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Chapter 3 Indices and surds

3G Example 17d

7 Express the following with positive indices. ( 3 )3 ( 2 )4 m 2x a b x3 4m5 e (3t –4 )2

i

(4y–2 )–2

c 2(x–7 )3

d 4(d –2 )3

f

5(x2 )–2

g (3x–5 )4

h –8(x5 )–3

j

(3h–3 )–4

k 7(j –2 )–4

l

FLUENCY

198

2(t –3 )–2

8 Express the following in simplest form with positive indices. a x2 y3 × x–3 y–4

b 4a–6 y4 × a3 y–2

e a3 b4 ÷ (a2 b7 )

f

p2 q3 ÷ (p7 q2 )

g

a4 b3 a2 b5

h

m3 n2 mn3

i

p2 q2 r4 pq4 r5

j

3x2 y 6xy2

k

4m3 n4 7m2 n7

l

12r4 s6 9rs–1

m

f 3 g–2 f –2 g3

n

r–3 s– 4 r3 s–2

o

3w–2 x3 6w–3 x–2

p

15c3 d 12c–2 d–3

9(½) Example 18

d 6a4 b3 × 3a–6 b

9–10(½)

9–10(½), 11

9 Simplify the following and express your answers with positive indices. a (a3 b2 )3 × (a2 b4 )–1

b (2p2 )4 × (3p2 q)–2

2a3 b2 2a2 b5 × a–3 b4 ( 2 3 )2 ( 4 )–2 a b ab g ÷ b–2 a2

(3rs2 )4 (2r2 s)2 × s7 r –3 s4 ( 4 –2 )2 ( –3 2 )2 m n m n h ÷ r3 r3

d

e

10 Evaluate without the use of a calculator. a 5–2 b 4–3 e 310 × (32 )–6 ( )–2 2 i 3

f

(42 )–5 × 4(4–3 )–3

j

(

–5 4

)–3

c 2 × 7–2 2 g 7–2 k

(4–2 )3 4–4

c 2(x2 y–1 )2 × (3xy4 )3 f

4(x–2 y4 )2 xy4 × x2 y –3 2x–2 y

i

3(x2 y–4 )2 (xy)–3 ÷ 2 2 2(xy ) (3x–2 y4 )2

PROBLEM-SOLVING

c 2a–3 b × 3a–2 b–3

d 5 × (–3– 4 ) –3 h 4–2 l

(10–4 )–2 (10–2 )–3

11 The width of a hair on a spider is approximately 3–5 cm. How many centimetres is this, correct to four decimal places?

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

12, 13

12 a

Simplify these numbers. ( )–1 ( )–1 2 5 i ii 3 7 ( )–1 a when expressed in simplest form? b What is b

a

+ 3–1

( )–1 3 d – 5(2 –2 ) 2

b

3–2

( )–2 ( ) 2–2 4 – e 5 3

Simple equations with negative indices

c 3x =

1 27

f

—

1 32 ( )x 4 3 d = 4 3

b 2x =

f

( )x 125 2 = 5 8

g

1 =8 2x

h

1 = 81 3x

i

1 =1 2x

j

5x – 2 =

l

10x – 5 =

1 9

3 2–2

) ( –1 )–1 2 – 3–2

16

1 25 1 1000

( )2x + 1 3 64 m = 4 27

( )3x – 5 2 25 n = 5 4

( )3x + 2 3 16 o = 2 81

( )1 – x 7 4 p = 4 7

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

(

—

( )x 9 2 = e 3 4

k 3x – 3 =

)–1

( )–1 ( )0 1 3 c – 4 2

+ 6–1

( )x 1 = 2–x giving reasons. 15 Prove that 2

16 Find the value of x. 1 a 2x = 4

2x y

1 . Explain the error made. 2x2

14 Evaluate the following by combining fractions. 2–1

(

3G

ENRICHMENT

13 A student simplifies 2x–2 and writes 2x–2 =

iii

14, 15

REASONING

12

199

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200

Chapter 3 Indices and surds

3H Scientific notation

CONSOLIDATING

Scientific notation is useful when working with very large or very small numbers. Combined with the use of significant figures, numbers can be written down with an appropriate degree of accuracy and without the need to write all the zeros that define the position of the decimal place. The approximate distance between the Earth and the Sun is 150 million kilometres or 1.5 × 108 km when written in scientific notation using two significant figures. Negative indices can be used for very small numbers, such as 0.0000382 g = 3.82 × 10–5 g.

Let’s start: Amazing facts large and small Think of an object, place or living thing that is associated with a very large or small number. • Give three examples using very large numbers. • Give three examples using very small numbers. • Can you remember how to write these numbers using scientific notation? • How are significant figures used when writing numbers with scientific notation?

Key ideas

A number written using scientific notation is of the form a × 10m , where 1 ≤ a < 10 or –10 < a ≤ –1 and m is an integer. • Large numbers: 24 800 000 = 2.48 × 107 9 020 000 000 = 9.02 × 109 • Small numbers: 0.00307 = 3.07 × 10–3 –0.0000012 = –1.2 × 10–6 Significant figures are counted from left to right, starting at the first non-zero digit. • When using scientific notation the digit to the left of the decimal point is the first significant figure. For example: 20 190 000 = 2.019 × 107 shows four significant figures. • The × 10n , EE or Exp keys can be used on calculators to enter numbers using scientific notation; e.g. 2.3E– 4 means 2.3 × 10– 4 .

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

201

Example 19 Converting from scientific notation to a basic numeral Write these numbers as a basic numeral. a 5.016 × 105

b 3.2 × 10–7

SO L U T I O N

EX P L A N A T I O N

a 5.016 × 105 = 501 600

Move the decimal point 5 places to the right.

b 3.2 × 10–7 = 0.00000032

Move the decimal point 7 places to the left.

Example 20 Converting to scientific notation using significant figures Write these numbers in scientific notation, using three significant figures. a 5 218 300 b 0.0042031 SO L U T I O N

EX P L A N A T I O N

a 5 218 300 = 5.22 × 106

Place the decimal point after the first non-zero digit. The digit following the third digit is at least 5, so round up.

b 0.0042031 = 4.20 × 10–3

Round down in this case, but retain the zero to show the value of the third significant figure.

3(½)

How many significant figures are showing in these numbers? a 2.12 × 107 b 5.902 × 104 c 1.81 × 10–3 e 461 f 91 g 0.0001

—

d 1.0 × 10–7 h 0.0000403

2 Write these numbers as powers of 10. a 1000

b 10 000 000

3 Convert to numbers using scientific notation. a 43 000 b 712 000 e 0.00078 f 0.00101

Example 19a

4 Write these numbers as a basic numeral. a 3.12 × 103 b 5.4293 × 104 4 e 5.95 × 10 f –8.002 × 105 i 2.105 × 108 j –5.5 × 104

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

1 1000

c 0.000001

d

c 901 200 g 0.00003

d 10 010 h 0.0300401

4–7(½)

c 7.105 × 105 g –1.012 × 104 k 2.35 × 109

4–8(½)

UNDERSTANDING

1

1–3(½)

4–8(½)

d 8.213 × 106 h 9.99 × 106 l 1.237 × 1012

FLUENCY

Exercise 3H

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Chapter 3 Indices and surds

3H Example 19b

5 Write these numbers as a basic numeral. a 4.5 × 10–3 b 2.72 × 10–2 –5 e –9.2 × 10 f 2.65 × 10–1 i 9.8 × 10–1 j –5.45 × 10–10

c 3.085 × 10– 4 g 1.002 × 10– 4 k 3.285 × 10–12

d 7.83 × 10–3 h –6.235 × 10–6 l 8.75 × 10–7

Example 20a

6 Write these numbers in scientific notation, using three significant figures. a 6241 b –572 644 c 30 248 d 423 578 e –10 089 f 34 971 863 g 72 477 h 356 088 i 110 438 523 j 909 325 k – 4 555 678 l 9 826 100 005

Example 20b

7 Write these numbers in scientific notation, using three significant figures. a 0.002423 b –0.018754 c 0.000125 d –0.0078663 e 0.0007082 f 0.11396 g 0.000006403 h 0.00007892 i 0.000129983 j 0.00000070084 k 0.000000009886 l –0.0004998

FLUENCY

202

8 Write in scientific notation, using the number of significant figures given in the brackets. a –23 900 (2) b 5 707 159 (3) c 703 780 030 (2) d 4875 (3) e 0.00192 (2) f –0.00070507 (3) g 0.000009782 (2) h –0.35708 (4) i 0.000050034 (3) 9–10(½)

9 Write the following numerical facts using scientific notation. a b c d e f

The area of Australia is about 7 700 000 km2 . The number of stones used to build the Pyramid of Khufu is about 2 500 000. The greatest distance of Pluto from the Sun is about 7 400 000 000 km. A human hair is about 0.01 cm wide. The mass of a neutron is about 0.000000000000000000000000001675 kg. The mass of a bacteria cell is about 0.00000000000095 g.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

10(½), 11

PROBLEM-SOLVING

9–10(½)

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10 Use a calculator to evaluate the following, giving the answers in scientific notation using three significant figures. a (2.31)–7 b (5.04)–4 c (2.83 × 102 )–3 d 5.1 ÷ (8 × 102 ) e (9.3 × 10–2 ) × (8.6 × 108 ) f (3.27 × 104 ) ÷ (9 × 10–5 ) √ √ 3 g 3.23 × 10–6 h p (3.3 × 107 )2 i 5.73 × 10–4

PROBLEM-SOLVING

Number and Algebra

203

3H

11 The speed of light is approximately 3 × 105 km/s and the average distance between Pluto and the Sun is about 5.9 × 109 km. How long does it take for light from the Sun to reach Pluto? Answer correct to the nearest minute. 12, 13(½)

13–14(½), 15

REASONING

12

12 Explain why 38 × 107 is not written using scientific notation. 13 Write the following using scientific notation. a 21 × 103 b 394 × 107 4 e 0.2 × 10 f 0.007 × 102 –2 i 0.4 × 10 j 0.0031 × 10–11

c 6004 × 10–2 g 0.01 × 109 k 210.3 × 10–6

d 179 × 10–6 h 0.06 × 108 l 9164 × 10–24

14 Combine your knowledge of index laws with scientific notation to evaluate the following and express using scientific notation. a (3 × 102 )2 b (2 × 103 )3 c (8 × 104 )2 –5 2 –3 –2 d (12 × 10 ) e (5 × 10 ) f (4 × 105 )–2 g (1.5 × 10–3 )2 h (8 × 10–8 )–1 i (5 × 10–2 ) × (2 × 10–4 ) –7 2 8 –11 j (3 × 10 ) × (4.25 × 10 ) k (15 × 10 ) × (12 × 10 ) l (18 × 105 ) ÷ (9 × 103 ) m (240 × 10–4 ) ÷ (3 × 10–2 ) n (2 × 10–8 ) ÷ (50 × 104 ) o (5 × 102 ) ÷ (20 × 10–3 ) 15 Rewrite 3 × 10–4 with a positive index and use this to explain why, when expressing 3 × 10–4 as a basic numeral, the decimal point is moved four places to the left. —

—

16

16 E = mc2 is a formula derived by Albert Einstein (1879–1955). The formula relates the energy (E joules) of an object to its mass (m kg), where c is the speed of light (approximately 3 × 108 m/s).

ENRICHMENT

E = mc 2

Use E = mc2 to answer these questions, using scientific notation. a

Find the energy, in joules, contained inside an object with these given masses. i 10 kg ii 26 000 kg iii 0.03 kg iv 0.00001 kg

b Find the mass, in kilograms, of an object that contains these given amounts of energy. Give your answer using three significant figures. i 1 × 1025 J ii 3.8 × 1016 J iii 8.72 × 104 J iv 1.7 × 10–2 J c

The mass of the Earth is about 6 × 1024 kg. How much energy does this convert to?

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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204

Chapter 3 Indices and surds

Progress quiz 38pt 3A

1

Express each number as a decimal and decide if it is rational or irrational. Then locate all the numbers on the same number line. √ 22 a 10 b c p d 315% 7

2

Simplify the following. √ √ √ 5 32 a 98 b 2 75 c 8 √ Express 8 3 as the square root of a positive integer.

10A

38pt 3A 10A

38pt 3A

3

d

√

125 16

10A

38pt 3B

4

Simplify the following. √ √ √ a 7 3–5 3+ 3 √ √ c 5 48 – 2 12

5

Simplify the following. √ √ a – 3× 5

10A

38pt 3C 10A

38pt 3C

6

√ √ √ √ b 4 2–3 5+2 2+5 5 √ √ √ √ d 7 45 – 243 – 2 20 + 27 √ √ b –5 21 × (– 14)

√ √ c 14 7 ÷ 21 35

√ √ √ Use the distributive law to expand 2 3( 6 + 5 24) and simplify the surds where necessary.

10A

38pt 3D

7

Expand and simplify. √ √ a (2 3 – 4)(5 3 + 1)

8

Rationalise the denominators. 3 a √ b 7

10A

38pt 3D 10A

38pt 3F

38pt 3G

38pt 3H

√ 2 3 √ 5

√ √ c ( 5 – 3)( 5 + 3) √ √ 6–3 5 √ c 2

9 Simplify, using index laws. a a3 × a2 b 4x2 y × 3xy3

c h6 ÷ h2 d 5m9 n4 ÷ (10m3 n) ( 4 3 )2 2p q e (a2 )3 f (3m5 )2 g h (2ab)0 + 5m0 7rt2 10 Simplify the following where possible and express your answers using positive indices. 7 4d –7 a x–3 b 2a–2 b4 c–3 c d m–2 5d –5 ( 3 )2 4k 20c–3 d2 e f (2a–2 )–3 g 6a–3 m4 × 2a–2 m–3 h 7 k 15c–1 d–3 11 a b

38pt 3G

√ √ b ( 3 – 2)2

Write these numbers as a basic numeral. i 7.012 × 106 ii 9.206 × 10–3 Write these numbers in scientific notation, using three significant figures. i 351 721 102 ii 0.0002104

12 Simplify the following and express your answers using positive indices. ( –4 )2 ( )2 ( –2 3 )3 (a3 b)5 a 3x2 2x c a × b ÷ 3 –2 2 –3 6 d 5a b c d b

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

3I Rational indices

205

10A

√ √ The square and cube roots of numbers, such as 81 = 9 and 3 64 = 4, can be written using fractional powers. 1 1 √ √ The following shows that 9 = 9 2 and 3 8 = 8 3 . Consider: √ √ 9× 9 = 3×3

1

1 1

1

92 × 92 = 92+2

and

=9

=9 ∴

√

1

9 = 92

Also: √ √ √ 3 3 3 8× 8× 8 = 2×2×2

1

1

1 1 1

1

83 × 83 × 83 = 83+3+3

and

=8

=8 ∴

√ 3

8=

If the volume of a cube is known then the cube root can be used to find its side length.

1 83

A rational index is an index that can be expressed as a fraction.

Let’s start: Making the connection For each part below use your knowledge of index laws and basic surds to simplify the numbers. Then √ discuss the connection that can be made between numbers that have a sign and numbers that have fractional powers. √

1 1 √ 5 × 5 and 5 2 × 5 2  1 2 √ • ( 5)2 and 5 2

•

1 1 1 √ √ 27 × 3 27 × 3 27 and 27 3 × 27 3 × 27 3  1 3 √ • ( 3 64)3 and 64 3

•

√ 3

1 √ an = n a √ • n a is the nth root of a. 1 √ 1 1 √ √ √ For example: 3 2 = 2 3 or 3, 5 3 = 3 5, 7 10 = 10 7

Key ideas

 1 m √ m m 1 √ a n = an = ( n a)m or a n = (am ) n = n am  1 2 2 2 1 For example: 8 3 = 8 3 or 8 3 = (82 ) 3 √ 3 = ( 8)2

1

= 22

= 64 3 √ 3 = 64

=4

=4

In most cases, the index laws apply to rational indices (i.e. fractional indices) just as they do for indices that are integers. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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206

Chapter 3 Indices and surds

Example 21 Writing in index form Express the following in index √ form. √ a 15 b 7x5 SO L U T I O N a

b

√

c

√ 4 3 x7

√ d 10 10

EX P L A N A T I O N √

1

15 = 15 2

means the square root or 1 √ Note: n a = a n .

√ 1 7x5 = (7x5 ) 2

Rewrite

√

1 5

= 72 x2

c

√ 4

7

= 3x 4

d

.

1 as power , then apply index laws 2

to simplify: 5 ×

√ 1 4 3 x7 = 3(x7 ) 4

√ 2

1 5 = . 2 2

1 means to the power of . 4

Apply index law 3 to multiply indices.

√ 1 10 10 = 10 × 10 2

1 Rewrite the square root as power and then 2 add indices for the common base 10. Recall 1 3 10 = 101 , so 1 + = . 2 2 1 √ √ An alternative answer is 100 × 10 = 1000 2 .

3

= 10 2

Example 22 Writing in surd form Express the following in surd form. a

1

2

b 53

35

SO L U T I O N

EX P L A N A T I O N

1 √ a 35 = 5 3

1 √ an = n a

b

2 ( 1 )2 53 = 53 √ 3 = ( 5)2

2 1 Use index law 3 whereby = × 2. 3 3 1 √ 53 = 3 5

Alternatively: 2

1

5 3 = (52 ) 3 √ 3 = 25

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

1 1 × 2 is the same as 2 × . 3 3

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Number and Algebra

207

Example 23 Evaluating numbers with fractional indices Evaluate the following without a calculator. a

1

1

c

b 16 4

16 2

SO L U T I O N a

1

27– 3

EX P L A N A T I O N

1 √ 16 2 = 16

1 √ 16 2 means 16.

=4 b

1 √ 4 16 4 = 16

1 √ 16 4 means 4 16 and 24 = 16.

=2 1

1

27– 3 =

Rewrite, using positive indices. Recall that 1 a–m = m . a 1 √ 27 3 means 3 27 and 33 = 27.

1

27 3 1 =√ 3 27 =

1 3

Exercise 3I 1

1–3(½)

Copy and complete each statement. √ a 2 = 8 and 3 8 = √ c 5 = 125 and 3 125 = √ e 3 = 81 and 81 = 3

2 Evaluate: √ a 9 √ 3 e 8 √ 4 i 16

b f j

√ 25 √ 3 27 √ 4 81

3

—

√ = 9 and 9 = √ d 2 = 32 and 32 = 2 √ f 10 = 100 000 and 100 000 = 10 b 3

√ 121 √ 3 g 125 √ 5 k 32

UNDERSTANDING

c

√ 625 √ 3 h 64 √ 5 l 100 000

c

d

3 Using a calculator, enter and evaluate each pair of numbers in their given form. Round your answer to two decimal places. 1 1 1 √ √ √ a 3 7, 7 3 b 5 10, 10 5 c 13 100, 100 13

Example 21a,b

4 Express the following in index form. √ √ a 29 b 3 35 √ √ 3 e 2a f 4t7

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

√ 5 2 x √ 5 g 10t2 c

4–7(½)

4–7(½)

√ 4 3 b √ 8 h 8m4 d

FLUENCY

4–7(½)

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208

Chapter 3 Indices and surds

Example 22

5 Express √ the following in index form. √ 3 a 7 x5 b 6 n7 √ √ 3 d 5 3 p2 r e 2 a4 b2 √ √ g 5 5 h 7 7 6 Express the following in surd form. 1

c 63

b 87

2

e 33

f

4

h 37

g 25

7 Evaluate without using a calculator.

1

1

b 27 3

a 36 2 1

c 64 3

1

d 49 2

e 16 4

1

1

h 32– 5

g 9– 2 j

d 11 10

3

2

73

1

1

1

1

a 25

Example 23

√ c 3 4 y12 √ f 2 4 g3 h5 √ i 43 4

1

1

1000– 3

k 400– 2

125 3

i

81– 4

l

10 000– 4

8–9(½)

8 Evaluate without using a calculator. 2

a 83 5

f

3

i

3

k

5 42

l

5

9 Use index laws to simplify the following. 1

3

3

e

(

3

b m2 × m2

)4 3 7 s2

f

(

g

10 Simplify the following. √ √ 3 4 a 25s b 27t6 1

e (x3 ) 3

f 1

(16a2 b8 ) 2 √ 25 m 49 i

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

4

1

(b12 ) 3

1

1 3

25– 2 10 3

5

(

) 2 0 t 11

√ 4 16t8 ( )12 1 g t4

3

d b4 ÷ b4  3 2

4

h a3  4

b3

c

(216m6 n3 ) 3 √ 3 3 8x n 27 j

7

c x3 ÷ x3

)1 1 3 y3

2

27– 3

100 2

92

a a2 × a2

8–10(½)

3

h 25– 2

2

1

c 36 2

3

2

j

3

e 16– 4

g 64– 3

1

8–9(½)

b 32 5

d 16 4

1

f

PROBLEM-SOLVING

Example 21c,d

FLUENCY

3I

√ 3 125t12 ( )10 1 h m5

d 1

k (32x10 y15 ) 5 ( )1 32 5 o x10

1

(343r9 t6 ) 3 ( 2 4 )14 10 x p 0.01 l

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Number and Algebra

11

11, 12

REASONING

11

5

11 As shown below, 16 4 can be evaluated in two ways. Method A 5 16 4

=

1 (165 ) 4

Method B   1 5 = 16 4 √ 4 = ( 16)5

5 16 4 1

= (1 048 576) 4 √ 4 = 1 048 576

209

3I

= 25 = 32

= 32 5

a If 16 4 is to be evaluated without a calculator, which method above would be preferable? b Use your preferred method to evaluate the following without a calculator. i

5

3

ii 36 2  3 1 2 vi 9

83 4

v 125 3

4

7

iii 16 4  5 4 2 vii 25

iv 27 3  4 27 3 viii 1000

√ 12 Explain why 6 64 is not a surd.

—

—

13

√ 13 We know that when y = x, where x < 0, y is not a real number. This is because the square of y cannot be negative; i.e. y2 π x since y2 is positive and x is negative. √ But we know that (–2)3 = –8 so 3 –8 = –2. a

Evaluate: √ i 3 –27

ii

√ 3

–1000

iii

√ 5

–32

b Decide if these are real numbers. √ √ √ i –5 ii 3 –7 iii 5 –16 √ c If y = n x and x < 0, for what values of n is y a real number?

iv

√ 7

–2187

iv

√ 4

–12

ENRICHMENT

Does it exist?

√ The square root of a negative value is not a real number. –1 = i and involves a branch of mathematics called complex numbers.

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210

Chapter 3 Indices and surds

3J Exponential growth and decay The population of a country increasing by 5% per year and an investment increasing, on average, by 12% per year are examples of exponential growth. When an investment grows exponentially, the increase per year is not constant. The annual increase is calculated on the value of the investment at that time, and this changes from year to year because of the added investment returns. The more money you have invested, the more interest you will make in a year. In the same way, a population can grow exponentially. A growth of 5% in a large population represents many more babies born in a year than 5% of a small population.

Population growth can be modelled using exponential equations.

Exponential relations will be studied in more detail in Chapter 7. Here we will focus on exponential growth and decay in general and compound interest will be studied in the next section.

Let’s start: A compound rule Imagine you have an antique car valued at $100 000 and you hope that it will increase in value at 10% p.a. The 10% increase is to be added to the value of the car each year. • Discuss how to calculate the value of the car after 1 year. • Discuss how to calculate the value of the car after 2 years. • Complete this table. Year Value ($)

0 100 000

1 100 000 × 1.1 =

2 100 000 × 1.1 × =

3 =

• Discuss how indices can be used to calculate the value of the car after the second year. • Discuss how indices can be used to calculate the value of the car after the tenth year. • What might be the rule connecting the value of the car ($A) and the time n years? • Repeat the steps above if the value of the car decreases by 10% p.a.

Key ideas

Per annum (p.a.) means ‘per year’. Exponential growth and decay can be modelled by the rule A = kat , where A is the amount, k is the initial amount and t is the time. • When a > 1, exponential growth occurs. • When 0 < a < 1, exponential decay occurs. r For a growth rate of r% p.a., the base ‘a’ is calculated using a = 1 + . 100 r For a decay rate of r% p.a., the base ‘a’ is calculated using a = 1 – . 100

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Number and Algebra

The basic exponential formula can be summarised as A = A0



r 1± 100

n

.

211

Key ideas

• The subscript zero is often used to indicate the initial value of a quantity (e.g. P0 is initial population).

Example 24 Writing exponential rules Form exponential rules for the following situations. a John has a painting that is valued at $100 000 and which is expected to increase in value by 14% per annum. b A city’s initial population of 50 000 is decreasing by 12% per year. SOL UTI ON

EX P L A NA TI ON

a

Define your variables.  n r . A = A0 1 ± 100

Let A = the value in $ of the painting at any time n = the number of years the painting is kept r = 14 A0 = 100 000 

n 14 A = 100 000 1 + 100

∴ A = 100 000(1.14)n b

Let P = the population at any time n = the number of years the population decreases r = 12 P0 = 50 000 

n 12 P = 50 000 1 – 100

∴ P = 50 000(0.88)n

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Substitute r = 14 and A0 = 100 000 and use ‘+’ since we have growth. Define your variables.  n r . P = P0 1 ± 100

Substitute r = 12 and P0 = 50 000 and use ‘–’ since we have decay.

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212

Chapter 3 Indices and surds

Example 25 Applying exponential rules House prices are rising at 9% per year and Zoe’s flat is currently valued at $600 000. a Determine a rule for the value of Zoe’s flat ($V) in n years’ time. b c

What will be the value of her flat: i next year?

ii in 3 years’ time?

Use trial and error to find when Zoe’s flat will be valued at $900 000, to one decimal place.

SO L U T I O N

EX P L A N A T I O N

a

n = number of years from now

Define ( your variables. )n r . V = V0 1 ± 100

r=9

Use ‘+’ since we have growth.

Let V = value of Zoe’s flat at any time V0 = starting value $600 000

V = V0 (1.09)n ∴ V = 600 000(1.09)n i When n = 1, V = 600 000(1.09)1

b

Substitute n = 1 for next year.

= 654 000 Zoe’s flat would be valued at $654 000 next year. ii When n = 3, V = 600 000(1.09)3

For 3 years, substitute n = 3.

= 777 017.40 In 3 years’ time Zoe’s flat will be valued at about $777 017. n V

4 5 4.6 4.8 4.7 846 949 923 174 891 894 907 399 899 613

Zoe’s flat will be valued at $900 000 in about 4.7 years’ time.

Exercise 3J 1

Try a value of n in the rule. If V is too low, increase your n value. If V is too high, decrease your n value. Continue this process until you get close to 900 000.

1–3

3

An antique ring is purchased for $1000 and is expected to grow in value by 5% per year. Round your answers to the nearest cent. a b c d e

Find the increase in value in the first year. Find the value of the ring at the end of the first year. Find the increase in value in the second year. Find the increase in value in the third year. Find the value of the ring at the end of the fifth year.

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—

UNDERSTANDING

c

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UNDERSTANDING

Number and Algebra

2 The mass of a limestone 5 kg rock exposed to the weather is decreasing at a rate of 2% per annum. a Find the mass of the rock at the end of the first year. b Write the missing numbers for the mass of the rock (M kg) after t years. )t M = 5(1 – c

213

3J

t = 5× Use your rule to calculate the mass of the rock after 5 years, correct to two decimal places.

3 Decide if the following represent exponential growth or exponential decay. a A = 1000 × 1.3t b A = 200 × 1.78t c A = 350 × 0.9t ( ( )t )t 3 7 t d P = 50 000 × 0.85 e P = P0 1 + f T = T0 1 – 100 100

Example 24

5–8

4 Define variables and form exponential rules for the following situations. a A flat is purchased for $200 000 and is expected to grow in value by 17% per annum. b A house initially valued at $530 000 is losing value at 5% per annum. c The value of a car, bought for $14 200, is decreasing at 3% per annum. d A population, which is initially 172 500, is increasing at 15% per year. e f g h

Example 25

4–7

FLUENCY

4–6

5

A tank with 1200 litres of water is leaking at a rate of 10% of the water in the tank every hour. A human cell of area 0.01 cm2 doubles its area every minute. An oil spill, initially covering an area of 2 square metres, is increasing at 5% per minute. A substance of mass 30 g is decaying at a rate of 8% per hour.

The value of a house purchased for $500 000 is expected to grow by 10% per year. Let $A be the value of the house after t years. a Write the missing number in the rule connecting A and t. t A = 500 000 × b Use your rule to find the expected value of the house after the following number of years. Round your answer to the nearest cent. i c

3 years

ii 10 years

iii 20 years

Use trial and error to estimate when the house will be worth $1 million. Round your answer to one decimal place.

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Chapter 3 Indices and surds

3J 6 A share portfolio, initially worth $300 000, is reduced by 15% p.a. over a number of years. Let $A be the share portfolio value after t years.

FLUENCY

214

a

Write the missing number in the rule connecting A and t. × 0.85t A= b Use your rule to find the value of the shares after the following number of years. Round your answer to the nearest cent. i 2 years ii 7 years iii 12 years c

Use trial and error to estimate when the share portfolio will be valued at $180 000. Round your answer to one decimal place.

7 A water tank containing 15 000 L has a small hole that reduces the amount of water by 6% per hour. a Determine a rule for the volume of water (V) left after t hours. b Calculate (to the nearest litre) the amount of water left in the tank after: i 3 hours ii 7 hours c How much water is left after two days? Round your answer to two decimal places. d Using trial and error, determine when the tank holds less than 500 L of water, to one decimal place. 8 Megan invests $50 000 in a superannuation scheme that has an annual return of 11%. a Determine the rule for the value of her investment (V) after n years. b How much will Megan’s investment be worth in: i 4 years? ii 20 years? Find the approximate time before her investment is worth $100 000. Round your answer to two decimal places.

9, 10

9 A certain type of bacteria grows according to the equation N = 3000(2.6)t , where N is the number of cells present after t hours. a How many bacteria are there at the start? b Determine the number of cells (round to the whole number) present after: i 1 hour ii 2 hours iii 4.6 hours c

9, 10

10, 11

PROBLEM-SOLVING

c

If 50 000 000 bacteria are needed to make a drop of serum, determine how long you will have to wait to make a drop (to the nearest minute).

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10 A car tyre has 10 mm of tread when new. It is considered unroadworthy when there is only 3 mm left. The rubber wears at 12.5% every 10 000 km. a Write an equation relating the depth of tread (D) for every 10 000 km travelled. b Using trial and error, determine when the tyre becomes unroadworthy, to the nearest 10 000 km. c If a tyre lasts 80 000 km, it is considered to be of good quality. Is this a good quality tyre?

PROBLEM-SOLVING

Number and Algebra

215

3J

11 A cup of coffee has an initial temperature of 90◦ C and the surrounding temperature is 0◦ C. a

If the temperature relative to surroundings reduces by 8% every minute, determine a rule for the temperature of the coffee (T ◦ C) after t minutes. b What is the temperature of the coffee (to one decimal place) after: i 90 seconds? ii 2 minutes? When is the coffee suitable to drink if it is best consumed at a temperature of 68.8◦ C? Give your answer to the nearest second. 12

12

12, 13

12 The monetary value of things can be calculated using different time periods. Consider a $1000 collector’s item that is expected to grow in value by 10% p.a. over 5 years. • If the increase in value is added annually then r = 10 and t = 5, so A = 1000(1.1)5 . 10 • If the increase in value is added monthly then r = and t = 5 × 12 = 60, so 12 ( )60 10 . A = 1000 1 + 1200 a

REASONING

c

If the increase in value is added annually, find the value of the collectors’ item, to the nearest cent, after: i 5 years ii 8 years iii 15 years

b If the increase in value is added monthly, find the value of the collectors’ item, to the nearest cent, after: i 5 years ii 8 years iii 15 years

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3J 13 You inherit a $2000 necklace that is expected to grow in value by 7% p.a. What will the necklace be worth, to the nearest cent, after 5 years if the increase in value is added: a annually? b monthly? c weekly (assume 52 weeks in the year)?

Half-life

—

—

14–16

Half-life is the period of time it takes for an object to decay by half. It is often used to compare the rate of decay for radioactive materials. 14 A 100 g mass of a radioactive material decays at a rate of 10% every 10 years. a

REASONING

Chapter 3 Indices and surds

ENRICHMENT

216

Find the mass of the material after the following time periods. Round your answer to one decimal place. i 10 years ii 30 years iii 60 years

b Estimate the half-life of the radioactive material (i.e. find how long it takes for the material to decay to 50 g). Use trial and error and round your answer to the nearest year. 15 An ice sculpture, initially containing 150 L of water, melts at a rate of 3% per minute. a

What will be the volume of the ice sculpture after half an hour? Round your answer to the nearest litre. b Estimate the half-life of the ice sculpture. Give your answer in minutes, correct to one decimal place.

16 The half-life of a substance is 100 years. Find the rate of decay per annum, expressed as a percentage correct to one decimal place.

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Number and Algebra

217

3K Compound interest For simple interest, the interest is always calculated on the principal amount. Sometimes, however, interest is calculated on the actual amount present in an account at each time period that interest is calculated. This means that the interest is added to the amount, then the next lot of interest is calculated again using this new amount. This process is called compound interest. Compound interest can be calculated using updated applications of the simple interest formula or by using the compound interest formula. It is a common example of exponential growth.

Let’s start: Investing using updated simple interest Consider investing $400 at 12% per annum. • Copy and complete the table below. Time (n) 1st year 2nd year 3rd year 4th year

• •

Amount (A ) $400 $448 $501.76

Interest (I ) $48 $53.76

New amount $448 $501.76

What is the balance at the end of 4 years if interest is added to the amount at the end of each year? Thinking about this as exponential growth, write a rule linking A with n. Compound interest can be found using updated applications of the simple interest formula. For example, $100 compounded at 10% p.a. for 2 years.

Key ideas

Year 1: 100 + 10% of 100 = $110 Year 2: 110 + 10% of 110 = $121, so compound interest = $21 The total amount in an account using compound interest for a given number of time periods is given by: n  r , where: A=P 1+ 100 • Principal (P) = the amount of money borrowed or invested. • Rate of interest (r) = the percentage applied to the principal per period of investment. • Periods (n) = the number of periods the principal is invested. • Amount (A) = the total amount of your investment. Interest = amount (A) – principal (P)

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218

Chapter 3 Indices and surds

Example 26 Converting rates and time periods Calculate the number of periods and the rates of interest offered per period for the following. a 6% p.a. over 4 years, paid monthly b

18% p.a. over 3 years, paid quarterly

SO L U T I O N a

b

n = 4 × 12

EX P L A N A T I O N r = 6 ÷ 12

= 48

= 0.5

n = 3×4

r = 18 ÷ 4

= 12

= 4.5

4 years is the same as 48 months, as 12 months = 1 year. 6% p.a. = 6% in 1 year. Divide by 12 to find the monthly rate. There are 4 quarters in 1 year.

Example 27 Using the compound interest formula Determine the amount after 5 years if $4000 is compounded annually at 8%. Round to the nearest cent. SO L U T I O N

EX P L A N A T I O N

P = 4000, n = 5, r = 8 ( )n r A=P 1+ 100 ( )5 8 = 4000 1 + 100

List the values for the terms you know. Write the formula and then substitute the known values.

= 4000(1.08)5

Simplify and evaluate.

= $5877.31

Write your answer to two decimal places (the nearest cent)

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Number and Algebra

219

Example 28 Finding compounded amounts using months Anthony’s investment of $4000 is compounded at 8.4% p.a. over 5 years. Determine the amount he will have after 5 years if the interest is paid monthly. Round to the nearest cent. SO L U T I O N

EX P L A N A T I O N

P = 4000

List the values of the terms you know. Convert the time in years to the number of periods (in this case, months); 60 months = 5 years. Convert the rate per year to the rate per period (months) by dividing by 12.

n = 5 × 12 = 60 r = 8.4 ÷ 12

r A=P 1+ 100

)n

Write the formula.

= 4000(1 + 0.007)60 = 4000(1.007)

Substitute the values, 0.7 ÷ 100 = 0.007.

60

= $6078.95

Simplify and evaluate, rounding to the nearest cent.

Exercise 3K 1

1–5

4, 5

Consider $500 invested at 10% p.a., compounded annually. a How much interest is earned in the first year? b What is the balance of the account once the first year’s interest is added? c How much interest is earned in the second year? d What is the balance of the account at the end of the second year? e Use your calculator to work out 500(1.1)2 .

—

UNDERSTANDING

= 0.7 (

2 Find the value of the following, correct to two decimal places. a $1000 × 1.05 × 1.05 b $1000 × 1.052 c $1000 × 1.05 × 1.05 × 1.05 d $1000 × 1.053 3 Fill in the missing numbers. a $700 invested at 8% p.a., compounded annually for 2 years. A=

(1.08)

b $1000 invested at 15% p.a., compounded annually for 6 years. )6

A = 1000(

c $850 invested at 6% p.a., compounded annually for 4 years. A = 850(

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)

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Chapter 3 Indices and surds

3K Example 26

4 Calculate the number of periods (n) and the rates of interest (r) offered per period for the following. (Round the interest rate to three decimal places where necessary.) a 6% p.a. over 3 years, paid bi-annually b 12% p.a. over 5 years, paid monthly c 4.5% p.a. over 2 years, paid fortnightly d 10.5% p.a. over 3.5 years, paid quarterly e 15% p.a. over 8 years, paid quarterly f 9.6% p.a. over 10 years, paid monthly

UNDERSTANDING

220

5 By considering an investment of $4000 at 5% p.a., compounded annually, copy and complete the table below. Year 1 2 3 4 5

Amount ($) 4000 4200

Interest ($) 200

Example 27

6–8(½)

6 Determine the amount after 5 years if: a $4000 is compounded annually at 5% b $8000 is compounded annually at 8.35% c $6500 is compounded annually at 16% d $6500 is compounded annually at 8%. 7 Determine the amount if $100 000 is compounded annually at 6% for: a 1 year b 2 years c 3 years d 5 years e 10 years

Example 28

7–8(½)

FLUENCY

6–8(½)

New amount ($) 4200

f

15 years

8 Calculate the value of the following investments if interest is compounded monthly. a $2000 at 6% p.a. for 2 years b $34 000 at 24% p.a. for 4 years c $350 at 18% p.a. for 8 years

1 d $670 at 6.6% p.a. for 2 years 2

e $250 at 7.2% p.a. for 12 years 9, 10

10, 11

9 An investment of $8000 is compounded at 12.6% over 3 years. Determine the amount the investor will have after 3 years if the interest is compounded monthly. 10 Darinia invests $5000 compounded monthly at 18% p.a. Determine the value of the investment after: a 1 month b 3 months c 5 months

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

PROBLEM-SOLVING

9, 10

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11 a For each rate below, calculate the amount of compound interest paid on $8000 at the end of 3 years. i 12% compounded annually ii 12% compounded bi-annually (i.e. twice a year) iii 12% compounded monthly iv 12% compounded weekly v 12% compounded daily

PROBLEM-SOLVING

Number and Algebra

221

3K

b What is the interest difference between annual and daily compounding in this case?

12

12(½), 13

REASONING

12

12 The following are expressions relating to compound interest calculations. Determine the principal (P), number of periods (n), rate of interest per period (r%), annual rate of interest (R%) and the overall time (t). a 300(1.07)12 , bi-annually b 5000(1.025)24 , monthly 65 c 1000(1.00036) , fortnightly d 3500(1.000053)30 , daily e 10 000(1.078)10 , annually 13 Paula must decide whether to invest her $13 500 for 6 years at 4.2% p.a. compounded monthly or 5.3% compounded bi-annually. Decide which investment would be the best choice for Paula.

—

—

14

14 You have $100 000 to invest and wish to double that amount. Use trial and error in the following. a Determine, to the nearest whole number of years, the length of time it will take to do this using the compound interest formula at rates of: i 12% p.a. ii 6% p.a. iii 8% p.a. iv 16% p.a. v 10% p.a. vi 20% p.a.

ENRICHMENT

Double your money

b If the amount of investment is $200 000 and you wish to double it, determine the time it will take using the same interest rates as above. c Are the lengths of time to double your investment the same in part a and part b?

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222

Chapter 3 Indices and surds

3L Comparing interest using technology In the following exercise, we compare compound and simple interest and look at their applications to the banking world. You are expected to use technology to its best advantage when solving the problems in this section.

Let’s start: Who earns the most? • • •

Key ideas

Ceanna invests $500 at 8% p.a., compounded monthly over 3 years. Huxley invests $500 at 10% p.a., compounded annually over 3 years. Loreli invests $500 at 15% p.a. simple interest over 3 years. – How much does each person have at the end of the 3 years? – Who earned the most? For either form of interest, you can calculate the total amount of your investment using technology. Graphics calculator To create programs for the two types of interest, enter the following data. This will allow you to calculate both types of interest for a given time period. If you invest $100 000 at 8% p.a. paid monthly for r 2 years, you will be asked for P, R = , t or n and the calculator will 100 do the work for you. Note: Some modifications may be needed for the CAS or other calculators. Spreadsheet Copy and complete the spreadsheets as shown, to compile a simple interest and compound interest sheet.

Fill in the principal in B3 and the rate per period in D3. For example, for $4000 invested at 0.054 5.4% p.a. paid monthly, B3 will be 4000 and D3 will be . 12 Prt Recall the simple interest formula from previous years: I = where I is the total amount of 100 interest, P is the initial amount or principal, r is percentage interest rate and t is the time. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

223

Example 29 Comparing simple and compound interest using technology Find the total amount of the following investments, using technology. a $5000 at 5% p.a., compounded annually for 3 years b

$5000 at 5% p.a. simple interest for 3 years

SO L U T I O N

EX P L A N A T I O N ) ( r n using P = 5000, A=P 1+ 100 r = 5 and n = 3. Alternatively, use a spreadsheet or computer program. Refer to the Key ideas.

a $5788.13

b $5750

Total = P +

Prt using P = 5000, r = 5 100

and t = 3.

1

1–3

2, 3

—

Which is better on an investment of $100 for 2 years: A simple interest calculated at 20% p.a. or B compound interest calculated at 20% p.a. and paid annually?

2 Write down the values of P, r and n for an investment of $750 at 7.5% p.a., compounded annually for 5 years.

UNDERSTANDING

Exercise 3L

3 Write down the values of I, P, r and t for an investment of $300 at 3% p.a. simple interest over 300 months.

Example 29

5

4 a Find the total amount of the following investments, using technology. i $6000 at 6% p.a., compounded annually for 3 years ii $6000 at 3% p.a., compounded annually for 5 years iii $6000 at 3.4% p.a., compounded annually for 4 years iv $6000 at 10% p.a., compounded annually for 2 years v $6000 at 5.7% p.a., compounded annually for 5 years

5

FLUENCY

4

b Which of the above yields the most interest? 5 a Find the total amount of the following investments, using technology where possible. i $6000 at 6% p.a. simple interest for 3 years ii $6000 at 3% p.a. simple interest for 6 years iii $6000 at 3.4% p.a. simple interest for 7 years iv $6000 at 10% p.a. simple interest for 2 years v $6000 at 5.7% p.a. simple interest for 5 years b Which of the above yields the most interest? Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter 3 Indices and surds

6

3L 6 a

6

6

Determine the total simple and compound interest accumulated in the following cases. i

$4000 at 6% p.a. payable annually for: I

II 2 years

1 year

III 5 years

IV 10 years

III 5 years

IV 10 years

III 5 years

IV 10 years

ii $4000 at 6% p.a. payable bi-annually for: I

II 2 years

1 year

PROBLEM-SOLVING

224

iii $4000 at 6% p.a. payable monthly for: I

c

Would you prefer the same rate of compound interest or simple interest if you were investing money? Would you prefer the same rate of compound interest or simple interest if you were borrowing money and paying off the loan in instalments?

7

7

8

7 a Copy and complete the following table when simple interest is applied. Principal $7000 $7000

$9000 $18 000

Rate

10% 10% 8% 8%

Overall time 5 years 5 years 3 years 3 years 2 years 2 years

Interest

Amount $8750 $10 500

REASONING

b

II 2 years

1 year

$990 $2400

b Explain the effect on the interest when we double the: i rate ii period iii overall time 8 Copy and complete the following table when compound interest is applied. You may need to use a calculator and trial and error to find some of the missing numbers. Principal $7000 $7000 $9000 $18 000

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Rate

8% 8%

Period annually annually fortnightly fortnightly

Overall time 5 years 5 years 2 years 2 years

Interest

Amount $8750 $10 500

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Number and Algebra

—

—

9, 10

9 If you invest $5000, determine the interest rate per annum (to two decimal places) if the total amount is approximately $7500 after 5 years and if: a interest is compounded annually b interest is compounded quarterly c interest is compounded weekly

ENRICHMENT

Changing the parameters

225

3L

Comment on the effect of changing the period for each payment on the rate needed to achieve the same total amount in a given time. 10 a Determine, to one decimal place, the equivalent simple interest rate for the following investments over 3 years. i $8000 at 4%, compounded annually ii $8000 at 8%, compounded annually b If you double or triple the compound interest rate, how is the simple interest rate affected?

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226

Chapter 3 Indices and surds

3M Exponential equations

10A

Equations can take many forms. For example, 2x – 1 = 5 and 5(a – 3) = –3(3a + 7) are both linear equations; x2 = 9 and 3x2 – 4x – 9 = 0 are quadratic equations; and 2x = 8 and 32x – 3x – 6 = 0 are exponential equations. Exponential equations contain a pronumeral within the index or indices of the terms in the equation. To solve for the unknown in exponential equations we use our knowledge of indices and surds and try to equate powers where possible.

Let’s start: 2 to the power of what number is 5? We know that 2 to the power of 2 is 4 and 2 to the power of 3 is 8, but 2 to the power of what number is 5? That is, what is x when 2x = 5? •

Use a calculator and trial and error to estimate the value of x when 2x = 5 by completing this table. x 2x Result

•

2 4 too small

3 8 too big

2.5 5.65... too big

2.1

Continue trying values until you find the answer, correct to three decimal places.

Key ideas

A simple exponential equation is of the form ax = b, where a > 0, b > 0 and a π 1. • There is only one solution to exponential equations of this form. Many exponential equations can be solved by expressing both sides of the equation using the same base. • We use this fact: if ax = ay then x = y.

Example 30 Solving exponential equations Solve for x in each of the following. a

2x = 16

b 3x =

SOL UTI ON a

4

2 =2 ∴ x=4 b

c

25x = 125

EX P L A NA TI ON

2x = 16 x

1 9

1 9 1 3x = 2 3 3x = 3–2 3x =

Rewrite 16 as a power, using the base 2. Equate powers using the result: if ax = ay then x = y. Rewrite 9 as a power of 3, then write using a negative index.

Equate powers with the same base.

∴ x = –2 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

25x = 125 ( )x 3 52 = 5 52x = 53

c

Since 25 and 125 are both powers of 5, rewrite both with a base of 5. Apply index law 3 to remove brackets and multiply indices, then equate powers and solve for x.

∴ 2x = 3 x=

227

3 2

Example 31 Solving exponential equations with a variable on both sides Solve 32x – 1 = 27x SO L U T I O N

EX P L A N A T I O N

32x – 1 = 27x 2x – 1

3

2x – 1

3

Rewrite 27 as a power of 3.

3 x

= (3 )

Apply index law 3 to remove brackets and then equate powers.

3x

=3

∴ 2x – 1 = 3x

Subtract 2x from both sides and answer with x as the subject.

–1 = x ∴ x = –1

1

a

1–3

Evaluate the following. i 22 ii 23

b Hence, state the value of x when: i 2x = 8 ii 2x = 32

3

iii 24

—

iv 25

iii 2x = 64

UNDERSTANDING

Exercise 3M

2 Complete these patterns, which involve powers. a b c d e

2, 4, 8, 3, 9, 27, 4, 16, 5, 25, 6, 36,

,

, ,

, , ,

, ,

, ,

, ,

, ,

,

3 Write these numbers in index form. For example, 32 = 25 . a 9 b 125 c 243 d 128

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e 729

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Chapter 3 Indices and surds

Example 30a

Example 30b

4 Solve for x in each of the following. a 3x = 27 b 2x = 8 x e 5 = 125 f 4x = 64 i 5x = 625 j 2x = 32

1 256

e 3x =

1 9

h 2–x =

g 3–x = Example 30c

c 6x = 36 g 3x = 81 k 10x = 10 000

5 Solve for x in each of the following. 1 1 a 7x = b 9x = 49 81 d 4x =

4–6(½)

6 Solve for x in each of the following. a 9x = 27 b 8x = 16 x e 81 = 9 f 216x = 6 –x i 7 = 49 j 4–x = 256

4–6(½)

FLUENCY

4–5(½)

3M

d 9x = 81 h 6x = 216 l 7x = 343

c 11x =

1 121

1 243

f

5–x =

1 125

1 64

i

7–x =

1 343

c 25x = 125 g 32x = 2 k 16–x = 64

d 16x = 64 h 10 000x = 10 l 25–x = 125

7, 8(½)

7, 8(½)

8(½), 9

7 The population of bacteria in a dish is given by the rule P = 2t , where P is the bacteria population and t is the time in minutes. a What is the initial population of bacteria; i.e. when t = 0? b What is the population of bacteria after: i 1 minute? ii 5 minutes? iii 1 hour? c

Example 31

iv 1 day?

PROBLEM-SOLVING

228

How long does it take for the population to reach: i 8? ii 256? iii more than 1000?

8 Solve for x in each of the following. a 2x + 1 = 8x b x + 3 2x d 5 = 25 e g 27x + 3 = 92x h 2x + 3 2x – 1 j 27 =9 k

32x + 1 = 27x 62x + 3 = 2162x 25x + 3 = 1253x 9x – 1 = 272x – 6

c f i l

7x + 9 = 492x 9x + 12 = 81x + 5 322x + 3 = 1282x 492x – 3 = 3432x – 1

9 Would you prefer $1 million now or 1 cent doubled every second for 30 seconds? Give reasons for your preference.

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Number and Algebra

10, 11(½)

11–13(½)

REASONING

10

10 Consider ax , where a = 1. a

Evaluate 1x when: i x=1

ii x = 3

b Are there any solutions to the equation

ax

iii x = 10 000 000

3M

= 2 when a = 1? Give a reason.

1 1 √ √ 11 Recall that x = x 2 and 3 x = x 3 . Now solve the following. √ √ √ a 3x = 81 b 5x = 25 c 6x = 3 36 √ √ √ e 2x = 4 32 f 3x = 9 27 g 25x = 5 125

12 a

229

Write these numbers as decimals. 1 i ii 2–3 22

iii

√ d 4x = 4 64 1 h 9x = √ 3 27  4 1 iv 5

10–3

b Write these decimal numbers as powers of prime numbers. i 0.04 ii 0.0625 iii 0.5

iv 0.0016

13 Show how you can use techniques from this section to solve these equations involving decimals. a 10x = 0.0001 b 2x = 0.015625 c 5x = 0.00032 1 d (0.25)x = 0.5 e (0.04)x = 125 f (0.0625)x + 1 = 2

14 Solve for n in the following. a 3n × 9n = 27 d 32n – 1 =

1 81

—

b 53n × 25–2n + 1 = 125 e 72n + 3 =

g 62n – 6 = 1 3n – 2 j =9 91 – n

1 49

h 113n – 1 = 11 53n – 3 k = 125 25n – 3

—

14

c 2–3n × 42n – 2 = 16 1 625

f

53n + 2 =

i

85n – 1 = 1 363 + 2n =1 6n

l

ENRICHMENT

Mixing index laws with equations

Many mathematicians work in scientific research and development, where index laws are often used in scientific formulas.

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230

Chapter 3 Indices and surds

Investigation True wealth Chances are that people who become wealthy have invested money in appreciating assets, such as property, shares and other businesses, rather than depreciating assets, such as cars, televisions and other electronic devices. Appreciating assets increase in value over time and usually earn income. Depreciating assets lose value over time. Appreciating or depreciating? Imagine you have $100 000 to invest or spend and you have these options. Option 1: Invest in shares and expect a return of 8% p.a. Option 2: Buy a car that depreciates at 8% p.a. a Find the value of the $100 000 share investment after: i 2 years ii 5 years

iii 10 years

b How long will it take for the share investment to double in value? c Find the value of the $100 000 car after: i 2 years ii 5 years

iii 10 years

d How long will it take for the value of the car to fall to half its original value? e Explain why people who want to create wealth might invest in appreciating assets. Buying residential property A common way to invest in Australia is to buy residential property. Imagine you have $500 000 to buy an investment property that is expected to grow in value by 10% p.a. Stamp duty and other buying costs total $30 000. Each year the property has costs of $4000 (e.g. land tax, rates and insurance) and earns a rental income of $1200 per month. a

What is the initial amount you can spend on a residential property after taking into account the stamp duty and other buying costs?

b

What is the total rental income per year?

c

What is the total net income from the property per year after annual expenses have been considered?

d

By considering only the property’s initial capital value, find the expected value of the property after: i 5 years ii 10 years iii 30 years

e

By taking into account the rise in value of the property and the net income, determine the total profit after: i 1 year ii 3 years iii 5 years

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Number and Algebra

231

Borrowing to invest Borrowing money to invest can increase returns but it can also increase risk. Interest has to be paid on the borrowed money, but this can be offset by the income of the investment. If there is a net loss at the end of the financial year, then negative gearing (i.e. borrowing) has occurred. This net loss can be used to reduce the amount of tax paid on other income, such as salary or other business income, under Australian taxation laws. Imagine that you take out a loan of $300 000 to add to your own $200 000 so you can spend $500 000 on the investment property. In summary: • The property is expected to grow in value by 10% p.a. • Your $300 000 loan is interest only at 7% p.a., meaning that only interest is paid back each year and the balance remains the same. • Property costs are $4000 p.a. • Rental income is $1200 per month. • Your taxable income tax rate is 30%. a

Find the amount of interest that must be paid on the loan each year.

b

Find the net cash loss for the property per year. Include property costs, rent and loan interest.

c

This loss reduces other income, so with a tax rate of 30% this loss is reduced by 30%. Now calculate the overall net loss, including this tax benefit.

d

Now calculate the final net gain of the property investment for the following number of years. You will need to find the value of the appreciating asset (which is initially $470 000) and subtract the net loss for each year from part c above. i 1 year ii 5 years iii 20 years

You now know how millions of Australians try to become wealthy using compounding investment returns with a little help from the taxman!

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232

Chapter 3 Indices and surds Up for a challenge? If you get stuck on a question, check out the 'Working with unfamiliar problems' poster at the end of the book to help you.

Problems and challenges 1

Write 3n – 1 + 3n – 1 + 3n – 1 as a single term with base 3.

2

Simplify: 256 × 54 a 1255

3

Solve 32x × 27x + 1 = 81.

4

Simplify: 2n + 1 – 2n + 2 a n–1 n–2 2 –2

5

b

8x × 3 x 6x × 9x

b

2a + 3 – 4 × 2 a 22a + 1 – 4a

√ A rectangular piece of paper has an area of 100 2 cm2 . The piece of paper is such that, when it is folded in half along the dashed line as shown, the new rectangle is similar (i.e. of the same shape) to the original rectangle. What are the dimensions of the piece of paper?

6

Simplify the following, leaving your answer with a rational denominator. √ 2 2 √ +√ 3+1 2 2+1

7

Simplify: 1

1

1 1

x 2 y– 2 – x– 2 y 2 a √ xy 8

9

1

1

1 1

x 2 y– 2 – x– 2 y 2 b x–1 y–1

Three circles, each of radius 1 unit, fit inside a square such that the two outer circles touch the middle circle and the sides of the square, as shown. Given the centres of the circle lie on the diagonal of the square, find the exact area of the square.

√ Given that 5x + 1 – 5x – 2 = 620 5, find the value of x.

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Negative indices a −m = 1m

Scientific notation Numbers (large and small) in scientific notation are of the form × 10-3 a × 10m, where 1 ≤ a < 10 or −10 < a ≤ −1 and m is an integer. e.g. 3.45 × 105 = 345 000 2.1 × 10−3 = 0.0021 e.g. 1475 231 in scientific notation, using three significant figures is 1.48 × 106.

a

1

= am

a −m

Distributive law (10A) e.g. 1

2 ( 2 3 + 5) = 2 6 + 5 2

2 ( 6 + 5)( 2 6 – 3) = 2 × 6 – 3 6 + 10 6 – 15 =7 6–3 3 ( 7 – 2 )( 7 + 2 ) = 7 + 14 – 14 – 2 =5

Rational Indices (10A) Index laws 1: am

an

1

a = a2

am + n

Law × = Law 2: am ÷ an = am − n

n

1

a = an

n m a =

Law 3: (a m )n = amn Law 4: (ab)m = ambm m m Law 5: a = a m b b Zero power: a0 = 1, a ≠ 0

m

Simplifying surds (10A)

an 1

3

e.g. 273 = 27 =3

Exponential equations (10A) If a x = a y, then x = y. e.g. 1 2x = 16 2x = 24 ∴ x=4 27x − 4 = 9x (33)x − 4 = (32)x 33x − 12 = 32x ∴ 3x − 12 = 2x x = 12

233

Chapter summary

Number and Algebra

Indices and surds

A surd uses symbol and as a decimal is infinite and non-recurring. Simplify surds: use the highest square factor. e.g. 1 20 = 4 × 5 = 4× 5 =2 5 2 4 27 = 4 × 9 × 3 = 12 3

2

Multiply/Divide (10A) x × y = xy

Exponential growth/decay

Rule: amount A = A0(1 ± ) A0 = initial amount r = rate of growth/decay, as a % n = time period Use + for growth. Use − for decay. r n 100

a x × b y = ab xy x x÷ y = y a x a x ÷ (b y ) = b y

Add/Subtract (10A) Like surds only e.g. 1 3 2 + 4 2 = 7 2 4 2+ 5 − 2 = 3 2+ 5 Simplify first e.g. 2 3 8 − 4 2 = 3× 4 × 2 − 4 2 = 6 2 −4 2 =2 2

Compound interest A = P (1 + 100 ) A (amount): total value P (principal): initial amount r (interest rate): per period n (period): number of periods r

n

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Rationalise the denominator (10A) Express with a whole number in the denominator 2 = 2 × 5 5 5 5 2 5 = 5

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Chapter review

234

Chapter 3 Indices and surds

Multiple-choice questions 38pt 3A

1

10A

38pt 3A 10A

38pt 3A 10A

38pt 3B 10A

38pt 3C/D 10A

38pt 3E 10A

38pt 3F

Which of the following is a surd? √ A 36 B p

38pt 3I 10A

38pt 3M 10A

38pt 3J

D

√ 3 8

√ 3 4 5 is equivalent to: √ √ A 100 B 80

√ C 2 10

D

√ √ 4 3 12 + 7 – 4 3 simplifies to: √ √ A 7+ 3 B 8 3+7

√ √ C 6 6+3 3

√ D 3 3

√ √ 5 The expanded form of 2 5(5 – 3 3) is: √ √ √ √ A 10 5 – 6 15 B 7 5 – 5 15 √ √ √ D 10 – 5 15 E 7 5–5 3 √ 2 5 6 √ is equivalent to: 6√ √ √ 2 30 5 6 A √ B C 2 5 3 6 7 The simplified form of

8

E 1.6˙

y4 2x6

B

(6xy3 )2 3x3 y2 × 4x4 y0

3y3 x10

√ 20

E

√ 40

√ E 2 3+7

√ √ C 10 5 – 12 2

√

30 3

D

√ 30 E 10

is:

C

y6 x5

D

3y4 x5

E

y6 2x6

E

3 4 7 a b 2

8a–1 b–2 expressed with positive indices is: 12a3 b–5 A

38pt 3H

√ 7

2 A square has an area of 75 square units. Its side length, in simplified form, is: √ √ √ √ A 3 5 B 8.5 C 25 3 D 5 3 E 6 15

A 38pt 3G

C

2a2 3b3

B

a2 b3 96

C

2b3 3a4

D –

2b7 3a2

9 The radius of the Earth is approximately 6 378 137 m. In scientific notation, using three significant figures, this is: A 6.378 × 106 m B 6.38 × 106 m C 6.4 × 105 m D 6.37 × 106 m E 6.36 × 106 m √ 10 8x6 in index form is: A 8x3

B 8x2

11 The solution to 32x – 1 = 92 is: 3 A x= B x=2 2

1

C 4x3

C x=

5 2

1

D 8 2 x3

E 8 2 x4

D x=6

E x=3

12 A rule for the amount $A in an account after n years for an initial investment of $5000 that is increasing at 7% per annum is: A A = 5000(1.7)n B A = 5000(0.93)n C A = 5000(0.3)n D A = 5000(1.07)n E A = 5000(0.7)n

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Short-answer questions 38pt 3A 10A

38pt 3B/C 10A

38pt 3D 10A

38pt 3E 10A

1 Simplify the following surds. √ √ a 24 b 72 √ √ 4 8 e f 49 9 2 Simplify the following. √ √ a 2 3+4+5 3 √ √ √ d 4 3 + 2 18 – 4 2 √ 2 15 g √ 3

√ d 4 54 √ 2 45 h 15

√ c 3 200 √ 5 28 g 2 √ √ √ √ b 6 5– 7–4 5+3 7 √ √ e 2 5× 6 √ 5 14 √ h 15 2

3 Expand and simplify. √ √ a 2(2 3 + 4) √ √ c (4 3 + 4)(5 – 2 3) √ √ e ( 10 – 2)( 10 + 2) √ 2 g (3 + 7)

b d f h

4 Rationalise the denominator. 1 10 b √ a √ 6 2 √ √ 3 3 5 5 √ e √ f 4 10 2 6

√ 6 3 c √ 2 √ 5+2 √ g 2

√ √ c 8+3 2 √ √ f –3 2 × 2 10 √ 27 √ i – 3 3

√ √ √ 2 3(2 15 – 3) √ √ √ (2 5 + 10)(6 – 3 2) √ √ √ √ ( 11 – 2 5)( 11 + 2 5) √ 2 √ (4 3 – 2 2)

d h

√ 4 7 √ 2 √ √ 4 2– 3 √ 3

38pt 3F/G

5 Simplify the following, expressing all answers with positive indices when required. a (5y3 )2 b 7m0 – (5n)0 c 4x2 y3 × 5x5 y7 ( ) 3x 2 x–5 3(a2 b–4 )2 (ab)–2 d 3x2 y–4 e f ÷ × 2 –3 2 2 y 6y (2ab ) (3a–2 b)2

38pt 3I

6 Express in index form. √ √ a 21 b 3x √ √ 3 e 10x3 f 2a9 b

10A

38pt 3I 10A

235

Chapter review

Number and Algebra

√ 3 m5 √ g 7 7

√ 5 2 a √ 3 h 4 4

c

d

7 Evaluate these without using a calculator. a

1 25 2 1

d 49– 2

b

( )1 1 3 c 8

1 64 3 1

e 100– 2

f

1

125– 3

38pt 3H

8 a Write the following numbers as a basic numeral. i 3.21 × 103 ii 4.024 × 106 iii 7.59 × 10–3 iv 9.81 × 10–5 b Write the following numbers in scientific notation, using three significant figures. i 0.0003084 ii 0.0000071753 iii 5 678 200 iv 119 830 000

38pt 3J

9 Form exponential rules for the following situations. a An antique bought for $800 is expected to grow in value by 7% per year. b A balloon with volume 3000 cm3 is leaking air at a rate of 18% per minute.

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Chapter review

236

Chapter 3 Indices and surds

38pt 3K/L

10 Determine the final amount after 4 years if: a $1000 is compounded annually at 5% b $3000 is compounded monthly at 4% c $5000 is compounded daily at 3%.

38pt 3M 10A

11 Solve the following exponential equations for x. a 3x = 27 b 7x = 49 c 42x + 1 = 64 1 81

f

5x =

73x – 4 = 49x

j

11x – 5 =

e 9x = i

1 125 1 121

d 2x – 2 = 16

g 36x = 216

h 8x + 1 = 32

k 100x – 2 = 1000x

l

93 – 2x = 27x + 2

Extended-response questions 10A

1

√ √ A small rectangular jewellery box has a base with dimensions 3 15 cm by (12 + 3) cm and √ a height of (2 5 + 4) cm. a

Determine the exact area of the base of the box, in expanded and simplified form.

b

What is the exact volume of the box?

c d

√ Julie’s earring boxes occupy an area of 9 5 cm2 . What is the exact number that would fit across the base of the jewellery box? Give your answer with a rational denominator. √ The surface of Julie’s rectangular dressing table has dimensions ( 2 – 1) m by √ ( 2 + 1) m. i

Find the area of the dressing table, in square centimetres.

ii

What percentage of the area of the dressing table does the jewellery box occupy? Give your answer to one decimal place.

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2 Georgia invests $10 000 in shares in a new company. She has been told that their value is expected to increase at 6.5% per year. a Write a rule for Georgia’s expected value, V dollars, in shares after n years. b Use your rule to find the value she expects the shares to be after: i 2 years ii 5 years c When her shares are valued at $20 000 Georgia plans to cash them in. According to this rule, how many years will it take to reach this amount? Give your answer to one decimal place. d After 6 years there is a downturn in the market and the shares start to drop, losing value at 3% per year. i

What is the value of Georgia’s shares prior to the downturn in the market? Give your answer to the nearest dollar.

ii

Using your answer from part d i, write a rule for the expected value, V dollars, of Georgia’s shares t years after the market downturn.

237

Chapter review

Number and Algebra

iii Ten years after Georgia initially invested in the shares the market is still falling at this rate. She decides it’s time to sell her shares. What is their value, to the nearest dollar? How does this compare with the original amount of $10 000 she invested?

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Chapter

4

Trigonometry

What you will learn

Australian curriculum

4A Trigonometric ratios 4B Finding angles 4C Applications in two dimensions 4D Bearings 4E Applications in three dimensions (10A) 4F Obtuse angles and exact values (10A) 4G The sine rule (10A) 4H The cosine rule (10A) 4I Area of a triangle (10A) 4J The four quadrants (10A) 4K Graphs of trigonometric functions (10A)

MEASUREMENT AND GEOMETRY

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Pythagoras and trigonometry Solve right-angled triangle problems, including those involving direction and angles of elevation and depression. (10A) Establish the sine, cosine and area rules for any triangle and solve related problems. (10A) Use the unit circle to define trigonometric functions, and graph them with and without the use of digital technologies. (10A) Solve simple trigonometric equations. (10A) Apply trigonometry to solving three-dimensional problems in right-angled triangles.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Online resources • Chapter pre-test • Videos of all worked examples • Interactive widgets • Interactive walkthroughs • Downloadable HOTsheets • Access to HOTmaths Australian Curriculum courses

Global positioning systems Global positioning systems, once a secret military technology, are now a commonplace navigational tool. The global positioning system (GPS) network relies on 24 satellites that orbit the Earth at about 3000 km/h, taking about 12 hours to complete one orbit. If a handheld or car GPS device receives radio signals from at

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

least three satellites, then a process of triangulation can be used to pinpoint the position of the receiver. Triangulation involves knowing the position of the satellites and the distance between the satellites and the receiver. Trigonometry is used in these calculations.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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2D applications • Draw right-angled triangle with key information. • Use trigonometry to find unknown.

Finding lengths

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter

W h a t y o u w ill le a r n

5

5 A Expanding expressions (Consolidating) 5 B Factorising expressions 5 C Factorising trinomials of the form x 2 + bx + c 5 D Factorising quadratic trinomials of the form ax 2 + bx + c (10A) 5 E Factorising by completing the square 5 F Solving quadratic equations 5 G Applications of quadratics 5 H Solving quadratic equations by completing the square 5 I The quadratic formula

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Quadratic equations

Au s t r a l i a n c u r r ic u lu m

NUMBER AND ALGEBRA

Patterns and algebra Factorise algebraic expressions by taking out a common algebraic factor. Expand binomial products and factorise monic quadratic expressions using a variety of strategies. Substitute values into formulas to determine an unknown. Linear and non-linear relationships Solve simple quadratic equations using a range of strategies. (10A) Factorise monic and non-monic quadratic expressions and 32x32from a variety 16x16of quadratic equations derived solve a wide range of contexts.

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O n lin e re s o u rc e s • • • • • •

Chapter pre-test Videos of all worked examples Interactive widgets Interactive walkthroughs Downloadable HOTsheets Access to HOTmaths Australian Curriculum courses

T a ilg a tin g w ith q u a d r a tic s Have you ever thought about why there are signs on highways that say ‘Don’t tailgate’? Tailgating is the act of driving too close to the car in front, and at high speeds this is very dangerous. It is important to leave enough space between you and the car in front just in case of an accident. The gap that is required depends on the type of car being driven and the speed at which it is travelling. The relationship between stopping distance and speed is non-linear and can be explored using quadratics. From the work of Galileo in the 17th century, and later Newton in the 20th century, we know that the velocity, v, of an object is given by v = u + at (1) where u is the initial velocity with constant acceleration a after time t. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

This then leads to the position of an object s, given by 1 s = ut + at 2 2

(2)

which is a quadratic equation. To find a car’s stopping distance using acceleration = –a (from a car’s brakes), we solve for t using v = 0 in equation (1) and u2 substitute into equation (2). This gives s = . 2a This is another quadratic equation that tells us that the stopping distance is proportional to the square of the initial velocity u. So if you double your speed (× 2), you should quadruple the distance (× 22 = 4) between you and the car in front.

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324

Chapter 5 Quadratic equations

5A Expanding expressions

CONSOLIDATING

You will recall that expressions that include numerals and pronumerals are central to the topic of algebra. Sound skills in algebra are essential for solving most mathematical problems and this includes the ability to expand expressions involving brackets. This includes binomial products, perfect squares and the difference of perfect squares. Exploring how projectiles fly subject to the Earth’s gravity, for example, can be modelled with expressions with and without brackets. The path of this soccer ball could be modelled with the use of algebra.

Let’s start: Five key errors Here are five expansion problems with incorrect answers. Discuss what error has been made and then find the correct answer. • –2(x – 3) = –2x – 6 • (x + 3)2 = x2 + 9 • (x – 2)(x + 2) = x2 + 4x – 4 • 5 – 3(x – 1) = 2 – 3x • (x + 3)(x + 5) = x2 + 8x + 8

Key ideas

Like terms have the same pronumeral part. • They can be collected (i.e. added and subtracted) to form a single term. For example: 7x – 11x = – 4x and 4a2 b – 7ba2 = –3a2 b The distributive law is used to expand brackets. • a(b + c) = ab + ac and a(b – c) = ab – ac • (a + b)(c + d) = ac + ad + bc + bd • (a + b)(c + d) is called a binomial product because each expression in the brackets has two terms. Perfect squares • (a + b)2 = (a + b)(a + b) = a2 + 2ab + b2 •

(a – b)2 = (a – b)(a – b) = a2 – 2ab + b2

Difference of perfect squares (DOPS) • (a + b)(a – b) = a2 – b2 By definition, a perfect square is an integer that is the square of an integer; however, the rules above also apply for a wide range of values for a and b, including all real numbers. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Number and Algebra

325

Example 1 Expanding simple expressions Expand and simplify where possible. a

b 2x(1 – x)

–3(x – 5)

c

2 (14x + 3) 7

d x(2x – 1) – x(3 – x)

SO L U T I O N

EX P L A N A T I O N

a –3(x – 5) = –3x + 15

Use the distributive law: a(b – c) = ab – ac. –3 × x = –3x and –3 × (–5) = 15

b 2x(1 – x) = 2x – 2x2

Recall that 2x × (–x) = –2x2 .

c

2 2 2 (14x + 3) = × 14x + × 3 7 7 7 = 4x +

d

When multiplying fractions cancel before multiplying numerators and denominators. 3 Recall that 3 = . 1

6 7

x(2x – 1) – x(3 – x) = 2x2 – x – 3x + x2 2

= 3x – 4x

Apply the distributive law to each set of brackets first, then simplify by collecting like terms. Recall that –x × (–x) = x2 .

Example 2 Expanding binomial products, perfect squares and difference of perfect squares Expand the following. a (x + 5)(x + 4)

b (x – 4)2

SO L U T I O N

c

(2x + 1)(2x – 1)

EX P L A N A T I O N

a (x + 5)(x + 4) = x2 + 4x + 5x + 20 2

= x + 9x + 20 b (x – 4)2 = (x – 4)(x – 4)

For binomial products use (a + b)(c + d) = ac + ad + bc + bd. Simplify by collecting like terms. Rewrite and expand using the distributive law.

2

= x – 4x – 4x + 16 = x2 – 8x + 16 Alternatively: (x – 4)2 = x2 – 2(x)(4) + 42

Alternatively for perfect squares (a – b)2 = a2 – 2ab + b2 . Here a = x and b = 4.

= x2 – 8x + 16 c (2x + 1)(2x – 1) = 4x2 – 2x + 2x – 1 = 4x2 – 1

Cancel the –2x and +2x terms.

Alternatively: (2x + 1)(2x – 1) = (2x)2 – (1)2 2

= 4x – 1 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Expand, recalling that 2x × 2x = 4x2 .

Alternatively for difference of perfect squares (a – b)(a + b) = a2 – b2 . Here a = 2x and b = 1.

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326

Chapter 5 Quadratic equations

Example 3 Expanding more binomial products Expand and simplify. a (2x – 1)(3x + 5)

b 2(x – 3)(x – 2)

SOL UTI ON

Expand using the distributive law and simplify.

= 6x2 + 7x – 5

Note: 2x × 3x = 2 × 3 × x × x = 6x2 .

2(x – 3)(x – 2) = 2(x2 – 2x – 3x + 6)

First expand the brackets using the distributive law, simplify and then multiply each term by 2.

= 2(x2 – 5x + 6) = 2x2 – 10x + 12 c

(x + 2)(x + 4) – (x – 2)(x – 5) 2

Expand each binomial product.

2

= (x + 4x + 2x + 8) – (x – 5x – 2x + 10) = (x + 6x + 8) – (x – 7x + 10)

Remove brackets in the last step before simplifying.

= x2 + 6x + 8 – x2 + 7x – 10

– (x2 – 7x + 10) = –1 × x 2 + (–1) × (–7x) + (–1) × 10

2

2

= –x 2 + 7x – 10

= 13x – 2

Exercise 5A 1

1–3(½)

Use each diagram to help expand the expressions. a x(x + 2) b (x + 3)(x + 1)

x

x

3(½)

—

c (x + 4)2

x

3

x

x2

x

x

2

2x

1

4

4

UNDERSTANDING

b

(x + 2)(x + 4) – (x – 2)(x – 5)

EX P L A NA TI ON

(2x – 1)(3x + 5) = 6x2 + 10x – 3x – 5

a

c

2 Write expressions for the following. a b c d e

The sum of a2 and ab. 5 less than the product of 2 and x. The difference between a2 and b2 , where a2 > b2 . The square of the sum of y and x. Half of the difference between a and b, where a > b.

3 Simplify these expressions. a 2 × 3x b – 4 × 5x e 5x ÷ 10 f 3x ÷ 9 i 3x – 21x j 12x – 5x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

c x × 2x g – 4x2 ÷ x k –3x + 8x

d –x × 4x h –6x2 ÷ (2x) l –5x – 8x

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Number and Algebra

4 Expand and simplify where possible. a 2(x + 5) b 3(x – 4) d – 4(x – 2) e 3(2x – 1) g –2(5x – 3) h –5(4x + 3) j x(3x – 1) k 2x(1 – x) m –2x(3x + 2) n –3x(6x – 2)

Example 2b,c

Example 3a

3 (8x – 5) 4

q

2 (10x + 4) 5

r

1 – (6x + 1) 3

t

1 – (4x – 3) 2

3 u – (24x – 1) 8

w

3x (3x + 8) 4

x

2 v – (9x + 7) 9

Example 2a

–5(x + 3) 4(3x + 1) x(2x + 5) 3x(2 – x) –5x(2 – 2x)

p – 4x(1 – 4x) s

Example 1d

c f i l o

4–8(½)

FLUENCY

Example 1a–c

4–8(½)

2x (7 – 3x) 5

5 Expand and simplify. a x(3x – 1) + x(4 – x) d 3x(2x + 4) – x(5 – 2x)

b x(5x + 2) + x(x – 5) e 4x(2x – 1) + 2x(1 – 3x)

c x(4x – 3) – 2x(x – 5) f 2x(2 – 3x) – 3x(2x – 7)

6 Expand the following. a (x + 2)(x + 8) d (x + 8)(x – 3) g (x – 7)(x + 3)

b (x + 3)(x + 4) e (x + 6)(x – 5) h (x – 4)(x – 6)

c (x + 7)(x + 5) f (x – 2)(x + 3) i (x – 8)(x – 5)

7 Expand the following. a (x + 5)2 d (x – 3)2 g (x + 4)(x – 4) j (3x + 4)(3x – 4)

b e h k

(x + 7)2 (x – 8)2 (x + 9)(x – 9) (4x – 5)(4x + 5)

c f i l

(x + 6)2 (x – 10)2 (2x – 3)(2x + 3) (8x – 7)(8x + 7)

8 Expand the following using the distributive law. a (2x + 1)(3x + 5) b (4x + 5)(3x + 2) d (3x + 2)(3x – 5) e (5x + 3)(4x – 2) g (4x – 5)(4x + 5) h (2x – 9)(2x + 9) j (7x – 3)(2x – 4) k (5x – 3)(5x – 6) 2 m (2x + 5) n (5x + 6)2

c f i l o

(5x + 3)(2x + 7) (2x + 5)(3x – 5) (5x – 7)(5x + 7) (7x – 2)(8x – 2) (7x – 1)2

9–10(½)

9 Write the missing number. a (x + ?)(x + 2) = x2 + 5x + 6 c (x + 7)(x – ?) = x2 + 4x – 21 e (x – 6)(x – ?) = x2 – 7x + 6

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

5A

10–11(½)

b (x + ?)(x + 5) = x2 + 8x + 15 d (x + 4)(x – ?) = x2 – 4x – 32 f (x – ?)(x – 8) = x2 – 10x + 16

10–11(½), 12

PROBLEM-SOLVING

4–7(½)

327

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Chapter 5 Quadratic equations

5A Example 3b

Example 3c

10 Expand the following. a 2(x + 3)(x + 4) d – 4(x + 9)(x + 2) g –3(a + 2)(a – 7) j 3(y – 4)(y – 5) m 3(2x + 3)(2x + 5) p 2(x + 3)2 s –3(y – 5)2

b e h k n q t

11 Expand and simplify the following. a (x + 1)(x + 3) + (x + 2)(x + 4) c (y + 3)(y – 1) + (y – 2)(y – 4) e (2a + 3)(a – 5) – (a + 6)(2a + 5) g (x + 5)2 – 7 i 3 – (2x – 9)2

c f i l o r u

3(x + 2)(x + 7) 5(x – 3)(x + 4) –5(a + 2)(a – 8) –2(y – 3)(y – 8) 6(3x – 4)(x + 2) 4(m + 5)2 3(2b – 1)2 b d f h j

–2(x + 8)(x + 2) 3(x + 5)(x – 3) 4(a – 3)(a – 6) –6(y – 4)(y – 3) –2(x + 4)(3x – 7) 2(a – 7)2 –3(2y – 6)2

PROBLEM-SOLVING

328

(x + 8)(x + 3) + (x + 4)(x + 5) (y – 7)(y + 4) + (y + 5)(y – 3) (4b + 8)(b + 5) – (3b – 5)(b – 7) (x – 7)2 – 9 14 – (5x + 3)2

12 Find an expanded expression for the area of the pictures centred in these rectangular frames. a b 5 cm 3 cm x cm picture = x cm picture 5 cm

(x + 30) cm

13–14(½)

15–16(½)

13 Prove the following by expanding the left-hand side. a (a + b)(a – b) = a2 – b2 b (a + b)2 = a2 + 2ab + b2 c (a – b)2 = a2 – 2ab + b2 d (a + b)2 – (a – b)2 = 4ab

REASONING

13

14 Use the distributive law to evaluate the following without the use of a calculator. For example: 4 × 102 = 4 × 100 + 4 × 2 = 408. a 6 × 103 b 4 × 55 c 9 × 63 d 8 × 208 e 7 × 198 f 3 × 297 g 8 × 495 h 5 × 696 15 Each problem below has an incorrect answer. Find the error and give the correct answer. a –x(x – 7) = –x2 – 7x b 3a – 7(4 – a) = – 4a – 28 2 2 c (2x + 3) = 4x + 9 d (x + 2)2 – (x + 2)(x – 2) = 0 16 Expand these cubic expressions. a (x + 2)(x + 3)(x + 1) b (x + 4)(x + 2)(x + 5) d (x – 4)(2x + 1)(x – 3) e (x + 6)(2x – 3)(x – 5)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

c (x + 3)(x – 4)(x + 3) f (2x – 3)(x – 4)(3x – 1)

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Number and Algebra

—

—

17 One of the ways to prove Pythagoras’ theorem is to arrange four congruent right-angled triangles around a square to form a larger square, as shown.

17

c b Find an expression for the total area of the four shaded triangles by multiplying the area of one triangle by 4. a b Find an expression for the area of the four shaded triangles by subtracting the area of the inner square from the area of the outer square. c By combining your results from parts a and b, expand and simplify to prove Pythagoras’ theorem: a2 + b2 = c2 .

ENRICHMENT

Expanding to prove

329

5A

a

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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330

Chapter 5 Quadratic equations

5B Factorising expressions A common and key step in the simplification and solution of equations involves factorisation. Factorisation is the process of writing a number or expression as a product of its factors. In this section we look at expressions in which each term has a common factor, expressions that are a difference of perfect squares and four-term expressions, which can be factorised by grouping.

Let’s start: But there are no common factors! An expression such as xy + 4x + 3y + 12 has no common factors across all four terms, but it can still be factorised. The method of grouping can be used. •

Complete this working to show how to factorise the expression. xy + 4x + 3y + 12 = x( ) + 3( ) =(

•

Now repeat with the expression rearranged. xy + 3y + 4x + 12 = y( ) + 4( ) =(

•

Key ideas

)(x + 3)

)(

)

Are the two results equivalent? Factorise expressions with common factors by ‘taking out’ the common factors. For example: –5x – 20 = –5(x + 4) and 4x2 – 8x = 4x(x – 2). Factorise a difference of perfect squares (DOPS) using a2 – b2 = (a + b)(a – b). • We use surds when a2 or b2 is not a perfect square, such as 1, 4, 9, … √ √ √ For example: x2 – 5 = (x + 5)(x – 5) using ( 5)2 = 5. Factorise four-term expressions if possible by grouping terms and factorising each pair. For example: x2 + 5x – 2x – 10 = x(x + 5) – 2(x + 5) = (x + 5)(x – 2)

Example 4 Taking out common factors Factorise by taking out common factors. a –3x – 12 b 20a2 + 30a

c

2(x + 1) – a(x + 1)

SOL UTI ON

EX P L A NA TI ON

a –3x – 12 = –3(x + 4)

–3 is common to both –3x and –12. Divide each term by –3 to determine the terms in the brackets. Expand to check.

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Number and Algebra

b 20a2 + 30a = 10a(2a + 3)

The HCF of 20a2 and 30a is 10a.

c 2(x + 1) – a(x + 1) = (x + 1)(2 – a)

(x + 1) is a common factor to both parts of the expression.

331

Example 5 Factorising difference of perfect squares Factorise the following difference of perfect squares. You may need to look for a common factor first. a x2 – 16 b 9a2 – 4b2 c 12y2 – 1200 d (x + 3)2 – 4 SO L U T I O N a

EX P L A N A T I O N

x2 – 16 = (x)2 – (4)2

Use a2 – b2 = (a + b)(a – b), where a = x and b = 4.

= (x + 4)(x – 4) b

9a2 – 4b2 = (3a)2 – (2b)2

9a2 = (3a)2 and 4b2 = (2b)2 .

= (3a + 2b)(3a – 2b) c

12y2 – 1200 = 12(y2 – 100) = 12(y + 10)(y – 10)

First, take out the common factor of 12. 100 = 102 , use a2 – b2 = (a – b)(a + b).

d

(x + 3)2 – 4 = (x + 3)2 – (2)2

Use a2 – b2 = (a – b)(a + b), where a = x + 3 and b = 2. Simplify each bracket.

= (x + 3 + 2)(x + 3 – 2) = (x + 5)(x + 1)

Example 6 Factorising DOPS using surds Factorise these DOPS using surds. a x2 – 10 b x2 – 24 SO L U T I O N a

√ 2 x2 – 24 = x2 – ( 24) √ √ = (x + 24)(x – 24) √ √ = (x + 2 6)(x – 2 6) √ c (x – 1)2 – 5 = (x – 1)2 – ( 5)2 √ √ = (x – 1 + 5)(x – 1 – 5)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

(x – 1)2 – 5

EX P L A N A T I O N

√ 2 x2 – 10 = x2 – ( 10) √ √ = (x + 10)(x – 10)

b

c

√ Recall that ( 10)2 = 10. √ Use ( 24)2 = 24. Simplify: √ √ √ √ √ 24 = 4 × 6 = 4 × 6 = 2 6 Use a2 – b2 = (a + b)(a – b), where a = x – 1 √ and b = 5.

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332

Chapter 5 Quadratic equations

Example 7 Factorisation by grouping Factorise by grouping x2 – x + ax – a. SO L U T I O N

EX P L A N A T I O N

x2 – x + ax – a = x(x – 1) + a(x – 1)

Factorise two pairs of terms, then take out the common binomial factor (x – 1).

1

Example 4a,b

1–2(½)

Determine the highest common factor of these pairs of terms. a 7x and 14 b 12x and 30 c –8y and 40 2 2 e 4a and 2a f 12a and 9a g –5a2 and –50a

2 Factorise by taking out the common factors. a 3x – 18 b 4x + 20 e –5x – 30 f – 4y – 2 i 4x2 + x j 5x2 – 2x 2 m 10a – 5a n 12x – 30x2 2 2 q ab – a b r 2x2 yz – 4xy

c g k o s

Example 5a,b

Example 5c,d

—

d –5y and –25 h –3x2 y and –6xy

7a + 7b –12a – 3 6b2 – 18b –2x – x2 –12m2 n – 12mn2

3–6(½) Example 4c

2(½)

d h l p t

3–7(½)

9a – 15 –2ab – bc 14a2 – 21a – 4y – 8y2 6xyz2 – 3z2

3–7(½)

3 Factorise, noting the common binomial factor. (Hint: For parts g-i, insert a 1 where appropriate.) a 5(x – 1) – a(x – 1) b b(x + 2) + 3(x + 2) c a(x + 5) – 4(x + 5) d x(x + 2) + 5(x + 2) e x(x – 4) – 2(x – 4) f 3(x + 1) – x(x + 1) g a(x + 3) + (x + 3) h x(x – 2) – (x – 2) i (x – 6) – x(x – 6) 4 Factorise the following difference of perfect squares. a x2 – 9 b x2 – 25 c y2 – 49 2 2 e 4x – 9 f 36a – 25 g 1 – 81y2 i 25x2 – 4y2 j 64x2 – 25y2 k 9a2 – 49b2

d y2 – 1 h 100 – 9x2 l 144a2 – 49b2

5 Factorise the following. a 2x2 – 32 e 3x2 – 75y2 i (x + 5)2 – 16 m (3x + 5)2 – x2

d h l p

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b f j n

5x2 – 45 3a2 – 300b2 (x – 4)2 – 9 (2y + 7)2 – y2

c g k o

6y2 – 24 12x2 – 27y2 (a – 3)2 – 64 (5x + 11)2 – 4x2

UNDERSTANDING

Exercise 5B

FLUENCY

= (x – 1)(x + a)

3y2 – 48 63a2 – 112b2 (a – 7)2 – 1 (3x – 5y)2 – 25y2

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Example 6

6 Factorise using surds and remember to simplify surds where possible. a x2 – 7 b x2 – 5 c x2 – 19 2 2 d x – 21 e x – 14 f x2 – 30 g x2 – 15 h x2 – 11 i x2 – 8 2 2 j x – 18 k x – 45 l x2 – 20 m x2 – 32 n x2 – 48 o x2 – 50 2 2 p x – 200 q (x + 2) – 6 r (x + 5)2 – 10 s (x – 3)2 – 11 t (x – 1)2 – 7 u (x – 6)2 – 15 2 2 v (x + 4) – 21 w (x + 1) – 19 x (x – 7)2 – 26

Example 7

7 Factorise by grouping. a x2 + 4x + ax + 4a d x2 + 2x – ax – 2a g x2 – ax – 4x + 4a

9 Factorise by first rearranging. a xy – 6 – 3x + 2y d xy + 12 – 3y – 4x 10 Factorise fully. a 5x2 – 120 e 2(x + 3)2 – 10

b 3x2 – 162 f 3(x – 1)2 – 21

7 16 g (x + 1)2 – 75 k 7x2 – 5 o –10 + 3x2

5 36 h (x – 7)2 – 40 l 6x2 – 11 p –7 + 13x2 d x2 –

c ax – 10 + 5x – 2a f 2ax – 20 + 8a – 5x

c 7x2 – 126 g 4(x – 4)2 – 48 11

8–10(½)

PROBLEM-SOLVING

8–9(½)

c x2 –

b ax – 12 + 3a – 4x e 2ax + 3 – a – 6x

5B

c x2 – 3x + ax – 3a f x2 + 3x – 4ax – 12a i 3x2 – 6ax – 7x + 14a

8(½)

8 Factorise fully and simplify surds. 3 2 a x2 – b x2 – 9 4 e (x – 2)2 – 20 f (x + 4)2 – 27 i 3x2 – 4 j 5x2 – 9 2 m –9 + 2x n –16 + 5x2

333

d 2x2 – 96 h 5(x + 6)2 – 90 11(½), 12

11(½), 13, 14

11 Evaluate the following, without the use of a calculator, by first factorising. a 162 – 142 b 182 – 172 c 132 – 102 d 152 – 112 e 172 – 152 f 112 – 92 g 272 – 242 h 522 – 382

REASONING

b x2 + 7x + bx + 7b e x2 + 5x – bx – 5b h x2 – 2bx – 5x + 10b

FLUENCY

Number and Algebra

12 a Show that 4 – (x + 2)2 = –x(x + 4) by factorising the left-hand side. b Now factorise the following. i 9 – (x + 3)2 ii 16 – (x + 4)2 iii 25 – (x – 5)2 2 2 iv 25 – (x + 2) v 49 – (x – 1) vi 100 – (x + 4)2

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334

Chapter 5 Quadratic equations

REASONING

5B 13 a Prove that, in general, (x + a)2 π x2 + a2 . b Are there any values of x for which (x + a)2 = x2 + a2 ? If so, what are they? 4 1 14 Show that x2 – = (3x + 2)(3x – 2) using two different methods. 9 9

—

—

15 Factorise and simplify the following without initially expanding the brackets. a (x + 2)2 – (x + 3)2 b (y – 7)2 – (y + 4)2 2 2 c (a + 3) – (a – 5) d (b + 5)2 – (b – 5)2 e (s – 3)2 – (s + 3)2 f (y – 7)2 – (y + 7)2 2 2 g (2w + 3x) – (3w + 4x) h (d + 5e)2 – (3d – 2e)2 i (4f + 3j)2 – (2f – 3j)2 j (3r – 2p)2 – (2p – 3r)2

15, 16

ENRICHMENT

Hidden DOPS

16 a Is it possible to factorise x2 + 5y – y2 + 5x? Can you show how? b Also try factorising: i x2 + 7x + 7y – y2 ii x2 – 2x – 2y – y2 iii 4x2 + 4x + 6y – 9y2 iv 25y2 + 15y – 4x2 + 6x

Factorising is a key component of the proof of Fermat’s last theorem, which states that there are no solutions to x n + y n = z n for n ≥ 3.

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Number and Algebra

335

Using calculators to expand and factorise 1 Expand and simplify 3(2x – 3)(5x + 1) – (4x – 1)2 . 2 Factorise: a ax2 + x – ax – 1

b 6x2 + 7x – 3

Using the TI-Nspire:

Using the ClassPad:

1

In a calculator page use b>Algebra>Expand and type as shown.

1 In the Main application, type and highlight the expression, then tap Interactive, Transformation, expand and type in as shown below.

2

In a Calculator page use b>Algebra>Factor and type as shown.

2 Use the VAR keyboard to type the expression as shown. Highlight the expression and tap Interactive, Transformation, factor.

Note: Use a multiplication sign between the a and x.

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336

Chapter 5 Quadratic equations

5C Factorising trinomials of the form x 2 + bx + c A quadratic trinomial of the form x2 + bx + c is called a monic quadratic because the coefficient of x2 is 1. Now consider: (x + m)(x + n) = x2 + xn + mx + mn = x2 + (m + n)x + mn We can see from this expansion that mn gives the constant term (c) and m + n is the coefficient of x. This tells us that to factorise a monic quadratic trinomial we should look for factors of the constant term (c) that add to give the coefficient of the middle term (b).

Let’s start: Factorising x 2 – 6x – 72 Discuss what is wrong with each of these statements when trying to factorise x2 – 6x – 72. • • • • •

Find factors of 72 that add to 6. Find factors of 72 that add to –6. Find factors of –72 that add to 6. –18 × 4 = –72 so x2 – 6x – 72 = (x – 18)(x + 4) –9 × 8 = –72 so x2 – 6x – 72 = (x – 9)(x + 8)

Can you write a correct statement that correctly factorises x2 – 6x – 72?

Key ideas

Monic quadratics have a coefficient of x2 equal to 1. Monic quadratics of the form x2 + bx + c can be factorised by finding the two numbers that multiply to give the constant term (c) and add to give the coefficient of x (i.e. b). mn = (x + m)(x + n) x2 + (m + n)x +    c b

Example 8 Factorising trinomials of the form x 2 + bx + c Factorise: a x2 + 8x + 15 c

b x2 – 5x + 6

2x2 – 10x – 28

d x2 – 8x + 16

SOL UTI ON

EX P L A NA TI ON

a x2 + 8x + 15 = (x + 3)(x + 5)

3 × 5 = 15 and 3 + 5 = 8 Check: (x + 3)(x + 5) = x2 + 5x + 3x + 15 = x2 + 8x + 15

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Number and Algebra

b

x2 – 5x + 6 = (x – 3)(x – 2)

337

–3 × (–2) = 6 and –3 + (–2) = –5 Check: (x – 3)(x – 2) = x2 – 2x – 3x + 6 = x2 – 5x + 6

c

2x2 – 10x – 28 = 2(x2 – 5x – 14) = 2(x – 7)(x + 2)

d

x2 – 8x + 16 = (x – 4)(x – 4) = (x – 4)

2

First, take out the common factor of 2. –7 × 2 = –14 and –7 + 2 = –5 – 4 × (– 4) = 16 and – 4 + (– 4) = –8 (x – 4)(x – 4) = (x – 4)2 is a perfect square.

Example 9 Simplifying algebraic fractions Use factorisation to simplify these algebraic fractions. x2 – 4x – 5 x2 – x – 6 x2 – 9 × a b x+2 2x – 6 x2 – 2x – 15

a

b

EX P L A N A T I O N

(x  + 2) 1 x2 – x – 6 (x – 3) = x+2 (x  + 2) 1  =x–3 x2 – 9 x2 – 4x – 5 × 2x – 6 x2 – 2x – 15 1     (x  – 5) 1 (x + 1) (x + 3)  (x – 3) 1   × =   1 1 (x  + 3) (x  – 5)  2 (x  – 3) 1  x+1 = 2

Exercise 5C 1

First, factorise x2 – x – 6 and then cancel (x + 2).

First, factorise all expressions in the numerators and denominators. Cancel to simplify where possible.

1, 2, 3(½)

3(½)

—

Find two integers that multiply to give the first number and add to give the second number. a 18, 11 b 20, 12 c –15, 2 d –12, 1 e –24, –5 f –30, –7 g 10, –7 h 36, –15

2 A number (except zero) divided by itself always equals 1. a1 2 (x  – 3) 1 (a  + 5) 1 1  For example:
1 = 1, = 2, =   1 1  2  a (x – 3) 2 (a + 5) 


UNDERSTANDING

SO L U T I O N

Invent some algebraic fractions that are equal to: a 1

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b 3

c –5

d

1 3

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5C 3 Simplify by cancelling common factors. For parts i to l, first factorise the numerator. 2x 7x 6a 4a a b c d 4 21 2a 20a e

3(x + 1) 9(x + 1)

f

2(x – 2) 8(x – 2)

g

15(x – 5) 3(x – 5)

h

8(x + 4) 12(x + 4)

i

x2 + x x

j

x2 – 2x x

k

x2 – 3x 2x

l

2x – 4x2 2x

4–6(½) Example 8a,b

Example 8c

Example 8d

Example 9a

4–7(½)

4 Factorise these quadratic trinomials. a x2 + 7x + 6 b x2 + 5x + 6 d x2 + 7x + 10 e x2 + 7x + 12 2 g x + 5x – 6 h x2 + x – 6 j x2 + 3x – 4 k x2 + 7x – 30 2 m x – 7x + 10 n x2 – 6x + 8 p x2 – 2x + 1 q x2 – 9x + 18 2 s x – 4x – 12 t x2 – x – 20 v x2 – x – 12 w x2 + 4x – 32

c f i l o r u x

x2 + 6x + 9 x2 + 11x + 18 x2 + 2x – 8 x2 + 9x – 22 x2 – 7x + 12 x2 – 11x + 18 x2 – 5x – 14 x2 – 3x – 10

5 Factorise by first taking out the common factor. a 2x2 + 14x + 20 b 3x2 + 21x + 36 2 d 5x – 5x – 10 e 4x2 – 16x – 20 g –2x2 – 14x – 24 h –3x2 + 9x – 6 2 j – 4x + 4x + 8 k –5x2 – 20x – 15

c f i l

2x2 + 22x + 36 3x2 – 9x – 30 –2x2 + 10x + 28 –7x2 + 49x – 42

6 Factorise these perfect squares. a x2 – 4x + 4 d x2 – 14x + 49 g 2x2 + 44x + 242 j –3x2 + 36x – 108

c f i l

x2 + 12x + 36 x2 – 20x + 100 5x2 – 50x + 125 – 4x2 – 72x – 324

b e h k

x2 + 6x + 9 x2 – 18x + 81 3x2 – 24x + 48 –2x2 + 28x – 98

UNDERSTANDING

Chapter 5 Quadratic equations

4–7(½)

FLUENCY

338

7 Use factorisation to simplify these algebraic fractions. In some cases, you may need to remove a common factor first. a d g

x2 – 3x – 54 x–9 x+2 x2 + 9x + 14 2(x + 12) 2 x + 4x – 96

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b e h

x2 + x – 12 x+4 x–3 x2 – 8x + 15 x2 – 5x – 36 3(x – 9)

c f i

x2 – 6x + 9 x–3 x+1 x2 – 5x – 6 x2 – 15x + 56 5(x – 8)

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Number and Algebra

Example 9b

8–9(½)

8–10(½)

PROBLEM-SOLVING

8(½)

8 Simplify by first factorising. a c e g

x2 – 4 5x – 15 × 2 2 x – x – 6 x + 4x – 12 x2 + 2x – 3 2x – 10 × x+3 x2 – 25 x2 – 4x + 3 4x + 4 × x2 + 4x – 21 x2 – 1 x2 – x – 6 x2 + 5x + 4 × x2 + x – 12 x2 – 1

b d f h

x2 + 3x + 2 x2 – 9 × x2 + 4x + 3 3x + 6 x2 – 9 4x – 8 × 2 2 x – 5x + 6 x + 8x + 15 x2 + 6x + 8 6x – 24 × 2 x2 – 4 x – 16 x2 – 4x – 12 x2 – 6x + 8 × 2 x2 – 4 x – 36

339

5C

9 Simplify these expressions that involve surds. x2 – 7 √ x+ 7 √ 5x + 3 d 5x2 – 9 a

g

b e

(x + 1)2 – 2 √ x+1+ 2

h

x2 – 10 √ x – 10 √ 3x – 4 3x2 – 16 (x – 3)2 – 5 √ x–3– 5

c

x2 – 12 √ x+2 3

f

7x2 – 5 √ √ 7x + 5

i

(x – 6)2 – 6 √ x–6+ 6

10 Simplify using factorisation. x2 + 2x – 3 3x – 3 ÷ 2x + 10 x2 – 25

b

x2 + 3x + 2 4x + 8 ÷ x2 + 4x + 3 x2 – 9

c

x2 – x – 12 x2 – 16 ÷ 3x + 12 x2 – 9

d

4x + 28 x2 – 49 ÷ x2 – 3x – 28 6x + 24

e

x2 + 5x – 14 x2 + 9x + 14 ÷ 2 x2 + 2x – 3 x +x–2

f

x2 + 8x + 15 x2 + 6x + 5 ÷ x2 + 5x – 6 x2 + 7x + 6 11

11 A businessman is showing off his new formula to determine the company’s profit, in millions of dollars, after t years. Profit =

11, 12

12–14

REASONING

a

t2 – 49 t2 – 5t – 24 × 5t – 40 2t2 – 8t – 42

Show that this is really the same as Profit =

t+7 . 10

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340

Chapter 5 Quadratic equations

REASONING

5C 12 Note that an expression with a perfect square can be simplified as shown. (x  + 3) 1 (x + 3)2 (x + 3) =   x+3 x + 31 =x+3 Use this idea to simplify the following. a d

x2 – 6x + 9 x–3 6x – 12 x2 – 4x + 4

b e

x2 + 2x + 1 x+1 4x + 20 x2 + 10x + 25

c f

x2 – 16x + 64 x–8 2 x – 14x + 49 5x – 35

a2 + 2ab + b2 a2 – b2 ÷ 2 = 1. a2 + ab a – ab b Make up your own expressions, like the one in part a, which equal 1. Ask a classmate to check them.

13 a

Prove that

14 Simplify. a2 + 2ab + b2 a2 – b2 ÷ 2 a(a + b) a – 2ab + b2

b

a2 – 2ab + b2 a2 – b2 ÷ 2 2 2 a –b a + 2ab + b2

c

a2 – b2 a2 – b2 ÷ a2 – 2ab + b2 a2 + 2ab + b2

d

a2 + 2ab + b2 a(a – b) ÷ 2 a(a + b) a – 2ab + b2

Addition and subtraction with factorisation

—

—

15

15 Factorisation can be used to help add and subtract algebraic fractions. Here is an example. x 3 x 3 + = + x – 2 x2 – 6x + 8 x – 2 (x – 2)(x – 4) 3(x – 4) x = + (x – 2)(x – 4) (x – 2)(x – 4) 3x – 12 + x = (x – 2)(x – 4) 4x – 12 = (x – 2)(x – 4) 4(x – 3) = (x – 2)(x – 4) Now simplify the following. x 2 + 2 a x + 3 x – x – 12 3 2x c – x + 4 x2 – 16 x+4 x–5 e – x2 – x – 6 x2 – 9x + 18 x+1 x–2 g – x2 – 25 x2 – 6x + 5

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b d f h

ENRICHMENT

a

3x 4 + 2 x + 2 x – 7x – 18 4 1 – 2 2 x – 9 x – 8x + 15 x+3 x – x2 – 4x – 32 x2 + 7x + 12 x+2 x+3 – x2 – 2x + 1 x2 + 3x – 4

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Number and Algebra

5D Factorising quadratic trinomials of the form ax 2 + bx + c

341

10A

There are a number of ways of factorising non-monic quadratic trinomials of the form ax2 + bx + c, where a π 1. The cross method, for example, uses lists of factors of a and c so that a correct combination can be found. For example, to factorise 4x2 – 4x – 15:

2x + 3 factors of 4

2

1

4

2

4

1

−15 1 −1 15

5 −3 −5

1 −15 15 −1 −3

5

3

3 −5

factors of −15

2x − 5 2 × (–5) + 2 × 3 = – 4, so choose (2x + 3) and (2x – 5). ∴ 4x2 – 4x – 15 = (2x + 3)(2x – 5) The method outlined in this section, however, uses grouping. Unlike the method above, grouping is more direct and involves less trial and error.

Let’s start: Does the order matter? To factorise the non-monic quadratic 4x2 – 4x – 15 using grouping, we multiply a by c, which is 4 × (–15) = –60. Then we look for numbers that multiply to give –60 and add to give – 4 (the coefficient of x). • •

What are the two numbers that multiply to give –60 and add to give – 4? Complete the following using grouping. 4x2 – 4x – 15 = 4x2 – 10x + 6x – 15 –10 × 6 = –60, –10 + 6 = – 4 = 2x(

) + 3(

= (2x – 5)( •

)

)

If we changed the order of the –10x and +6x do you think the result would change? Copy and complete to find out. 4x2 – 4x – 15 = 4x2 + 6x – 10x – 15 6 × (–10) = –60, 6 + (–10) = – 4 = 2x( =(

) – 5( )(

) )

To factorise a non-monic trinomial of the form ax2 + bx + c, follow these steps: • Find two numbers that multiply to give a × c and add to give b. For 15x2 – x – 6, a × c = 15 × (–6) = –90. The factors of –90 that add to –1 (b) are –10 and 9. • Use the two numbers shown in the example above to split bx, then factorise by grouping. 15x2 – x – 6 = 15x2 – 10x + 9x – 6

Key ideas

= 5x(3x – 2) + 3(3x – 2) = (3x – 2)(5x + 3) There are other valid methods that can be used to factorise non-monic trinomials. The cross method is illustrated in the introduction. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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342

Chapter 5 Quadratic equations

Example 10 Factorising non-monic quadratics Factorise: a 6x2 + 19x + 10

b 9x2 + 6x – 8

SO L U T I O N

EX P L A N A T I O N

6x2 + 19x + 10 = 6x2 + 15x + 4x + 10

a

= 3x(2x + 5) + 2(2x + 5)

a × c = 6 × 10 = 60; choose 15 and 4 since 15 × 4 = 60 and 15 + 4 = 19 (b).

= (2x + 5)(3x + 2)

Factorise by grouping.

9x2 + 6x – 8 = 9x2 + 12x – 6x – 8

b

a × c = 9 × (–8) = –72; choose 12 and –6 since 12 × (–6) = –72 and 12 + (–6) = 6 (b).

= 3x(3x + 4) – 2(3x + 4) = (3x + 4)(3x – 2)

Example 11 Simplifying algebraic fractions involving quadratic expressions Simplify

4x2 – 9 25x2 – 10x + 1 × . + 13x – 3 10x2 – 17x + 3

10x2

SO L U T I O N

EX P L A N A T I O N

4x2 – 9 25x2 – 10x + 1 × + 13x – 3 10x2 – 17x + 3     (5x– 1) 1  (5x– 1) 1 (2x+ 3) 1  (2x– 3) 1   × =     (2x+ 3) 1  (5x– 1) 1  (2x– 3) 1  (5x– 1) 1 

10x2

First, use the range of factorising techniques to factorise all quadratics. Cancel to simplify.

=1

1

1, 2(½)

2(½)

Complete this table. ax 2 + bx + c

a × c

6x 2 + 13x + 6

36

2

8x + 18x + 4 12x 2 + x – 6

Two numbers that multiply to give a × c and add to give b 9 and

—

UNDERSTANDING

Exercise 5D

32 –8 and

2

10x – 11x – 6 21x 2 – 20x + 4

–6 and

2

15x – 13x + 2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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x2 + 4x + 6x + 24 x2 – 3x – 4x + 12 3x2 – 12x + 2x – 8 10x2 + 12x – 15x – 18

c f i l

3(½) Example 10

3 Factorise the following. a 3x2 + 10x + 3 d 3x2 – 5x + 2 g 3x2 – 11x – 4 j 2x2 – 9x + 7 m 2x2 – 9x – 5 p 8x2 – 14x + 5 s 6x2 + 13x + 6 v 8x2 – 26x + 15

b e h k n q t w

4 Factorise the following. a 18x2 + 27x + 10 d 30x2 + 13x – 10 g 24x2 – 38x + 15

b 20x2 + 39x + 18 e 40x2 – x – 6 h 45x2 – 46x + 8

2x2 + 3x + 1 2x2 – 11x + 5 3x2 – 2x – 1 3x2 + 2x – 8 13x2 – 7x – 6 6x2 + x – 12 4x2 – 5x + 1 6x2 – 13x + 6

3–4(½)

c f i l o r u x

3–4(½)

3x2 + 8x + 4 5x2 + 2x – 3 7x2 + 2x – 5 2x2 + 5x – 12 5x2 – 22x + 8 10x2 + 11x – 6 8x2 – 14x + 5 9x2 + 9x – 10

c 21x2 + 22x – 8 f 28x2 – 13x – 6 i 25x2 – 50x + 16

5(½), 7

5 Factorise by first taking out the common factor. a 6x2 + 38x + 40 b 6x2 – 15x – 36 d 32x2 – 88x + 60 e 16x2 – 24x + 8 2 g –50x – 115x – 60 h 12x2 – 36x + 27

x2 + 3x + 7x + 21 x2 – 5x + 3x – 15 8x2 – 4x + 6x – 3 12x2 – 6x – 10x + 5

5D

FLUENCY

b e h k

343

5–6(½), 7

5–6(½), 7

c 48x2 – 18x – 3 f 90x2 + 90x – 100 i 20x2 – 25x + 5

PROBLEM-SOLVING

2 Factorise by grouping pairs. a x2 + 2x + 5x + 10 d x2 – 7x – 2x + 14 g 6x2 – 8x + 3x – 4 j 5x2 + 20x – 2x – 8

UNDERSTANDING

Number and Algebra

6 Simplify by first factorising. a d

6x2 – x – 35 3x + 7 10x – 2 15x2 + 7x – 2

b e

8x2 + 10x – 3 2x + 3 4x + 6 14x2 + 17x – 6

c f

9x2 – 21x + 10 3x – 5 20x – 12 10x2 – 21x + 9

g

2x2 + 11x + 12 6x2 + 11x + 3

h

12x2 – x – 1 8x2 + 14x + 3

i

10x2 + 3x – 4 14x2 – 11x + 2

j

9x2 – 4 15x2 + 4x – 4

k

14x2 + 19x – 3 49x2 – 1

l

8x2 – 2x – 15 16x2 – 25

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter 5 Quadratic equations

5D 7 A cable is suspended across a farm channel. The height (h), in metres, of the cable above the water surface is modelled by the equation h = 3x2 – 19x + 20, where x metres is the distance from one side of the channel. a

Factorise the right-hand side of the equation. b Determine the height of the cable when x = 3. Interpret this result. c Determine where the cable is at the level of the water surface.

Example 11

8–9(½)

8–9(½), 10

REASONING

8(½)

8 Combine all your knowledge of factorising to simplify the following. a

9x2 – 16 x2 + x – 12 × 2 – 6x + 9 3x + 8x – 16

x2

b

4x2 – 1 9x2 – 4 × 2 6x – x – 2 8x – 4

c

1 – x2 25x2 + 30x + 9 × 15x + 9 5x2 + 8x + 3

d

20x2 + 21x – 5 16x2 – 24x + 9 × 16x2 + 8x – 15 25x2 – 1

e

2x2 – 7x + 3 100x2 – 25 ÷ 2x2 – 9x – 5 5x2 – 40x + 75

f

3x2 – 12 2x2 – 3x – 2 ÷ 30x + 15 4x2 + 4x + 1

g

9x2 – 6x + 1 9x2 – 1 ÷ 6x2 – 11x + 3 6x2 – 7x – 3

h

16x2 – 25 4x2 – 17x + 15 ÷ 4x2 – 7x – 15 16x2 – 40x + 25

PROBLEM-SOLVING

344

9 Find a method to show how –12x2 – 5x + 3 factorises to (1 – 3x)(4x + 3). Then factorise the following. a –8x2 + 2x + 15 b –6x2 + 11x + 10 c –12x2 + 13x + 4 d –8x2 + 18x – 9 e –14x2 + 39x – 10 f –15x2 – x + 6 10 Make up your own complex expression like those in Question 8, which simplifies to 1. Check your expression with your teacher or a classmate. —

—

11

11 Factorise the quadratics in the expressions and then simplify using a common denominator. 2 x 3 x a + 2 b – 2 2x – 3 8x – 10x – 3 3x – 1 6x + 13x – 5 c e g

4x x + 2x – 5 8x2 – 18x – 5

d

2 1 + 2 – 1 6x – x – 2

f

4x2

4 2 – 8x2 – 18x – 5 12x2 – 5x – 2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

h

ENRICHMENT

Non-monics with addition and subtraction

4x 3x – 12x2 – 11x + 2 3x – 2 9x2

2 3 – 2 – 25 9x + 9x – 10

1 2 + 10x2 – 19x + 6 4x2 + 8x – 21

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Number and Algebra

345

5E Factorising by completing the square Consider the quadratic expression x2 + 6x + 1. We cannot factorise this using the methods we have established in the previous exercises because there are no factors of 1 that add to 6.

1× 1=1

We can, however, use our knowledge of perfect squares and the difference of perfect squares to help find factors using surds.

But 1 + 1 ≠ 6

Let’s start: Make a perfect square x

3

x

x2

3x

3

3x

9

This diagram is a square. Its sides are x + 3 and its area is given by x2 + 6x + 9 = (x + 3)2 . Use a similar diagram to help make a perfect square for the following and determine the missing number for each. • x2 + 8x + ? • x2 + 12x + ? Can you describe a method for finding the missing number without drawing a diagram?  2 b To complete the square for x2 + bx, add . 2  2  2 b b • x2 + bx + = x+ 2 2 To factorise by completing the square:  2  2 b b • Add and balance by subtracting . 2 2 • •

Factorise the perfect square and simplify. Factorise using DOPS: a2 – b2 = (a + b)(a – b); surds can be used.

Key ideas  2  2 6 6 x + 6x + 1 = x + 6x + – +1 2 2  2 6 = x+ –8 2 √ = (x + 3)2 – ( 8)2 √ √ = (x + 3 + 8)(x + 3 – 8) √ √ = (x + 3 + 2 2)(x + 3 – 2 2) 2

2

Not all quadratic expressions factorise. This will be seen when you end up with expressions such as (x + 3)2 + 6, which is not a difference of two perfect squares. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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346

Chapter 5 Quadratic equations

Example 12 Completing the square Decide what number must be added to these expressions to complete the square. Then factorise the resulting perfect square. a x2 + 10x b x2 – 7x SO L U T I O N

EX P L A N A T I O N

( )2 10 a = 52 = 25 2

b

x2

+ 10x + 25 =

(

–7 2

)2

For

( )2 b + bx, add . 2

( )2 b . Here b = 10, and evaluate 2

(x + 5)2

x2

( )2 ( )2 b b + bx + = x+ 2 2

( )2 b In x2 – 7x, b = –7 and evaluate . 2

49 = 4

49 x2 – 7x + = 4

x2

(

7 x– 2

)2

Factorise the perfect square.

Example 13 Factorising by completing the square Factorise the following by completing the square if possible. a x2 + 8x – 3 b x2 – 2x + 8 SO L U T I O N a

b

(

( )2 ) ( )2 8 8 – x2 + 8x – 3 = x2 + 8x + –3 2 2 ( )2 8 = x+ – 16 – 3 2 = (x + 4)2 – 19 √ = (x + 4)2 – ( 19)2 √ √ = (x + 4 – 19)(x + 4 + 19)  ( )2  ( )2 2  2 x2 – 2x + 8 = x2 – 2x + – +8 2 2 =

(

)2 2 x– +7 2

= (x – 1)2 + 7 ∴ x2 – 2x + 8 cannot be factorised.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

EX P L A N A T I O N ( )2 b Add to complete the square and balance 2 ( )2 b also. Factorise the by subtracting 2 resulting perfect square and simplify. √ Express 19 as ( 19)2 to set up a DOPS. Apply a2 – b2 = (a + b)(a – b) using surds. ( )2 2 Add = (1)2 to complete the square and 2 balance by subtracting (1)2 also. Factorise the perfect square and simplify. (x – 1)2 + 7 is not a difference of perfect squares.

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Number and Algebra

347

Example 14 Factorising with fractions 1 Factorise x2 + 3x + . 2 EX P L A N A T I O N

)2 9 1 3 – + = x+ 2 4 2 ( )2 3 7 = x+ – 2 4 ( )2 (√ )2 7 3 – = x+ 2 4 ( √ )( √ ) 7 7 3 3 = x+ – x+ + 2 2 2 2 ( √ )( √ ) 3– 7 3+ 7 = x+ x+ 2 2 (

Exercise 5E 1

( )2 3 Add to complete the square and balance 2 ( )2 3 . Leave in fraction form. by subtracting 2 Factorise the perfect square and simplify. 9 1 9 2 7 – + =– + =– 4 2 4 4 4

Recall that

1–3(½)

√ √ 7 7 7 =√ = and use DOPS. 4 2 4

3(½)

—

( )2 b These expressions are of the form + bx. Evaluate for each one. 2 a x2 + 6x b x2 + 12x c x2 + 2x 2 2 d x – 4x e x – 8x f x2 – 10x g x2 + 5x h x2 + 3x i x2 – 9x x2

2 Factorise these perfect squares. a x2 + 4x + 4 b x2 + 8x + 16 2 d x – 12x + 36 e x2 – 6x + 9

c x2 + 10x + 25 f x2 – 18x + 81

3 Factorise using surds. Recall that a2 – b2 = (a + b)(a – b). a (x + 1)2 – 5 b (x + 2)2 – 7 d (x – 3)2 – 11 e (x – 6)2 – 22

c (x + 4)2 – 10 f (x – 5)2 – 3

4–5(½) Example 12

√

UNDERSTANDING

( )2  ( )2 1 1 3  3 x2 + 3x + = x2 + 3x + + – 2 2 2 2 

4–6(½)

4–6(½)

4 Decide what number must be added to these expressions to complete the square. Then factorise the resulting perfect square. a x2 + 6x b x2 + 12x c x2 + 4x d x2 + 8x 2 2 2 e x – 10x f x – 2x g x – 8x h x2 – 12x i x2 + 5x j x2 + 9x k x2 + 7x l x2 + 11x 2 2 2 m x – 3x n x – 7x o x –x p x2 – 9x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

FLUENCY

SO L U T I O N

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5E

Example 13b

5 Factorise by completing the square. a x2 + 4x + 1 b x2 + 6x + 2 2 e x – 8x + 13 f x2 – 12x + 10

c x2 + 2x – 4 g x2 – 4x – 3

d x2 + 10x – 4 h x2 – 8x – 5

6 Factorise, if possible. a x2 + 6x + 11 e x2 + 10x + 3 i x2 – 12x + 2

c x2 + 8x + 1 g x2 – 10x + 30 k x2 – 8x – 1

d x2 + 4x + 2 h x2 – 6x + 6 l x2 – 4x + 6

b x2 + 4x + 7 f x2 + 4x – 6 j x2 – 2x + 2

7(½) Example 14

7 Factorise the following. a x2 + 3x + 1 b x2 + 7x + 2 e x2 – 3x +

1 2

f

x2 – 5x +

1 2

7–8(½)

c x2 + 5x – 2 g x2 – 5x –

7–9(½)

d x2 + 9x – 3

3 2

h x2 – 9x –

8 Factorise by first taking out the common factor. a 2x2 + 12x + 8 b 3x2 + 12x – 3 d 3x2 – 24x + 6 e –2x2 – 4x + 10 2 g – 4x – 16x + 12 h –2x2 + 16x + 4

c 4x2 – 8x – 16 f –3x2 – 30x – 3 i –3x2 + 24x – 15

9 Factorise the following. a 3x2 + 9x + 3 d 4x2 – 28x + 12 g – 4x2 + 12x + 20

c 2x2 – 10x + 4 f –2x2 – 14x + 8 i –2x2 + 10x + 8

b 5x2 + 15x – 35 e –3x2 – 21x + 6 h –3x2 + 9x + 6

10

10

5 2

10, 11

10 A student factorises x2 – 2x – 24 by completing the square. a Show the student’s working to obtain the factorised form of x2 – 2x – 24. b Now that you have seen the answer from part a, what would you suggest is a better way to factorise x2 – 2x – 24? 11 a Explain why x2 + 9 cannot be factorised using real numbers. b Decide whether the following can or cannot be factorised. i x2 – 25 ii x2 – 10 iii x2 + 6 2 2 v (x + 1) + 4 vi (x – 2) – 8 vii (x + 3)2 – 15 c

PROBLEM-SOLVING

Example 13a

FLUENCY

Chapter 5 Quadratic equations

REASONING

348

iv x2 + 11 viii (2x – 1)2 + 1

For what values of m can the following be factorised, using real numbers? i x2 + 4x + m ii x2 – 6x + m iii x2 – 10x + m

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

—

—

2 12 A non-monic quadratic such as ) 2x – 5x + 1 can be factorised in the following way. ( 5 1 2x2 – 5x + 1 = 2 x2 – x + 2 2   ( )2 ( )2 5 5 1 5 5 5 = 2x2 – x + – + ; Note: ÷ 2 = 2 4 4 2 2 4  ( )2 5 25 8 – +  = 2 x – 4 16 16 (  )2 5 17 = 2 x – –  4 16 ( √ )( √ ) 17 17 5 5 =2 x– + x– – 4 4 4 4

Factorise these using a similar technique. a 2x2 + 5x – 12 b 3x2 + 4x – 3 d 3x2 – 2x + 6 e –2x2 – 3x + 4 2 g – 4x + 11x – 24 h –2x2 + 3x + 4 j 3x2 + 4x – 5 k –2x2 – 3x + 5

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

c f i l

12

ENRICHMENT

Non-monic quadratics and completing the square

349

5E

4x2 – 7x – 16 –3x2 – 7x – 3 2x2 + 5x – 7 –3x2 – 7x – 4

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350

Chapter 5 Quadratic equations

5F Solving quadratic equations The result of multiplying a number by zero is zero. Consequently, if an expression equals zero then at least one of its factors must be zero. This is called the Null Factor Law and it provides us with an important method that can be utilised to solve a range of mathematical problems involving quadratic equations.

Parabolic arches (like this one supporting the weight of a bridge) can be modelled by quadratic equations.

Let’s start: Does factorisation beat trial and error? Set up two teams. Team A: Trial and error Team B: Factorisation Instructions: •

Team A must try to find the two solutions of 3x2 – x – 2 = 0 by guessing and checking values for x that make the equation true. • Team B must solve the same equation 3x2 – x – 2 = 0 by first factorising the left-hand side.

Which team was the first to find the two solutions for x? Discuss the methods used.

Key ideas

The Null Factor Law states that if the product of two numbers is zero, then either or both of the two numbers is zero. • If a × b = 0, then either a = 0 or b = 0. To solve a quadratic equation, write it in standard form (i.e. ax2 + bx + c = 0) and factorise. Then use the Null Factor Law. • If the coefficients of all the terms have a common factor, then first divide by that common factor.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

351

Example 15 Solving quadratic equations using the Null Factor Law Solve the following quadratic equations. a x2 – 2x = 0 b x2 – 15 = 0 SO L U T I O N a

2x2 = 50

E X P LA N A T I O N

x2 – 2x = 0

Factorise by taking out the common factor x. Apply the Null Factor Law: if a × b = 0, then a = 0 or b = 0. Solve for x. Check your solutions by substituting back.

x(x – 2) = 0 ∴ x = 0 or x – 2 = 0 ∴ x = 0 or x = 2 b

c

x2 – 15 = 0 √ √ (x + 15)(x – 15) = 0 √ √ ∴ x + 15 = 0 or x – 15 = 0 √ √ ∴ x = – 15 or x = 15

Factorise a2 – b2 = (a – b)(a + b) using surds. Alternatively, add 15 to both sides to give x2 = 15, then take the positive and negative square root. √ So x = ± 15.

2x2 = 50

c

First, write in standard form (i.e. ax2 + bx + c = 0). Take out the common factor of 2 and then factorise using a2 – b2 = (a + b)(a – b).

2x2 – 50 = 0 2(x2 – 25) = 0

Alternatively, divide first by 2 to give x2 = 25 and x = ± 5.

2(x + 5)(x – 5) = 0 ∴ x + 5 = 0 or x – 5 = 0 ∴ x = –5 or x = 5

Example 16 Solving ax2 + bx + c = 0 Solve the following quadratic equations. a

x2 – 5x + 6 = 0

SO L U T I O N a

b x2 + 2x + 1 = 0

10A

c

10x2 – 13x – 3 = 0

E X P LA N A T I O N x2 – 5x + 6 = 0

(x – 3)(x – 2) = 0 ∴ x – 3 = 0 or x – 2 = 0

Factorise by finding two numbers that multiply to 6 and add to –5 : –3 × (–2) = 6 and –3 + (–2) = –5. Apply the Null Factor Law and solve for x.

∴ x = 3 or x = 2 b

x2 + 2x + 1 = 0 (x + 1)(x + 1) = 0 (x + 1)2 = 0

1 × 1 = 1 and 1 + 1 = 2 (x + 1)(x + 1) = (x + 1)2 is a perfect square. This gives one solution for x.

∴ x+1=0 ∴ x = –1

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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352

Chapter 5 Quadratic equations

10x2 – 13x – 3 = 0

c 2

10x – 15x + 2x – 3 = 0

First, factorise using grouping or another method. 10 × (–3) = –30, –15 × 2 = –30 and –15 + 2 = –13.

5x(2x – 3) + (2x – 3) = 0 (2x – 3)(5x + 1) = 0 ∴ 2x – 3 = 0 or 5x + 1 = 0

Solve using the Null Factor Law.

∴ 2x = 3 or 5x = –1 ∴ x=

1 3 or x = – 2 5

Check your solutions by substitution.

Example 17 Solving disguised quadratics Solve the following by first writing in the form ax2 + bx + c = 0. a

x2 = 4(x + 15)

b

SO L U T I O N a

x+6 =x x

E X P LA N A T I O N x2 = 4(x + 15) x2 = 4x + 60

x2 – 4x – 60 = 0 (x – 10)(x + 6) = 0

First expand and then write in standard form by subtracting 4x and 60 from both sides. Factorise and apply the Null Factor Law: –10 × 6 = –60 and –10 + 6 = – 4.

∴ x – 10 = 0 or x + 6 = 0 ∴ x = 10 or x = –6 x+6 =x x x + 6 = x2 0 = x2 – x – 6 0 = (x – 3)(x + 2) ∴ x – 3 = 0 or x + 2 = 0 ∴ x = 3 or x = –2

Exercise 5F 1

First multiply both sides by x and then write in standard form. Factorise and solve using the Null Factor Law. Check your solutions.

1–3(½)

3(½)

—

Write the solutions to these equations, which are already in factorised form. a x(x + 1) = 0 b x(x – 5) = 0 c 2x(x – 4) = 0 d (x – 3)(x + 2) = 0 e (x + 5)(x – 4) = 0 f (x + 1)(x – 1) = 0 √ √ √ √ √ √ g (x + 3)(x – 3) = 0 h (x + 5)(x – 5) = 0 i (x + 2 2)(x – 2 2) = 0 j (2x – 1)(3x + 7) = 0 k (4x – 5)(5x + 2) = 0 l (8x + 3)(4x + 3) = 0

2 Rearrange to write in standard form ax2 + bx + c = 0. Do not solve. a x2 + 2x = 3 b x2 – 3x = 10 c 2 2 d 5x = 2x + 7 e 3x = 14x – 8 f g x(x + 1) = 4 h 2x(x – 3) = 5 i j 2 = x(x – 3) k – 4 = x(3x + 2) l Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

x2 – 5x = –6 4x2 = 3 – 4x x2 = 4(x – 3) x2 = 3(2 – x)

UNDERSTANDING

b

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3 How many different solutions for x will these equations have? a (x – 2)(x – 1) = 0 b (x + 7)(x + 3) = 0 √ √ e (x + 2)(x – 2) = 0 d (x – 3)(x – 3) = 0 g (x + 2)2 = 0 h (x + 3)2 = 0

c (x + 1)(x + 1) = 0 √ f (x + 8)(x – 5) = 0 i 3(2x + 1)2 = 0

Example 15

Example 16a,b

4–6(½)

5F

4–6(½)

4 Solve the following quadratic equations. a x2 – 4x = 0 b x2 – 3x = 0 d 3x2 – 12x = 0 e 2x2 – 10x = 0 2 g x –7=0 h x2 – 11 = 0 j x2 = 2x k x2 = –5x 2 m 5x = 20 n 3x2 = 27

c f i l o

x2 + 2x = 0 4x2 + 8x = 0 3x2 – 15 = 0 7x2 = –x 2x2 = 72

5 Solve the following quadratic equations. a x2 + 3x + 2 = 0 b x2 + 5x + 6 = 0 2 d x – 7x + 10 = 0 e x2 + 4x – 12 = 0 g x2 – x – 20 = 0 h x2 – 5x – 24 = 0 2 j x + 4x + 4 = 0 k x2 + 10x + 25 = 0 m x2 – 14x + 49 = 0 n x2 – 24x + 144 = 0

c f i l o

x2 – 6x + 8 = 0 x2 + 2x – 15 = 0 x2 – 12x + 32 = 0 x2 – 8x + 16 = 0 x2 + 18x + 81 = 0

6 Solve the following quadratic equations. a 2x2 + 11x + 12 = 0 b 4x2 + 16x + 7 = 0 10A 2 d 2x – 23x + 11 = 0 e 3x2 – 4x – 15 = 0 g 6x2 + 7x – 20 = 0 h 7x2 + 25x – 12 = 0

353

FLUENCY

4–5(½)

UNDERSTANDING

Number and Algebra

Example 16c

c 2x2 – 17x + 35 = 0 f 5x2 – 7x – 6 = 0 i 20x2 – 33x + 10 = 0

7 Solve by first taking out a common factor. a 2x2 + 16x + 24 = 0 b 2x2 – 20x – 22 = 0 10A e d 5x2 – 20x + 20 = 0 –8x2 – 4x + 24 = 0 Example 17a

Example 17b

7–8(½)

10A

8 Solve the following by first writing in the form ax2 + bx + c = 0. a x2 = 2(x + 12) b x2 = 4(x + 8) d x2 + 7x = –10 e x2 – 8x = –15 g 2x – 16 = x(2 – x) h x2 + 12x + 10 = 2x + 1 j x2 – 5x = –15x – 25 k x2 – 14x = 2x – 64 10A 10A m 2x(x – 2) = 6 n 3x(x + 6) = 4(x – 2)

c 3x2 – 18x + 27 = 0 f 18x2 – 57x + 30 = 0 c f i l o

9 Solve the following by first writing in the form ax2 + bx + c = 0. 5x + 84 9x + 70 a =x b =x c x x 20 – 3x 6x + 8 10A d 10A e 10A f = 2x =x x 5x 3 1 10A h g =x+2 = 3 – 2x i x x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

7–9(½)

x2 = 3(2x – 3) x(x + 4) = 4x + 9 x2 + x – 9 = 5x – 4 x(x + 4) = 4(x + 16) 4x(x + 5) = 6x – 4x2 – 3

PROBLEM-SOLVING

8(½)

18 – 7x =x x 7x + 10 = 3x 2x 4 =x+1 x–2

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Chapter 5 Quadratic equations

10

5F 10 a

10, 11

11, 12

REASONING

354

Write down the solutions to the following equations. i 2(x – 1)(x + 2) = 0 ii (x – 1)(x + 2) = 0

b What difference has the common factor of 2 made to the solutions in the first equation? c Explain why x2 – 5x – 6 = 0 and 3x2 – 15x – 18 = 0 have the same solutions. 11 Explain why x2 + 16x + 64 = 0 has only one solution. 12 When solving x2 – 2x – 8 = 7 a student writes the following. x2 – 2x – 8 = 7 (x – 4)(x + 2) = 7 x – 4 = 7 or x + 2 = 7 x = 11 or x = 5 Discuss the problem with this solution and then write a correct solution.

—

—

13 Solve these equations by first multiplying by an appropriate expression. a x+3=–

2 x

b –

1 =x–2 x

c –

5 = 2x – 11 x

d

x2 – 48 =2 x

e

x2 + 12 = –8 x

f

2x2 – 12 = –5 x

g

x–5 6 = 4 x

h

x–2 5 = 3 x

i

x–4 2 =– 2 x

j

x+4 3 – =1 2 x–3

k

x x+1 – =1 x–2 x+4

l

1 1 1 – = x–1 x+3 3

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

13

ENRICHMENT

More algebraic fractions with quadratics

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Number and Algebra

355

Using calculators to solve quadratic equations 1 Solve: a 2x2 – 41x = 0

b 4x2 + 4x – 15 = 0

2 Solve ax2 + bx + c = 0.

Using the TI-Nspire:

Using the ClassPad:

1 In a Calculator page use b>Algebra>Solve and type as shown ending with: ,x.

1 In the Main application, type and highlight the equation then tap Interactive, Advanced, Solve, OK.

Note: if your answers are decimal then change the Calculation Mode to Auto in Settings on the Home screen. 2

Use b>Algebra>Solve and type as shown.

2 Use the VAR keyboard to type the equation as shown. Highlight the equation and tap Interactive, Advanced, Solve, OK. This gives the general quadratic formula studied in section 5I.

Note: use a multiplication sign between a and x 2 in ax 2 and b and x in bx. This gives the general quadratic formula studied in section 5I

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356

Chapter 5 Quadratic equations

Progress quiz 38pt 5A

38pt 5B

38pt 5C

1

2

3

Expand brackets and simplify where possible. 2x a – (12x – 5) b 3 c (m + 2)(m + 5) d e (3m – 2)(3m + 2) f g 5(x – 4)(x – 3) h

a(3a – 2) – a(5 – a) (k – 3)2 (4h + 7)(2h – 5) (p + 5)(p + 4) – (p – 2)(p – 8)

Factorise the following. a 4a – 20 c 4(x + 5) – x(x + 5) e 16a2 – 121b2 g (k + 2)2 – 49 i x2 – 20 (use surds) k x2 + 5x + ax + 5a

b d f h j l

Factorise. a x2 + x – 20

c 3k2 – 21k – 54

b a2 – 10a + 21

–12m2 + 18m a2 – 81 5m2 – 125 (x – 1)2 – 4 (h + 3)2 – 7 (use surds) 4x2 – 8mx – 5x + 10m d m2 – 12m + 36

38pt 5C

4

Use factorisation to simplify these algebraic fractions. x2 + 3x – 28 x2 + 2x – 15 x2 – 25 × a b x+5 2x + 14 x2 – 9x + 20

38pt 5E

5

Complete the square and factorise, if possible. a x2 + 8x + 3

b x2 – 12x + 26

c x2 + 14x + 50

d x2 + 5x –

1 2

38pt 5F

6 Solve the following quadratic equations by first writing in the form ax2 + bx + c = 0 where necessary. a x2 – 5x = 0 b x2 – 10 = 0 2 c 3x = 48 d x2 – 7x + 12 = 0 e x2 + 6x + 9 = 0 f 2x2 – 12x – 110 = 0 7x – 10 g x2 = 4(8 – x) h =x x

38pt 5D

7

10A

38pt 5D

8

Factorise. a 6a2 + 19a + 10 Simplify

b 8m2 – 6m – 9

c 15x2 – 22x + 8

d 6k2 – 11k – 35

9x2 – 49 2x2 + 7x + 5 × 2 . 2 3x – 4x – 7 6x + 5x – 21

10A

38pt 5F 10A

9 Solve the following quadratic equations by first writing in the form ax2 + bx + c = 0 where necessary. 19x + 4 a 6x2 – x – 35 = 0 b 2x(2x + 5) = 2x2 + 7(x + 5) c 6x = 5x

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Number and Algebra

357

5G Applications of quadratics Defining variables, setting up equations, solving equations and interpreting solutions are all important elements of applying quadratic equations to problem solving. The area of a rectangular paddock, for example, that can be fenced off using a limited length of fencing can be found by setting up a quadratic equation, solving it and then interpreting the solutions.

Let’s start: The 10 cm2 triangle There are many base and height measurements for a triangle that give an area of 10 cm2 .

height base •

Draw three different triangles that have a 10 cm2 area. Include the measurements for the base and the height. • Do any of your triangles have a base length that is 1 cm more than the height? Find the special triangle with area 10 cm2 that has a base 1 cm more than its height by following these steps. – Let x cm be the height of the triangle. –

Write an expression for the base length.

–

Write an equation if the area is 10 cm2 .

–

Solve the equation to find two solutions for x.

–

Which solution is to be used to describe the special triangle? Why?

x cm

When applying quadratic equations, follow these steps. • Define a variable; i.e. ‘Let x be …’. • Write an equation. • Solve the equation. • Choose the solution(s) that solves the equation and answers the question in the context in which it was given. Check that the solutions seem reasonable.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Key ideas

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358

Chapter 5 Quadratic equations

Example 18 Finding dimensions using quadratics The area of a rectangle is fixed at 28 m2 and its length is 3 metres more than its width. Find the dimensions of the rectangle. SOL UTI ON

EX P L A NA TI ON

Let x m be the width of the rectangle. Length = (x + 3) m

Draw a diagram to help.

xm

x2 + 3x – 28 = 0 (x + 7)(x – 4) = 0 x + 7 = 0 or x – 4 = 0 ∴ x = –7 or x = 4 x > 0 so, choose x = 4. Rectangle has width 4 m and length 7 m.

Exercise 5G Example 18

1

(x + 3) m 28 m2

Write an equation using the given information. Then write in standard form and solve for x. Disregard x = –7 because x > 0. Answer the question in full. Note: Length is 4 + 3 = 7.

1–2

2

—

A rectangle has an area of 24 m2 . Its length is 5 m longer than its width. x m 24 m2 a Copy this sentence: ‘Let x m be the width of the rectangle.’ b Write an expression for the rectangle’s length. c Write an equation using the rectangle’s area. d Write your equation from part c in standard form (i.e. ax2 + bx + c = 0) and solve for x. e Find the dimensions of the rectangle.

UNDERSTANDING

x(x + 3) = 28

2 Repeat all the steps in Question 1 to find the dimensions of a rectangle with the following properties. a Its area is 60 m2 and its length is 4 m more than its width. b Its area is 63 m2 and its length is 2 m less than its width. c Its area is 154 mm2 and its length is 3 mm less than its width. 3–5

4–6

3 Find the height and base lengths of a triangle that has an area of 24 cm2 and height 2 cm more than its base.

FLUENCY

3–5

4 Find the height and base lengths of a triangle that has an area of 7 m2 and height 5 m less than its base. 5 The product of two consecutive numbers is 72. Use a quadratic equation to find the two sets of numbers. 6 The product of two consecutive, even positive numbers is 168. Find the two numbers.

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Number and Algebra

7–10

8–11

PROBLEM-SOLVING

7, 8

7 A 100 m2 hay shed is to be extended to give 475 m2 of floor space in total, as shown. All angles are right angles. Find the value of x.

100 m2 10 m

359

5G

=

10 m x m= xm

8 Solve for x in these right-angled triangles, using Pythagoras’ theorem. a b c x+2 x−1

x

10

x

5

x+1

x−2 9

9 A square hut of side length 5 m is to be surrounded by a veranda of width x metres. Find the width of the veranda if its area is to be 24 m2 . 10 A father’s age is the square of his son’s age (x). In 20 years’ time the father will be three times as old as his son. What are the ages of the father and son? 11 A rectangular painting is to have a total area (including the frame) of 1200 cm2 . If the painting is 30 cm long and 20 cm wide, find the width of the frame. 12, 13

13, 14

1 12 The sum of the first n positive integers is given by n(n + 1). 2 a Find the sum of the first 10 positive integers (i.e. use n = 10). b Find the value of n if the sum of the first n positive integers is: i 28 ii 91 iii 276

REASONING

12

13 A ball is thrust vertically upwards from a machine on the ground. The height (h metres) after t seconds is given by h = t(4 – t). a Find the height after 1.5 seconds. b Find when the ball is at a height of 3 metres. c Why are there two solutions to part b? d Find when the ball is at ground level. Explain. e Find when the ball is at a height of 4 metres. f Why is there only one solution for part e? g Is there a time when the ball is at a height of 5 metres? Explain.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter 5 Quadratic equations

REASONING

5G 14 The height (h metres) of a golf ball is given by h = –x2 + 100x, where x metres is the horizontal distance from where the ball was hit.

a b c

Find the values of x when h = 0. Interpret your answer from part a. Find how far the ball has travelled horizontally when the height is 196 metres.

Fixed perimeter and area

—

—

15, 16

15 A small rectangular block of land has a perimeter of 100 m and an area of 225 m2 . Find the dimensions of the block of land. 16 A rectangular farm has perimeter 700 m and area 30 000 m2 . Find its dimensions.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

ENRICHMENT

360

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Number and Algebra

361

5H Solving quadratic equations by completing the square In section 5E we saw that some quadratics cannot be factorised using integers but instead could be factorised by completing the square. Surds were also used to complete the factorisation. We can use this method to solve many quadratic equations.

√ Let’s start: Where does 6 come in? Consider the equation x2 – 2x – 5 = 0 and try to solve it by discussing these points. • Are there any common factors that can be taken out? • Are there any integers that multiply to give –5 and add to give –2? • Try completing the square on the left-hand side. Does this help and how? √ • Show that the two solutions contain the surd 6.

To solve quadratic equations of the form ax2 + bx + c = 0 for which you cannot factorise using integers: • Complete the square for the quadratic expression. • Solve the quadratic equation using the Null Factor Law.

Key ideas

Expressions such as x2 + 5 and (x – 1)2 + 7 cannot be factorised further and therefore give no solutions as they cannot be expressed as a difference of two squares.

Example 19 Solving quadratic equations by completing the square Solve these quadratic equations by first completing the square. a x2 – 4x + 2 = 0 b x2 + 6x – 11 = 0 SOL UTI ON

c

x2 – 3x + 1 = 0

EX P L A NA TI ON x2 – 4x + 2 = 0

a 2

x – 4x + 4 – 4 + 2 = 0 (x – 2)2 – 2 = 0 √ (x – 2 + 2)(x – 2 – 2) = 0 √ √ ∴ x – 2 + 2 = 0 or x – 2 – 2 = 0 √ √ ∴ x = 2 – 2 or x = 2 + 2 √

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Complete the square:



–4 2

2

= 4.

x2 – 4x + 4 = (x – 2)(x – 2) = (x – 2)2 Use a2 – b2 = (a + b)(a – b). Apply the Null Factor Law and solve for x. √ √ For x – 2 + 2 = 0, add 2 and subtract 2 from both sides. √ The solutions can also be written as 2 ± 2.

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362

Chapter 5 Quadratic equations

( )2 6 = 9. Complete the square: 2

2

x + 6x – 11 = 0 2

x + 6x + 9 – 9 – 11 = 0 (x + 3)2 – 20 = 0 √ √ (x + 3 – 20)(x + 3 + 20) = 0 √ √ (x + 3 – 2 5)(x + 3 + 2 5) = 0 √ √ ∴ x + 3 – 2 5 = 0 or x + 3 + 2 5 = 0 √ √ ∴ x = –3 + 2 5 or x = –3 – 2 5 √ Alternatively, x = –3 ± 2 5. x2 – 3x + 1 = 0

c

9 9 x2 – 3x + – + 1 = 0 4 4 ( )2 3 5 x– – =0 2 4 ( √ )( √ ) 5 5 3 3 x– + x– – =0 2 2 4 4 √ √ 5 5 3 3 x– + = 0 or x – – =0 2 2 4 4 √ 5 3 – 2 2

√ 5 3 or x = + 2 2

√ 3– 5 x= 2 √ 3± 5 So x = . 2

√ 3+ 5 or x = 2

∴ x=

Apply the Null Factor Law and solve for x. Alternatively, write solutions using the ± symbol. ( )2 9 3 = – 2 4

a2 – b2 = (a + b)(a – b) Use the Null Factor Law.

Recall that

√

√ √ 5 5 5 =√ = . 4 2 4

Combine using the ± symbol.

Exercise 5H 1

Use difference of perfect squares with surds. √ √ √ Recall that 20 = 4 × 5 = 2 5.

1–3(½)

3(½)

What number must be added to the following expressions to form a perfect square? a x2 + 2x b x2 + 6x c x2 + 20x d x2 + 50x 2 2 2 e x – 4x f x – 10x g x + 5x h x2 – 3x

2 Factorise using surds. a x2 – 3 = 0 d (x + 1)2 – 5 = 0

b x2 – 7 = 0 e (x + 3)2 – 11 = 0

3 Solve these equations. √ √ a (x – 2)(x + 2) = 0 √ √ c (x – 10)(x + 10) = 0 √ √ e (x – 4 + 6)(x – 4 – 6) = 0

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

—

UNDERSTANDING

b

c x2 – 10 = 0 f (x – 1)2 – 2 = 0

√ √ b (x – 7)(x + 7) = 0 √ √ d (x – 3 + 5)(x – 3 – 5) = 0 √ √ f (x + 5 + 14)(x + 5 – 14) = 0

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Number and Algebra

Example 19a

Example 19b

4–6(½)

4–6(½)

4 Solve by first completing the square. a x2 + 6x + 3 = 0 b x2 + 4x + 2 = 0 2 d x + 4x – 2 = 0 e x2 + 8x – 3 = 0 g x2 – 8x – 1 = 0 h x2 – 12x – 3 = 0 2 j x – 10x + 18 = 0 k x2 – 6x + 4 = 0 m x2 + 6x – 4 = 0 n x2 + 20x + 13 = 0

c f i l o

5 Solve by first completing the square. a x2 + 8x + 4 = 0 b x2 + 6x + 1 = 0 2 d x – 4x – 14 = 0 e x2 – 10x – 3 = 0 g x2 – 2x – 31 = 0 h x2 + 12x – 18 = 0

c x2 – 10x + 5 = 0 f x2 + 8x – 8 = 0 i x2 + 6x – 41 = 0

x2 + 10x + 15 = 0 x2 + 6x – 5 = 0 x2 – 2x – 16 = 0 x2 – 8x + 9 = 0 x2 – 14x – 6 = 0

FLUENCY

4–5(½)

363

5H

6 Decide how many solutions there are to these equations. Try factorising the equations if you are unsure. a x2 – 2 = 0 b x2 – 10 = 0 c x2 + 3 = 0 2 2 d x +7=0 e (x – 1) + 4 = 0 f (x + 2)2 – 7 = 0 g (x – 7)2 – 6 = 0 h x2 – 2x + 6 = 0 i x2 – 3x + 10 = 0 2 2 j x + 2x – 4 = 0 k x + 7x + 1 = 0 l x2 – 2x + 17 = 0

Example 19c

7 Solve by first completing the square. a x2 + 5x + 2 = 0 b x2 + 3x + 1 = 0 2 d x – 3x – 2 = 0 e x2 – x – 3 = 0 g x2 – 7x + 2 = 0 h x2 – 9x + 5 = 0 3 j x2 + 9x + 9 = 0 k x2 – 3x – = 0 4

7–9(½), 10

7–9(½), 11

c x2 + 7x + 5 = 0 f x2 + 5x – 2 = 0 i x2 + x – 4 = 0 5 l x2 + 5x + = 0 4

PROBLEM-SOLVING

7(½), 10

8 Solve the following, if possible, by first factoring out the coefficient of x2 and then completing the square. a 2x2 – 4x + 4 = 0 b 4x2 + 20x + 8 = 0 c 2x2 – 10x + 4 = 0 2 2 d 3x + 27x + 9 = 0 e 3x + 15x + 3 = 0 f 2x2 – 12x + 8 = 0 9 Solve the following quadratic equations, if possible. a x2 + 3x = 5 b x2 + 5x = 9 d x2 – 8x = –11 e x2 + 12x + 10 = 2x + 5 10 A rectangle’s length is 3 cm more than its width. Find the dimensions of the rectangle if its area is 20 cm2 .

c x2 + 7x = –15 f x2 + x + 9 = 5x – 3

x cm 20 cm2 (x + 3) cm

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364

Chapter 5 Quadratic equations

PROBLEM-SOLVING

5H 11 The height, h km, of a ballistic missile launched from a submarine at sea level is given x(400 – x) , where x km is the horizontal distance travelled. by h = 20 000 a Find the height of a missile that has travelled the following horizontal distances. i 100 km ii 300 km b Find how far the missile has travelled horizontally when the height is: i 0 km ii 2 km Find the horizontal distance the missile has travelled when its height is 1 km. (Hint: Complete the square.) 12

12, 13

14(½), 15

REASONING

c

12 Complete the square to show that the following have no (real) solutions. a x2 + 4x + 5 = 0 b x2 – 3x = –3 13 A friend starts to solve x2 + x – 30 = 0 by completing the square but you notice there is a much quicker way. What method do you describe to your friend? 14 A slightly different way to solve by completing the square is shown here. Solve the following using this method. x2 – 4x + 1 = 0 a x2 – 6x + 2 = 0 x2 – 4x + 4 – 4 + 1 = 0 b x2 + 8x + 6 = 0 2 (x – 2) – 3 = 0 c x2 + 4x – 7 = 0 (x – 2)2 = 3 d x2 – 2x – 5 = 0 √ ∴x–2= ± 3 e x2 – 8x + 4 = 0 √ x = 2± 3 f x2 + 10x + 1 = 0 15 This rectangle is a golden rectangle. • ABEF is a square. • Rectangle BCDE is similar to rectangle ACDF.

b

1 a . = 1 a–1 Find the exact value of a (which will give you the golden ratio) by completing the square. —

D

A

B a

C

1

Show that

Completing the square with non-monics

E

—

16

16 In the Enrichment section of Exercise 5E we looked at a method to factorise non-monic quadratics by completing the square. It involved taking out the coefficient of x2 . Dividing both sides by that number is possible in these equations and this makes the task easier. Use this technique to solve the following equations. a 2x2 + 4x – 1 = 0 b 3x2 + 6x – 12 = 0 c –2x2 + 16x – 10 = 0 2 2 d 3x – 9x + 3 = 0 e 4x + 20x + 8 = 0 f 5x2 + 5x – 15 = 0

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

ENRICHMENT

a

F

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Number and Algebra

365

5I The quadratic formula A general formula for solving quadratic equations can be found by completing the square for the general case. Consider ax2 + bx + c = 0, where a, b, c are constants and a π 0. Start by dividing both sides by a. c b x2 + x + = 0 a a  2  2 b b b c x2 + x + – + =0 a 2a 2a a   b 2 b2 c x+ – 2 + =0 2a a 4a    2  b 2 b – 4ac x+ =0 – 2a 4a2   b 2 b2 – 4ac x+ = 2a 4a2  b2 – 4ac b = ± x+ 2a 4a2 √ b2 – 4ac b x=– ± 2a √ 2a 2 –b ± b – 4ac = 2a

The path of an object under the influence of gravity can be modelled by a quadratic equation.

This formula now gives us a mechanism to solve quadratic equations and to determine how many solutions the equation has. The expression under the root sign, b2 – 4ac, is called the discriminant (∆) and helps us to identify the number of solutions. A quadratic equation can have 0, 1 or 2 solutions.

Let’s start: How many solutions? Complete this table to find the number of solutions for each equation. 2

ax + bx + c = 0

a

b

c

2

b – 4ac

–b +



b 2 – 4ac 2a

–b –



b 2 – 4ac 2a

2x 2 + 7x + 1 = 0 9x 2 – 6x + 1 = 0 x 2 – 3x + 4 = 0

Discuss under what circumstances a quadratic equation has: • 2 solutions • 1 solution • 0 solutions

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366

Chapter 5 Quadratic equations

Key ideas

If ax2 + bx + c = 0 (where a, b, c are constants and a π 0), then √ √ –b + b2 – 4ac –b – b2 – 4ac or x = x= 2a 2a • This is called the quadratic formula. • The quadratic formula is useful when a quadratic cannot be factorised easily. The discriminant is ∆ = b2 – 4ac. √ • When ∆ < 0, the quadratic equation has 0 real solutions (since ∆ is undefined when ∆ is negative).   b . • When ∆ = 0, the quadratic equation has 1 real solution x = – 2a   √ –b ± b2 – 4ac • When ∆ > 0, the quadratic equation has 2 real solutions x = . 2a

Example 20 Using the discriminant Determine the number of solutions to the following quadratic equations using the discriminant. a x2 + 5x – 3 = 0 b 2x2 – 3x + 4 = 0 c x2 + 6x + 9 = 0 SOL UTI ON

EX P L A NA TI ON

a

State the values of a, b and c in ax2 + bx + c = 0. Calculate the value of the discriminant by substituting values.

a = 1, b = 5, c = –3 2

∆ = b – 4ac = (5)2 – 4(1)(–3) = 25 + 12 = 37 ∆ > 0, so there are 2 solutions. b

Interpret the result with regard to the number of solutions. State the values of a, b and c and substitute to evaluate the discriminant. Recall that (–3)2 = –3 × (–3) = 9.

a = 2, b = –3, c = 4 2

∆ = b – 4ac = (–3)2 – 4(2)(4) = 9 – 32 = –23 ∆ < 0, so there are no solutions. c

a = 1, b = 6, c = 9

Interpret the result. Substitute the values of a, b and c to evaluate the discriminant and interpret the result.

2

∆ = b – 4ac = (6)2 – 4(1)(9) = 36 – 36

Note: x2 + 6x + 9 = (x + 3)2 is a perfect square.

=0 ∆ = 0, so there is 1 solution.

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Number and Algebra

367

Example 21 Solving quadratic equations using the quadratic formula Find the exact solutions to the following using the quadratic formula. a x2 + 5x + 3 = 0 b 2x2 – 2x – 1 = 0 SO L U T I O N

EX P L A N A T I O N

a

a = 1, b = 5, c = 3 √ –b ± b2 – 4ac x= √2a –5 ± (5)2 – 4(1)(3) = 2(1) √ –5 ± 25 – 12 = √2 –5 ± 13 = 2

Determine the values of a, b and c in ax2 + bx + c = 0. Write out the quadratic formula and substitute the values.

a = 2, b = –2, c = –1 √ – b ± b2 – 4ac x= 2a √ –(–2) ± (–2)2 – 4(2)(–1) = 2(2) √ 2± 4 + 8 = √4 2 ± 12 = 4√ 2±2 3 = 4 √ 1± 3 = 2

Determine the values of a, b and c.

√ √ –5 – 13 –5 + 13 , . Two solutions: x = 2 2

Simplify:

12 =

√ √ 4 × 3 = 2 3.

Cancel: √ √ 2 ± 2 3 2(1 ± 3) = 4 √4 1± 3 = 2

Exercise 5I 1

√

1–3

3

For these quadratic equations in the form ax2 + bx + c = 0, state the values of a, b and c. a 3x2 + 2x + 1 = 0 b 2x2 + x + 4 = 0 c 5x2 + 3x – 2 = 0 d 4x2 – 3x + 2 = 0 e 2x2 – x – 5 = 0 f –3x2 + 4x – 5 = 0

2 Find the value of the discriminant (b2 – 4ac) for each part in Question 1 above. 3 State the number of solutions of a quadratic that has: a b2 – 4ac = 0 b b2 – 4ac < 0

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

—

UNDERSTANDING

b

Simplify.

c b2 – 4ac > 0

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Chapter 5 Quadratic equations

4–6(½)

4–6(½)

Example 20

4 Using the discriminant, determine the number of solutions for these quadratic equations. a x2 + 5x + 3 = 0 b x2 + 3x + 4 = 0 c x2 + 6x + 9 = 0 2 2 d x + 7x – 3 = 0 e x + 5x – 4 = 0 f x2 + 4x – 4 = 0 g 4x2 + 5x + 3 = 0 h 4x2 + 3x + 1 = 0 i 2x2 + 12x + 9 = 0 2 2 j –x – 6x – 9 = 0 k –2x + 3x – 4 = 0 l – 4x2 – 6x + 3 = 0

Example 21a

5 Find the exact solutions to the following quadratic equations, using the quadratic formula. a x2 + 3x – 2 = 0 b x2 + 7x – 4 = 0 c x2 – 7x + 5 = 0 d x2 – 8x + 16 = 0 e –x2 – 5x – 4 = 0 f –x2 – 8x – 7 = 0 2 2 g 4x + 7x – 1 = 0 h 3x + 5x – 1 = 0 i 3x2 – 4x – 6 = 0 j –2x2 + 5x + 5 = 0 k –3x2 – x + 4 = 0 l 5x2 + 6x – 2 = 0

Example 21b

6 Find the exact solutions to the following quadratic equations, using the quadratic formula. a x2 + 4x + 1 = 0 b x2 – 6x + 4 = 0 c x2 + 6x – 2 = 0 2 2 d –x – 3x + 9 = 0 e –x + 4x + 4 = 0 f –3x2 + 8x – 2 = 0 2 2 g 2x – 2x – 3 = 0 h 3x – 6x – 1 = 0 i –5x2 + 8x + 3 = 0

7, 8(½)

8(½), 10

7 A triangle’s base is 5 cm more than its height of x cm. Find its height if the triangle’s area is 10 cm2 .

x cm

8 Solve the following using the quadratic formula. a 3x2 = 1 + 6x b 2x2 = 3 – 4x 3 1 d 2x – 5 = – e = 3x + 4 x x 2x + 2 3x + 4 g 5x = h x= x 2x

FLUENCY

4–5(½)

5I

8(½), 9, 11

PROBLEM-SOLVING

368

c 5x = 2 – 4x2 5 f – =2–x x 10x – 1 i 3x = 2x

9 Two positive numbers differ by 3 and their product is 11. Find the numbers. 10 Find the exact perimeter of this right-angled triangle.

x+3

x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

x+4

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11 A rectangular pool measuring 6 m by 3 m is to have a path surrounding it. If the total area of the pool and path is to be 31 m2 , find the width (x m) of the path, correct to the nearest centimetre.

6m

PROBLEM-SOLVING

Number and Algebra

369

5I

xm

12

12, 13

13, 14

REASONING

3m

√ –b ± b2 – 4ac gives only one solution when b2 – 4ac = 0. 12 Explain why the rule x = 2a 13 Make up three quadratic equations that have: a no solutions b 1 solution c 2 solutions 14 For what two values of k does x2 + kx + 9 = 0 have only one solution? —

—

15 The discriminant for x2 + 2x + k = 0 is 4 – 4k, so there: • are no solutions for 4 – 4k < 0, ∴ k > 1 • is 1 solution for 4 – 4k = 0, ∴ k = 1 • are 2 solutions for 4 – 4k > 0, ∴ k < 1 a

For what values of k does x2 + 4x + k = 0 have: i no solutions? ii 1 solution?

b For what values of k does i no solutions?

kx2

c

x2

For what values of k does i no solutions?

+ 3x + 2 = 0 have: ii 1 solution?

+ kx + 1 = 0 have: ii 1 solution?

d For what values of k does 3x2 + kx – 1 = 0 have: i no solutions? ii 1 solution?

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

15

ENRICHMENT

k determines the number of solutions

iii 2 solutions? iii 2 solutions? iii 2 solutions? iii 2 solutions?

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370

Chapter 5 Quadratic equations

Investigation Binomial expressions Blaise Pascal and expansion Blaise Pascal (1623–1662) was a French mathematician and philosopher. By the age of 16 he had proved many theorems in geometry and by 17 he had invented and made what is regarded as the first calculator. One of his mathematical investigations involved exploring the properties and patterns of numbers in a triangular arrangement that is known today as Pascal’s triangle. The triangle has many applications in mathematics, including algebraic expansion and probability. The diagram below shows part of this triangle. Pascal’s triangle row 0 row 1 row 2 row 3 row 4 row 5 row 6 row 7 row 8 row 9 row 10

1 1 1 1 1 1

1 2

3 4

5

1 3

6 10

1 4

10

1 5

1

1 1

Expanding the triangle a b c

Observe and describe the pattern of numbers shown in rows 0 to 4. State a method that might produce the next row in the triangle. Complete the triangle to row 10.

Expanding brackets Consider the expansions of binomial expressions. If you look closely, you can see how the coefficients in each term match the values in the triangle you produced in the triangle above. (x + y)0 = 1 (x + y)1 = 1x + 1y (x + y)2 = (x + y)(x + y) = 1x2 + 2xy + 1y2 (x + y)3 = (x + y)(x + y)2 = (x + y)(x2 + 2xy + y2 ) = x3 + 2x2 y + xy2 + yx2 + 2xy2 + y3 = 1x3 + 3x2 y + 3xy2 + 1y3 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

371

Expand (x + y)4 , (x + y)5 , (x + y)6 , (x + y)7 and (x + y)8 by completing the triangle below. (x + y)0 (x + y)1 (x + y)2 (x + y)3 (x + y)4 (x + y)5 (x + y)6 (x + y)7 (x + y)8

1 1x + 1y 1x2 + 2xy + 1y2 3 1x + 3x2 y + 3xy2 + 1y3

Factorials and combinations Another way of generating Pascal’s triangle is by using the formula for combinations. In general, the number(of)ways to select r objects from a group of n objects is given by n! nC = n = , where n! = n × (n – 1) × (n – 2) × (n – 3) × . . . × 1 and 0! = 1. r r!(n – r)! r Consider a group of two objects from which you wish to choose 0, 1 or 2 objects. ( ) 2 2! You can choose 0 objects = 1 way = 0!(2 – 0)! 0 ( ) 2 2! 1 object = 2 ways = 1!(2 – 1)! 1 ( ) 2 2! 2 objects = = 1 way 2!(2 – 2)! 2 Use the combinations formula to copy and complete the triangle below for the number of ways of selecting objects from a group. 0 objects 1 object 2 objects

( ) 2 =1 0

( ) 1 =1 0

( ) 0 =1 0

( ) 2 =2 1

( ) 1 =1 1

( ) 2 =1 2

3 objects 4 objects Expanding binomial expressions By noting the patterns above, expand the following. a (x + y)5

b (x + y)10

c (x + 1)3

d (x + 3)5

e (2x + 3)4

f (3x + 1)5

g (1 – x)4

h (3 – 2x)5

i (x2 – 1)6

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372

Chapter 5 Quadratic equations

Up for a challenge? If you get stuck on a question, check out the 'Working with unfamiliar problems' poster at the end of the book to help you.

Problems and challenges 1

√ √ Find the monic quadratic in the form x2 + bx + c = 0 with solutions x = 2 – 3 and x = 2 + 3.

2

If x +

3

Find all the solutions to each equation. (Hint: Consider letting a = x2 in each equation.) a x4 – 5x2 + 4 = 0 b x4 – 7x2 – 18 = 0

4

Make a substitution as you did in Question 3 to obtain a quadratic equation to help you solve the following. a 32x – 4 × 3x + 3 = 0 b 4 × 22x – 9 × 2x + 2 = 0

5

Quadrilateral ABCD has a perimeter of 64 cm with measurements as shown. What is the area of the quadrilateral?

1 1 = 7, what is x2 + 2 ? x x

B 6 cm

8 cm

A

C

D 6

A cyclist in a charity ride rides 300 km at a constant average speed. If the average speed had been 5 km/h faster, the ride would have taken 2 hours less. What was the average speed of the cyclist?

7

Find the value of x, correct to one decimal place, in this diagram if the area is to be 20 square units.

x x + 10

8

Prove that x2 – 2x + 2 > 0 for all values of x.

9

A square has the same perimeter as a rectangle of length x cm and width y cm. Determine a simplified expression for the difference in their areas and, hence, show that when the perimeters are equal the square has the greatest area.

10 The equation x2 + wx + t = 0 has solutions a and b , where the equation x2 + px + q = 0 has solutions 3a and 3b . Determine the ratios w : p and t : q. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Factorising x 2 + bx + c

Factorising non-monic quadratics ax 2 + bx + c (10A) Use grouping, split up bx using two numbers that multiply to a × c and add to give b : e.g. 1 6x 2 − 5x − 4 a × c = 6 × (− 4) = −24 −8 × 3 = −24 −8 + 3 = −5 2 6x 2 + 3x − 8x − 4 = 3x (2x + 1) − 4(2x + 1) = (2x + 1)(3x − 4)

Two numbers × to give c. Two numbers + to give b. e.g. x 2 − 7x − 18 = (x − 9)(x + 2) −9 × 2 = −18 −9 + 2 = −7

Factorising by completing the square e.g. x 2 + 4x − 3 = (x 2 + 4x + ( 42 ) 2 ) − ( 42 )2 − 3 = (x + 42 )2 − 4 − 3 = (x + 2)2 − 7 = (x + 2 − 7)(x + 2 + 7) Note, for example, (x + 2)2 + 5 cannot be factorised.

Quadratic equations

Factorising and DOPS Always take out common factors first. Difference of perfect squares a 2 − b 2 = (a − b)(a + b) e.g. 4x 2 − 9 = (2x)2 − (3)2 = (2x − 3)(2x + 3) x 2 − 7 = (x − 7)(x + 7)

Expanding brackets a (b + c) = ab + ac (a + b)(c + d ) = ac + ad + bc + bd (a + b)(a − b ) = a 2 − b 2 (a + b)2 = a 2 + 2ab + b 2 (a − b)2 = a 2 − 2ab + b 2

373

Chapter summary

Number and Algebra

Solving quadratic equations Null factor law: If ab = 0 then a = 0 or b = 0. Write each quadratic in standard form ax 2 + bx + c = 0, factorise then apply the Null Factor Law to solve. e.g. 1 x 2 − 4x = 0 x (x − 4) = 0 x = 0 or x − 4 = 0 x = 0 or x = 4 2 x 2 = 3x − 10 x 2 − 3x + 10 = 0 (x − 5)(x + 2) = 0 x − 5 = 0 or x + 2 = 0 x = 5 or x = −2

If

Quadratic formula (10A) + bx + c = 0, then

ax 2

b 2 − 4ac 2a The discriminant Δ = b 2 − 4ac tells us how many solutions: Δ>0 2 solutions Δ=0 1 solution Δ 0 2 D b – 4ac π 0 E b2 – 4ac < 0

C b2 – 4ac ≤ 0

Short-answer questions 38pt 5A

38pt 5A

1

Expand the following and simplify where possible. a 2(x + 3) – 4(x – 5) b (x + 5)(3x – 4) 2 d (x – 6) e (x + 4)2 – (x + 3)(x – 2)

c (5x – 2)(5x + 2) f (3x – 2)(4x – 5)

2 Write, in expanded form, an expression for the shaded areas. All angles are right angles. a b c x−1 x+2 xm

6m

3m

375

Chapter review

Number and Algebra

xm

x+4 38pt 5B

3 Factorise the following difference of perfect squares. Remember to look for a common factor first. a x2 – 49 b 9x2 – 16 c 4x2 – 1 d 3x2 – 75 e 2x2 – 18 f x2 – 11 2 2 g –2x + 40 h (x + 1) – 16 i (x – 3)2 – 10

38pt 5C

4 Factorise these quadratic trinomials. a x2 – 8x + 12 b x2 + 10x – 24

38pt 5D 10A

38pt 5C/D

38pt 5E

5 Factorise these non-monic quadratic trinomials. a 3x2 + 17x + 10 b 4x2 + 4x – 15 6 Simplify. x2 – 1 12x × a x2 + 2x – 3 6x + 6

10A

b

c 12x2 – 16x – 3

d 12x2 – 23x + 10

8x + 12 4x2 – 9 ÷ 2x2 + x – 6 x2 – 2x + 8

7 Factorise the following by completing the square. a x2 + 8x + 10 b x2 + 10x – 4 d x2 + 3x – 2

c –3x2 + 21x – 18

e x2 + 5x + 3

c x2 – 6x – 3 f

x2 + 7x +

9 2

38pt 5F

8 Solve these quadratic equations by factorising and applying the Null Factor Law. a x2 + 4x = 0 b 3x2 – 9x = 0 c x2 – 25 = 0 2 2 d x – 10x + 21 = 0 e x – 8x + 16 = 0 f x2 + 5x – 36 = 0 10A g 10A h 10A i 2x2 + 3x – 2 = 0 6x2 + 11x – 10 = 0 18x2 + 25x – 3 = 0

38pt 5F

9 Solve the following quadratic equations by first writing them in standard form. a 3x2 = 27 b x2 = 4x + 5 3x + 18 d =x c 2x2 – 28 = x(x – 3) x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter review

376

Chapter 5 Quadratic equations

38pt 5G

10 A rectangular sandpit is 2 m longer than it is wide. If it occupies an area of 48 m2 , determine the dimensions of the sandpit by solving a suitable equation.

38pt 5H

11 Solve these quadratic equations by first completing the square. a x2 + 4x – 3 = 0 b x2 – 6x + 1 = 0 c x2 – 3x – 2 = 0 d x2 + 5x – 5 = 0

38pt 5I

12 For each quadratic equation, determine the number of solutions by finding the value of the discriminant. a x2 + 2x + 1 = 0 b x2 – 3x – 3 = 0 c 2x2 – 4x + 3 = 0 d –3x2 + x + 5 = 0

10A

38pt 5I 10A

13 Use the quadratic formula to give exact solutions to these quadratic equations. a x2 + 3x – 6 = 0 b x2 – 2x – 4 = 0 c 2x2 – 4x – 5 = 0 d –3x2 + x + 3 = 0

Extended-response questions 1

A zoo enclosure for a rare tiger is rectangular in shape and has a trench of width x m all the way around it to ensure the tiger doesn’t get far if it tries to escape. The dimensions are as shown. a Write an expression in terms of x for: i

the length of the enclosure and trench combined

ii

the width of the enclosure and trench combined.

trench enclosure 12 m

xm

15 m xm

b Use your answers from part a to find the area of the overall enclosure and trench, in expanded form. c Hence, find an expression for the area of the trench alone. d Zoo restrictions state that the trench must have an area of at least 58 m2 . By solving a suitable equation, find the minimum width of the trench.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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377

Chapter review

Number and Algebra

2 The surface area S of a cylindrical tank with a hemispherical top is given by the equation S = 3p r2 + 2p rh, where r is the radius and h is the height of the cylinder.

r

h

a If the radius of a tank with height 6 m is 3 m, determine its exact surface area. b If the surface area of a tank with radius 5 m is 250 m2 , determine its height, to two decimal places. c The surface area of a tank of height 6 m is found to be 420 m2 .

10A

i

Substitute the values and rewrite the equation in terms of r only.

ii

Rearrange the equation and write it in the form ar2 + br + c = 0.

iii Solve for r using the quadratic formula and round your answer to two decimal places.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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378

Semester review 1

Semester review 1 Linear relations Multiple-choice questions 1

The simplified form of 2x(4 – 3y) – 3(3x – 4xy) is: A 6xy – x B 2xy C x – 18xy

2 The point that is not on the line y = 3x – 2 is: A (–1, –5) B (1, 1) C (–2, – 4)

D 12xy – 7x

E 2x – 3y – 4xy

D (4, 10)

E (0, –2)

3 The length, d, and midpoint, M, of the line segment joining the points (–2, 4) and (3, –2) are: √ √ √ A d = 5, M = (0.5, 1) B d = 61, M = (2.5, 3) C d = 29, M = (1, 1) √ √ D d = 61, M = (0.5, 1) E d = 11, M = (1, 2) 4 The equation of the line that is perpendicular to the line with equation y = –2x – 1 and passes through the point (1, –2) is: 1 3 A y=– x+ B y = 2x – 2 C y = –2x – 4 2 2 5 1 E y= x– D y=x–2 2 2 Ext

5 The graph of 3x + 2y < 6 is: A B y

2 3

x

y

D

y

C

2

O

O

3

x

3 O

2

x

y

E

6

3 −2

y

O

x

O

x

3

Short-answer questions 1

Simplify: 12 – 8x a 4 3 2 c – 4 a

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

5x – 10 12 × 3 x–2 4 5 d + x+2 x–3 b

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Semester review 1

2 a

b

3 a b

Solve these equations for x. i 2 – 3x = 14 x–3 iii =5 2

379

ii 2(2x + 3) = 7x 3x – 2 2x + 1 iv = 4 5

Solve these inequalities for x and graph their solutions on a number line. x i 3x + 2 ≤ 20 ii 2 – > 1 3 Find the gradient and y-intercept for these linear relations and sketch each graph. i y = 3x – 2 ii 4x + 3y = 6 Sketch by finding the x- and y-intercepts where applicable. i y = 2x – 6 ii 3x + 5y = 15 iii x = 3

4 Find the equation of the straight lines shown. a b y

iv y = –2x

y

3 O

(3, 3) x

x

O

(4, -1) (-2, -5)

5 Find the value(s) of a in each of the following when: a The lines y = ax – 3 and y = –3x + 2 are parallel. b The gradient of the line joining the points (3, 2) and (5, a) is –3. √ c The distance between (3, a) and (5, 4) is 13. 1 d The lines y = ax + 4 and y = x – 3 are perpendicular. 4 6 Solve these pairs of simultaneous equations. a y = 2x – 1 b 2x – 3y = 8 c y = 5x + 8

y=x–2

2x + y = 2

d

5x + 3y = 7

3x – 2y = 19 4x + 3y = –3

7 At a fundraising event, two hot dogs and three cans of soft drink cost $13, and four hot dogs and two cans of soft drink cost $18. What are the individual costs of a hot dog and a can of soft drink? Ext

8 Sketch the half planes for these linear inequalities. a y ≥ 3 – 2x b 3x – 2y < 9

c y > –3

Extended-response question A block of land is marked on a map with coordinate axes and with boundaries given by the equations y = 4x – 8 and 3x + 2y = 17. a Solve the two equations simultaneously to find their point of intersection. b Sketch each equation on the same set of axes, labelling axis intercepts and the point of intersection.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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380

Semester review 1

The block of land is determined by the intersecting region x ≥ 0, y ≥ 0, y ≥ 4x – 8 and 3x + 2y ≤ 17. Ext

c Shade the area of the block of land (i.e. the intersecting region on the graph in part b). d Find the area of the block of land if 1 unit represents 100 metres.

Geometry Multiple-choice questions 1

The value of a in the diagram shown is: A 40 B 25 C 30 D 50 E 45

70°

2 The values of x and y in these similar figures are: A x = 2.6, y = 5 B x = 4, y = 4.5 C x = 4, y = 7.5 D x = 3, y = 6 E x = 3.5, y = 4.5 10A

Ext

Ext

3

4

5

65°

100°

85° a° 6 cm

3 cm

7.5 cm

5 cm y cm

x cm

The values of the pronumerals in this diagram are: A a = 17, b = 56 B a = 34, b = 73 C a = 68, b = 56 D a = 34, b = 34 E a = 68, b = 34

34° a° b°

The value of angle a in this diagram is: A 115 B 165 C 140 D 130 E 155

a°

The value of x in this diagram is: A 5.5 B 0.75 C 3

25°

1.5 D 4

E 8

4

x

8

Short-answer questions 1

Prove the following congruence statements, giving reasons, and use this to find the value of the pronumerals. a △ABC ≡ △DEF b △ABC ≡ △ADC

E

B 85° A

60°

C

D

B 3 cm A

60° a°

x cm F

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

C D

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Semester review 1

381

2 Use congruence to prove that a parallelogram (with opposite parallel sides) has equal opposite sides. 3 Find the value of the pronumeral, given these pairs of triangles are similar. a b 4.5 5 6 x x 1.4 9 3.5

10A

4 Use the chord and circle theorems to find the value of each pronumeral. a b c x cm a°

83°

x 10

8 cm d

a°

e

50°

65°

b° f

b° a°

c°

x° 30°

110°

Ext

20°

5 Use tangent properties to find the value of the pronumerals. a b c 8m x° 70° 2m

a°

xm

Ext

10A

6 Find the value of x in each figure. a b 6 3 5 12 x 4 x 5

55° 63°

b°

c

5

x

8

Extended-response question A logo for a car manufacturing company is silver and purple and shaped as shown, with O indicating the centre of the circle.

A

C O

D

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

B

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382

Semester review 1

The radius of the logo is 5 cm and chord AB is 6 cm. Given the two chords are equidistant from the centre of the circle, complete the following. a What is the length of CD? Give a reason. b Hence, prove that △OAB ≡ △OCD. c By first finding the length of OM, where M is the point such that OM ⊥ AB, find the area of △OAB. d Hence, determine what percentage of the logo is occupied by the silver portion, given the area of a circle is p r2 . Answer correct to one decimal place. e Given that ∠OCD = 53.1◦ , what is the angle between the two triangles (i.e. ∠BOD)?

Indices and surds Multiple-choice questions 10A

10A

10A

√ The simplified form of 2 45 is: √ √ √ A 90 B 6 5 C 10 3 √ √ √ √ 2 7 3 – 4 2 + 12 + 2 simplifies to: √ √ √ √ A 7 3–3 2 B 5 3+3 2 √ √ √ √ D 9 3–3 2 E 3 3+5 2 √ √ 3 The expanded form of (2 6 + 1)(3 – 4 6) is: √ √ A 3 B 6 – 33 C 6 6+3

1

√ D 18 5 C

√ E 6 15 √ √ 3–3 2

√ D 2 6

√ E 2 6 – 45

12(a3 )–2 , when written using positive indices, is: (2ab)2 × a2 b–1 3 3 6a B 3a2 C D 2 3 E b a b a10 b

4 The simplified form of A

6 a2 b

5 0.00032379 in scientific notation, using three significant figures, is: A 3.23 × 10–4 B 3.24 × 104 C 3.24 × 10–4 D 32.4 × 103 E 0.324 × 10–5 Short-answer questions 10A

1

Simplify: √ a 54 d g

10A

10A

√

√ 5× 2

√

√ 15 ÷ 5

√ b 4 75 √ √ e 3 7× 7 √ 3 30 √ h 9 6

2 Simplify fully. √ √ √ √ √ √ a 2 5+3 7+5 5–4 7 b 20 – 2 5 3 Expand and simplify these expressions. √ √ a 2 3( 5 – 2) √ √ √ √ c ( 7 – 3)( 7 + 3)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

√ 3 24 c 2 √ √ f 3 6×4 8 √ 200 i 49 c

√ √ √ 18 – 4 + 6 2 – 2 50

√ √ b (3 5 – 2)(1 – 4 5) √ 2 √ d (3 3 + 4 2)

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Semester review 1

10A

4 Rationalise the denominator. 3 a √ b 2

√ 2 3 √ 5 6

383

√ 2– 5 √ 5

c

5 Use the index laws to simplify the following. Express all answers with positive indices. ( )2 2b10 3a × a (2x2 )3 × 3x4 y2 b 4 b 6(2a5 )0 c 3a–5 b2

d

6 Convert: a to a basic numeral i 3.72 × 104 b

10A

7 a b

10A

ii 4.9 × 10–6

to scientific notation, using three significant figures i 0.000072973 ii 4 725 400 000 Express in index form. √ √ i 10 ii 7x6 Express in surd form. i

10A

4x–2 y3 10x –4 y6

√ 5 iii 4 x3 1

1

3

ii 20 5

62

√ iv 15 15 iii 7 4

8 Evaluate without using a calculator. a 5–1

b 2–4

1

c 81 4

1

d 8– 3

9 Solve these exponential equations for x. 1 49 d 55x + 1 = 125x

a 4x = 64

b 7–x =

c 9x = 27 10 Determine the final amount after 3 years if: a $2000 is compounded annually at 6% b $7000 is compounded annually at 3%. Extended-response question

Lachlan’s share portfolio is rising at 8% per year and is currently valued at $80 000. a Determine a rule for the value of Lachlan’s share portfolio (V dollars) in n years’ time. b What will be the value of the portfolio, to the nearest dollar: i next year? ii in 4 years’ time? c Use trial and error to find when, to two decimal places, the share portfolio will be worth $200 000. d After 4 years, however, the market takes a downwards turn and the share portfolio begins losing value. Two years after the downturn, Lachlan sells his shares for $96 170. If the market was declining in value at a constant percentage per year, what was this rate of decline, to the nearest percentage?

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384

Semester review 1

Trigonometry Multiple-choice questions 1

x

The value of x in the triangle shown is approximately: A 7.6 B 12.0 C 10.4 D 6.5 E 8.2

11.2 47°

2 A bird 18 m up in a tree spots a worm on the ground 12 m from the base of the tree. The angle of depression from the bird to the worm is closest to: A 41.8◦ B 56.3◦ C 61.4◦ D 33.7◦ E 48.2◦ 3 A walker travels due south for 10 km and then on a true bearing of 110◦ for 3 km. The total distance south from the starting point to the nearest kilometre is: A 11 km B 1 km C 9 km D 13 km E 15 km 4 The area of the triangle shown is closest to:

10A

8 cm 68° 7 cm

A 10A

69 cm2

B 52 cm2

5 Choose the incorrect statement. A q=

290◦

is in quadrant 4

C cos 110◦ = – cos 20◦ E sin 230◦ = – sin 50◦

C 28 cm2

D 26 cm2

E 10 cm2

√ 3 B = 2 D tan q is positive for 200◦ < q < 250◦ . sin 120◦

Short-answer questions 1

Find the value of the pronumeral in each right-angled triangle, correct to one decimal place. a b 9m

xm

ym 13 m

θ

52° 12 m

2 For the following bearings, give the true bearing of: i A from O ii O from A

N

a

N

b

A W

O S

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

30° A

E

W 40°

O

E

S Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Semester review 1

3 Two wires reach from the top of an antenna to points A and B on the ground, as shown. Point A is 25 m from the base of the antenna, and the wire from point B is 42 m long and makes an angle of 50◦ with the ground. a Find the height of the antenna, to three decimal places. b Find the angle the wire at point A makes with the ground, to one decimal place.

10A

385

42 m

50° A

25 m

B

4 Find the value of the pronumeral, correct to one decimal place. a b q is obtuse

10A

25 cm

43° 78°

36° θ

14 m

18 cm

xm 5 Find the largest angle, correct to one decimal place, in a triangle with side lengths 8 m, 12 m and 15 m.

10A

10A

6 a b

c

10A

If q = 223◦ , state which of sin q, cos q and tan q are positive? Choose the angle q to complete each statement. i sin 25◦ = sin q, where q is obtuse. ii tan 145◦ = – tan q, where q is acute. iii cos 318◦ = cos q, where q is the reference angle. State the exact value of: i cos 60◦

ii sin 135◦

iii tan 330◦

7 Use the graph of sin q below to answer the following. a Estimate the value of sin q for q = 160◦ . b Estimate the two values of q for which sin q = –0.8. c Is sin 40◦ < sin 120◦ ? sinθ 1

O

90°

180°

270°

360°

θ

–1

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386

Semester review 1

Extended-response question A group of walkers set out on a trek to get to the base of a mountain range. The mountains have two peaks, which are 112 m and 86 m above ground level from the base. The angle of elevation from the peak of the smaller mountain to the peak of the taller mountain is 14◦ . a Find the horizontal distance between the two mountain peaks, correct to one decimal place. 10A

To get to the base of the mountain range, the walkers set out from the national park entrance on a bearing of 52◦ T for a distance of 13 km and then turn on a bearing of 340◦ T for the last 8 km of the trek. b Draw a diagram representing the trek. Label all known measurements. c If the walkers are able to trek directly from their start location to their endpoint, what distance would they cover? Round your answer to three decimal places. d After they have explored the mountains, the group will be taken by bus back along the direct path from their end location to the park entrance. Determine the true bearing on which they will travel. Round your answer to the nearest degree.

Quadratic equations Multiple-choice questions 1

The expanded form of 2(2x – 3)(3x + 2) is: A 12x2 – 5x – 6 B 12x2 – 12 2 D 24x – 20x – 24 E 12x2 – x – 6

C 12x2 – 10x – 12

2 The factorised form of 25y2 – 9 is: A (5y – 3)2 B (5y – 3)(5y + 3) D (5y – 9)(5y + 1) E 5(y + 1)(y – 9)

C (25y – 3)(y + 3)

3

x2 – 4x + 3 x2 – 4 × simplifies to: 4x – 8 x2 – x – 6 x+3 x2 + 1 B C A x–2 12 2x

D

x–1 4

4 The solution(s) to the quadratic equation x2 – 4x + 4 = 0 is/are: A x = 0, 4 B x=2 C x = 1, 4 D x = 2, –2 10A

E

x2 x–2

E x = –1, 4

5 A quadratic equation ax2 + bx + c = 0 has a discriminant equal to 17. This tells us that: A The equation has a solution x = 17. B The equation has no solutions. C a + b + c = 17 D The equation has two solutions. E The equation has one solution.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Semester review 1

387

Short-answer questions 1

Expand and simplify. a (3x + 1)(3x – 1)

b (2x – 5)2

c (2x + 3)(x + 5) – (3x – 5)(x – 4)

2 Factorise fully these quadratics. Remember to take out any common factors first. a 4x2 – y2 b (x + 2)2 – 7 c 3x2 – 48 d x2 + 5x – 14 e x2 – 10x + 25 f 2x2 – 16x + 24 10A

3 Factorise these non-monic quadratics. a 3x2 – 2x – 8 b 6x2 + 7x – 3

c 10x2 – 23x + 12

4 Solve these quadratic equations using the Null Factor Law. a 2x(x – 3) = 0 b (x + 4)(2x – 1) = 0 2 c x + 5x = 0 d x2 – 16 = 0 2 e x –7=0 f x2 – 4x + 4 = 0 10A h g x2 – 5x – 24 = 0 3x2 + 5x – 2 = 0 5 Solve these quadratic equations by first writing them in standard form. x + 20 a x2 = 40 – 3x b x(x – 6) = 4x – 21 c =x x 6 a b

10A

Factorise by completing the square. i x2 – 6x + 4 ii x2 + 4x + 7

iii x2 + 3x + 1

Use your answers to part a to solve these equations, if possible. i x2 – 6x + 4 = 0 ii x2 + 4x + 7 = 0 iii x2 + 3x + 1 = 0

7 Solve these quadratic equations using the quadratic formula. Leave your answers in exact surd form. a 2x2 + 3x – 6 = 0 b x2 – 4x – 6 = 0 Extended-response question A rectangular backyard swimming pool, measuring 12 metres by 8 metres, is surrounded by a tiled path of width x metres, as shown. a Find a simplified expression for the area of the tiled path.

xm 12 m pool

8m

xm

tiles

b If x = 1, what is the tiled area? c Solve an appropriate equation to determine the width, x metres, if the tiled area is 156 m2 . 10A

d Find the width, x metres, if the tiled area is 107.36 m2 . Use the quadratic formula.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter

6

Measurement

What you will learn

Australian curriculum

6A Review of length (Consolidating) 6B Pythagoras’ theorem 6C Area (Consolidating) 6D Surface area – prisms and cylinders 6E Surface area – pyramids and cones (10A) 6F Volume – prisms and cylinders 6G Volume – pyramids and cones (10A) 6H Spheres (10A) 6I Limits of accuracy (Extending)

MEASUREMENT AND GEOMETRY

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Using units of measurement Solve problems involving surface area and volume for a range of prisms, cylinders and composite solids. (10A) Solve problems involving surface area and volume of right pyramids, right cones, spheres and related composite solids. Pythagoras and trigonometry Solve right-angled triangle problems. 32x32 (10A) Apply Pythagoras’ theorem to 16x16 solve three-dimensional problems in right-angled triangles.

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Online resources • Chapter pre-test • Videos of all worked examples • Interactive widgets • Interactive walkthroughs • Downloadable HOTsheets • Access to HOTmaths Australian Curriculum courses

Monolithic domes Monolithic domes are round one-piece structures with a smooth spherical-like surface. They offer excellent protection from earthquakes, bushfires and cyclones because of their shape and durability. They are extremely energy efficient because of the minimal surface area for the volume contained within the structure. The first type of monolithic domes used were igloos, which are very strong and provide good insulation

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

in freezing conditions. The minimal surface area of the dome means that there is less surface for heat to be transferred to the outside air. A cube, for example, containing the same volume of air as the dome has about 30% more surface area exposed to the outside air. Volumes and surface areas of spheres and other solids can be calculated using special formulas.

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390

Chapter 6 Measurement

6A Review of length

CONSOLIDATING

Length measurements are common in many areas of mathematics, science and engineering, and are clearly associated with the basic measures of perimeter and circumference, which will be studied here.

Engineers at a refinery checking measurements.

Let’s start: The simple sector This sector looks simple enough but can you describe how to find its perimeter? Discuss these points to help.

100°

Converting between metric units of length

km

× 100 m

× 10 cm

mm

÷ 1000 ÷ 100 ÷ 10 Perimeter is the distance around the outside of a closed shape. The circumference of a circle is the distance around the circle. • C = 2p r = p d, where d = 2r. Perimeter of a sector q × 2p r • P = 2r + 360

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

θ

)

× 1000

us (r

Key ideas

10 cm

Recall the rule for the circumference of a circle. What is a definition of perimeter? What fraction of a circle is this sector? Find the perimeter using both exact and rounded numbers.

radi

• • • •

diameter (d)

r

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Measurement and Geometry

391

Example 1 Finding the perimeter of polygons 4.5 m

Consider the given two-dimensional shape. a Find the perimeter of the shape when x = 2.6. b

Find x when the perimeter is 11.9 m.

c

Write an expression for x in terms of the perimeter, P.

xm

2.1 m 3.4 m

SOL UTI ON

EX P L A NA TI ON

a

Simply add the lengths of all four sides using x = 2.6.

Perimeter = 4.5 + 2.1 + 3.4 + 2.6 = 12.6 m

b

Add all four sides and set equal to the perimeter 11.9. Simplify and solve for x by subtracting 10 from both sides.

11.9 = 4.5 + 2.1 + 3.4 + x = 10 + x ∴ x = 1.9

c

Use P for the perimeter and add all four sides.

P = 4.5 + 2.1 + 3.4 + x

Simplify and rearrange to make x the subject.

= 10 + x ∴ x = P – 10

Example 2 Using the formula for the circumference of a circle If a circle has radius r cm, find the following, rounding the answer to two decimal places where necessary. a The circumference of a circle when r = 2.5. b A rule for r in terms of the circumference, C.

r cm

c The radius of a circle with a circumference of 10 cm. SOL UTI ON

EX P L A NA TI ON

a

Write the rule for circumference and substitute r = 2.5, then evaluate and round as required.

Circumference = 2p r = 2p (2.5) = 15.71 cm (to 2 d.p.)

b

C = 2p r C ∴ r= 2p

c

r=

C 2p 10 = 2p = 1.59 cm (to 2 d.p.)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Write the rule for circumference, then divide both sides by 2p to make r the subject.

Substitute C = 10 into the rule from part b and evaluate.

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392

Chapter 6 Measurement

Example 3 Finding perimeters of sectors This sector has a radius of 3 cm. a Find the sector’s exact perimeter.

3 cm

b Find the perimeter, correct to one decimal place.

SOL UTI ON

EX P L A NA TI ON

q × 2p r 360 240 = 2×3 + ×2×p ×3 360 2 = 6 + ×2×p ×3 3 = 6 + 4p cm

P = 2r +

b

The perimeter of a sector consists of two   240 2 radii and a fraction = of the 360 3 circumference of a circle. 6 + 4p is the exact value.

P = 6 + 4p

Round to the required one decimal place, using a calculator.

= 18.6 cm (to 1 d.p.)

Exercise 6A 1

1–2(½), 3

2

—

UNDERSTANDING

a

240°

Convert the following length measurements to the units given in brackets. a 4 cm (mm) b 0.096 m (cm) c 0.001 km (m) d 300 m (km) e 800 cm (m) f 297 m (km) g 0.0127 m (cm) h 5102 mm (cm) i 0.0032 km (m)

2 What fraction of a circle (in simplest form) is shown in these sectors? a b c

d

e

f

30°

105°

40°

3 Find the perimeter of these shapes. a b

0.5 km

220 m

12 cm

c

34 cm

32 cm

2.6 km

185 m

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

Example 1

4–7, 8(½)

4 Consider the given two-dimensional shape. a Find the perimeter of the shape when x = 8. b Find x when the perimeter is 33.7 cm. c Write an expression for x in terms of the perimeter, P.

4–7, 8(½)

x cm 12 cm

FLUENCY

4–7

393

6A

6.2 cm 10.4 cm

5 Consider the given two-dimensional shape. a Find the perimeter of the shape when x = 5. b Find x when the perimeter is 20 m. c Write an expression for x in terms of the perimeter, P.

8.4 m

xm 6 Find the circumference of these circles, correct to two decimal places. a b c 1.2 m 7 cm mm 28.4

d

1.1 km Example 2

7

If a circle has radius r m, find the following, rounding to two decimal places, where necessary. a The circumference of a circle when r = 12. b A rule for r in terms of the circumference, C. c The radius of a circle with a circumference of 35 m.

Example 3

rm

8 Find the perimeter of these sectors: i using exact values ii by rounding the answer to one decimal place. a

b

c

2m

60°

1 km

4m d

e

f

60° 6 cm

80° 5 mm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

205° 3 cm

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Chapter 6 Measurement

9(½), 10

6A

9 Find the value of x for these shapes with the given perimeters. a b 2

9(½), 10, 11(½)

11, 12

c

5

2x

x 7

x

Perimeter = 0.072 5.3 Perimeter = 22.9

Perimeter = 17 d

2x

2.8

PROBLEM-SOLVING

394

11.61

e

3.72 7.89

Perimeter = 16.2

1.5x

f

x

3x

2x Perimeter = 10.8

Perimeter = 46.44 10 A rectangular rose garden of length 15 m and width 9 m is surrounded by a path of width 1.2 m. Find the distance around the outside of the path.

11 Find the perimeter of these composite shapes, correct to two decimal places. a b c 7.9

1.5 0.3

3.6 d

e

3.6

f

6 10

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

12

2.2

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PROBLEM-SOLVING

Measurement and Geometry

12 Find the perimeter of these shapes, giving your answers as exact values. a b c 3m 5m

6A

1 km

13

13, 14

15, 16

13 A bicycle has wheels with diameter 64 cm. Find how far, correct to the nearest centimetre, the bicycle moves when the wheels turn: i one rotation ii five rotations

REASONING

2.5 m

a

395

b How many rotations are required for the bike to travel 10 km? Round your answer to the nearest whole number. c Find an expression for the number of rotations required to cover 10 km if the wheel has a diameter of d cm.

14 A square of side length n has the same perimeter as a circle. What is the radius of the circle? Give an expression in terms of n. 15 A square of side length x just fits inside a circle. Find the exact circumference of the circle in terms of x.

x

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6A l

16 Consider a rectangle with perimeter P, length l and width w. a b c d

Express l in terms of w and P. Express l in terms of w when P = 10. If P = 10, state the range of all possible values of w. If P = 10, state the range of all possible values of l.

Rotating circles

w

—

17 When a circle rolls around the outside of another circle it will rotate by a certain angle. For these problems the fixed circle will have radius r. Given the following conditions, by how many degrees will the moving circle rotate if it rolls around the fixed circle once?

—

REASONING

Chapter 6 Measurement

17

rotating circle

r

fixed circle

r

ENRICHMENT

396

a

Assume the rotating circle has radius r (shown). 1 b Assume the rotating circle has radius r. 2 c Assume the rotating circle has radius 2r. 1 d Assume the rotating circle has radius r. 3

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Measurement and Geometry

397

6B Pythagoras’ theorem You will recall that for any right-angled triangle we can link the length of the three sides using Pythagoras’ theorem. When given two of the sides, we can work out the length of the remaining side. This has applications in all sorts of two- and three-dimensional problems.

Let’s start: President Garfield’s proof Five years before he became president of the United States of America in 1881, James Garfield discovered a proof of Pythagoras’ theorem. It involves arranging two identical   right-angled triangles 1 and 2 to form a trapezium, as shown.

US President Garfield discovered a proof of Pythagoras’ theorem.

a 1 c

b

3 a

c 2 b

•

Use the formula for the area of a trapezium



 1 (a + b)h to find an expression for the area of the 2

entire shape. • Explain why the third triangle 3 is right-angled. • Find an expression for the sum of the areas of the three triangles. • Hence, prove that c2 = a2 + b2 .

Pythagoras’ theorem states that: The sum of the squares of the two shorter sides of a right-angled triangle equals the square of the hypotenuse. 2

2

a +b =c

c

a

2

To write an answer using an exact value, use a square root sign where possible (e.g.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Key ideas

√

b 3).

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398

Chapter 6 Measurement

Example 4 Finding side lengths using Pythagoras’ theorem Find the length of the unknown side in these right-angled triangles, correct to two decimal places. a b 1.1 m x cm 5 cm ym 1.5 m 9 cm SOL UTI ON

EX P L A NA TI ON

c2 = a2 + b2

a

∴ x =5 +9 2

2

x cm is the length of the hypotenuse.

2

Substitute the two shorter sides a = 5 and b = 9 (or a = 9 and b = 5).

= 106 √ ∴ x = 106 = 10.30 (to 2 d.p.) The length of the unknown side is 10.30 cm. b

a2 + b2 = c2 2

2

y + 1.1 = 1.5

Find the square root of both sides and round your answer as required.

Substitute the shorter side b = 1.1 and the hypotenuse c = 1.5.

2

Subtract 1.12 from both sides.

y2 = 1.52 – 1.12 = 2.25 – 1.21 = 1.04 √ ∴ y = 1.04

Find the square root of both sides and evaluate.

= 1.02 (to 2 d.p.) The length of the unknown side is 1.02 m.

For centuries builders, carpenters and landscapers have used Pythagoras’ theorem to construct right angles for their foundations and plots. The ancient Egyptians used three stakes joined by a rope to make a triangular shape with side lengths of 3, 4 and 5 units, which form a right angle when the rope is taut. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

399

Example 5 Using Pythagoras’ theorem in 3D Consider a rectangular prism ABCDEFGH with the side lengths AB = 7, AE = 4 and EH = 2. Find: a BE, leaving your answer in exact form

10A

H

G C

D

E

F

b BH, correct to two decimal places.

A SOL UTI ON a

B

EX P L A NA TI ON

E

Draw the appropriate right-angled triangle with two known sides.

4 7

A

B

c2 = a2 + b2 ∴ BE = 4 + 7 2

2

Substitute a = 4 and b = 7.

2

Solve for BE exactly.

= 65 √ ∴ BE = 65 b

Leave intermediate answers in surd form to reduce the chance of accumulating errors in further calculations.

H

Draw the appropriate triangle.

2 E

√65

B √ Substitute HE = 2 and EB = 65. √ √ √ Note ( 65)2 = 65 × 65 = 65.

BH2 = HE2 + EB2 √ = 22 + ( 65)2 = 4 + 65 = 69 √ ∴ BE = 69 = 8.31 (to 2 d.p.)

1

1, 2

2

—

Solve for a in these equations, leaving your answer in exact form using a square root sign. Assume a > 0. a a2 + 32 = 82 b a2 + 52 = 62 c 22 + a2 = 92 d a2 + a2 = 22 e a2 + a2 = 42 f a2 + a2 = 102

2 Write an equation connecting the pronumerals in these right-angled triangles. a b c y d

x

z

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b

a

UNDERSTANDING

Exercise 6B

c

x

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Chapter 6 Measurement

3–5(½)

6B Example 4a

3–6(½)

3–6(½)

3 Use Pythagoras’ theorem to find the length of the hypotenuse for these right-angled triangles. Round your answers to two decimal places where necessary. a b c 10 m 7 km 3 cm 15 km 5m

FLUENCY

400

4 cm 0.2 mm

d

e 0.37 km

f

0.21 km

72.1 cm

1.8 mm

Example 4b

27.3 cm 4 Find the length of the unknown side in these right-angled triangles, correct to two decimal places. a b c 0.3 m 12 m

5m

2m

0.7 m

9m d

e

0.71 cm

0.14 cm

f

19.3 cm 24.2 cm

1.32 cm Example 5 10A

0.11 cm

5 For each of these rectangular prisms, find: i BE, leaving your answer in exact form

H

a

G C

D

5

b H

D G

2 F

E

A

ii BH, correct to two decimal places.

B

C

3

14

A

E 8

H

c

G 3

E

F

D

C

F 4 B H

d

D

G C 9

7 E

A 1 B

F 5

A Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

8

B

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6 Use Pythagoras’ theorem to help decide whether these triangles are right-angled. They may not be drawn to scale. a b c 3 8 2 13 6 12 4 11

FLUENCY

Measurement and Geometry

401

6B

5 9

e

10

f

√5 1

12

12

6

2

15

7, 8

8–9(½)

9–11

7 Use Pythagoras’ theorem to find the distance between points A and B in these diagrams, correct to two decimal places. a b c B 3 cm

A

35 m

1.7 cm A

1.2 m 2.6 m

3.5 cm B

49 m

1.9 cm

26 m A

5.3 cm

d

e

A

B 0.5 km

A 2.1 km

5.3 cm

1.8 km

B

f

17.2 km

B 10A

PROBLEM-SOLVING

d

19.7 km

12.3 km

A B

8 A 20 cm drinking straw sits diagonally in a glass of radius 3 cm and height 10 cm. What length of straw protrudes from the glass? Round your answer to one decimal place.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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402

Chapter 6 Measurement

10A

PROBLEM-SOLVING

6B 9 Find the value of x, correct to two decimal places, in these three-dimensional solids. a b c x mm xm

9.3 mm

3.7 m

d

5 cm

x cm 15.5 m

11.4 mm

xm

2m

1.5 m 3m

3 cm

x km

e

f

6.7 km

2.93 cm 4.04 cm

5.31 cm

6.2 km

x cm 8.2 km y

10 Find the exact distance between these pairs of points on a number plane.

8 6 4 2

a (0, 0) and (4, 6) b (–2, 3) and (2, –1) c (–5, –3) and (4, 7)

−8 −6 −4 −2−2O −4 −6 −8 11 a

x

Find the length of the longest rod that will fit inside these objects. Give your answer correct to one decimal place. i a cylinder with diameter 10 cm and height 20 cm ii a rectangular prism with side lengths 10 cm, 20 cm and 10 cm

b Investigate the length of the longest rod that will fit in other solids, such as triangular prisms, pentagonal prisms, hexagonal prisms and truncated rectangular pyramids. Include some three-dimensional diagrams. 12

12 Two joining chords in a semicircle have lengths 1 cm and 2 cm, as shown. Find the exact radius, r cm, of the semicircle. Give reasons.

12, 13

13, 14

2 cm

1 cm

REASONING

10A

2 4 6 8

r cm

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REASONING

Measurement and Geometry

13 The diagonals of a rectangle are 10 cm long. Find the exact dimensions of the rectangle if: a the length is twice the width b the length is three times the width c the length is ten times the width. 14 Streamers are used to decorate the interior of a rectangular room that is 4.5 m long, 3.5 m wide and 3 m high, as shown. a

Find the length of streamer, correct to two decimal places, required to connect from: i A to H ii E to B iii A to C iv A to G via C v E to C via D vi E to C directly

A

6B

B

3m D

E 4.5 m

C

F H 3.5 m

G

b Find the shortest length of streamer required, correct to two decimal places, to reach from A to G if the streamer is not allowed to reach across open space. (Hint: Consider a net of the prism.)

How many proofs?

—

—

15

15 There are hundreds of proofs of Pythagoras’ theorem. a Research some of these proofs using the internet and pick one you understand clearly. b Write up the proof, giving full reasons. c Present your proof to a friend or the class. Show all diagrams, algebra and reasons.

ENRICHMENT

10A

403

Mathematical proofs are vital to the subject and are considered to be among its greatest achievements. Most mathematicians consider a good proof to be a thing of beauty.

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404

Chapter 6 Measurement

6C Area

CONSOLIDATING

Area is a measure of surface and is expressed as a number of square units. By the inspection of a simple diagram like the one shown, a rectangle with side lengths 2 m and 3 m has an area of 6 square metres or 6 m2 .

3m 2m Area = 6 m2

The appropriate unit for the area of a desert is the square kilometre.

For rectangles and other basic shapes, we can use area formulas to help us calculate the number of square units. Some common metric units for area include square kilometres (km2 ), square metres (m2 ), square centimetres (cm2 ) and square millimetres (mm2 ).

Let’s start: Pegs in holes Discuss, with reasons relating to the area of the shapes, which is the better fit:

Key ideas

a square peg in a round hole? a round peg in a square hole?

Conversion of units of area

× 10002 km2

× 1002 m2

÷ 10002

cm2 ÷ 1002

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

1 m = 100 cm

× 102 mm2 ÷ 102

1 m2

1 m = 100 cm

• •

1 m2 = 100 cm × 100 cm = 1002 cm2

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Measurement and Geometry

The area of a two-dimensional shape can be defined as the number of square units contained within its boundaries. Some common area formulas are given. square

rectangle

w

h b

l

1 Area = bh 2

Area = lw

rhombus

y

parallelogram

trapezium a

y

1 Area = (a + b)h 2

Area = bh

The rule for the area of a circle is: Area = p r2 , where r is the radius. The rule for the area of a sector is A =

x

b

b 1 Area = xy 2

kite

h

h

x

Key ideas

triangle

l

Area = l2

405

1 Area = xy 2

θ

q p r2 . 360

r

Example 6 Converting between units of area Convert these areas to the units shown in the brackets. a 2.5 cm2 (mm2 ) b 2 000 000 cm2 (km2 ) SOL UTI ON a

EX P L A NA TI ON

2.5 cm2 = 2.5 × 102 mm2 = 2.5 × 100 mm2

10 mm

1 cm2

= 250 mm2

10 mm 1 cm2 = 10 × 10 mm2 = 100 mm2 b

2 000 000 cm2 = 2 000 000 ÷ 1002 m2 = 200 m

2

= 200 ÷ 10002 km2 = 0.0002 km

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

2

km2 ÷ 10002

m2

cm2 ÷ 1002

1002 = 10 000 and 10002 = 1 000 000

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406

Chapter 6 Measurement

Example 7 Finding the area of basic shapes Find the area of these basic shapes, correct to two decimal places where necessary. a b c 3 cm 1.06 km 3.3 m

2 cm

5.8 m

5 cm

SOL UTI ON a

b

c

EX P L A NA TI ON

1 (a + b)h 2 1 = (3 + 5)2 2 = 8 cm2

A=

The shape is a trapezium, so use this formula. Substitute a = 3, b = 5 and h = 2. Simplify and include the correct units.

1 bh 2 1 = (5.8)(3.3) 2 = 9.57 m2

The shape is a triangle.

A=

Substitute b = 5.8 and h = 3.3. Simplify and include the correct units.

A = p r2

The shape is a circle. 2

= p (0.53)

The radius, r, is half the diameter; i.e. 1.06 ÷ 2 = 0.53

= 0.88 km2 (to 2 d.p.)

Evaluate using a calculator and round your answer to the required number of decimal places.

Example 8 Using area to find unknown lengths Find the value of the pronumeral for these basic shapes, rounding to two decimal places where necessary. a b 1.3 m

Area = 11 2.3 mm

mm2

Area = 0.5 m2 am

0.4 m

l mm

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Measurement and Geometry

SOL UTI ON

EX P L A NA TI ON

a

Use the rectangle area formula.

A = lw 11 = l × 2.3 11 ∴ l= 2.3 = 4.78 (to 2 d.p.)

b

407

Substitute A = 11 and w = 2.3. Divide both sides by 2.3 to solve for l.

1 (a + b)h 2 1 0.5 = (a + 1.3) × 0.4 2 0.5 = 0.2(a + 1.3)

Use the trapezium area formula.

A=

Substitute A = 0.5, b = 1.3 and h = 0.4.   1 Simplify × 0.4 = 0.2 , then divide both sides 2 by 0.2 and solve for a.

2.5 = a + 1.3 ∴ a = 1.2

Example 9 Finding areas of sectors and composite shapes Find the area of this sector and composite shape. Write your answer as an exact value and as a decimal, correct to two decimal places. a b

80° 3 cm

5m SOL UTI ON a

EX P L A NA TI ON

q × p r2 360 280 × p × 32 = 360 = 7p

A=

Write the formula for the area of a sector. Sector angle = 360◦ – 80◦ = 280◦ .

= 21.99 cm2 (to 2 d.p.) b

1 × p × 52 4 25p = 50 – 4 = 30.37 m2 (to 2 d.p.)

A = 2 × 52 –

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Simplify to express as an exact value (7p ), then round as required.

The area consists of two squares minus a quarter circle with radius 5 m. 50 –

25p is the exact value. 4

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408

Chapter 6 Measurement

1

Write the formula for the area of these shapes. a circle b sector e kite f trapezium i parallelogram j semicircle

2(½)

c square d rectangle g triangle h rhombus k quadrant (quarter circle)

2 Convert the following area measurements to the units given in brackets. a 3000 mm2 (cm2 ) b 29 800 cm2 (m2 ) c 205 000 m2 (km2 ) d 0.5 m2 (cm2 ) e 5 km2 (m2 ) f 0.0001 km2 (m2 ) 2 2 2 2 g 0.023 m (cm ) h 537 cm (mm ) i 0.0027 km2 (m2 ) j 10 m2 (mm2 ) k 0.00022 km2 (cm2 ) l 145 000 000 mm2 (km2 )

3–4(½) Example 7

—

3–5(½)

3–5(½)

FLUENCY

Example 6

1–2(½)

UNDERSTANDING

Exercise 6C

3 Find the area of these basic shapes, rounding to two decimal places where necessary. a b c 1.3 km 5.2 m

5 cm

2.8 km 10.5 m

d

e

0.3 mm

f

7m 0.1 mm

10 cm

0.2 mm g

64 m

23 m

h

0.4 mm

i

15 cm

2 3

1 4

0.25 mm

j

k

0.82 m

√3 km 1.37 m

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

km

l

km

√10 cm

√2 cm

√6 cm

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Example 8

4 Find the value of the pronumeral for these basic shapes, rounding to two decimal places where necessary. c a b 5.2 cm 1.8 km

Area = 15 cm2 d

e

5.34 mm

Area = 80 m2

xm h

r cm

i

dm

dm Area = 26 km2

Area = 0.21 m2

Area = 5 cm2 Example 9a

18 m

Area = 10.21 mm2

am Area = 2.5 m2

g

f

h mm

1.3 m

6C

Area = 1.3 km2

Area = 206 m2

2.8 m

409

h km

lm

w cm

FLUENCY

Measurement and Geometry

5 Find the area of each sector. Write your answer as an exact value and as a decimal rounded to two places. a b c 10 m

120° 6 cm d

7m e

100° 2m

f

6 km

3 mm

150° 325°

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Chapter 6 Measurement

6, 7

6C 6

7(½), 8

7(½), 8

A lawn area is made up of a semicircular region with diameter 6.5 metres and a triangular region of length 8.2 metres, as shown. Find the total area of lawn, to one decimal place.

6.5 m

PROBLEM-SOLVING

410

8.2 m

Example 9b

7 Find the area of these composite shapes. Write your answers as exact values and as decimals, correct to two decimal places. a b c 7m 1.7 m 4m 1.8 m 10 m 5 cm

1.6 m d

28 km

e

f

0.25 m 18 km

4.2 mm

26 km

0.3 m 0.7 m

8 An L-shaped concrete slab being prepared for the foundation of a new house is made up of two rectangles with dimensions 3 m by 2 m and 10 m by 6 m. a Find the total area of the concrete slab. b If two bags of cement are required for every 5 m2 of concrete, how many whole bags of cement will need to be purchased for the job?

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Measurement and Geometry

9, 10

10, 11

REASONING

9

1 square miles (1 mile ≈ 1.61 km). 9 1 hectare (1 ha) is 10 000 m2 and an acre is 640 Find how many: a b c d

411

6C

hectares in 1 km2 square metres in 20 hectares hectares in 1 acre (round to one decimal place) acres in 1 hectare (round to one decimal place).

10 Consider a trapezium with area A, parallel side lengths a and b and height h.

a h

A=

1 (a + b)h 2

b a Rearrange the area formula to express a in terms of A, b and h. b Hence, find the value of a for these given values of A, b and h. i A = 10, b = 10, h = 1.5 ii A = 0.6, b = 1.3, h = 0.2 iii A = 10, b = 5, h = 4 c

Sketch the trapezium with the dimensions found in part b iii above. What shape have you drawn?

11 Provide a proof of the following area formulas, using only the area formulas for rectangles and triangles. a parallelogram b kite c trapezium

—

—

12 Find, correct to one decimal place, the percentage areas for these situations. a The largest square inside a circle. b The largest circle inside a square.

c The largest square inside a right isosceles triangle.

12

ENRICHMENT

Percentage areas

d The largest circle inside a right isosceles triangle.

x

x

x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

x

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412

Chapter 6 Measurement

6D Surface area – prisms and cylinders Knowing how to find the area of simple shapes combined with some knowledge about three-dimensional objects helps us to find the total surface area of a range of solids. A cylindrical can, for example, has two circular ends and a curved surface that could be rolled out to form a rectangle. Finding the sum of the two circles and the rectangle will give the total surface area of the cylinder. You will recall the following information about prisms and cylinders.  A prism is a polyhedron with a uniform cross-section and two congruent ends. • A prism is named by the shape of the cross-section. • The remaining sides are parallelograms.  A cylinder has a circular cross-section. • A cylinder is similar to a prism in that it has a uniform cross-section and two congruent ends. right triangular prism oblique pentagonal prism Let’s start: Drawing nets Drawing or visualising a net can help when finding the total surface area of a solid. Try drawing a net for these solids.

square prism

cylinder

By labelling the dimensions, can you come up with a formula for the total surface area of these solids?

Key ideas

The total surface area (TSA) of a three-dimensional object can be found by finding the sum of the areas of each of the shapes that make up the surface of the object. A net is a two-dimensional illustration of all the surfaces of a solid object. Given below are the net and surface area of a cylinder. Diagram r

Net

TSA = 2 circles + 1 rectangle = 2p r2 + 2p rh

r

= 2p r(r + h)

h 2πr

h

r

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Measurement and Geometry

413

Key ideas

Composite solids are solids made up of two or more basic solids. • To find a total surface area do not include any common faces. – In this example, the top circular face area of the cylinder is equal to the common face area, so the TSA = surface area of prism + curved surface area of cylinder.

common face

Example 10 Finding the surface area of prisms and cylinders Find the total surface area of this rectangular prism and cylinder. Round your answer to two decimal places where necessary. a b 3 cm 5.3 m 5 cm 8 cm

1.7 m SOL UTI ON

EX P L A NA TI ON

a

Draw the net of the solid if needed to help you. Sum the areas of the rectangular surfaces.

TSA = 2 × (8 × 3) + 2 × (5 × 3) + 2 × (8 × 5) = 48 + 30 + 80 = 158 cm2

5 cm 3 cm

3 cm

8 cm

5 cm b

TSA = 2p r2 + 2p rh 2

= 2p (1.7) + 2p (1.7) × 5.3

Write the formula and substitute the radius and height. 1.7 m

= 74.77 m2 (to 2 d.p.)

2πr

5.3 m

1.7 m

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414

Chapter 6 Measurement

Example 11 Finding the surface area of composite solids A composite object consists of a square-based prism and a cylinder, as shown. Find the total surface area, correct to one decimal place.

6 cm 8 cm 10 cm

20 cm EX P L A NA TI ON

TSA = 4 × (20 × 10) + 2 × (20 × 20) + 2 × p × 6 × 8 2

2

+ p (6) – p (6) = 1600 + 96p

= 1901.6 cm2 (to 1 d.p.)

Exercise 6D 1

The common circular area, which should not be included, is added back on with the top of the cylinder. So the total surface area of the prism is added to only the curved area of the cylinder.

1, 2

Draw an example of these solids. a cylinder b rectangular prism

2 Draw a net for each of these solids. a b 4 cm

—

c triangular prism

c

3 cm

10 cm 2 cm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

2

UNDERSTANDING

SOL UTI ON

1 cm 2 cm

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Measurement and Geometry

Example 10

3–4(½), 5

3–4(½), 5, 6

3 Find the total surface area of these solids. Round your answers to two decimal places where necessary. a b c 2.1 cm 3 cm

3 cm

6 cm

4.5 cm

1.3 mm 8m

d

12.8 m

e

6D

7 cm

2.7 mm

5.1 mm

FLUENCY

3–4(½)

415

5.2 m

f

15 m 9.2 m

20 m

6.4 m

17 m 4 Find the total surface area of these solids. a

b 5 mm

7 mm

3 mm 4 mm

1.2 cm c

d

25 m 10 m

5 mm

8.66 m

16 mm

e 0.5 cm 1.2 cm

1.4 m

f

2.1 m

1.5 cm 0.76 cm

6 mm

12 mm

0.8 cm

1.9 m

4.8 m 2.3 m

5 Find the total surface area, in square metres, of the outer surface of an open pipe with radius 85 cm and length 4.5 m, correct to two decimal places. 6 What is the minimum area of paper required to wrap a box with dimensions 25 cm wide, 32 cm long and 20 cm high?

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Chapter 6 Measurement

7–8(½)

6D

7–8(½)

8(½), 9, 10

7 The cross-sections of these solids are sectors. Find the total surface area, rounding to one decimal place. a b 1.6 m

c

2m

10 cm

3 cm

PROBLEM-SOLVING

416

d

4m

20 cm 40°

6m 40 cm 8 Use Pythagoras’ theorem to determine any unknown side lengths and find the total surface area of these solids, correct to one decimal place. a b c 3.2 1.8 7

10A

1.4

8

7

2

10

3 d

e

8 10

f

5.2 1.8

33 26

Example 11

9 Find the total surface area of these composite solids. Answer correct to one decimal place. a b 4 cm

8m

4 cm 2 cm 6 cm c

10 m 10 cm

d

20 cm

15 m

20 m 40 m

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PROBLEM-SOLVING

Measurement and Geometry

10 Find the total surface area of this triangular prism, correct to one decimal place.

6D

5m

11

11, 12

12–14

11 Find a formula for the total surface area of these solids, using the given pronumerals. a b c

a

b

REASONING

2m

417

x c

d

h h

d

r

12 Find the exact total surface area for a cylinder with the given dimensions. Your exact answer will be in terms of p . 1 a r = 1 and h = 2 b r = and h = 5 2 13 The total surface area of a cylinder is given by the rule: Total surface area = 2p r(r + h) Find the height, to two decimal places, of a cylinder that has a radius of 2 m and a total surface area of: a 35 m2 b 122 m2 14 Can you find the exact radius of the base of a cylinder if its total surface area is 8p cm2 and its height is 3 cm?

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6D

Chapter 6 Measurement

Deriving formulas for special solids

—

—

15

15 Derive the formulas for the total surface area of the following solids. a a cylinder with its height equal to its b a square-based prism with square side radius r length x and height y

ENRICHMENT

418

x

r r

y

c a half cylinder with radius r and height h

r

d a solid with a sector cross-section, radius r, sector angle q and height h

r

θ

h

h

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Measurement and Geometry

6E Surface area – pyramids and cones

419

10A

Pyramids and cones are solids for which we can also calculate the total surface area by finding the sum of the areas of all the outside surfaces. The total surface area of a pyramid involves finding the sum of the areas of the base and its triangular faces. The rule for the surface area of a cone can be developed after drawing a net that includes a circle (base) and sector (curved surface), as you will see in the following examples.

Let’s start: The cone formula A Use a pair of compasses to construct a large sector. Use any sector angle q that you like. Cut out the sector and join the s points A and B to form a cone of radius r. θ • Give the rule for the area of the base of the cone. s r B • Give the rule for the circumference of the base of the cone. • Give the rule for the circumference of a circle with radius s. • Use the above to find an expression for the area of the base of the cone as a fraction of the area p s2 . • Hence, explain why the rule for the surface area of a cone is given by: Total surface area = p r2 + p rs. A cone is a solid with a circular base and a curved surface that reaches from the base to a point called the apex. • A right cone has its apex directly above the centre of the base. • The pronumeral s is used for the slant height and r is the radius of the base. • Cone total surface area is given by:

s s

r

Key ideas

apex s r right cone

and

r

2πr Area (base) = p r2

Area =

2p r × p s2 = p rs 2p s

∴ TSA (cone) = p r2 + p rs = p r(r + s)

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420

Chapter 6 Measurement

Key ideas

A pyramid has a base that is a polygon and its remaining faces are triangles that meet at the apex. • A pyramid is named by the shape of its base. • A right pyramid has its apex directly above the centre of the base.

apex

right squarebased pyramid

Example 12 Finding the total surface area of a cone and pyramid Find the total surface area of these solids, using two decimal places for part a. a A cone with radius 2 m and slant height 4.5 m.

4.5 m

b A square-based pyramid with square-base length 25 mm and triangular face height 22 mm.

2m SOL UTI ON a

EX P L A NA TI ON

TSA = p r2 + p rs 2

= p (2) + p (2) × (4.5)

The cone includes the circular base plus the curved part. Substitute r = 2 and s = 4.5.

= 40.84 m2 (to 2 d.p.) b

1 TSA = l2 + 4 × bh 2 1 2 = 25 + 4 × × 25 × 22 2 = 1725 mm2

22 mm 25 mm

Example 13 Finding the slant height and vertical height of a cone A cone with radius 3 cm has a curved surface area of 100 cm2 . a Find the slant height of the cone, correct to one decimal place. 10A

b

Find the height of the cone, correct to one decimal place.

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Measurement and Geometry

SOL UTI ON a

EX P L A NA TI ON

Surface area = p rs

Substitute the given information into the rule for the curved surface area of a cone and solve for s.

100 = p × 3 × s 100 s= 3p = 10.6 cm (to 1 d.p.) b

421

h 2 + r2 = s 2   100 2 h2 + 32 = 3p   100 2 2 h = –9 3p    100 2 h= –9 3p

h

s

r Identify the right-angled triangle within the cone and use Pythagoras’ theorem to find the height h. Use the exact value of s from part a to avoid accumulating errors.

= 10.2 cm (to 1 d.p.)

1

1–3(a,b)

2(½)

—

UNDERSTANDING

Exercise 6E Write the rule for the following.

a area of a triangle b surface area of the base of a cone with radius r c surface area of the curved part of a cone with slant height s and radius r 10A

2 Find the exact slant height for these cones, using Pythagoras’ theorem. Answer exactly, using a square root sign. a b c 5m 5 cm 6 cm

10 cm

14 m 2 cm 3 Draw a net for each of these solids. a b

c

3 cm 2 cm

4 cm

2 cm

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Chapter 6 Measurement

4–6(a,b)

6E Example 12a

4 Find the total surface area of these cones, correct to two decimal places. a b c 9 mm 0.5 m

12 mm Example 12b

4–6(a,c), 7

11 km

0.8 m 15 km

5 Find the total surface area of these square-based pyramids. a b 8 cm 6m

4m

4–7(b)

FLUENCY

422

c

0.2 m

0.3 m

5 cm

6 For each cone, find the area of the curved surface only, correct to two decimal places. a b c 1.1 cm 26 mm 10 m 48 mm 1.5 cm 2m 7 A cone has height 10 cm and radius 3 cm.

10A

a

Use Pythagoras’ theorem to find the slant height of the cone, rounding your answer to two decimal places. b Find the total surface area of the cone, correct to one decimal place.

Example 13 10A

9–11

10, 11, 12(½)

PROBLEM-SOLVING

8, 9

8 A cone with radius 5 cm has a curved surface area of 400 cm2 . a Find the slant height of the cone, correct to one decimal place. b Find the height of the cone, correct to one decimal place. 9 A cone with radius 6.4 cm has a curved surface area of 380 cm2 . a Find the slant height of the cone, correct to one decimal place. b Find the height of the cone, correct to one decimal place.

10A

10 This right square-based pyramid has base side length 4 m and vertical height 6 m. a b

6m

Find the height of the triangular faces, correct to one decimal place. Find the total surface area, correct to one decimal place.

4m

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11 Party hats A and B are in the shape of open cones with no base. Hat A has radius 7 cm and slant height 25 cm, and hat B has radius 9 cm and slant height 22 cm. Which hat has the greater surface area?

PROBLEM-SOLVING

Measurement and Geometry

423

6E

12 Find the total surface area of these composite solids, correct to one decimal place as necessary. a b 3 cm 4 cm

2 cm 3 cm

5 cm c

d

4.33 cm

10 m

8 cm

4.33 cm

3m 8m

5 cm e

f

9 mm

5m

6 mm 10 mm 2m

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

7m

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Chapter 6 Measurement

13

6E

13, 14

14, 15

13 Explain√why the surface area of a cone with radius r and height h is given by the expression p r(r + r2 + h2 ).

REASONING

424

14 A cone has a height equal to its radius (i.e. h = r). Show that its total surface area is given by the √ expression p r2 (1 + 2).

Carving pyramids from cones

—

10 cm 10 cm

—

16

16 A woodworker uses a rotating lathe to produce a cone with radius 4 cm and height 20 cm. From that cone the woodworker then cut slices off the sides of the cone to produce a square-based pyramid of the same height. a Find the exact slant height of the cone. b Find the total surface area of the cone, correct to two decimal places. c Find the exact side length of the base of the square-based pyramid. d e f

20 cm

ENRICHMENT

15 There is enough information in this diagram to find the total surface area, although the side length of the base and the height of the triangular faces are not given. Find the total surface area, correct to one decimal place.

8 cm

Find the height of the triangular faces of the pyramid, correct to three decimal places. Find the total surface area of the pyramid, correct to two decimal places. Express the total surface area of the pyramid as a percentage of the total surface area of the cone. Give the answer correct to the nearest whole percentage.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

425

Progress quiz In the following questions round all answers to two decimal places, where necessary. 38pt 6A

1

Find the perimeter of these shapes. (Note: In part c, all angles are right angles.) a b 11 cm

4 cm

7 cm

5 cm 13 cm c

3 cm 8 cm 38pt 6B

2

For each of these shapes, find: i the value of x ii the perimeter iii the area. a

b

13 cm

x cm

9 cm x cm

12 cm

c

8 cm x cm

5 cm 40 cm 12 cm

38pt 6A/C

3 Calculate the circumference and area of these circles. a b

4 mm

38pt 6C

4

17.6 mm

Find the areas of these sectors. a

b

210°

5 cm 80° 5 cm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

4 cm

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426

Chapter 6 Measurement

38pt 6C

5

Convert 4.5 cm2 to the following units. a mm2 b m2

38pt 6C

6

Find the shaded (purple) area of these shapes. a b 4 cm 7 cm 6 cm

38pt 6D/E

7

Calculate the surface area of these solids. a

c

10 cm

1.1 m

b

5 cm

50 cm

3 cm 8 cm c

d

12 cm

8 cm

7.2 cm 38pt 6E

8

6 cm

A cone is made from a sector of a circle with radius 7.5 cm and central angle 120◦ .

120°

7.5 cm

O a Calculate the curved surface area of the cone. b Calculate the radius of the circular base of the cone. c 38pt 6B

9

Calculate the perpendicular height of the cone.

Find the diagonal length of a rectangular prism with dimensions 10 cm by 8 cm by 5 cm.

10A

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

427

6F Volume – prisms and cylinders Volume is the amount of space contained within the outer surfaces of a three-dimensional object and is measured in cubic units. The common groups of objects considered in this section are the prisms and the cylinders.

Let’s start: Right and oblique prisms Recall that to find the volume of a right prism, you would first find the area of the base and multiply by the height. Here is a right rectangular prism and two oblique rectangular prisms with the same side lengths.

10

3 • • • • •

6

10

3

6

10

3

6

What is the volume of the right rectangular prism? (The one on the left.) Do you think the volume of the two oblique prisms would be calculated in the same way as that for the right rectangular prism and using exactly the same lengths? Would the volume of the oblique prisms be equal to or less than that of the rectangular prism? Discuss what extra information is required to find the volume of the oblique prisms. How does finding the volume of oblique prisms (instead of a right prism) compare with finding the area of a parallelogram (instead of a rectangle)?

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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428

Chapter 6 Measurement

Key ideas

Metric units for volume include cubic kilometres (km3 ), cubic metres (m3 ), cubic centimetres (cm3 ) and cubic millimetres (mm3 ).

× 10003 km3

Units for capacity include megalitres (ML), kilolitres (kL), litres (L) and millilitres (mL). • 1 cm3 = 1 mL

× 1003 m3

× 103 cm3

mm3

÷ 10003

÷ 1003

÷ 103

× 1000

× 1000

× 1000

ML

L

kL ÷ 1000

÷ 1000

mL ÷ 1000

For right and oblique prisms and cylinders the volume is given by: V = Ah, where • A is the area of the base • h is the perpendicular height.

h h

w

l

right rectangular prism V = Ah

h r

x x oblique square prism V = Ah

right cylinder V = Ah

= x2 h

= lwh

= p r2 h

Example 14 Finding the volume of right prisms and cylinders Find the volume of these solids, rounding to two decimal places for part b. a b 2 cm

4m 6 cm 2m

3m

SOL UTI ON

EXPLA NA TI ON

a

Write the volume formula for a rectangular prism.

V = lwh = 2×3×4

Substitute l = 2, w = 3 and h = 4.

= 24 m3 b

V = p r2 h

The prism is a cylinder with base area p r2 .

= p (2)2 × 6 3

Substitute r = 2 and h = 6.

= 75.40 cm (to 2 d.p.) Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Evaluate and round your answer as required.

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Measurement and Geometry

429

Example 15 Finding the volume of an oblique prism Find the volume of this oblique prism.

8 cm

3 cm 5 cm SOL UTI ON

EX P L A NA TI ON

1 bh × 8 2 1 = ×5×3×8 2 = 60 cm3

V=

The base is a triangle, so multiply the area of the base triangle by the perpendicular height.

Example 16 Finding the volume of a composite solid Find the volume of this composite solid, correct to one decimal place.

2 cm

2 cm

6 cm

SOL UTI ON Radius of cylinder =

EX P L A NA TI ON 6 = 3 cm 2

V = lwh + p r2 h = 6 × 6 × 2 + p × 32 × 2 = 72 + 18p

First, find the radius length, which is half the side length of the square base. Add the volume of the square-based prism and the volume of the cylinder.

= 128.5 cm3 (to 1 d.p.)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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430

Chapter 6 Measurement

1

1, 2(½)

2(½)

Find the volume of these solids with the given base areas. a b

—

UNDERSTANDING

Exercise 6F

2m

10 cm A = 16 m2 A = 8 cm2 c

12 mm

A = 9 mm2 2 Convert these volume measurements to the units given in brackets. (Refer to the Key ideas for help.) a 2 cm3 (mm3 ) b 0.2 m3 (cm3 ) c 0.015 km3 (m3 ) d 5700 mm3 (cm3 ) e 28 300 000 m3 (km3 ) f 762 000 cm3 (m3 ) 3 3 3 3 g 0.13 m (cm ) h 0.000001 km (m ) i 2.094 cm3 (mm3 ) j 2.7 L (mL) k 342 kL (ML) l 35 L (kL) m 5.72 ML (kL) n 74 250 mL (L) o 18.44 kL (L)

Example 14a

3

Find the volume of each rectangular prism. a b

2 cm 4 cm

3–5(a, c), 6(½)

3–5(b), 6(½)

FLUENCY

3–5(a, b)

c

5 cm

7 mm

35 m

3.5 mm

30 m

10.6 mm

10 m Example 14b

4 Find the volume of each cylinder, correct to two decimal places. a b 2 cm

10 m

1.5 cm

c

12 m 14 m

5m

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Example 15

FLUENCY

Measurement and Geometry

5 Find the volume of these oblique solids. Round to one decimal place for part b. a b 5m

3 cm

431

6F

20 m

2 cm c

1.2 mm 2.4 mm 1.7 mm

6 Find the volume of these solids, rounding your answers to three decimal places where necessary. a b c 3m 8 cm

2m

2 km

10 km

3.5 cm

3 km

7 cm

d

e

5

cm2

3m

7m

f

21.2 cm

2 cm 0.38 m

18.3 cm

0.12 m

0.5 m

g

24.5 cm

h

i

6m 2m

12 cm

3m

0.8 m 2m

12 cm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter 6 Measurement

7, 8

6F

8, 9(½)

9(½), 10

PROBLEM-SOLVING

432

7 How many containers holding 1000 cm3 (1 L) of water are needed to fill 1 m3 ? 8 How many litres of water are required to fill a rectangular fish tank that is 1.2 m long, 80 cm wide and 50 cm high?

Example 16

9 Find the volume of these composite objects, rounding to two decimal places where necessary. a b c

5m

5 cm

1.1 m

2.6 m

2 cm 1.6 m

8 cm

2.3 m

5m d

e

5 mm 8 mm

11 mm

4 cm

f

5 cm 4 cm

9 cm

6 cm

3 cm 10 Find the exact volume of a cube if its surface area is: a 54 cm2 b 18 m2 11, 12

12–14

11 Use the rule V = p r2 h to find the height of a cylinder, to one decimal place, with radius 6 cm and volume 62 cm3 . 12 A fellow student says the volume of this prism is given by V = 4 × 3 × 5. Explain his error.

5 4

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

REASONING

11

3

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13 Find a formula for the volume of a cylindrical portion with angle q, radius r and height h, as shown.

r

REASONING

Measurement and Geometry

433

6F

θ h

14 Decide whether there is enough information in this diagram of a triangular prism to find its volume. If so, find the volume, correct to one decimal place.

Concrete poles

10 m

—

—

15

15 A concrete support structure for a building is made up of a cylindrical base and a square-based prism as the main column. The cylindrical base is 1 m in diameter and 1 m high, and the square prism is 10 m long and sits on the cylindrical base as shown.

ENRICHMENT

4m

10 m

1m

1m a Find the exact side length of the square base of the prism. b Find the volume of the entire support structure, correct to one decimal place.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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434

Chapter 6 Measurement

6G Volume – pyramids and cones

10A

The volume of a cone or pyramid is a certain fraction of the volume of a prism with the same base area. This particular fraction is the same for both cones and pyramids and will be explored in this section.

The volume of sand needed to construct these garden features can be calculated from the formula for the volume of a cone.

Let’s start: Is a pyramid half the volume of a prism? Here is a cube and a square pyramid with equal base side lengths and equal heights.

• •

Discuss whether or not you think the pyramid is half the volume of the cube. Now consider this diagram of the cube with the pyramid inside. The cube is black. The pyramid is green. The triangular prism is blue.

• Compared to the cube, what is the volume of the triangular prism (blue)? Give reasons. • Is the volume of the pyramid (green) more or less than the volume of the triangular prism (blue)? • Do you know what the volume of the pyramid is as a fraction of the volume of the cube?

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

Key ideas

1 For pyramids and cones the volume is given by V = Ah, 3 where A is the area of the base and h is the perpendicular height. right square pyramid

right cone

435

oblique cone

h

h

h x

r

x

r

1 V = Ah 3 1 = p r2 h 3

1 V = Ah 3 1 = x2 h 3

1 Ah 3 1 = p r2 h 3

V=

Example 17 Finding the volume of pyramids and cones Find the volume of this rectangular-based pyramid and cone. Give the answer for part b, correct to two decimal places. a b 23 mm 1.3 m

29 mm

1.4 m 1.2 m SOL UTI ON a

b

EX P L A NA TI ON

1 Ah 3 1 = (l × w) × h 3 1 = (1.4 × 1.2) × 1.3 3 = 0.728 m3

V=

The pyramid has a rectangular base with area l × w.

Substitute l = 1.4, w = 1.2 and h = 1.3.

1 Ah 3 1 = p r2 h 3 1 = p (11.5)2 × 29 3 = 4016.26 mm3 (to 2 d.p.)

V=

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

The cone has a circular base of area p r2 .

Substitute r =

23 = 11.5 and h = 29. 2

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436

Chapter 6 Measurement

1

1–3

3(a, c)

—

A cylinder has volume 12 cm3 . What will be the volume of a cone with the same base area and perpendicular height?

2 A pyramid has volume 5 m3 . What will be the volume of a prism with the same base area and perpendicular height? 3 Find the volume of these solids with the given base areas. a b

UNDERSTANDING

Exercise 6G

c

6m 4 cm

A = 2 cm2

A = 25 mm2

4–5(½) Example 17a

4–5(½)

4–5(½)

4 Find the volume of the following pyramids. For the oblique pyramids (i.e. parts d, e, f) use the given perpendicular height. a

b

3 cm

c

FLUENCY

A = 5 m2

7 mm

5 km

18 m

12 km 15 m

5 km

2 cm 13 m

d

2 cm

e

4 cm

f

1.4 mm 0.6 mm

3 cm

Perpendicular height = 2 cm Perpendicular height = 4 cm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Perpendicular height = 1.2 mm

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Example 17b

FLUENCY

Measurement and Geometry

5 Find the volume of the following cones, correct to two decimal places. a b c 1.6 mm

437

6G

20 m 60 m

2.6 m

3.5 mm

1.1 m d

e

0.6 m

f

20 cm

6m

5 cm 1.3 m

2m

6, 7(½)

7(½), 8

6 A medicine cup is in the shape of a cone with base radius 3 cm and height 5 cm. Find its capacity in millilitres, correct to the nearest millilitre. 7 Find the volume of these composite objects, rounding to two decimal places where necessary. a b 6m 5 cm

8m

PROBLEM-SOLVING

6, 7(½)

3m 6 cm

c

d

4m 2m

4.8 mm

3m 6m

1.5 mm 5.2 mm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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438

Chapter 6 Measurement

e

PROBLEM-SOLVING

6G 10 m

f

2.6 m 3.96 m 2.8 m

8

7.1 m

3 cm

The volume of ice-cream in the cone is half the volume of the cone. The cone has a 3 cm radius and 6 cm height. What is the depth of the ice-cream, correct to two decimal places?

ice-cream

6 cm

depth of ice-cream

9, 10

10, 11

9 A wooden cylinder is carved to form a cone that has the same base area and the same height as the original cylinder. What fraction of the wooden cylinder is wasted? Give a reason.

REASONING

9

10 A square-based pyramid and a cone are such that the diameter of the cone is equal to the length of the side of the square base of the pyramid. They also have the same height. a

Using x as the side length of the pyramid and h as its height, write a rule for: i the volume of the pyramid in terms of x and h ii the volume of the cone in terms of x and h.

b Express the volume of the cone as a fraction of the volume of the pyramid. Give an exact answer. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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1 2 p r h to find the base radius of a cone, to one decimal place, with height 3 23 cm and volume 336 cm3 . 1 b Rearrange the rule V = p r2 h to write: 3 Use the rule V =

i

h in terms of V and r

439

6G

ii r in terms of V and h

Truncated cones

—

—

12

12 A truncated cone is a cone that has its apex cut off by an intersecting plane. In this example, the top has radius r2 , the base has radius r1 and the two circular ends are parallel.

h2

ENRICHMENT

11 a

REASONING

Measurement and Geometry

h1

r2 r1 r1 h1 = . r2 h2 b Find a rule for the volume of a truncated cone. c Find the volume, to one decimal place, of a truncated cone when r1 = 2 cm, h1 = 5 cm and h2 equals: 1 2 h1 h i ii 2 3 1 a

Give reasons why

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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440

Chapter 6 Measurement

6H Spheres

10A

Planets are spherical in shape due to the effects of gravity. This means that we can describe a planet’s size using only one measurement – its diameter or radius. Mars, for example, has a diameter of about half that of the Earth, which is about 12 756 km. The Earth’s volume is about 9 times that of Mars and this is because the volume of a sphere varies with the cube of the radius. The surface area of the Earth is about 3.5 times that of Mars because the surface of a sphere area varies with the square of the radius. Mars is approximately spherical in shape.

Let’s start: What percentage of a cube is a sphere? A sphere of radius 1 unit just fits inside a cube. • First, guess the percentage of space occupied by the sphere. • Draw a diagram showing the sphere inside the cube. • •

Key ideas

4 3 pr . 3 Now calculate the percentage of space occupied by the sphere. How close was your guess?

Calculate the volume of the cube and the sphere. For the sphere, use the formula V =

The surface area of a sphere depends on its radius, r, and is given by: Surface area = 4p r2

r

The volume of a sphere depends on its radius, r, and is given by: 4 Volume = p r3 3

Example 18 Finding the surface area and volume of a sphere Find the surface area and volume of a sphere of radius 7 cm, correct to two decimal places. SOL UTI ON

EX P L A NA TI ON

TSA = 4p r2

Write the rule for the surface area of a sphere and substitute r = 7.

= 4p (7)2 = 615.75 cm2 (to 2 d.p.) 4 V = p r3 3 4 = p (7)3 3 = 1436.76 cm3 (to 2 d.p.) Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Evaluate and round the answer. Write the rule for the volume of a sphere and substitute r = 7. Evaluate and round the answer.

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Measurement and Geometry

441

Example 19 Finding the radius of a sphere Find the radius of a sphere with volume 10 m3 , correct to two decimal places.

rm

SOL UTI ON

EX P L A NA TI ON

4 3 pr 3 4 10 = p r3 3 30 = 4p r3 V=

Substitute V = 10 into the formula for the volume of a sphere. Solve for r3 by multiplying both sides by 3 and then dividing both sides by 4p . Simplify 15 30 = . 4p 2p

15 = r3 2p 

15 2p = 1.34 (to 2 d.p.)

∴ r=

3

Take the cube root of both sides to make r the subject and evaluate.

∴ The radius is 1.34 m.

Example 20 Finding the surface area and volume of composite solids with spherical portions This composite object includes a hemisphere and cone, as shown. a Find the surface area, rounding to two decimal places. b Find the volume, rounding to two decimal places.

SOL UTI ON a

5 cm

7 cm

EX P LA NA TI ON

2

2

2

s =5 +2 s2 = 29 √ s = 29

Calculate the slant height, s, of the cone using Pythagoras’ theorem.

1 × 4p r2 + p rs 2 √ 1 = × 4p (2)2 + p (2)( 29) 2 √ = 8p + 2 29p

TSA =

= 58.97 cm2 (to 2 d.p.)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

2 cm

First find the radius, r cm, of the hemisphere.

Radius r = 7 – 5 = 2

5 cm

s cm

Write the rules for the surface area of each component and note that the top shape is a hemisphere (i.e. half sphere). Only the curved surface of the cone is required. Substitute r = 2 and h = 5. Simplify and then evaluate, rounding as required.

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442

Chapter 6 Measurement

1 4 3 1 2 × pr + pr h 2 3 3 1 1 4 = × p (2)3 + p (2)2 (5) 2 3 3 16p 20p + = 3 3 36p = 3

b

1 Volume (sphere) + Volume (cone) 2 Substitute r = 2 and h = 5.

V=

Volume (object) =

Simplify and then evaluate, rounding as required.

= 12p = 37.70 cm3 (to 2 d.p.)

1

1–4

4

—

Evaluate and round your answer to two decimal places. a 4 × p × 52 d

b 4 × p × 2.22

4 × p × 23 3

e

c 4×p ×

4 × p × 2.83 3

f

4p (7)3 3

UNDERSTANDING

Exercise 6H

 2 1 2

2 Rearrange 12 = 4p r2 to write r as the subject. 4 3 p r to write r as the subject. 3

4 What fraction of a sphere is shown in these diagrams? a b

c

5–6(½) Example 18

5–7(½)

5–7(½)

5 Find the surface area and volume of the following spheres, correct to two decimal places. a b c 0.5 m 2 cm

FLUENCY

3 Rearrange 8 =

38 mm d

e

18 cm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

f

1.36 m

0.92 km

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6 Find the total surface area and volume of a sphere with the given dimensions. Give the answer correct to two decimal places. a radius 3 cm b radius 4 m c radius 7.4 m √ √ d diameter 5 mm e diameter 7 m f diameter 2.2 km Example 19

7

a

FLUENCY

Measurement and Geometry

443

6H

Find the radius of these spheres with the given volumes, correct to two decimal places. i ii iii

r cm

r km

r cm

V = 15 cm3 V = 180 cm3 V = 0.52 km3 b Find the radius of these spheres with the given surface area, correct to two decimal places. i ii iii

r cm

S = 10 m2

S = 120 cm2 8–10

8

A box with dimensions 30 cm long, 30 cm wide and 30 cm high holds 50 tennis balls of radius 3 cm. Find: a the volume of one tennis ball, correct to two decimal places b the volume of 50 tennis balls, correct to one decimal place c the volume of the box not taken up by the tennis balls, correct to one decimal place.

r mm

S = 0.43 mm2 11–13, 14(½)

14-15(½), 16, 17

PROBLEM-SOLVING

rm

9 An expanding spherical storage bag has 800 cm3 of water pumped into it. Find the diameter of the bag, correct to one decimal place, after all the water has been pumped in.

10 A sphere just fits inside a cube. What is the surface area of the sphere as a percentage of the surface area of the cube? Round your answer to the nearest whole percentage.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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444

Chapter 6 Measurement

11 Find the volume of these portions of a sphere, correct to two decimal places. a b c 4 cm 5 cm

PROBLEM-SOLVING

6H 1m

12 Two sports balls have radii 10 cm and 15 cm. Find the difference in their total surface areas, correct to one decimal place. 13 A monolithic structure has a cylindrical base of radius 4 m and height 2 m and a hemispherical top. a What is the radius of the hemispherical top? b Find the total volume of the entire monolithic structure, correct to one decimal place.

Example 20a

2m

4m

14 Find the total surface area for these solids, correct to two decimal places. a b 5m 0.7 cm

1.3 cm

c

d

8 mm 4 mm

3m

5m

1m e

f

3 cm

2 cm 2m √2 cm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Example 20b

PROBLEM-SOLVING

Measurement and Geometry

15 Find the volume of the following composite objects, correct to two decimal places. a b

1m

13 m

3m

445

6H

19 m

c

d

28 cm

2 cm 20 cm e

hollow

f

2m 2m 1 cm

2m

16 A spherical party balloon is blown up to help decorate a room. a

Find the volume of air, correct to two decimal places, needed for the balloon to be: i 10 cm wide ii 20 cm wide iii 30 cm wide

b If the balloon pops when the volume of air reaches 120 000 cm3 , find the diameter of the balloon at that point, correct to one decimal place. 17 A hemisphere sits on a cone and two height measurements are given as shown. Find: a the radius of the hemisphere b the exact slant height of the cone in surd form c the total surface area of the solid, correct to one decimal place.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

10 cm

15 cm

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Chapter 6 Measurement

18

6H

18, 19

20, 21

REASONING

446

18 a Find a rule for the radius of a sphere with surface area S. b Find a rule for the radius of a sphere with volume V. 19 A ball’s radius is doubled. a By how much does its surface area change? b By how much does its volume change? 20 Show that the volume of a sphere is given by V =

1 3 p d , where d is the diameter. 6

21 A cylinder and a sphere have the same radius, r, and volume, V. Find a rule for the height of the cylinder in terms of r.

—

—

22

ENRICHMENT

Comparing surface areas

22 Imagine a cube and a sphere that have the same volume. a

If the sphere has volume 1 unit3 , find: i the exact radius of the sphere ii the exact surface area of the sphere iii the value of x (i.e. the side length of the cube) iv the surface area of the cube

x r x

x

v the surface area of the sphere as a percentage of the surface area of the cube, correct to one decimal place. b Now take the radius of the sphere to be r units. Write: i the rule for the surface area of the sphere ii the rule for x in terms of r, given the volumes are equal iii the surface area of the cube in terms of r. c

Now write the surface area of the sphere as a fraction of the surface area of the cube, using your results from part b.  p Simplify to show that the result is 3 . 6 d Compare your answers from part a v with that of part c (i.e. as a percentage).

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

6I Limits of accuracy

447

EXTENDING

Humans and machines measure many different things, such as the time taken to swim a race, the length of timber needed for a building and the volume of cement needed to lay a concrete path around a swimming pool. The degree or level of accuracy required usually depends on the intended purpose of the measurement. All measurements are approximate. Errors can happen as a result of the equipment being used or the person using the measuring device. Accuracy is the measure of how true the measure is to the ‘real’ one, whereas precision is the ability to obtain the same result over and over again.

During many sporting events, a high degree of accuracy is required for time-keeping.

Let’s start: Rounding a decimal • A piece of timber is measured to be 86 cm, correct to the nearest centimetre. a What is the smallest decimal that it could be rounded from? •

b What is the largest decimal that is recorded as 86 cm when rounded to the nearest whole? If a measurement is recorded as 6.0 cm, correct to the nearest millimetre, then: a What units were used when measuring? b What is the smallest decimal that could be rounded to this value?

•

c What is the largest decimal that would have resulted in 6.0 cm? Consider a square with side length 7.8941 cm. a What is the perimeter of the square if the side length is: i used with the four decimal places? ii rounded to one decimal place? iii truncated (i.e. chopped off) at one decimal place (i.e. 7.8)? b

What is the difference between the perimeters if the decimal is rounded to two decimal places or truncated at two decimal places or written with two significant figures? The limits of accuracy tell you what the upper and lower boundaries are for the true measurement. • Usually, it is ± 0.5 × the smallest unit of measurement. For example, when measuring to the nearest centimetre, 86 cm has limits from 85.5 cm up to (but not including) 86.5 cm.

85.5 85

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Key ideas

86.5 86.0

87

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448

Chapter 6 Measurement

Key ideas

• When measuring to the nearest millimetre, the limits of accuracy for 86.0 cm are 85.95 cm to 86.05 cm. 85.95 86.05

85.8

85.9

86.0

86.1

86.2

Example 21 Finding limits of accuracy Give the limits of accuracy for these measurements. a 72 cm b 86.6 mm SOL UTI ON

E X P LA NA TI ON

a

72 ± 0.5 × 1 cm

Smallest unit of measurement is one whole cm.

= 72 – 0.5 cm to 72 + 0.5 cm

Error = 0.5 × 1 cm

= 71.5 cm to 72.5 cm

This error is subtracted and added to the given measurement to find the limits of accuracy.

86.6 ± 0.5 × 0.1 mm

Smallest unit of measurement is 0.1 mm.

= 86.6 ± 0.05 mm

Error = 0.5 × 0.1 mm = 0.05 mm

= 86.6 – 0.05 mm to 86.6 + 0.05 mm

This error is subtracted and added to the given measurement to find the limits of accuracy.

b

= 86.55 mm to 86.65 mm

Example 22 Considering the limits of accuracy Janis measures each side of a square as 6 cm. Find: a the upper and lower limits for the sides of the square b the upper and lower limits for the perimeter of the square c the upper and lower limits for the square’s area. SOL UTI ON

E X P LA NA TI ON

a

6 ± 0.5 × 1 cm

Smallest unit of measurement is one whole cm.

= 6 – 0.5 cm to 6 + 0.5 cm

Error = 0.5 × 1 cm

= 5.5 cm to 6.5 cm b

Lower limit P = 4 × 5.5 = 22 cm Upper limit P = 4 × 6.5

The lower limit for the perimeter uses the lower limit for the measurement taken and the upper limit for the perimeter uses the upper limit of 6.5 cm.

= 26 cm c

Lower limit = 5.52 = 30.25 cm2 Upper limit = 6.52

The lower limit for the area is 5.52 , whereas the upper limit will be 6.52 .

= 42.25 cm2 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

1

1–3

3

—

UNDERSTANDING

Exercise 6I

449

Write down a decimal that gives 3.4 when rounded from two decimal places.

2 Write down a measurement of 3467 mm, correct to the nearest: a centimetre b metre 3 What is the smallest decimal that could result in an answer of 6.7 when rounded to one decimal place?

Example 21

4–5(½), 6, 7(½), 8

4–5(½), 8, 9

4 For each of the following: i Give the smallest unit of measurement (e.g. 0.1 cm is the smallest unit in 43.4 cm). ii Give the limits of accuracy. a 45 cm e 56.8 g i 12.34 km

b 6.8 mm f 10 m j 0.987 km

c 12 m g 673 h k 1.65 L

5 Give the limits of accuracy for the following measurements. a 5m b 8 cm c 78 mm e 2 km f 34.2 cm g 3.9 kg i 457.9 L j 18.65 m k 7.88 km

FLUENCY

4–5(½), 6

d 15.6 kg h 9.84 m l 9.03 mL

d 5 mL h 19.4 kg l 5.05 s

6 What are the limits of accuracy for the amount $4500 when it is written: a to two significant figures? b to three significant figures? c to four significant figures? 7 Write the following as a measurement, given that the lower and upper limits of these measurements are as follows. a 29.5 m to 30.5 m b 140 g to 150 g c 4.55 km to 4.65 km d 8.95 km to 9.05 km e 985g to 995 g f 989.5 g to 990.5 g 8 Martha writes down the length of her fabric as 150 cm. As Martha does not give her level of accuracy, give the limits of accuracy of her fabric if it was measured correct to the nearest: a centimetre b 10 centimetres c millimetre

9 A length of copper pipe is given as 25 cm, correct to the nearest centimetre. a What are the limits of accuracy for this measurement? b If 10 pieces of copper, each with a given length of 25 cm, are joined end to end, what is the minimum length that it could be? c What is the maximum length for the 10 pieces of pipe in part b?

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Chapter 6 Measurement

10

6I Example 22

10

10, 11

PROBLEM-SOLVING

450

10 The side of a square is recorded as 9.2 cm, correct to two significant figures. a b c d

What is the minimum length that the side of this square could be? What is the maximum length that the side of this square could be? Find the upper and lower boundaries for this square’s perimeter. Find the upper and lower limits for the area of this square.

11 The side of a square is recorded as 9.20 cm, correct to three significant figures. What is the minimum length that the side of this square could be? What is the maximum length that the side of this square could be? Find the upper and lower boundaries for this square’s perimeter. Find the upper and lower limits for the area of this square. How has changing the level of accuracy from 9.2 cm (see Question 10) to 9.20 cm affected the calculation of the square’s perimeter and area?

12

12

13

12 Cody measures the mass of an object to be 6 kg. Jacinta says the same object is 5.8 kg and Luke gives his answer as 5.85 kg. a Explain how all three people could have different answers for the same measurement. b Write down the level of accuracy being used by each person. c Are all their answers correct? Discuss.

REASONING

a b c d e

13 Write down a sentence explaining the need to accurately measure items in our everyday lives and the accuracy required for each of your examples. Give three examples of items that need to be measured correct to the nearest: a kilometre b millimetre c millilitre d litre

—

—

14

To calculate the percentage error of any measurement, the error (i.e. ± the smallest unit of measurement) is compared to the given or recorded measurement and then converted to a percentage. For example: 5.6 cm Error = ± 0.5 × 0.1 = ± 0.05 ± 0.05 Percentage error = × 100% 5.6 = ± 0.89% (to two significant figures)

ENRICHMENT

Percentage error

14 Find the percentage error for each of the following. Round to two significant figures. a 28 m b 9 km c 8.9 km d 8.90 km e 178 mm f $8.96 g 4.25 m h 700 mL

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Measurement and Geometry

451

Investigation Similar objects Enlarging a rectangular prism The larger rectangular prism shown here is an enlargement of the smaller prism by a factor of 1.5.

3

2 4

9

6

6

a

Write down the ratios of the side lengths.

b

Find the surface areas of the prisms and write these as a ratio in simplest form.

c

Find the volumes of the prisms and write these as a ratio in simplest form.

d

What do you notice about the ratios for length, area and volume? Explain.

e

Can you write all three ratios using indices with the same base?

Using similarity From the previous section, we note that if two similar objects have a length ratio a : b, then the following apply. Length ratio = a : b Area ratio = a2 : b2 Volume ratio = a3 : b3 a

b a b2 Scale factor = 2 a b3 Scale factor = 3 a Scale factor =

A sphere of radius 1 cm is enlarged by a length scale factor of 3. i Find the ratio of the surface areas of the two spheres. ii Find the ratio of the volumes of the two spheres. iii Use your ratio to find the surface area of the larger sphere. iv Use your ratio to find the volume of the larger sphere.

b

1 This rectangular prism is enlarged by a factor of , which means that 2 the prism is reduced in size. i

Find the ratio, large : small, of the surface areas of the prisms.

2m 6m

10 m

ii Find the ratio, large : small, of their volumes. iii Use your ratios to find the surface area of the smaller prism. iv Use your ratios to find the volume of the smaller prism. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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452

Chapter 6 Measurement

Density Density is defined as the mass or weight of a substance per cubic unit of volume. mass or Mass = density × volume Density = volume Finding mass Find the total mass of these objects with the given densities, correct to one decimal place where necessary. a

b

4m

1m 2m Density = 50 kg per m3

1m

c

6 cm

2m 10 cm Density = 100 kg per m3

5 cm Density = 0.05 kg per cm3

Finding density a

Find the density of a compound with the given mass and volume measurements, rounding to two decimal places where necessary. i mass 30 kg, volume 0.4 m3 ii mass 10 g, volume 2 cm3 iii mass 550 kg, volume 1.8 m3

b

The density of a solid depends somewhat on how its molecules are packed together. Molecules represented as spheres are tightly packed if they are arranged in a triangular form. The following relates to this packing arrangement.

1 cm

A

C

i

Find, the length AC for three circles, each of radius 1 cm, as shown. Use exact values.

ii Find the total height of four spheres, each of radius 1 cm, if they are packed to form a triangular-based pyramid. Use exact values. First, note that AB = 2BC for an equilateral triangle (shown at right). Pythagoras’ theorem can be used to prove this, but this is trickier.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

A

B

C

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Measurement and Geometry

453

Concentration Concentration is associated with the purity of dissolved substances and will be considered here using percentages. Concentration (%) =

volume of substance 100 × total volume 1

Finding concentration Find the concentration of acid as a percentage when 10 cm3 of pure acid is mixed into the given containers that are half full (i.e. half the volume) of water. Give your answers correct to two decimal places. a

b

c

6 cm

12 cm

10 cm 5 cm

8 cm 2 cm

Finding volumes a

Find the volume of pure acid required to give the following concentrations and total volumes. i concentration 25%, total volume 100 cm3 ii concentration 10%, total volume 90 m3 iii concentration 32%, total volume 1.8 L

b

A farmer mixes 5 litres of a chemical herbicide into a three-quarter full tank of water. The tank is cylindrical with diameter 1 metre and height 1.5 metres. Find: i the number of litres of water in the tank, correct to three decimal places ii the concentration of the herbicide after it has been added to the water, correct to two decimal places iii the possible diameter and height measurements of a tank that would make the concentration of herbicide 1%.

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454

Chapter 6 Measurement

Problems and challenges

Up for a challenge? If you get stuck on a question, check out the 'Working with unfamiliar problems' poster at the end of the book to help you.

1

A cube has a surface area that has the same value as its volume. What is the side length of this cube?

2

The wheels of a truck travelling at 60 km/h make 4 revolutions per second. What is the diameter of each wheel in metres, correct to one decimal place?

3

A sphere fits exactly inside a cylinder, and just touches the top, bottom and curved surface. a

Show that the surface area of the sphere equals the curved surface area of the cylinder.

b

What percentage of the volume of the cylinder is taken up by the sphere? Round your answer to the nearest whole percentage.

4

A sphere and cone with the same radius, r, have the same volume. Find the height of the cone in terms of r.

5

Four of the same circular coins of radius r are placed such that they are just touching, as shown. What is the area of the shaded region enclosed by the coins in terms of r?

6

Find the exact ratio of the equator to the distance around the Earth at latitude 45◦ north. (Assume that the Earth is a perfect sphere.)

45°

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Area Units of area × 10002 km2

× 1002

m2

÷ 10002

circle

square

r

A=

A=

pr 2

A=

y

h b

l

A = lw

A=

a h

Length Units of length

1 bh 2

× 1000 × 100 × 10 km m cm mm

kite

parallelogram trapezium

x

÷ 1000

y

h

b

1 × pr 2 A = 2 xy

triangle w

x

÷ 100

÷ 10

Perimeter is the distance around the outside of a closed shape.

b

A = bh

c2 = a2 + b2

a Can occur in 3D shapes

mm2 ÷ 102

Formulas rectangle

l2

rhombus

θ

θ 360

cm2 ÷ 1002

c

b

× 102

l

sector r

Pythagoras’ theorem

455

Chapter summary

Measurement and Geometry

Circumference of a circle C = 2pr = pd

A = 12 (a + b)h A = 12 xy

Perimeter of a sector r

P = 2r +

θ

Measurement

θ × 2pr 360

Surface area For prisms and pyramids draw the net and add the areas of all the faces.

Units of volume

e.g.

km3

× 10003

× 1003

m3

÷ 10003

h

2pr

× 1000

h

× 1000

h l V = lwh

TSA = prs + pr 2 curved + base TSA = 4pr 2

For composite solids consider which surfaces are exposed.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

× 1000

For right and oblique prisms and cylinders V = Ah, where A is the area of the base h is the perpendicular height. rectangular prism cylinder

s

r

mm3 ÷ 103

megalitres kilolitres litres millilitres (ML) (kL) (L) (mL) ÷ 1000 ÷ 1000 ÷ 1000

Limits of accuracy Usually ± 0.5 × the smallest unit of measurement e.g. 72 cm is 71.5 cm to 72.5 cm.

r

Sphere (10A)

÷ 1003

1 cm3 = 1 mL

r TSA = 2prh + 2pr 2 curved base + surface ends Cone (10A)

× 103 cm3

Units of capacity

TSA = 4 × triangles + square base Cylinder

Volume

h r V = pr 2h

w

For pyramids and cones: (10A) 1 V = 3 Ah, where A is the area of the base. Cone: V =

1 pr 2h 3

Sphere: V =

4 pr 3 3

h (10A)

r

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Chapter review

456

Chapter 6 Measurement

Multiple-choice questions 38pt 6A

1

A B C D E 38pt 6A

38pt 6C

38pt 6C

12.8 6.5 5.7 6.4 3.6

7.1 cm 10.4 cm

2 The perimeter of the sector shown, rounded to one decimal place, is: A B C D E

38pt 6B

x cm

If the perimeter of this shape is 30.3 cm, then the value of x is:

8.7 m 22.7 m 18.7 m 56.7 m 32.7 m

5m 100°

3 The value of x in this triangle is closest to: A 4.9 B 3.5 C 5.0 D 4.2 E 3.9

7 km

4 The exact shaded (purple) area in square units is: A 32 – 72p B 120 – 36p C 32 + 6p D 120 – 18p E 48 + 18p 5 0.128 m2 is equivalent to: A 12.8 cm2 B 128 mm2

C 1280 cm2

x km

10 12

D 0.00128 cm2 E 1280 mm2

38pt 6D

6 A cube has a total surface area of 1350 cm2 . The side length of the cube is: A 15 cm B 11 cm C 18 cm D 12 cm E 21 cm

38pt 6D

7 A cylindrical tin of canned food has a paper label glued around its curved surface. If the can is 14 cm high and has a radius of 4 cm, the area of the label is closest to: A 452 cm2 B 352 cm2 C 126 cm2 D 704 cm2 E 235 cm2

38pt 6E

8 The exact surface area of a cone of diameter 24 cm and slant height 16 cm is: A 216p cm2 B 960p cm2 C 528p cm2 D 336p cm2 E 384p cm2

10A

38pt 6B 10A

9 A cone has a radius of 7 m and a slant height of 12 m. The cone’s exact height, in metres, is: √ √ √ A 52 B 193 C 85 √ √ D 137 E 95

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

7m

12 m

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38pt 6H 10A

38pt 6F

10 cm

10 The volume of the hemisphere shown, correct to the nearest cubic centimetre, is: A 1257 cm3 B 628 cm3 C 2094 cm3 3 3 D 1571 cm E 4189 cm 11 The volume of this oblique square-based prism is: A 72 cm3 B 48 cm3 C 176 cm3 3 3 D 144 cm E 120 cm

9 cm 4 cm

38pt 6H 10A

12 The volume of air in a sphere is 100 cm3 . The radius of the sphere, correct to two decimal places, is: A 1.67 cm B 10.00 cm C 2.82 cm D 23.87 cm E 2.88 cm

457

Chapter review

Measurement and Geometry

Short-answer questions 6A/C/F 38pt

38pt 6B

1 Convert the given measurements to the units in the brackets. a 0.23 m (cm) b 270 mm2 (cm2 ) d 8.372 litres (mL) e 638 250 mm2 (m2 )

c 2.6 m3 (cm3 ) f 0.0003 km2 (cm2 )

2 Find the perimeter of these shapes, correct to one decimal place where necessary. Note: Pythagoras’ theorem may be required in part b. a b c

8m

9m 4m

8m 38pt 6A/C

80°

6 cm

7m

3 A floral clock at the Botanic Gardens is in the shape of a circle and has a circumference of 14 m. a Find the radius of the clock, in exact form. b Hence, find the area occupied by the clock. Answer to two decimal places.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter review

458

Chapter 6 Measurement

38pt 6B 10A

H

4 For the rectangular prism with dimensions as shown, use Pythagoras’ theorem to find:

2

a AF, leaving your answer in exact form b AG, to two decimal places.

C

D A

38pt 6C

G

4

E

F 7

B

5 Find the area of these shapes. Round to two decimal places where necessary. a b c

12 m

6.2 m

13 m

11 m 7m

18 m 5.2 m d

e

13 m

f

10 cm

8.1 m 4.2 m

6.4 m

7m

9.8 m 38pt 6C

6 A backyard deck, as shown, has an area of 34.8 m2 .

9.2 m

a Find the width, w metres, of the deck. b Calculate the perimeter of the deck, correct to two decimal places. Pythagoras’ theorem will be required to calculate the missing length. 38pt 6D–H

7

wm

deck 5.3 m

For each of these solids find, correct to two decimal places where necessary: i the total surface area ii the volume a b

6m 5m

8m c

8 cm

5 cm

4 cm 10 cm

6 cm 10A

d

13 cm

12 cm

20 cm 10 cm 10 cm 10A

e

10A

10 cm

f

4m 12 cm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

13 cm

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38pt 6E

8 A cone has a radius of 6 cm and a curved surface area of 350 cm2 . a Find the slant height of the cone, in exact form. b Find the height of the cone, correct to one decimal place.

10A

38pt 6E/G

9 A papier mâché model of a square-based pyramid with base length 30 cm has volume 5400 cm3 . a What is the height of the pyramid? b Use Pythagoras’ theorem to find the exact height of the triangular faces. c Hence, find the total surface area of the model, correct to one decimal place.

10A

38pt 6F 10A

10 A cylinder and a cone each have a base radius of 1 m. The cylinder has a height of 4 m. Determine the height of the cone if the cone and cylinder have the same volume.

38pt 6F/G

11 For each of the following composite solids find, correct to two decimal places where necessary: i the total surface area ii the volume a b c 5m 3 cm 4m 3 cm 0.9 mm 7m 2.5 mm 10 cm 1.6 mm

38pt 6G

12 For each of the following composite solids find, correct to two decimal places where necessary: i the total surface area ii the volume a b 3.9 cm c

10A

459

Chapter review

Measurement and Geometry

3 cm

5 cm

3.2 cm

2 cm 5 cm 4 cm

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

1.8 cm

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Chapter review

460

Chapter 6 Measurement

38pt 6G/H

13 A water buoy is in the shape shown. Find:

10A

38pt 6I

36 cm

a the volume of air inside the buoy, in exact form b the surface area of the buoy, in exact form.

14 Give the limits of accuracy for these measurements. a 8m b 10.3 kg

39 cm 51 cm

c 4.75 L

Extended-response questions 1

A waterski ramp consists of a rectangular flotation container and a triangular angled section, as shown.

1m 2m

cm

8m

6m a b

What volume of air is contained within the entire ramp structure? Find the length of the angled ramp (c metres), in exact surd form.

The entire structure is to be painted with a waterproof paint costing $20 per litre. One litre of paint covers 25 square metres. c d

Find the total surface area of the ramp, correct to one decimal place. Find the number of litres and the cost of paint required for the job. Assume you can purchase only 1 litre tins of paint.

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2 A circular school oval of radius 50 metres is marked with spray paint to form a square pitch, as shown. a State the diagonal length of the square. b Use Pythagoras’ theorem to find the side length of the square, in exact surd form. c Find the area of the square pitch. d Find the percentage area of the oval that is not part of the square pitch. Round your answer to the nearest whole percentage. Two athletes challenge each other to a one-lap race around the oval. Athlete A runs around the outside of the oval at an average rate of 10 metres per second. Athlete B runs around the outside of the square at an average rate of 9 metres per second. Athlete B’s average running speed is less because of the need to slow down at each corner.

461

Chapter review

Measurement and Geometry

e Find who comes first and the difference in times, correct to the nearest hundredth of a second.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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7

Chapter

Parabolas and other graphs

W h a t y o u w ill le a r n

Au s t r a l i a n c u r r ic u lu m

7 A Exploring parabolas 7 B Sketching parabolas with transformations 7 C Sketching parabolas using factorisation 7 D Sketching by completing the square 7 E Sketching using the quadratic formula (10A) 7 F Applications involving parabolas 7 G Intersection of lines and parabolas (10A) 7 H Graphs of circles 7 I Graphs of exponentials 7 J Graphs of hyperbolas 7 K Further transformations of graphs (10A)

NUMBER AND ALGEBRA

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Patterns and algebra Factorise algebraic expressions by taking out a common algebraic factor. Expand binomial products and factorise monic quadratic expressions using a variety of strategies. Substitute values into formulas to determine an unknown. Linear and non-linear relationships Explore the connection between algebraic and graphical representations of relations such as simple quadratics, circles and exponentials, using digital technology as appropriate. Solve simple quadratic equations using a range of strategies. (10A) Describe, interpret and sketch parabolas, hyperbolas, circles and exponential functions and their transformations. (10A) Factorise monic and non-monic quadratic expressions 32x32 16x16 and solve a wide range of quadratic equations derived from a variety of contexts.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

O n lin e re s o u rc e s • • • • • •

Chapter pre-test Videos of all worked examples Interactive widgets Interactive walkthroughs Downloadable HOTsheets Access to HOTmaths Australian Curriculum courses

S o la r r e fle c to r s Solar power is one of the world’s developing renewable energy sources. Collecting solar power involves reflecting the Sun’s rays and capturing and transferring the heat generated. One of the most effective types of solar power plant uses parabolic reflectors. The parabolic shape of these reflectors can be described by a quadratic relation. Consequently they direct all the rays reflected by the mirror to the same place.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

In a parabolic disc system the Sun’s rays are concentrated onto a glass tube receiver. This receiver is positioned at the focal point of the parabola. The fluid in the receiver is heated up and transferred to an engine and converted to electricity. Other types of solar power plants also use parabolic trough reflectors and collect the Sun’s rays to generate electricity.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

464

Chapter 7 Parabolas and other graphs

7A Exploring parabolas One of the simplest and most important non-linear graphs is the parabola. When a ball is thrown or water streams up and out from a garden hose, the path followed has a parabolic shape. The parabola is the graph of a quadratic relation with the basic rule y = x2 . Quadratic rules, such as y = (x – 1)2 , y = 2x2 – x – 3 and y = (x + 4)2 – 7, also give graphs that are parabolas and are transformations of the graph of y = x2 .

Let’s start: To what effect? To see how different quadratic rules compare to the graph of y = x2 , complete this table and plot the graph of each equation on the same set of axes. x

–3

–2

y1 = x 2

9

4

y2 = –x 2

–1

0

1

2

–9

y3 = (x – 2)2 y4 = x 2 – 3

•

•

For all the graphs, find such features as the: – turning point – axis of symmetry – y-intercept – x-intercepts. Discuss how each of the graphs of y2 , y3 and y4 compare to the graph of y = x2 . Compare the rule with the position of the graph.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

y

3

9 8 7 6 5 4 3 2 1 −3 −2 −1−1O −2 −3 −4 −5 −6 −7 −8 −9

1 2 3

x

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Number and Algebra

9 8 7 6 5 4 3 2 1

Simple transformations of the graph of y = x2 include: • dilation

• reflection

y

y y = 2x2

1

y = x2

−3 −2 −1−1O

y = x2

(1, 2) y = x2 2 y = 12 x2 1 1 1 (1, 2 ) (−1, 2 ) x −1−1O 1

Key ideas

y

A parabola is the graph of a quadratic relation. The basic parabola has the rule y = x2 . • The vertex (or turning point) is (0, 0). • It is a minimum turning point. • Axis of symmetry is x = 0. • y-intercept is 0. • x-intercept is 0.

465

x

1 2 3

(1, 1) x 1 (1, −1) y = −x2

−1−1O −2 −3

y = −2x2

• translation

y y = (x + 3)2

4

y = x2 + 2 y = x2

2

−3

y = (x − 2)2

−1 O −3

2

4

x

y = x2 − 3

Example 1 Identifying key features of parabolas Determine the following key features of each of the given graphs. i turning point and whether it is a maximum or minimum ii axis of symmetry iii x-intercepts iv y-intercept

y

a

y

b

−1−1O

3

x

−2

O

x

−4 −3 (1, −4)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

466

Chapter 7 Parabolas and other graphs

SOL UTI ON

EX P L A NA TI ON

a i Turning point is a minimum at (1, – 4).

Lowest point of graph is at (1, – 4). Line of symmetry is through the x-coordinate of the turning point. x-intercepts lie on the x-axis ( y = 0) and the y-intercept on the y-axis (x = 0).

ii Axis of symmetry is x = 1. iii x-intercepts are –1 and 3. iv y-intercept is –3. b i Turning point is a maximum at (–2, 0).

Graph has a highest point at (–2, 0). Line of symmetry is through the x-coordinate of the turning point. Turning point is also the one x-intercept.

ii Axis of symmetry is x = –2. iii x-intercept is –2. iv y-intercept is – 4.

Example 2 Transforming parabolas Copy and complete the table for the following graphs. a b y y = 4x2 y y = (x + 2)2

y

c

y = x2

y = x2

3

y = x2

2 x

O

−2

x

O

x

O

y = −x2 + 3 Maximum or minimum

Formula a

y = 4x 2

b

y = (x + 2)2

c

y = –x 2 + 3

Reflected in the x-axis (yes/no)

Turning point

y value when x = 1

SOL UTI ON

Formula

Wider or narrower than y = x 2

EX P L A NA TI ON y value Wider or narrower Maximum or Reflected in the Turning when x-axis (yes/no) point x = 1 than y = x 2 minimum

a y = 4x 2

minimum

no

(0, 0)

4

narrower

b y = (x + 2)2

minimum

no

(–2, 0)

9

same

c y = –x 2 + 3 maximum

yes

(0, 3)

2

same

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Read features from graphs and consider the effect of each change in equation on the graph.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Number and Algebra

1

1, 2, 3(½)

3(½)

—

Complete this table and grid to plot the graph of y = x2 . x y

–3 9

–2

–1

0

1

2

UNDERSTANDING

Exercise 7A

y

3

9 8 7 6 5 4 3 2 1 −3 −2 −1 O

1 2 3

(−2, 4)

The parabola has a (maximum or minimum). The coordinates of the turning point are . The y-intercept is . The x-intercepts are and . The axis of symmetry is .

2

−5

Example 1

3 Determine these key features of the following graphs. i turning point and whether it is a maximum or minimum ii axis of symmetry iii x-intercepts

y

a

b

−1 O

4

−3

x

x

y

2 Write the missing features for this graph. a b c d e

467

−2

c

2

1

x

iv y-intercept

y

O −1

O

x

y 6 4 O

2

x

(2, −5)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

468

Chapter 7 Parabolas and other graphs

e

4

3

O

x

x

1 3 (2, −2)

4–6

4 Copy and complete the table below for the following graphs. a b y y y = x2

y = x2

8 6 4 2

8 6 4 2

y = 3x2

−4 −2 O

x

2 4 y

d

4 3 2 1 O −3 −2 −1 −1

−2 −3 −4

Formula a b

y = 2x 2

d

y = – 4x 2 1 y = – x2 3

f

x

y=

−4x2

Maximum or minimum

O −3 −2 −1 −1

−2 −3 −4

Reflected in the x-axis (yes/no)

y = 2x2 2 4 y

f

4 3 2 1

y = x2

1 2 3

x

O −3 −2 −1 −1

−2 −3 −4

1

y = − 3 x2 Turning point

x

y value when x = 1

y = x2

1 2 3

x

y = −2x2

Wider or narrower than y = x 2

y = 3x 2 1 y = x2 2

c

e

4 3 2 1

y = x2

y = x2

−4 −2 O

y

e

1 2 3

2 4

x

y 8 6 4 2

x

3

4–6(½)

c

y = 12 x2

−4 −2 O

O

−3

4–6 Example 2a

y

6

O

−3

f

y

FLUENCY

y

d

UNDERSTANDING

7A

y = –2x 2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Example 2b

5 Copy and complete the table below for the following graphs. a b y

y = x2

8 6 4 2

y = (x + 3)2

−8 −6 −4 −2 O c

x

y = (x − 2)2

−4 −2 O

Formula

y

Turning point

Axis of symmetry

y = x2

8 6 4 2

y = (x + 4)2

x

2 4

7A

x

2 4

d

8 6 4 2

469

y = (x − 1)2

−4 −2 O

y y = x2

Example 2c

2 4

y 8 6 4 2

y = x2

FLUENCY

Number and Algebra

x

−8 −6 −4 −2 O

2 4

y-intercept (x = 0)

x-intercept

2

a

y = (x + 3)

b

y = (x – 1)2

c

y = (x – 2)2

d

y = (x + 4)2

6 Copy and complete the table for the following graphs. a b c y y

d

y

y = x2

y = x2 y = x2 − 1

y = x2 + 3

4 O

2

x

Formula

y

O −1

Turning point

y= 2

x2

y = x2 + 2

2 x

y-intercept (x = 0)

O

2

x

O −4

y = x2 x

2 y = x2 − 4

y value when x = 1

2

a

y =x +3

b

y = x2 – 1

c

y = x2 + 2

d

y = x2 – 4

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter 7 Parabolas and other graphs

7–9(½)

7A

7–9(½), 10

7–8(½), 10

PROBLEM-SOLVING

470

7 Write down the equation of the axis of symmetry for the graphs of these rules. a y = x2 b y = x2 + 7 c y = –2x2 2 2 d y = –3x e y=x –4 f y = (x – 2)2 g y = (x + 1)2 h y = –(x + 3)2 i y = –x2 – 3 1 j y = x2 + 2 k y = x2 – 16 l y = –(x + 4)2 2 8 Write down the coordinates of the turning point for the graphs of the equations in Question 7. 9 Find the y-intercept (i.e. when x = 0) for the graphs of the equations in Question 7. 10 Match each of the following equations to one of the graphs below. a y = 2x2 b y = x2 – 6 c y = (x + 2)2 1 e y = (x – 3)2 f y = x2 g y = x2 + 4 4 i ii iii y y

d y = –5x2 h y = (x – 2)2 + 3

y

9 (1, 5)

4

O iv

x

3

y

v

O

x

O

−6 vi

y

x

(1, −5)

y

7

(1, 5)

(2, 1)

viii

y

(2, 3) x

O vii

x

O

O

x

y

4 2

−2

O

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

x

O

(1, 2) x

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Number and Algebra

11–13

11–13(b), 14

11 a Using technology, plot the following pairs of graphs on the same set of axes for –5 ≤ x ≤ 5 and compare their tables of values. 1 ii y = x2 and y = x2 i y = x2 and y = 4x2 3 1 iii y = x2 and y = 6x2 iv y = x2 and y = x2 4 2 v y = x2 and y = 7x2 vi y = x2 and y = x2 5

REASONING

11–13

471

7A

b Suggest how the constant a in y = ax2 transforms the graph of y = x2 . 12 a

Using technology, plot the following sets of graphs on the same set of axes for –5 ≤ x ≤ 5 and compare the turning point of each. i

y = x2 , y = (x + 1)2 , y = (x + 2)2 , y = (x + 3)2

ii y = x2 , y = (x – 1)2 , y = (x – 2)2 , y = (x – 3)2 b Explain how the constant h in y = (x + h)2 transforms the graph of y = x2 . 13 a

Using technology, plot the following sets of graphs on the same set of axes for –5 ≤ x ≤ 5 and compare the turning point of each. i

y = x2 , y = x2 + 1, y = x2 + 2, y = x2 + 3

ii y = x2 , y = x2 – 1, y = x2 – 3, y = x2 – 5 b Explain how the constant k in y = x2 + k transforms the graph of y = x2 . 14 Write down an example of a quadratic equation whose graph has: a two x-intercepts b one x-intercept c no x-intercepts

—

—

15 Find a quadratic rule that satisfies the following information. a b c d e f g h

turning point (0, 2) and another point (1, 3) turning point (0, 2) and another point (1, 1) turning point (–1, 0) and y-intercept 1 turning point (2, 0) and y-intercept 4 turning point (0, 0) and another point (2, 8) turning point (0, 0) and another point (–1, –3) turning point (–1, 2) and y-intercept 3 turning point (4, –2) and y-intercept 0

15, 16

ENRICHMENT

Finding the rule

16 Plot a graph of the parabola x = y2 for –3 ≤ y ≤ 3 and describe its features.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

472

Chapter 7 Parabolas and other graphs

7B Sketching parabolas with transformations Previously we have explored simple transformations of the graph of y = x2 and plotted these on a number plane. We will now formalise these transformations and sketch graphs showing key features without the need to plot every point.

Let’s start: So where is the turning point? Consider the quadratic rule y = –(x – 3)2 + 7. • • • •

•

Key ideas

Discuss the effect of the negative sign in y = –x2 compared with y = x2 . Discuss the effect of –3 in y = (x – 3)2 compared with y = x2 . Discuss the effect of +7 in y = x2 + 7 compared with y = x2 . Now for y = –(x – 3)2 + 7, find: – the coordinates of the turning point – the axis of symmetry – the y-intercept. What would be the coordinates of the turning point in these quadratics? – y = (x – h)2 + k – y = –(x – h)2 + k

The gateway arch in St Louis, USA, is a parabola 192 m high.

Sketch parabolas by drawing a parabolic curve and labelling key features including: • turning point • axis of symmetry • y-intercept (substitute x = 0) For y = ax2 , a dilates the graph of y = x2 . • Turning point is (0, 0). • y-intercept and x-intercept are both 0. • Axis of symmetry is x = 0. • When a > 0, the parabola is upright. • When a < 0, the parabola is inverted. For y = (x – h)2 , h translates the graph of y = x2 horizontally. • When h > 0, the graph is translated h units to the right. • When h < 0, the graph is translated h units to the left. For y = x2 + k, k translates the graph of y = x2 vertically. • When k > 0, the graph is translated k units up. • When k < 0, the graph is translated k units down. The turning point form of a quadratic is y = a(x – h)2 + k. • The turning point is (h, k). • The axis of symmetry is x = h.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Number and Algebra

473

Example 3 Sketching with transformations Sketch graphs of the following quadratic relations, labelling the turning point and the y-intercept. a y = 3x2 b y = –x2 + 4 c y = (x – 2)2 SOL UTI ON

EXPLA NA TI ON

a

y = 3x2 is upright and narrower than y = x2 . Turning point and y-intercept are at the origin (0, 0). Substitute x = 1 to label a second point.

y

(1, 3)

x

O

y = –x2 + 4 is inverted (i.e. has a maximum) and is translated 4 units up compared with y = –x2 . Turning point is at (0, 4) and y-intercept (i.e. when x = 0) is 4.

y

b

4 (1, 3)

x

O

c

y = (x – 2)2 is upright (i.e. has a minimum) and is translated 2 units right compared with y = x2 . Thus, the turning point is at (2, 0). Substitute x = 0 for y-intercept: y = (0 – 2)2

y 4

= (–2)2 =4

O

2

x

Example 4 Using turning point form Sketch the graphs of the following, labelling the turning point and the y-intercept. a y = (x – 3)2 – 2 b y = –(x + 1)2 + 4

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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474

Chapter 7 Parabolas and other graphs

SOL UTI ON

EX P L A NA TI ON

a y = (x – 3)2 – 2

In y = a(x – h)2 + k, h = 3 and k = –2, so the vertex is (3, –2). Substitute x = 0 to find the y-intercept: y = (0 – 3)2 – 2

y

7

=9–2 =7

x

O (3, −2) b y = –(x + 1)2 + 4

The graph is inverted since a = –1. h = –1 and k = 4, so the vertex is (–1, 4). When x = 0: y = –(0 + 1)2 + 4

y (−1, 4)

= –1 + 4 =3

3

x

O

Example 5 Finding a rule from a simple graph Find a rule for this parabola with turning point (0, 1) and another point (1, 2).

y

(1, 2) 1 O

SOL UTI ON y = ax2 + 1 When x = 1, y = 2 so: 2 = a(1)2 + 1

x

EX P L A NA TI ON In y = a(x – h)2 + k, h = 0 and k = 1 so the rule is y = ax2 + 1. We need y = 2 when x = 1, so a = 1.

∴ a=1 So y = x2 + 1. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Number and Algebra

1

1–3(½)

3(½)

—

UNDERSTANDING

Exercise 7B

475

Give the coordinates of the turning point for the graphs of these rules. a y = x2 b y = –x2 c y = x2 + 3 d y = –x2 – 3 e y = –x2 + 7 f y = (x – 2)2 1 i y = – x2 g y = (x + 5)2 h y = 2x2 3

2 Substitute x = 0 to find the y-intercept of the graphs with these equations. a y = x2 + 3 b y = –x2 – 3 c y = 2x2 – 3 2 2 d y = –6x – 1 e y = (x – 1) f y = (x + 2)2 2 2 g y = –(x + 4) h y = –(x – 5) i y = –2(x + 1)2 j y = (x + 1)2 + 1 k y = (x – 3)2 – 3 l y = –(x + 7)2 – 14 3 Choose the word: left, right, up or down to suit. Compared with the graph of y = x2 , the graph of y = x2 + 3 is translated Compared with the graph of y = x2 , the graph of y = (x – 3)2 is translated Compared with the graph of y = x2 , the graph of y = (x + 1)2 is translated Compared with the graph of y = x2 , the graph of y = x2 – 6 is translated Compared with the graph of y = –x2 , the graph of y = –x2 – 2 is translated Compared with the graph of y = –x2 , the graph of y = –(x + 3)2 is translated Compared with the graph of y = –x2 , the graph of y = –(x – 2)2 is translated Compared with the graph of y = –x2 , the graph of y = –x2 + 4 is translated 4–6(½) Example 3

4–6(½)

. . . . . . . . 4–6(½)

4 Sketch graphs of the following quadratics, labelling the turning point and the y-intercept. If the turning point is on the y-axis, also label a point where x = 1 1 a y = 2x2 b y = –3x2 c y = x2 2 1 2 2 d y=– x e y=x +2 f y = x2 – 4 3 g y = –x2 + 1 h y = –x2 – 3 i y = (x + 3)2 2 2 j y = (x – 1) k y = –(x + 2) l y = –(x – 3)2

FLUENCY

a b c d e f g h

5 State the coordinates of the turning point for the graphs of these rules. a y = (x + 3)2 + 1 b y = (x + 2)2 – 4 c y = (x – 1)2 + 3 2 2 d y = (x – 4) – 2 e y = (x – 3) – 5 f y = (x – 2)2 + 2 g y = –(x – 3)2 + 3 h y = –(x – 2)2 + 6 i y = –(x + 1)2 + 4 2 2 j y = –(x – 2) – 5 k y = –(x + 1) – 1 l y = –(x – 4)2 – 10 Example 4

6 Sketch graphs of the following quadratics, labelling the turning point and the y-intercept. a y = (x + 1)2 + 1 b y = (x + 2)2 – 1 c y = (x + 3)2 + 2 d y = (x – 1)2 + 2 e y = (x – 4)2 + 1 f y = (x – 1)2 – 4 2 2 g y = –(x – 1) + 3 h y = –(x – 2) + 1 i y = –(x + 3)2 – 2 j y = –(x – 2)2 + 1 k y = –(x – 4)2 – 2 l y = –(x + 2)2 + 2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter 7 Parabolas and other graphs

7–8(½)

7B

7–8(½), 9

8(½), 9

PROBLEM-SOLVING

476

7 Write the rule for each graph when y = x2 is transformed by the following. a b c d e f g h i Example 5

reflected in the x-axis translated 2 units to the left translated 5 units down translated 4 units up translated 1 unit to the right reflected in the x-axis and translated 2 units up reflected in the x-axis and translated 3 units left translated 5 units left and 3 units down translated 6 units right and 1 unit up

8 Determine the rule for the following parabolas. a b y y

c

y 9

(1, 6)

(1, 5)

4 x

O

d

x

O

y

e

y

O

x

3

y

f

2 −2

O

(1, 1)

x

−4

(1, O

g

h

y

1 ) 2

x

y

i

y

O

−1−1 O 1

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

x

x

O

x

1 O

1

x

(1, −7)

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

PROBLEM-SOLVING

Number and Algebra

9 The path of a basketball is given by y = –(x – 5)2 + 25, where y metres is the height and x metres is the horizontal distance. a

Is the turning point a maximum or a minimum? b What are the coordinates of the turning point? c What is the y-intercept? d What is the maximum height of the ball? e What is the height of the ball at these horizontal distances? i x=3 ii x = 7 iii x = 10 10–11(½)

11(½), 12

10 Recall that y = (x – h)2 + k and y = a(x – h)2 + k both have the same turning point coordinates. State the coordinates of the turning point for the graphs of these rules. y = 2(x – 1)2 y = 3x2 – 4 y = 6(x + 4)2 – 1 y = – 4(x + 2)2 + 3

b e h k

y = 3(x + 2)2 y = 5x2 – 2 y = 2(x + 2)2 + 3 y = –2(x + 3)2 – 5

11 Describe the transformations that take y = x2 to: a y = (x – 3)2 b y = (x + 2)2 2 d y=x +7 e y = –x2 g y = (x – 5)2 + 8 h y = –(x + 3)2 12 For y = a(x – h)2 + k, write: a the coordinates of the turning point Sketching with many transformations

c f i l

y = – 4(x + 3)2 y = –2x2 + 5 y = 3(x – 5)2 + 4 y = –5(x – 3)2 – 3

c y = x2 – 3 f y = (x + 2)2 – 4 i y = –x2 + 6 b the y-intercept. —

—

13 Sketch the graph of the following, showing the turning point and y-intercept. a y = 2(x – 3)2 + 4 b y = 3(x + 2)2 + 5 c y = –2(x – 3)2 + 4 1 1 d y = –2(x + 3)2 – 4 e y = (x – 3)2 + 4 f y = – (x – 3)2 + 4 2 2 g y = 4 – x2 h y = –3 – x2 i y = 5 – 2x2 1 j y = 2 + (x – 1)2 k y = 1 – 2(x + 2)2 l y = 3 – 4(x – 2)2 2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

13

ENRICHMENT

a d g j

7B

REASONING

10(½)

477

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478

Chapter 7 Parabolas and other graphs

Using calculators to sketch parabolas 1

Sketch the graph of the family y = x2 + k, using k = {–3, –1, 0, 1, 3}.

2

Sketch a graph of y = x2 – 4x – 12 and show the x-intercepts and the turning point. Using the TI-Nspire:

Using the ClassPad:

1 In a Graphs page, type the rule in f 1(x ) using the given symbol (l) which is accessed using /=.

1 In the Graph&Table application enter the rule y 1 = x 2 + { –3, –1, 0, 1, 3} followed by EXE. to see the graph. Tap

Enter the rule f 1(x ) = x 2 – 4x –12. Change the scale using the window settings. Use b>Trace>Graph Trace and scroll along the graph to show significant points.

2 Enter the rule y 1 = x 2 – 4x – 12. Tap and set an appropriate scale. Tap Analysis, G-Solve, root to locate the x -intercepts. Tap Analysis, G-Solve, Min to locate the turning point.

2

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479

Number and Algebra

7C Sketching parabolas using factorisation A quadratic relation written in the form y = x2 + bx + c differs from that of turning point form, y = a(x – h)2 + k, and so the transformations of the graph of y = x2 to give y = x2 + bx + c are less obvious. To address this, we have a number of options. We can first try to factorise to find the x-intercepts and then use symmetry to find the turning point or, alternatively, we can complete the square and express the quadratic relation in turning point form. The second of these methods will be studied in the next section.

y

x-intercepts x

O

y-intercept

turning point

Let’s start: Why does the turning point of y = x 2 – 2x – 3 have coordinates (1, – 4)? • Factorise y = x2 – 2x – 3. • Hence, find the x-intercepts. • Discuss how symmetry can be used to locate the turning point. • Hence, confirm the coordinates of the turning point. To sketch a graph of y = x2 + bx + c: • Find the y-intercept by substituting x = 0. • Find the x-intercept(s) by substituting y = 0. Factorise where possible and use the Null Factor Law. Once the x-intercepts are known, the turning point can be found using symmetry. • The axis of symmetry (also the x-coordinate of the turning point) lies halfway between 1+5 the x-intercepts: x = = 3. 2 y x=3

O

1

5

Key ideas

x

• Substitute this x-coordinate into the rule to find the y-coordinate of the turning point.

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480

Chapter 7 Parabolas and other graphs

Example 6 Using the x-intercepts to find the turning point Sketch the graph of the quadratic y = x2 – 6x + 5 and determine the coordinates of the turning point, using symmetry. SOL UTI ON

EX P L A NA TI ON

y-intercept at x = 0: y = 5 x-intercepts at y = 0: 0 = x2 – 6x + 5

Identify key features of the graph: y-intercept (when x = 0), x-intercepts (when y = 0), then factorise and solve by applying the Null Factor Law. Recall that if a × b = 0, then a = 0 or b = 0.

0 = (x – 5)(x – 1) x – 5 = 0 or x – 1 = 0 ∴ x = 5, x = 1 x-intercepts are 1 and 5. Turning point at x = 2

1+5 = 3. 2

y = (3) – 6 × (3) + 5 = 9 – 18 + 5 = –4 Turning point is a minimum at (3, – 4).

y

Using symmetry the x-coordinate of the turning point is halfway between the x-intercepts. Substitute x = 3 into y = x2 – 6x + 5 to find the y-coordinate of the turning point. It is a minimum turning point since the coefficient of x2 is positive.

5

O 1

5

x

Label key features on the graph and join points in the shape of a parabola.

(3, −4)

Computer-controlled milling machines like this can be programmed with equations to create complex shapes.

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Number and Algebra

481

Example 7 Sketching a perfect square Sketch the graph of the quadratic y = x2 + 6x + 9. SOL UTI ON

EX P L A NA TI ON

y-intercept at x = 0: y = 9 x-intercepts at y = 0: 0 = x2 + 6x + 9

For y-intercept substitute x = 0. For x-intercepts substitute y = 0 and factorise: (x + 3)(x + 3) = (x + 3)2 . Apply the Null Factor Law to solve for x.

0 = (x + 3)2 x+3=0 ∴ x = –3 x-intercept is –3. Turning point is at (–3, 0).

As there is only one x-intercept, it is also the turning point.

y 9 Label key features on graph.

−3 O

x

Example 8 Finding a turning point from a graph The equation of this graph is of the form y = (x + a)(x + b). Use the x-intercepts to find the values of a and b, then find the coordinates of the turning point. y

−3

O

1

x

SOL UTI ON

EX P L A NA TI ON

a = 3 and b = –1. y = (x + 3)(x – 1)

Using the Null Factor Law, (x + 3)(x – 1) = 0 gives x = –3 and x = 1, so a = 3 and b = –1. Find the average of the two x-intercepts to find the x-coordinate of the turning point.

x-coordinate of the turning point is

–3 + 1 = –1. 2

y-coordinate is (–1 + 3)(–1 – 1) = 2 × (–2) = – 4. Turning point is (–1, – 4).

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Substitute x = –1 into the rule to find the y value of the turning point.

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482

Chapter 7 Parabolas and other graphs

1

1–3(½), 4

4

—

Use the Null Factor Law to find the x-intercepts (y = 0) for these factorised quadratics. a y = (x + 1)(x – 2) b y = (x – 3)(x – 4) c y = (x + 1)(x + 5) d y = x(x – 3) e y = 2x(x – 5) f y = –3x(x + 2) √ √ √ √ √ √ g y = (x + 5)(x – 5) h y = (x + 7)(x – 7) i y = (x + 2 2)(x – 2 2)

UNDERSTANDING

Exercise 7C

2 Factorise and use the Null Factor Law to find the x-intercepts for these quadratics. a y = x2 + 3x + 2 b y = x2 + 2x – 8 c y = x2 – 8x + 16 2 2 d y = x – 4x e y = x – 6x f y = 2x2 + 10x g y = x2 – 9 h y = x2 – 25 i y = x2 – 10 3 Find the y-intercept for the quadratics in Question 2. 4 A parabola has the rule y = x2 – 2x – 48. Factorise x2 – 2x – 48. Find the x-intercepts of y = x2 – 2x – 48. Hence, state the equation of the axis of symmetry. Find the coordinates of the turning point. 5–8(½) Example 6

5 Sketch the graphs of the following quadratics. a y = x2 – 6x + 8 c y = x2 + 8x + 15 e y = x2 – 2x – 8 g y = x2 + 8x + 7

b d f h

y = x2 – 8x + 12 y = x2 – 6x – 16 y = x2 – 4x – 21 y = x2 – 12x + 20

6 Sketch graphs of the following quadratics. a y = x2 – 9x + 20 c y = x2 – 13x + 12 e y = x2 + 5x + 4 g y = x2 – 4x – 12 i y = x2 – 5x – 14 k y = x2 + 7x – 30

b d f h j l

y = x2 – 5x + 6 y = x2 + 11x + 30 y = x2 + 13x + 12 y = x2 – x – 2 y = x2 + 3x – 4 y = x2 + 9x – 22

7 Sketch by first finding x-intercepts. a y = x2 + 2x b y = x2 + 6x d y = x2 – 5x e y = x2 + 3x Example 7

5–9(½)

8 Sketch graphs of the following perfect squares. a y = x2 + 4x + 4 c y = x2 – 10x + 25

5–9(½)

FLUENCY

a b c d

c y = x2 – 4x f y = x2 + 7x b y = x2 + 8x + 16 d y = x2 + 20x + 100

9 Sketch graphs of the following quadratics that include a difference of perfect squares. a y = x2 – 9 b y = x2 – 16 c y = x2 – 4

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Number and Algebra

10–13(½)

10 Determine the turning points of the following quadratics. a y = 2(x2 – 7x + 10) b y = 3(x2 – 7x + 10) 2 d y = 4x + 24x + 32 e y = 4(x2 – 49) g y = 3x2 – 6x + 3 h y = 5x2 – 10x + 5 Example 8

11, 12–13(½), 14

PROBLEM-SOLVING

10(½), 11

c y = 3x2 + 18x + 24 f y = – 4(x2 – 49) i y = 2x2 – 4x + 10

483

7C

11 The equations of these graphs are of the form y = (x + a)(x + b). Use the x-intercepts to find the values of a and b, and then find the coordinates of the turning point. a b c y y y

−5 O

1

3

O

1

x

−2 O

6

x

x

12 State the x-intercepts and turning point for these quadratics. a y = x2 – 2 b y = x2 – 11

c y = x2 – 50

13 Sketch a graph of these quadratics. a y = 9 – x2 b y = 1 – x2 d y = 3x – x2 e y = –x2 + 2x + 8

c y = 4x – x2 f y = –x2 + 8x + 9

14 If the graph of y = a(x + 2)(x – 4) passes through the point (2, 16), determine the value of a and the coordinates of the turning point for this parabola.

15, 16

15 Explain why y = (x – 3)(x – 5) and y = 2(x – 3)(x – 5) both have the same x-intercepts. 16 a Explain why y = x2 – 2x + 1 has only one x-intercept. b Explain why y = x2 + 2 has zero x-intercepts.

17, 18

REASONING

15

17 Consider the quadratics y = x2 – 2x – 8 and y = –x2 + 2x + 8. a Show that both quadratics have the same x-intercepts. b Find the coordinates of the turning points for both quadratics. c Compare the positions of the turning points. 18 A quadratic has the rule y = x2 + bx. Give: a the y-intercept b the x-intercepts c the coordinates of the turning point.

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7C

Chapter 7 Parabolas and other graphs

More rules from graphs

—

—

19

19 Determine the equation of each of these graphs in factorised form; for example, y = 2(x – 3)(x + 2). a b c y y

y

x

4

O

x

2 (1, −1)

(2, −4) d

−9 (−3, −9) y

e

y

O

-3

x

3

O

−2

y

f

x

2

-4

y

h

y

i

5

2

y −1

x

O

1

O

√5

-5

−4

-9

x

O

-√5

g

x

O

−6 O

ENRICHMENT

484

5

3

O

x

x −3

-8 j

y

y

k

y

l

(2, 4) 12

O

4

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

x

−2

O

10

6

x

−√10

O

√10

x

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Number and Algebra

485

7D Sketching by completing the square We have learnt previously that the turning point of a parabola can be read directly from a rule in the form y = a(x – h)2 + k. This form of quadratic can be obtained by completing the square.

Let’s start: I forgot how to complete the square!  2 6 ) since x2 + 6x + 9 = (x + 3)2 . So to complete 2  2  2 6 6 the square for x2 + 6x + 2 we write x2 + 6x + – + 2 = (x + 3)2 – 7. 2 2 To make x2 + 6x a perfect square we need to add 9 (from

• Discuss the rules for completing the square and explain how x2 + 6x + 2 becomes (x + 3)2 – 7. • What does the turning point form of x2 + 6x + 2 tell us about its graph? • How can you use the turning point form of x2 + 6x + 2 to help find the x-intercepts of y = x2 + 6x + 2? By completing the square, all quadratics in the form y = ax2 + bx + c can be expressed in turning point form; i.e. y = a(x – h)2 + k. To sketch a quadratic in the form y = a(x – h)2 + k, follow these steps. • Determine the coordinates of the turning point (h, k). When a is positive, the parabola has a minimum turning point. When a is negative, the parabola has a maximum turning point. • Determine the y-intercept by substituting x = 0. • Determine the x-intercepts, if any, by substituting y = 0 and factorising to solve the equation. Use the Null Factor Law to help solve the equation.

Key ideas

Example 9 Finding key features of quadratics in turning point form For y = – 4(x – 1)2 + 16: a Determine the coordinates of its turning point and state whether it is a maximum or minimum. b

Determine the y-intercept.

c

Determine the x-intercepts (if any).

SOL UTI ON

E X P L A NA TI ON

a Turning point is a maximum at (1, 16).

For y = a(x – h)2 + k the turning point is at (h, k). As a = – 4 is negative, the parabola has a maximum turning point.

b y-intercept at x = 0: y = – 4(0 – 1)2 + 16

Substitute x = 0 to find the y-intercept. Recall that (0 – 1)2 = (–1)2 = 1.

= – 4 + 16 = 12 ∴ y-intercept is 12. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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486

Chapter 7 Parabolas and other graphs

c x-intercepts at y = 0: 0 = – 4(x – 1)2 + 16

Substitute y = 0 for x-intercepts. Divide both sides by – 4. Use a2 – b2 = (a – b)(a + b) with a = x – 1 and b = 2 to write in factorised form. Simplify and apply the Null Factor Law to solve for x. Note: Check that the x-intercepts are evenly spaced either side of the turning point. Alternatively, solve (x – 1)2 – 4 = 0

0 = (x – 1)2 – 4 0 = (x – 1)2 – (2)2 0 = (x – 1 – 2)(x – 1 + 2) 0 = (x – 3)(x + 1) ∴ x – 3 = 0 or x + 1 = 0 ∴ x = 3, x = –1 x-intercepts are –1 and 3.

(x – 1)2 = 4 x – 1 = ±2 x = –1, 3

Example 10 Sketching by completing the square Sketch these graphs by completing the square, giving the x-intercepts in exact form. a y = x2 + 6x + 15 b y = x2 – 3x – 1 SOL UTI ON

EX P L A NA TI ON

a Turning point form: y = x2 + 6x + 15  2  2 6 6 2 = x + 6x + – + 15 2 2

To change the equation into turning point form, complete the square by adding and  2 6 subtracting = 9. 2

= (x + 3)2 + 6 Turning point is a minimum at (–3, 6). y-intercept at x = 0: y = (0)2 + 6(0) + 15 = 15 ∴ y-intercept is 15. x-intercepts at y = 0: 0 = (x + 3)2 + 6 There is no solution and there are no x-intercepts. y

15

Read off the turning point, which is a minimum, as a = 1 is positive. For the y-intercept, substitute x = 0 into the original equation. For the x-intercepts, substitute y = 0 into the turning point form. This cannot be expressed as a difference of perfect squares and, hence, there are no factors and no x-intercepts. Note also that the turning point was a minimum with lowest y-coordinate of 6, telling us there are no x-intercepts. Sketch the graph, showing the key points.

(−3, 6) O

x

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Number and Algebra

b Turning point form: y = x2 – 3x – 1  2  2 3 3 = x2 – 3x + – –1 2 2   3 2 13 = x– – 2 4   3 13 ,– . Turning point is a minimum at 2 4 y-intercept at x = 0: y = (0)2 – 3(0) – 1 = –1 ∴ y-intercept is –1. x-intercepts at y = 0:   3 2 13 0= x– – 2 4 2 √ 2  13 3 – 0= x– 2 2  √  √  13 13 3 3 x– + 0= x– – 2 2 2 2 √ √ 3 – 13 3 + 13 ,x= x= 2 2 y

3 − √13 2

O −1

3 + √13 2

487

Complete the square to write in turning point  2 9 9 4 13 3 9 form: = and – – 1 = – – = – . 2 4 4 4 4 4

Substitute x = 0 to find the y-intercept.

Substitute y = 0 to find the x-intercepts. Factorise using difference of perfect squares.  √ √ 13 13 13 = √ = 4 2 4 Solve for x using the Null Factor Law: √ √ 3 13 13 3 = 0 or x – + =0 x– – 2 2 2 2 The solutions can be also expressed as √ 3 ± 13 x= . 2 Label key features on graph, using exact values.

x

) 32 , − 134 )

The study of quadratic equations, started by ancient Greek mathematicians looking at the properties of circles and curves, remained principally a part of pure mathematics until around 1600, when it was realised they could be applied to the description of motion, including the orbits of the planets. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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488

Chapter 7 Parabolas and other graphs

1

1–3(½)

3(½)

—

UNDERSTANDING

Exercise 7D

Copy and complete the square, then state the coordinates of the turning point (TP). a y = x2 + 2x – 5 b y = x2 + 4x – 1 = x2 + 2x + =( TP = ( c

–

2

,

) – )

=( TP = (

y = x2 – 6x + 10 = x2 – 6x + = TP = (

= x2 + 4x +

–

,

–1

,

) – )

y = x2 – 3x – 7

d –

–

2

=

+

= TP = (

)

,

)

2 Solve these equations for x, using DOPS and giving exact answers. a x2 – 9 = 0 b x2 – 3 = 0 c (x – 1)2 – 4 = 0 2 2 d (x + 1) – 16 = 0 e (x + 4) – 2 = 0 f (x – 6)2 – 5 = 0 3 State whether the turning points of the following are a maximum or a minimum and give the coordinates. a y = 2(x – 3)2 + 5 b y = –2(x – 1)2 + 3 2 c y = – 4(x + 1) – 2 d y = 6(x + 2)2 – 5 e y = 3(x + 5)2 + 10 f y = – 4(x – 7)2 + 2 2 g y = –5(x – 3) + 8 h y = 2(x – 3)2 – 7

4–7(½) Example 9b

Example 9c

4–8(½)

4–8(½)

4 Determine the y-intercept of each of the following. a y = (x + 1)2 + 5 b y = (x + 2)2 – 6 d y = (x – 4)2 – 7 e y = –(x + 5)2 + 9 2 g y = x + 6x + 3 h y = x2 + 5x + 1 j y = x2 + x – 8 k y = x2 – 5x + 13

c f i l

y = (x – 3)2 – 2 y = –(x – 7)2 – 6 y = x2 + 7x – 5 y = x2 – 12x – 5

5 Determine the x-intercepts (if any) of the following. a y = (x – 3)2 – 4 b y = (x + 4)2 – 9 2 d y = 2(x + 2) – 10 e y = –3(x – 1)2 + 30 g y = (x – 4)2 h y = (x + 6)2 2 j y = –2(x – 3) – 4 k y = –(x – 2)2 + 5

c f i l

y = (x – 3)2 – 36 y = (x – 5)2 – 3 y = 2(x – 7)2 + 18 y = –(x – 3)2 + 10

FLUENCY

Example 9a

6 Determine the x-intercepts (if any) by first completing the square and rewriting the equation in turning point form. Give exact answers. a y = x2 + 6x + 5 b y = x2 + 6x + 2 c y = x2 + 8x – 5 2 2 d y = x + 2x – 6 e y = x – 4x + 14 f y = x2 – 12x – 5

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7 Sketch the graphs of the following. a y = (x – 2)2 – 4 b d y = (x – 3)2 – 4 e g y = (x – 2)2 + 1 h j y = –(x + 4)2 – 9 k

Label the turning point and intercepts. y = (x + 4)2 – 9 c y = (x + 4)2 – 1 2 y = (x + 8) + 16 f y = (x + 7)2 + 2 y = (x – 3)2 + 6 i y = –(x – 5)2 – 4 2 y = –(x + 9) + 25 l y = –(x – 2)2 + 4

7D

8 Sketch these graphs by completing the square. Label the turning point and intercepts. a y = x2 + 4x + 3 b y = x2 – 2x – 3 c y = x2 + 6x + 9 d y = x2 – 8x + 16 e y = x2 – 2x – 8 f y = x2 – 2x – 15 2 2 g y = x + 8x + 7 h y = x + 6x + 5 i y = x2 + 12x 9(½)

Example 10b

489

9–10(½)

9–11(½)

9 Sketch these graphs by completing the square. Label the turning point and intercepts with exact values. a y = x2 + 4x + 1 b y = x2 + 6x – 5 c y = x2 – 2x + 6 2 2 d y = x – 8x + 20 e y = x + 4x – 4 f y = x2 – 3x + 1 g y = x2 + 5x + 2 h y = x2 – x – 2 i y = x2 + 3x + 3

PROBLEM-SOLVING

Example 10a

FLUENCY

Number and Algebra

10 Complete the square and decide if the graphs of the following quadratics will have zero, one or two x-intercepts. a y = x2 – 4x + 2 b y = x2 – 4x + 4 c y = x2 + 6x + 9 d y = x2 + 2x + 6 e y = x2 – 3x + 4 f y = x2 – 5x + 5 11 Take out a common factor and complete the square to find the x-intercepts for these quadratics. a y = 2x2 + 4x – 10 b y = 3x2 – 12x + 9 c y = 2x2 – 12x – 14 d y = 4x2 + 16x – 24 e y = 5x2 + 20x – 35 f y = 2x2 – 6x + 2 12(½), 13

13, 14

12 To sketch a graph of the form y = –x2 + bx + c we can complete the square by taking out a factor of –1. Here is an example. y = –x2 – 2x + 5

REASONING

12(½)

= –(x2 + 2x – 5)  ( )2 ( )2  2 2 = –x2 + 2x + – – 5 2 2

= –((x + 1)2 – 6)

= –(x + 1)2 + 6 So the turning point is a maximum at (–1, 6). Sketch the graph of these quadratics using the technique above. a y = –x2 – 4x + 3 b y = –x2 + 2x + 2 d y = –x2 + 8x – 8 e y = –x2 – 3x – 5

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

c y = –x2 + 6x – 4 f y = –x2 – 5x + 2

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Chapter 7 Parabolas and other graphs

REASONING

7D 13 For what values of k will the graph of y = (x – h)2 + k have: a zero x-intercepts? b one x-intercept? c two x-intercepts? 14 Show that

x2

+ bx + c =



b x+ 2

2

–

b2 – 4c . 4

Completing the square with non-monics

—

—

15 This example shows how to complete the square with non-monic quadratics of the form y = ax2 + bx + c. y = 3x2 + 6x + 1   1 2 = 3 x + 2x + 3    2  2 2 2 1 2 = 3 x + 2x + – + 2 2 3   2 = 3 (x + 1)2 – 3

Key features:

15

ENRICHMENT

490

The turning point is (–1, –2). y-intercept is 1. x-intercepts:

0 = 3(x + 1)2 – 2 2 (x + 1)2 = 3  = 3(x + 1)2 – 2 2 –1 x= ± 3 Use this technique to sketch the graphs of these non-monic quadratics. a d g j

y = 4x2 + 8x + 3 y = 2x2 + x – 3 y = 6x2 + 5x + 9 y = 7x2 + 10x

b e h k

y = 3x2 – 12x + 10 y = 2x2 – 7x + 3 y = 5x2 – 3x + 7 y = –3x2 – 9x + 2

c f i l

y = 2x2 + 12x + 1 y = 4x2 – 8x + 20 y = 5x2 + 12x y = – 4x2 + 10x – 1

One of the most important applications of quadratic equations is in modelling acceleration, first formulated by Galileo in the early 1600s, and shown in action here in the present day.

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Number and Algebra

7E Sketching using the quadratic formula

491

10A

So far we have found x-intercepts for parabolas by factorising (and using the Null Factor Law) and by completing the square. An alternative method is to use the quadratic formula, which states that √ 2 –b ± b – 4ac if ax2 + bx + c = 0, then x = . 2a The discriminant ∆ = b2 – 4ac determines the number of solutions to the equation. If ∆ = 0, i.e. b2 – 4ac = 0.

If ∆ > 0, i.e. b2 – 4ac > 0.

If ∆ < 0, i.e. b2 – 4ac < 0.

The solution to the equation b becomes x = – . 2a There is one solution and one

The solution to the equation  –b ± b2 – 4ac . becomes x = 2a There are two solutions and two

Square roots exist for positive

x-intercept.

x-intercepts.

numbers only. There are no solutions or x-intercepts. y

y

y

x-intercepts O

x

O

x O

x-intercept

x

Let’s start: No working required Three students set to work to find the x-intercepts for y = x2 – 2x + 3:

• • • •

Student A finds the intercepts by factorising. Student B finds the intercepts by completing the square. Student C uses the discriminant in the quadratic formula. Try the method for student A. What do you notice? Try the method for student B. What do you notice? What is the value of the discriminant for student C? What does this tell them about the number of x-intercepts for the quadratic? What advice would student C give students A and B? To sketch the graph of y = ax2 + bx + c, find the following points. • y-intercept at x = 0 : y = a(0)2 + b(0) + c = c • x-intercepts when y = 0 : For 0 = ax2 + bx + c, use the quadratic formula: √ √ –b + b2 – 4ac –b – b2 – 4ac or x= 2a 2a √ 2 –b ± b – 4ac . Alternatively, x = 2a

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Key ideas

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492

Chapter 7 Parabolas and other graphs

Key ideas

• To determine if there are zero, one or two solutions/x-intercepts, use the discriminant ∆ = b2 – 4ac. If ∆ < 0 → no x-intercepts. If ∆ = 0 → one x-intercept. If ∆ > 0 → two x-intercepts. b • Turning point: The x-coordinate lies halfway between the x-intercepts, so x = – . The 2a y-coordinate is found by substituting the x-coordinate into the original equation. b • x = – is the axis of symmetry. 2a y

x=–b 2a x

O –b –√b2 – 4ac 2a

–b +√b2 – 4ac 2a c

Example 11 Using the discriminant and using x = –

b to find the turning point 2a

Consider the parabola given by the quadratic equation y = 3x2 – 6x + 5. a Determine the number of x-intercepts. b c

Determine the y-intercept. b Use x = – to determine the turning point. 2a

SOL UTI ON a

EX P L A NA TI ON

∆ = b2 – 4ac

Use the discriminant ∆ = b2 – 4ac to find the number of x-intercepts. In 3x2 – 6x + 5, a = 3, b = –6 and c = 5.

= (–6)2 – 4(3)(5) = –24 ∆ < 0, so there are no x-intercepts. b y-intercept is 5. c

b 2a (–6) =1 =– 2(3)

For the x-coordinate of the turning point use b x = – with a = 3 and b = –6, as above. 2a

x=–

y = 3(1)2 – 6(1) + 5 =2 ∴ Turning point is at (1, 2). Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Interpret the result. Substitute x = 0 for the y-intercept.

Substitute the x-coordinate into y = 3x2 – 6x + 5 to find the corresponding y-coordinate of the turning point.

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Number and Algebra

493

Example 12 Sketching graphs using the quadratic formula Sketch the graph of the quadratic y = 2x2 + 4x – 3, labelling all significant points. Round the x-intercepts to two decimal places. SOL UTI ON

EX P LA NA TI ON

y-intercept is –3. x-intercepts (y = 0): 2x2 + 4x – 3 = 0 √ –b ± b2 – 4ac x= 2a – 4 ± 42 – 4(2)(–3) = 2(2) √ – 4 ± 40 = 4 √ –2 – 4 ± 21 10  = 2 √4 –2 ± 10 = 2

Identify key features; i.e. x- and y-intercepts and the turning point. Substitute x = 0 for the y-intercept. Use the quadratic formula to find the x-intercepts. For y = 2x2 + 4x – 3, a = 2, b = 4 and c = –3.

√ √ √ √ √ Simplify 40 = 4 × 10 = 4 × 10 = 2 10, then cancel the common factor of 2.

x = 0.58, –2.58 (to 2 d.p.)

Use a calculator to round to two decimal places.

∴ x-intercepts are 0.58, –2.58. b Turning point is at x = – 2a (4) = –1 =– 2(2) and ∴ y = 2(–1)2 + 4(–1) – 3 = –5. ∴ Turning point is at (–1, –5).

Substitute x = –1 into y = 2x2 + 4x – 3 to find the y-coordinate of the turning point.

y Label the key features on the graph and sketch.

−2.58

O 0.58

x

−3 (−1, −5)

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494

Chapter 7 Parabolas and other graphs

1–3

3

—

UNDERSTANDING

Exercise 7E

A graph has the rule y = ax2 + bx + c. Determine the number of x-intercepts it will have if: a b2 – 4ac > 0 b b2 – 4ac < 0 c b2 – 4ac = 0 √ –b ± b2 – 4ac when: 2 Give the exact value of 2a a a = 1, b = 2, c = –1 b a = –2, b = 3, c = 5 c a = –2, b = –1, c = 2 d a = 3, b = 6, c = –2

1

3 For the following graphs, state whether the discriminant of these quadratics would be zero, positive or negative. a b c y y y

x

x

O

4–6(½)

x

O

4–7(½)

4–7(½)

Example 11a

4 Use the discriminant to determine the number of x-intercepts for the parabolas given by the following quadratics. a y = x2 + 4x + 4 b y = x2 – 3x + 5 c y = –x2 + 4x + 2 2 2 d y = 3x – 4x – 2 e y = 2x – x + 2 f y = 2x2 – 12x + 18 g y = 3x2 – 2x h y = 3x2 + 5x i y = –3x2 – 2x 2 2 j y = 3x + 5 k y = 4x – 2 l y = –5x2 + x

Example 11b

5 Determine the y-intercept for the parabolas given by the following quadratics. a y = x2 + 2x + 3 b y = x2 – 4x + 5 c y = 4x2 + 3x – 2 d y = 5x2 – 2x – 4 e y = –2x2 – 5x + 8 f y = –2x2 + 7x – 10 2 2 g y = 3x + 8x h y = – 4x – 3x i y = 5x2 – 7

Example 11c

6 Use x = –

FLUENCY

O

b to determine the coordinates of the turning points for the parabolas defined by 2a the following quadratics. a y = x2 + 2x + 4 b y = x2 + 4x – 1 c y = x2 – 4x + 3 d y = –x2 + 2x – 6 e y = –x2 – 3x + 4 f y = –x2 + 7x – 7 2 2 g y = 2x + 3x – 4 h y = 4x – 3x i y = – 4x2 – 9 j y = – 4x2 + 2x – 3 k y = –3x2 – 2x l y = –5x2 + 2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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7 Sketch the graph of these quadratics, labelling all significant points. two decimal places. a y = 2x2 + 8x – 5 b y = 3x2 + 6x – 2 c 2 2 d y = 2x – 4x – 9 e y = 2x – 8x – 11 f g y = –3x2 + 6x + 8 h y = –2x2 – 4x + 7 i j y = –2x2 – x + 12 k y = –3x2 – 2x l 8–9(½)

Round the x-intercepts to

495

7E

y = 4x2 – 2x – 3 y = 3x2 + 9x – 10 y = – 4x2 + 8x + 3 y = –5x2 – 10x – 4 8–9(½)

8–9(½), 10

8 Sketch the graphs of these quadratics. a y = 4x2 + 12x + 9 b y = 9x2 – 6x + 1 c y = – 4x2 – 20x – 25 d y = –9x2 + 30x – 25 2 2 e y = –2x + 8x – 11 f y = –3x + 12x – 16 g y = 4x2 + 4x + 3 h y = 3x2 + 4x + 2 9 Give the exact x-intercepts of the graphs of these parabolas. Simplify any surds. a y = 3x2 – 6x – 1 b y = –2x2 – 4x + 3 c y = – 4x2 + 8x + 6 d y = 2x2 + 6x – 3 e y = 2x2 – 8x + 5 f y = 5x2 – 10x – 1 10 Find a rule in the form y = ax2 + bx + c that matches this graph.

PROBLEM-SOLVING

Example 12

FLUENCY

Number and Algebra

y

−1 − √6

O

−1 + √6

x

(−1, −6)

11 Write down two rules in the form y = ax2 + bx + c that have: a two x-intercepts b one x-intercept

11, 12

12, 13

REASONING

11

c no x-intercepts

12 Explain why the quadratic formula gives only one solution when the discriminant b2 – 4ac = 0. 13 Write down the quadratic formula for monic quadratic equations (i.e. where a = 1). —

—

14, 15

b into y = ax2 + bx + c to find the general rule for the y-coordinate of the 2a turning point in terms of a, b and c. √ – b ± b2 – 4ac by solving ax2 + bx + c = 0. 15 Prove the formula x = 2a

14 Substitute x = –

ENRICHMENT

Some proof

(Hint: Divide both sides by a and complete the square.)

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496

Chapter 7 Parabolas and other graphs

7F Applications involving parabolas Quadratic equations and their graphs can be used to solve a range of practical problems. These could involve, for example, the path of a projectile or the shape of a bridge’s arch. We can relate quantities with quadratic rules and use their graphs to illustrate key features. For example, x-intercepts show where one quantity (y) is equal to zero, and the turning point is where a quantity is a maximum or minimum.

Let’s start: The civil engineer Belinda, a civil engineer, designs a model for the curved cable of a 6 m suspension bridge using the equation h = (d – 3)2 + 2, where h metres is the height of the hanging cables above the 6m road for a distance d metres from the left pillar. • What are the possible values for d? • Sketch the graph of h = (d – 3)2 + 2 for appropriate values of d. • What is the height of the pillars above the road? • What is the minimum height of the cable above the road? • Discuss how key features of the graph have helped to answer the questions above.

Key ideas

road

Applying quadratics to solve problems may involve: • defining variables • forming equations • solving equations • deciding on a suitable range of values for the variables • sketching graphs showing key features • finding the maximum or minimum turning point.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

497

Example 13 Applying quadratics in problems A piece of wire measuring 100 cm in length is bent into the shape of a rectangle. Let x cm be the width of the rectangle. a Use the perimeter to write an expression for the length of the rectangle in terms of x. b Write an equation for the area of the rectangle (A cm2 ) in terms of x. c Decide on the suitable values of x. d Sketch the graph of A versus x for suitable values of x. e Use the graph to determine the maximum area that can be formed. f

What will be the dimensions of the rectangle to achieve its maximum area?

SOL UTI ON

EX P L A NA TI ON

a 2 × length + 2x = 100 2 × length = 100 – 2x

P = 100 cm

x cm

∴ Length = 50 – x 100 cm of wire will form the perimeter. Length is half of (100 – 2 × width). b A = x(50 – x)

Area of a rectangle = length × width.

c Length and width must be positive, so we require: x > 0 and 50 – x > 0 i.e. x > 0 and 50 > x i.e. 0 < x < 50

Require each dimension to be positive, solve for x.

d

A (25, 625)

O

50

x

Sketch the graph, labelling the intercepts and turning point, which has x-coordinate halfway between the x-intercepts; i.e. x = 25. Substitute x = 25 into the area formula to find the maximum area: A = 25(50 – 25) = 625. Note open circles at x = 0 and x = 50 as these points are not included in the range of x values.

e The maximum area that can be formed is 625 cm2 .

Read from the graph. The maximum area is the y-coordinate of the turning point.

f Maximum occurs when width x = 25 cm, so Length = 50 – 25

From turning point, x = 25 gives the maximum area. Substitute to find the corresponding length. Length = 50 – x.

= 25 cm Dimensions that give maximum area are 25 cm by 25 cm, which is, in fact, a square.

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498

Chapter 7 Parabolas and other graphs

1

1–3

3

—

A ball is thrown upwards from ground level and reaches a height of h metres after t seconds, given by the formula h = 20t – 5t2 . Sketch a graph of the rule for 0 ≤ t ≤ 4 by finding the t-intercepts (x-intercepts) and the coordinates of the turning point. b What maximum height does the ball reach? c How long does it take the ball to return to ground level? a

UNDERSTANDING

Exercise 7F

2 The path of a javelin thrown by Jo is given by the formula 1 h = – (d – 10)2 + 9, where h metres is the height of the 16 javelin above the ground and d metres is the horizontal distance travelled. Sketch the graph of the rule for 0 ≤ d ≤ 22 by finding the intercepts and the coordinates of the turning point. b What is the maximum height the javelin reaches? c What horizontal distance does the javelin travel? a

1 2 x – 27, where d cm is the depth 3 of the bowl and x cm is the distance from the centre of the bowl.

3 A wood turner carves out a bowl according to the formula d =

a Sketch a graph for –9 ≤ x ≤ 9, showing x-intercepts and the turning point. b What is the width of the bowl? c What is the maximum depth of the bowl? 4–6

5–7

FLUENCY

4, 5

4 The equation for the arch of a particular bridge is 1 given by h = – (x – 100)2 + 20, where h m is the 500 height above the base of the bridge and x m is the distance from the left side. a b c d e

Determine the coordinates of the turning point of the graph. Determine the x-intercepts of the graph. Sketch the graph of the arch for appropriate values of x. What is the span of the arch? What is the maximum height of the arch?

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Richmond Bridge, Tasmania, is the oldest bridge in Australia that is still in use.

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Example 13

5 A 20 cm piece of wire is bent to form a rectangle. Let x cm be the width of the rectangle. a Use the perimeter to write an expression for the length of the rectangle in terms of x. b Write an equation for the area of the rectangle (A cm2 ) in terms of x. c Decide on suitable values of x. d Sketch the graph of A versus x for suitable values of x. e Use the graph to determine the maximum area that can be formed. f What will be the dimensions of the rectangle to achieve its maximum area? 6 A farmer has 100 m of fencing to form a rectangular paddock with a river on one side (that does not require fencing), as shown.

499

7F

river

a Use the perimeter to write an expression for the length of the paddock in terms of the width, x metres. b Write an equation for the area of the paddock (A m2 ) in terms of x. c d e f

FLUENCY

Number and Algebra

xm

Decide on suitable values of x. Sketch the graph of A versus x for suitable values of x. Use the graph to determine the maximum paddock area that can be formed. What will be the dimensions of the paddock to achieve its maximum area?

7 The sum of two positive numbers is 20 and x is the smaller number. a b c d

Write the second number in terms of x. Write a rule for the product, P, of the two numbers in terms of x. Sketch a graph of P vs x. Find the values of x when: i P=0 ii P is a maximum.

e What is the maximum value of P?

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Chapter 7 Parabolas and other graphs

8, 9

7F

8, 9

9, 10

1 (x – 20)2 , where h m is the distance below 40 the base of a bridge and x m is the distance from the left side.

8 The equation for a support span is given by h = – a b c d

Determine the coordinates of the turning point of the graph. Sketch a graph of the equation using 0 ≤ x ≤ 40. What is the width of the support span? What is the maximum height of the support span?

PROBLEM-SOLVING

500

9 Jordie throws a rock from the top of a 30 metre high cliff and its height (h metres) above the sea is given by h = 30 – 5t2 , where t is in seconds. a

Find the exact time it takes for the rock to hit the water. b Sketch a graph of h vs t for appropriate values of t. c What is the exact time it takes for the rock to fall to a height of 20 metres? 10 A bird dives into the water to catch a fish. It follows a path given by h = t2 – 8t + 7, where h is the height in metres above sea level and t is the time in seconds. a Sketch a graph of h vs t, showing intercepts and the turning point. b Find the time when the bird: i enters the water ii exits the water iii reaches a maximum depth. c What is the maximum depth to which the bird dives? d At what times is the bird at a depth of 8 metres?

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Number and Algebra

11–13

13, 14

REASONING

11

11 The height, h metres, of a flying kite is given by the rule h = t2 – 6t + 10 for t seconds. a Find the minimum height of the kite during this time. b Does the kite ever hit the ground during this time? Give reasons.

501

7F

12 The sum of two numbers is 64. Show that their product has a maximum of 1024. 13 A rectangular framed picture has a total length and width of 20 cm and 10 cm, respectively. The frame has width x cm. a Find the rule for the area (A cm2 ) of the picture inside. b What are the minimum and maximum values of x? c Sketch a graph of A vs x using suitable values of x. d Explain why there is no turning point for your graph, using suitable values of x. e Find the width of the frame if the area of the picture is 144 cm2 .

x cm 10 cm 20 cm

1 14 A dolphin jumping out of the water follows a path described by h = – (x2 – 10x + 16), where h 2 is the vertical height, in metres, and x metres is the horizontal distance travelled. a How far horizontally does the dolphin travel out of the water? b Does the dolphin ever reach a height of 5 metres above water level? Give reasons.

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7F

Chapter 7 Parabolas and other graphs

The highway and the river and the lobbed ball

—

—

15, 16

1 x(x – 100) and all units are given in metres. 10 A highway is to be built near or over the river on the line y = c.

15 The path of a river is given by the rule y =

ENRICHMENT

502

a Sketch a graph of the path of the river, showing key features. b For the highway with equation y = c, decide how many bridges will need to be built if: i c=0 ii c = –300 c

Locate the coordinates of the bridge, correct to one decimal place, if: i c = –200 ii c = –10

d Describe the situation when c = –250. 16 A tennis ball is lobbed from ground level and must cover a horizontal distance of 22 m if it is to land just inside the opposite end of the court. If the opponent is standing 4 m from the baseline and he can hit any ball less than 3 m high, what is the lowest maximum height the lob must reach to win the point?

lic path of b parabo all

3m

tennis net

4m 22 m

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Number and Algebra

503

Progress quiz 38pt 7B

1

38pt 7B

2 Find a rule for this parabola with turning point (0, 2) and another point (1, 5).

Sketch graphs of the following quadratic relations, labelling the turning point and the y-intercept. (The x-intercepts are not required.) a y = 2x2 b y = –x2 + 3 c y = (x – 3)2 d y = –(x + 2)2 – 1

y (1, 5)

2 x

O 38pt 7C

3 Sketch graphs of the following quadratics, label the x- and y-intercepts and determine the coordinates of the turning point, using symmetry. a y = x2 – 2x – 3 b y = x2 – 4x + 4

38pt 7C

4 The equation of this graph is of the form y = (x + a)(x + b). Use the x-intercepts to find the values of a and b, then find the coordinates of the turning point.

y O −2

38pt 7D

38pt 7D

5

6

4

x

For y = –2(x – 3)2 + 8 determine the: a

coordinates of the turning point and state whether it is a maximum or minimum

b

y-intercept

c

x-intercepts (if any).

Sketch these graphs by completing the square to write the equation in turning point form. Label the exact x- and y-intercepts and turning point on the graph. a

y = x2 – 4x + 3

b

y = x2 – 2x – 6

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504

Chapter 7 Parabolas and other graphs

38pt 7E

7

For the parabolas given by the following quadratic equations: i

10A

Use the discriminant to determine the number of x-intercepts.

ii

Determine the y-intercept. b iii Use x = – to determine the turning point. 2a a y = x2 – 4x + 5 b y = x2 + 6x – 7 38pt 7E 10A

38pt 7F

c y = –x2 – 8x – 16

8 Sketch the graph of the quadratic y = 2x2 – 8x + 5, labelling all significant points. Give the x-intercepts, rounded to two decimal places.

9 A farmer has 44 metres of fencing to build three sides of a rectangular animal pen with the wall of the farm shed forming the other side. The width of the pen is x metres. a

Write an equation for the area of the pen (A m2 ) in terms of x.

b

Sketch the graph of A versus x for suitable values of x.

c

Determine the maximum area that can be formed and state the dimensions.

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Number and Algebra

7G Intersection of lines and parabolas

505

10A

We have seen previously when solving a pair of linear equations simultaneously that there is, at most, one solution. Graphically, this represents the point of intersection for the two straight lines. When the lines are parallel there will be no solution; that is, there is no point of intersection. For the intersection of a parabola and a line we can have either zero, one or two points of intersection. As we have done for linear simultaneous equations, we can use the method of substitution to solve a linear equation and a non-linear equation simultaneously.

Let’s start: How many times does a line cut a parabola? • Use computer graphing software to plot a graph of y = x2 . • By plotting lines of the form x = h, determine how many points of intersection a vertical line will have with the parabola. • By plotting lines of the form y = c, determine how many points of intersection a horizontal line could have with the parabola. • By plotting straight lines of the form y = 2x + k for various values of k, determine the number of possible intersections between a line and a parabola. • State some values of k for which the line above intersects the parabola: – twice •

– never

Can you find the value of k for which the line intersects the parabola exactly once?

Key ideas

When solving a pair of simultaneous equations involving a parabola and a line, we may obtain zero, one or two solutions. Graphically, this represents zero, one or two points of intersection between the parabola and the line. • A line that intersects a curve twice is called a secant. • A line that intersects a curve in exactly one place is called a tangent. y y y

x

O

zero points of intersection

O

one point of intersection

x

O

x

two points of intersection

The method of substitution is used to solve the equations simultaneously. • Substitute one equation into the other. • Rearrange the resulting equation into the form ax2 + bx + c = 0. • Solve for x by factorising and applying the Null Factor Law or use the quadratic formula √ –b ± b2 – 4ac . x= 2a • Substitute the x values into one of the original equations to find the corresponding y value. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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506

Key ideas

Chapter 7 Parabolas and other graphs

After substituting the equations and rearranging, we arrive at an equation of the form ax2 + bx + c = 0. Hence, the discriminant, b2 – 4ac, can be used to determine the number of solutions (i.e points of intersection) of the two equations. y y y

O

x

zero solutions b2 − 4ac < 0

x

O

one solution b2 − 4ac = 0

O

x

two solutions b2 − 4ac > 0

Example 14 Finding points of intersection of a parabola and a horizontal line Find any points of intersection of these parabolas and lines. a y = x2 – 3x b y = x2 + 2x + 4 y=4

y = –2

SOL UTI ON

EX P L A NA TI ON

a By substitution: x2 – 3x = 4

x – 4 = 0 or x + 1 = 0

Substitute y = 4 from the second equation into the first equation. Write in the form ax2 + bx + c = 0 by subtracting 4 from both sides. Factorise and apply the Null Factor Law to solve for x.

x = 4 or x = –1 ∴ The points of intersection are at (4, 4) and (–1, 4).

As the points are on the line y = 4, the y-coordinate of the points of intersection is 4.

x2 – 3x – 4 = 0 (x – 4)(x + 1) = 0

b By substitution: x2 + 2x + 4 = –2

Substitute y = –2 into the first equation.

x2 + 2x + 6 = 0 √ –b ± b2 – 4ac x= 2a –2 ± (2)2 – 4(1)(6) x= 2(1) √ –2 ± –20 = 2 ∴ There are no points of intersection.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Apply the quadratic formula to solve x2 + 2x + 6 = 0, where a = 1, b = 2 and c = 6. √

–20 has no real solutions.

The parabola y = x2 + 2x + 4 and the line y = –2 do not intersect.

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Number and Algebra

507

Example 15 Solving simultaneous equations involving a line and a parabola Solve the following equations simultaneously. a y = x2 b y = – 4x2 – x + 6 y = 2x

c

y = 3x + 7

y = x2 + 1 2x – 3y = – 4

SO L U T I O N

EX P L A N A T I O N

a By substitution: x2 = 2x

Substitute y = 2x into y = x2 .

x2 – 2x = 0

Rearrange the equation so that it is equal to 0.

x(x – 2) = 0

Factorise by removing the common factor x. Apply the Null Factor Law to solve for x.

x = 0 or x – 2 = 0 x = 0 or x = 2 When x = 0, y = 2 × (0) = 0. When x = 2, y = 2 × (2) = 4.

Substitute the x values into y = 2x to obtain the corresponding y value. Alternatively, the equation y = x2 can be used to find the y values or it can be used to check the y values.

∴ The solutions are x = 0, y = 0 and x = 2, y = 4.

The points (0, 0) and (2, 4) lie on both the line y = 2x and the parabola y = x2 .

b By substitution: – 4x2 – x + 6 = 3x + 7 –x + 6 = 4x2 + 3x + 7 6 = 4x2 + 4x + 7 0 = 4x2 + 4x + 1 ∴ (2x + 1)(2x + 1) = 0

Substitute y = 3x + 7 into the first equation. When rearranging the equation equal to 0, gather the terms on the side that makes the coefficient of x2 positive, as this will make the factorising easier. Hence, add 4x2 to both sides, then add x to both sides and subtract 6 from both sides. Factorise and solve for x.

2x + 1 = 0 1 x=– 2 ( ) 1 1 When x = – , y = 3 × – +7 2 2 11 1 = or 5 2 2 1 11 ∴ The only solution is x = – , y = . 2 2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Substitute the x value into y = 3x + 7 (or y = – 4x2 – x + 6 but y = 3x + 7 is a simpler equation). Finding only one solution indicates that this line is a tangent to the parabola.

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508

Chapter 7 Parabolas and other graphs

Replace y in 2x – 3y = – 4 with x2 + 1, making sure you include brackets.

c By substitution: 2x – 3(x2 + 1) = – 4 2x – 3x2 – 3 = – 4 2

2x – 3 = 3x – 4

Expand the brackets and then rearrange into the form ax2 + bx + c = 0.

2x = 3x2 – 1 ∴ 3x2 – 2x – 1 = 0 Factorise and solve for x.

(3x + 1)(x – 1) = 0 3x + 1 = 0 or x – 1 = 0 1 or x = 1 3 ( )2 1 1 When x = – , y = – +1 3 3 x=–

Substitute the x values into one of the two original equations to solve for y.

1 = +1 9 =

10 9

When x = 1, y = (1)2 + 1 =2

The line and parabola intersect in two places.

1 10 ∴ The solutions are x = – , y = 3 9 and x = 1, y = 2.

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Number and Algebra

509

Example 16 Solving simultaneous equations with the quadratic formula Solve the equations y = x2 + 5x – 5 and y = 2x simultaneously. Round your values to two decimal places. SO L U T I O N

EX P LA N A T I O N

By substitution: x2 + 5x – 5 = 2x

Rearrange into standard form. x2 + 3x – 5 does not factorise with whole numbers. 2 Quadratic √formula: If ax + bx + c = 0, then 2 –b ± b – 4ac . x= 2a Here, a = 1, b = 3 and c = –5.

x2 + 3x – 5 = 0 Using the√quadratic formula: –3 ± (3)2 – 4(1)(–5) x= 2(1) √ –3 ± 9 + 20 = √2 –3 ± 29 = 2 = 1.19258 . . . or – 4.19258 . . . ( √ ) –3 ± 29 In exact form, y = 2x = 2 × 2 √ = –3 ± 29 ∴ The solutions are x = 1.19, y = 2.39 and x = – 4.19, y = –8.39 (to 2 d.p.).

√ –3 – 29 Use a calculator to evaluate and 2 √ –3 + 29 . Recall that if the number under the 2 square root is negative, then there will be no real solutions. Substitute exact x values into y = 2x. Round your values to two decimal places, as required.

Example 17 Determining the number of solutions of simultaneous equations Determine the number of solutions (points of intersection) of the following pairs of equations. a y = x2 + 3x – 1 b y = 2x2 – 3x + 8 y=x–2

y = 5 – 2x

SO L U T I O N

EX P LA N A T I O N

a By substitution: x2 + 3x – 1 = x – 2 x2 + 2x + 1 = 0 Using the discriminant: b2 – 4ac = (2)2 – 4(1)(1)

Once the equation is in the form ax2 + bx + c = 0, the discriminant b2 – 4ac can be used to determine the number of solutions. Here, a = 1, b = 2 and c = 1.

=4–4 =0 ∴ There is one solution to the pair of equations.

Recall: b2 – 4ac > 0 means two solutions. b2 – 4ac = 0 means one solution. b2 – 4ac < 0 means no solutions.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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510

Chapter 7 Parabolas and other graphs

b By substitution: 2x2 – 3x + 8 = 5 – 2x

Substitute and rearrange into the form ax2 + bx + c = 0.

2x2 – x + 3 = 0 Using the discriminant: b2 – 4ac = (–1)2 – 4(2)(3)

Calculate the discriminant. Here, a = 2, b = –1 and c = 3.

= 1 – 24 b2 – 4ac < 0 means no solutions.

Exercise 7G 1

1–5

4, 5

—

UNDERSTANDING

= –23 < 0 ∴ There is no solution to the pair of equations.

y

By considering the graph shown: a

How many points of intersection would a vertical line have with this parabola? b How many possible points of intersection would a horizontal line have with this parabola?

x

O

2 a

Find the coordinates of the point where the vertical line x = 2 intersects the parabola y = 2x2 + 5x – 6. b Find the coordinates of the point where the vertical line x = –1 intersects the parabola y = x2 + 3x – 1.

y

3 The graph of a parabola is shown. a Add the line y = x + 2 to this graph. b Use the grid to write down the coordinates of the points of intersection of the parabola and the line. c The parabola has equation y = x2 – x – 1. Complete the following steps to verify algebraically the coordinates of the points of intersection. x2 – x – 1 = x2 –

–1=2

x2 – 2x – (x –

6 5 4 3 2 1 −2 −1−1O

1 2 3

x

=0

)(x + 1) = 0

= 0 or x + 1 = 0 or x = x= When x = ,y= +2= When x = ,y= +2=

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

. .

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UNDERSTANDING

Number and Algebra

4 Rearrange the following into the form ax2 + bx + c = 0, where a > 0. a x2 + 5x = 2x – 6 b x2 – 3x + 4 = 2x + 1 c x2 + x – 7 = –2x + 5 5 What do we know about the discriminant, b2 – 4ac, of the resulting equation from solving equations simultaneously which correspond to these graphs. a b c y y y

x

O

O

x

6–9(½) Example 14

y = 2x2 + 7x + 1 y = –2

e

6–10(½)

y = 4x2 – 12x + 9 y=0

f

y = 3x2 + 2x + 9 y=5

7 Solve these simultaneous equations using substitution. a

y = x2 y = 3x

b

y = x2 y = –2x

c

y = x2 y = 3x + 18

d

y = x2 – 2x + 5

e

y = –x2 – 11x + 4 y = –3x + 16

f

y = x2 + 3x – 1 y = 4x + 5

y = x2 – 2x – 4 y = –2x – 5

h

y = –x2 + 3x – 5

i

y = 2x2 + 4x + 10

y = 3x2 – 2x – 20 y = 2x – 5

k

y=x+5 g j

Example 15c

x

6 Find the points of intersection of these parabolas and horizontal lines. a y = x2 + x b y = x2 – 4x c y = x2 + 3x + 6 y=6 y = 12 y=1 d

Example 15a,b

6–10(½)

7G

FLUENCY

O

511

y = 3x – 1 y = –x2 – 4x + 3 y = 2x + 12

8 Solve these simultaneous equations by first substituting. a y = x2 b y = x2 2x + y = 8 d y = x2 + 3 5x + 2y = 4

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

y = 1 – 7x l

c y = x2

x – y = –2 e y = x2 + 2x 2x – 3y = –4

y = x2 + x + 2 y=1–x

2x + 3y = 1 f

y = –x2 + 9 6x – y = 7

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512

Chapter 7 Parabolas and other graphs

Example 16

9 Solve the following simultaneous equations, making use of the quadratic formula. a

Give your answers to one decimal place where necessary. i y = 2x2 + 3x + 6 ii y = x2 + 1 y=x+4

y = 2x + 3

2

iv y = 2x2 + 4x + 5 y = 3 – 2x

iii y = –2x + x + 3 y = 3x + 2 b Give your answers in exact surd form. i y = x2 + 2x – 5

ii

y=x

y = x2 – x + 1 y = 2x

iii y = –x2 – 3x + 3 y = –2x Example 17

FLUENCY

7G

iv y = x2 + 3x – 3 y = 2x + 1

10 Determine the number of solutions to the following simultaneous equations. a y = x2 + 2x – 3 b y = 2x2 + x c y = 3x2 – 7x + 3 y=x+4 d

2

y = x + 5x + 1 y = 2x – 3

y = 3x – 1 e

y = 1 – 2x

2

f

y = –x y = 2x + 1

y = 3x – 1

11 Ben, a member of an indoor cricket team, playing a match in a gymnasium, hits a ball that follows a path given by y = –0.1x2 + 2x + 1, where y is the height above ground, in metres, and x is the horizontal distance travelled by the ball, in metres. The ceiling of the gymnasium is 10.6 metres high. Will this ball hit the roof? Explain.

c

y = (x – 2)2 + 7 y=9–x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

11, 12(a, c)

12, 13

PROBLEM-SOLVING

11, 12(a)

12 Solve the following equations simultaneously. a y = x2 + 2x – 1 x–3 y= 2

y = –x2 + 2x

b

d

y = x(x – 4) 1 y= x–5 2 8 – x2 y= 2 y = 2(x – 1)

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13 A train track is to be constructed over a section of a lake. On a map, the edge of the lake that the train track will pass over is modelled by the equation y = 6 – 2x2 . The segment of train track is modelled by the equation y = x + 5. The section of track to be constructed will start and end at the points at which this track meets the lake.

PROBLEM-SOLVING

Number and Algebra

513

7G

a Determine the location (i.e. coordinates) of the points on the map where the framework for the track will start and end. b If 1 unit represents 100 metres, determine the length of track that must be built over the lake, correct to the nearest metre.

14, 15

15, 16

14 Consider the parabola with equation y = x2 – 6x + 5. a Use any suitable method to determine the coordinates of the turning point of this parabola. b Hence, state for which values of c the line y = c will intersect the parabola: i twice ii once iii not at all

REASONING

14

15 Consider the parabola with equation y = x2 and the family of lines y = x + k. a

Determine the discriminant, in terms of k, obtained when solving these equations simultaneously. b Hence, determine for which values of k the line will intersect the parabola: i twice ii once iii not at all 16 a Use the discriminant to show that the line y = 2x + 1 does not intersect the parabola y = x2 + 3. b Determine for which values of k the line y = 2x + k does intersect the parabola y = x2 + 3.

—

—

17

17 The line y = mx is a tangent to the parabola y = x2 – 2x + 4 (i.e the line touches the parabola at just one point). a Find the possible values of m. b Can you explain why there are two possible values of m? (Hint: A diagram may help.) c If the value of m is changed so that the line now intersects the parabola in two places, what is the set of possible values for m?

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

ENRICHMENT

Multiple tangents?

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514

Chapter 7 Parabolas and other graphs

7H Graphs of circles We know the circle as a common shape in geometry, but we can also describe a circle using an equation and as a graph on the Cartesian plane.

Let’s start: Plotting the circle A graph has the rule x2 + y2 = 9. • When x = 0, what are the two values of y? • When x = 1, what are the two values of y? • When x = 4, are there any values of y? Discuss. • Complete this table of values. x y

• • •

Key ideas

–3

–2 √ ± 5

–1

0

1

2

3

Now plot all your points on a number plane and join them to form a smooth curve. What shape have you drawn and what are its features? How does the radius of your circle relate to the equation? The Cartesian equation of a circle with centre (0, 0) and radius r is given by x2 + y2 = r2 .

y

Making x or y the subject: √ • y = ± r2 – x2  • x = ± r2 – y2

To find the intersection points of a circle and a line, use the method of substitution.

r P(x, y) r -r

(0, 0) O

x

y r

x

-r Using Pythagoras’ theorem, a 2 + b 2 = c 2 gives x 2 + y 2 = r 2 .

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

515

Example 18 Sketching a circle For the equation x2 + y2 = 4, complete the following. a State the coordinates of the centre. b State the radius. c

Find the values of y when x = 1.

e

Sketch a graph showing intercepts.

d Find the values of x when y =

1 . 2

SOL UTI ON

EX P L A NA TI ON

a (0, 0)

(0, 0) is the centre for all circles x2 + y2 = r2 .

b r=2

x2 + y2 = r2 , so r2 = 4.

c

x2 + y2 = 4 12 + y2 = 4

Substitute x = 1 and solve for y.

2

y =3

√ √ Recall that ( 3)2 and (– 3)2 both equal 3.

√ y= ± 3

d

x2 +

 2 1 =4 2 1 x2 + = 4 4 15 x2 = 4

1 . 2 1 16 1 15 – = 4– = 4 4 4 4

Substitute y =



15 x= ± 4 √ 15 x= ± 2

e

y

Draw a circle with centre (0, 0) and radius 2. Label intercepts.

2

−2

O

√ √ 15 15 15 = √ = 4 2 4

2

x

−2

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516

Chapter 7 Parabolas and other graphs

Example 19 Intersecting circles and lines Find the coordinates of the points where x2 + y2 = 4 intersects y = 2x. Sketch a graph showing the exact intersection points. SOL UTI ON

E X P L A NA TI ON

x2 + y2 = 4 and y = 2x

Substitute y = 2x into x2 + y2 = 4 and solve for x.

x2 + (2x)2 = 4

Recall that (2x)2 = 22 x2 = 4x2 .

x2 + 4x2 = 4 5x2 = 4 4 x2 = 5 x= ±



4 5

2 x = ±√ 5

If y = 2x: 2 4 2 x = √ gives y = 2 × √ = √ 5 5 5   2 2 4 x = –√ gives y = 2 × –√ = –√ 5 5 5

y 2

)−√52 , −√54 )

O

For x2 + y2 = 4, r = 2. Mark the intersection points and sketch y = 2x.

)√52 , √54 ) 2

x

−2

Exercise 7H 1

Substitute both values of x into y = 2x to find the y-coordinate.

1, 2, 3(½)

3(½)

Draw a circle on the Cartesian plane with centre (0, 0) and radius 2.

2 Solve these equations for the unknown variable. There are two solutions for each. a x 2 + 22 = 9 b x2 + 32 = 25 c x2 + 7 = 10 d 52 + y2 = 36 e 82 + y2 = 121 f 3 + y2 = 7

—

UNDERSTANDING

−2

√ 2 4 4 =√ =√ 5 5 5

3 A circle has equation x2 + y2 = r2 . Complete these sentences. a The centre of the circle is b The radius of the circle is

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Number and Algebra

Example 18

4(½), 5, 6–8(½), 9

5, 6–9(½)

FLUENCY

4(½), 5, 6–8(½)

4 A circle has equation x2 + y2 = 9. Complete the following. a State the coordinates of the centre. b State the radius. c Find the values of y when x = 2. 3 d Find the values of x when y = . 2 e Sketch a graph showing intercepts.

517

7H

5 For the equation x2 + y2 = 25 complete the following. a State the coordinates of the centre. b State the radius. 9 c Find the values of y when x = . 2 d Find the values of x when y = 4. e Sketch a graph showing intercepts. 6 Give the radius of the circles with these equations. a x2 + y2 = 36 b x2 + y2 = 81 d x2 + y2 = 5 e x2 + y2 = 14

c x2 + y2 = 144 f x2 + y2 = 20

7 Write the equation of a circle with centre (0, 0) and the given radius. a 2 b 7 c 100 √ √ e 6 f 10 g 1.1

d 51 h 0.5

8 For the circle with equation x2 + y2 = 4 find the exact coordinates where: a x=1 d y=–

1 2

1 2 y=0

b x = –1

c x=

e y = –2

f

9 Without showing any working steps, write down the x- and y-intercepts of these circles. a x2 + y2 = 1 b x2 + y2 = 16 c x2 + y2 = 3 d x2 + y2 = 11

10 Write down the radius of these circles. a x2 + y2 – 8 = 0 b x2 – 4 = –y2 d 10 – y2 – x2 = 0 e 3 + x2 + y2 = 15 Example 19

10(½), 11–14

12–15

c y2 = 9 – x2 f 17 – y2 = x2 – 3

11 Find the coordinates of the points where x2 + y2 = 9 intersects y = x. Sketch a graph showing the intersection points.

PROBLEM-SOLVING

10(½), 11

12 Find the coordinates of the points where x2 + y2 = 10 intersects y = 3x. Sketch a graph showing the intersection points. 1 13 Find the coordinates of the points where x2 + y2 = 6 intersects y = – x. Sketch a graph showing 2 the intersection points.

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Chapter 7 Parabolas and other graphs

7H 14 Determine the exact length of the chord formed by the intersection of y = x – 1 and x2 + y2 = 5. Sketch a graph showing the intersection points and the chord. 10A

15 For the circle x2 + y2 = 4 and the line y = mx + 4, determine the exact values of the gradient, m, so that the line:

PROBLEM-SOLVING

518

a is a tangent to the circle b intersects the circle in two places c does not intersect the circle. 16

16 Match equations a–f with graphs A–F. a x2 + y2 = 7 b y=x–1 2 d y=x –1 e y = –x(x + 2)

y

A

B

17, 18

REASONING

16, 17

c y = –2x + 2 f x2 + y2 = 9 C

y

y

3 O −1

x

1

O

−3

3

x

x

−1−1 O 1

−3 D

y

E

y

y

F

√7 2 −2 −√7

O

√7

x

O

1

O

x

x

−√7 √ 17 a Write x2 + y2 = 16 in the form y = ± r2 – x2 . b Write x2 + y2 = 3 in the form x = ± r2 – y2 .

18 a Explain why the graphs of y = 3 and x2 + y2 = 4 do not intersect. b Explain why the graphs of x = –2 and x2 + y2 = 1 do not intersect.

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Number and Algebra

—

—

19

ENRICHMENT

Half circles

√ 19 When we write x2 + y2 = 9 in the form y = ± 9 – x2 , we define two circle halves. y y

3 O

−3 −3

O

−3 y = −√9 − x2

b Write the rules for these half circles. i ii y

√ iii y = – 1 – x2  vi x = – 36 – y2  ix x = 12 – y2

y

O

−4 O

y

iii

5

iv

x

3

Sketch √ the graphs of these half circles. √ i y = 4 – x2 ii y = 25 – x2 √ √ iv y = – 10 – x2 v y = 16 – x2   vii x = – 7 – y2 viii x = 5 – y2

−5

7H

x

3

y = √9 − x2 a

519

4

2

x

O

x

5

−4 v

y

2

x

−2 y

vi

y √3

1 O

−1

x

−√3

x

O

√3

y

y

viii

y

ix

2√2

√10

−3√2 O

x

√5

−√5

−1 vii

O

−√5

√10

−√10

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

x

O

−2√2

O

3√2

x

x −3√2

−2√2

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520

Chapter 7 Parabolas and other graphs

Using calculators to graph circles and other graphs Sketch a graph of x2 + y2 = r2 using r = {1, 3}. 1 1 2 Sketch a graph of y = and y = – + 3 on the same set of axes. x x 1

Using the TI-Nspire:

Using the ClassPad:

1 In a Graphs page, use b >Graph Entry/Edit> Equation>Circle>centre form and enter (x – 0)2 + (y – 0)2 = 32 for r = 3.

1 In the Graph&Table application enter the rule y 1 = p({1, 3}2 – x 2 ) followed by EXE. Enter the rule y 2 = –p({1, 3}2 – x 2 ) followed by EXE. Tap to see the graph. Select Zoom, Square.

2

2 Enter the rules as y 1 and y 2. Change the scale by tapping . Use the Analysis, G-Solve menu to show significant points. Arrow up or down to toggle between graphs.

Enter the rules as f 1(x ) and f 2(x ). Change the scale using the window settings. Arrow up or down to toggle between graphs.

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Number and Algebra

521

7I Graphs of exponentials We saw in Chapter 3 that indices can be used to describe some special relations. The population of the world, for example, or the balance of an investment account can be described using exponential rules that include indices. The rule A = 100 000(1.05)t describes the account balance of $100 000 invested at 5% p.a. compound interest for t years.

Let’s start: What do y = 2x , y = –2x and y = 2–x all have in common? Complete this table and graph all three relations on the same set of axes before discussing the points below. y x y1 = 2 x

–3 1 8

–2

–1

0 1

1

2

8 6 4 2

3

4

y2 = –2x y3 = 2–x

-4 -3 -2 -1-2O -4 -6 -8

• Discuss the shape of each graph. • Where does each graph cut the y-axis? • Do the graphs have x-intercepts? Why not? • What is the one feature they all have in common?

1 2 3 4

y = 2x , y = (0.4)x , y = 3 × (1.1)x are examples of exponential relations. An asymptote is a line that a curve approaches, by getting closer and closer to it, but never reaching.

x

Key ideas

y y = 2x (1, 2) 1 O

x

asymptote ( y = 0)

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522

Key ideas

Chapter 7 Parabolas and other graphs

A simple exponential rule is of the form y = ax , where a > 0 and a π 1. • y-intercept is 1. • y = 0 is the equation of the asymptote. The graph of y = – ax is the reflection of the graph of y = ax in the x-axis. Note: y = – ax means y = –1 × ax . The graph of y = a–x is the reflection of the graph of y = ax in the y-axis. To find the intersection points of a simple exponential and a horizontal line, use the method of substitution and equate powers after expressing both sides of the equation using the same base. For example, for y = 2x and y = 16, solve 2x = 16

y y = 2-x

y = 2x 1 O -1

x y = -2x

2x = 24 ∴x=4

Example 20 Sketching graphs of exponentials Sketch the graph of the following on the same set of axes, labelling the y-intercept and the point where x = 1. a y = 2x b y = 3x c y = 4x SOL UTI ON

EX P L A NA TI ON

y = 4x y 4 3 2 1 −2 −1 O

a0 = 1, so all y-intercepts are at 1.

y = 3x y = 2x (1, 4)

y = 4x is steeper than y = 3x , which is steeper than y = 2x .

(1, 3) (1, 2)

Substitute x = 1 into each rule to obtain a second point to indicate the steepness of each curve.

1 2

x

Example 21 Sketching with reflections Sketch the graphs of these exponentials on the same set of axes. a y = 3x b y = –3x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

c

y = 3–x

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Number and Algebra

SOL UTI ON

EX P L A NA TI ON The graph of y = –3x is a reflection of the graph of y = 3x in the x-axis.

y y = 3x (-1, 3)

(1, 3) 1 O -1

523

Check: x = 1, y = –31 = –3 The graph of y = 3–x is a reflection of the graph of y = 3x in the y-axis. 1 Check: x = 1, y = 3–1 = 3 x = –1, y = 31 = 3

y = 3–x x y = -3x (1, -3)

Example 22 Solving exponential equations Find the intersection of the graphs of y = 2x and y = 8. SOL UTI ON

EX P L A NA TI ON

y = 2x

Set y = 8 and write 8 with base 2.

x

8=2

Since the bases are the same, equate the powers.

x

2 =2

y

x=3 ∴ Intersection point is (3, 8).

8

1 O

Exercise 7I 1

x

?

1–4

3, 4

—

UNDERSTANDING

3

Consider the exponential rule y = 2x . a Complete this table.

x y

–2 –1 1 2

0

1

2

1

b Plot the points in the table to form the graph of y = 2x .

y 4 3 2 1 −2 −1 O

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

1 2

x

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Chapter 7 Parabolas and other graphs

7I 2 Consider the exponential rule y = 3x . a Complete this table. x

–2 –1 1 3

y

0

1

2

b Plot the points in the table to form the graph of y = 3x .

y

1

9

-2 -1 O

–2 –1

0

1

2

x

y

x

y1 = 2

4 2

y2 = –2x y3 = 2

1 2

b Plot the graphs of y = 2x , y = –2x and y = 2–x on the same set of axes.

3 a Complete the table. x

UNDERSTANDING

524

–x

−2 −1−2O −4

1 2

x

Explain the difference between a–2 and –a2 . 1 b True or false: –32 = 2 ? Explain why. 3 1 1 1 c Write with negative indices: 3 , 2 , . 5 3 2 d Simplify: –32 , –53 , –2–2 .

4 a

5–7(a, c), 8

5–7(b), 8(½)

Example 20

5 Sketch the graph of the following on the same set of axes, labelling the y-intercept and the point where x = 1. a y = 2x b y = 4x c y = 5x

Example 21a,b

6 Sketch the graph of the following on the same set of axes, labelling the y-intercept and the point where x = 1. a y = –2x b y = –5x c y = –3x

Example 21c

7 Sketch the graph of the following on the same set of axes, labelling the y-intercept and the point where x = 1. a y = 2–x b y = 3–x c y = 6–x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

FLUENCY

5–7(a, b)

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Find the coordinates on the graph of y = 3x , where: i x=0 ii x = –1 iii y = 1

b Find the coordinates on the graph of y = i x=4 ii x = –1

–2x ,

c

4–x ,

Find the coordinates on the graph of y = i

x=1

ii x = –3

where: iii y = –1

9 a b c d

7I

iv y = –4

where: iii y = 1

9, 10 Example 22

iv y = 9

525

iv y =

9–11

1 4 9(½), 10–12

Find the intersection of the graphs of y = 2x and y = 4. Find the intersection of the graphs of y = 3x and y = 9. Find the intersection of the graphs of y = – 4x and y = – 4. Find the intersection of the graphs of y = 2–x and y = 8.

10 A study shows that the population of a town is modelled by the rule P = 2t , where t is in years and P is in thousands of people.

PROBLEM-SOLVING

8 a

FLUENCY

Number and Algebra

a State the number of people in the town at the start of the study. b State the number of people in the town after: i 1 year ii 3 years c

When is the town’s population expected to reach: i 4000 people? ii 16 000 people?

11 A single bacterium divides into two every second, so one cell becomes 2 in the first second and in the next second two cells become 4 and so on. a Write a rule for the number of bacteria, N, after t seconds. b How many bacteria will there be after 10 seconds? c How long does it take for the population to exceed 10 000? Round to the nearest second. 12 Use trial and error to find x when 2x = 5. Give the answer correct to three decimal places.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter 7 Parabolas and other graphs

13

7I

13, 14

13 Match equations a–f with graphs A–F. a y = –x – 2 b y = 3x x d y = –2 e y = (x + 2)(x – 3)

y

A

c y = 4 – x2 f x2 + y2 = 1

y

B

14–16

REASONING

526

y

C

1 (1, 3) −1

1 x

O

1

x

x

−2

y

E

O

−2

−1

y

D

O

F

y

4 x

O −1 (1, −2) −2

O

2

O

−2

3

x

x

14 Explain why the point (2, 5) does not lie on the curve with equation y = 2x . 15 Describe and draw the graph of the line with equation y = ax when a = 1. 16 Explain why 2x = 0 is never true for any value of x.  x 1 2

—

—

17

 x 1 . 2 a Using –3 ≤ x ≤ 3, sketch graphs of the rules on the same set of axes. What do you notice? b Write the following rules in the form y = ax , where 0 < a < 1. i y = 3–x ii y = 5–x iii y = 10–x

17 Consider the exponential rules y = 2–x and y =

Write the following rules in the form y = a–x , where a > 1.  x  x 1 1 i y= ii y = 4 7  x 1 d Prove that = a–x , for a > 0. a c

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

iii y =



1 11

ENRICHMENT

y = 2–x and y =

x

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527

Number and Algebra

7J Graphs of hyperbolas 1 A simple rectangular hyperbola is the graph of the equation y = . These types of equations are x common in many mathematical and practical situations. When two stones are thrown into a pond, the resulting concentric ripples intersect at a set of points that together form the graph of a hyperbola. In a similar way, when signals are received from two different satellites, a ship’s navigator can map the hyperbolic shape of the intersecting signals and help to determine the ship’s position.

Let’s start: How many asymptotes? 1 Consider the rule for the simple hyperbola y = . First, complete the table and graph, and then discuss the x points below. x

–4

–2

–1

–

1 2

–

1 4

1 4

1 2

1

2

4

y

• •

Discuss the shape of the graph of the hyperbola. What would the values of y approach as x increases to infinity or negative infinity? • What would the values of y approach as x decreases to zero from the left or from the right? 1 • What are the equations of the asymptotes for y = ? x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

y 4 3 2 1 −4 −3 −2 −1−1O −2 −3 −4

1 2 3 4

x

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528

Chapter 7 Parabolas and other graphs

Key ideas

A hyperbola has two asymptotes. Recall that an asymptote is a straight line that a curve approaches more and more closely but never quite reaches.

y asymptotes x=0 y=0

y=

4 3 2 1

(1, 1)

−4 −3 −2 −1−1O (−1, −1) −2 −3 −4 A rectangular hyperbola is the graph of the a rule y = , a π 0. x 1 • y = is the basic rectangular hyperbola. x • x = 0 (y-axis) and y = 0 (x-axis) are its asymptotes. • For a > 1 the hyperbola will be further out from the asymptotes. • For 0 < a < 1 the hyperbola will be closer in to the asymptotes. a a The graph of y = – is a reflection of the graph of y = x x in the x- (or y-) axis.

1 x

1 2 3 4

x

y 2 x 1 (−1,− 3 )

y=

−4 −3 −2 −1

4 3 2 1 O

(−1, −1) (−1, −2)

(1, 2) (1, 1) 1 2 3 4 −1 1 −2 (1, 3) 1 y= −3 3x −4

x

To find the intersection points of a hyperbola and a line, use the method of substitution.

Example 23 Sketching a hyperbola Sketch the graphs of each hyperbola, labelling the points where x = 1 and x = –1. 2 3 1 a y= b y= c y=– x x x SOL UTI ON

EX P L A NA TI ON

y

a

(1,1) O

x

Draw the basic shape of a rectangular 1 hyperbola y = . x Substitute x = 1 and x = –1 to find the two points.

(−1, −1)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

2 = 2. 1 2 = –2. For x = –1, y = –1

y

b

529

For x = 1, y =

(1, 2) x

O (−1, −2)

3 3 y = – is a reflection of y = in either the x x x- or y-axis. 3 When x = 1, y = – = –3. 1 3 = 3. When x = –1, y = – –1

y

c

(−1, 3)

x

O (1, −3)

Example 24 Intersecting with hyperbolas Find the coordinates of the points where y = a

y=3

SOL UTI ON a

1 and y = 3 x 1 3= x 3x = 1 1 x= 3

EX P L A NA TI ON

y=

∴ Intersection point is

1 intersects these lines. x b y = 4x

Substitute y = 3 into y =

y

( ) 1 , 3



 1 ,3 . 3

1 and solve. x

3

y=3

(1, 1) O

x

(−1, −1)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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530

Chapter 7 Parabolas and other graphs

1 and y = 4x x 1 4x = x 4x2 = 1 1 x2 = 4 1 x= ± 2    1 1 = 2 and y = 4 × – = –2 y = 4× 2   2 1 ∴ Intersection points are ,2 2   1 and – , –2 . 2 y=

Exercise 7J 1

Substitute and solve  by√multiplying both sides 1 1 1 by x. Note that =√ = . 4 4 2 Find the corresponding y values.

y ( 1 , 2) 2

x

O 1

(− 2 , −2)

1–4

4

—

1 A hyperbola has the rule y = . x a Complete this table. x

–2

–1

–

1 2

1 2

1

2

1 b Plot to form the graph of y = . x y

UNDERSTANDING

b

2 1

y

−2 −1−1O −2

1 2

x

3 2 A hyperbola has the rule y = . x a Complete this table. x

–3

–1

–

y

1 3

1 3

1

3

3 b Plot to form the graph of y = . x y

9 6 3 −3 −2 −1−3O −6 −9

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

1 2 3

x

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

UNDERSTANDING

Number and Algebra

2 3 A hyperbola has the rule y = – . x a Complete this table. x

–4

–2

–1

–

1 2

–

1 4

1 4

1 2

1

2

4

531

7J

y

2 b Plot to form the graph of y = – . x

y 8 6 4 2 −4 −3 −2 −1−2O −4 −6 −8

1 2 3 4

x

4 a

List in ascending order: 1 ÷ 0.1, 1 ÷ 0.001, 1 ÷ 0.01, 1 ÷ 0.00001. 1 b For y = , which of the following x values will give the largest value of y: x 1 1 1 1 , , or ? 5 10 2 100 1 c For y = , calculate the difference in the y values for x = 10 and x = 1000. x 1 1 1 d For y = , calculate the difference in the y values for x = – and x = – . x 2 1000

Example 23

5–8(½)

5–8(½)

5 Sketch the graphs of these hyperbolas, labelling the points where x = 1 and x = –1. 1 2 3 4 a y= b y= c y= d y= x x x x 1 2 3 4 e y=– f y=– g y=– h y=– x x x x 2 6 Find the coordinates on the graph of y = , where: x a x=2 b x=4 c x = –1

d x = –6

5 7 Find the coordinates on the graph of y = – , where: x a x = 10 b x = –4 c x = –7

d x=9

3 8 Find the coordinates on the graph of y = , where: x a y=3 b y=1 c y = –2

d y = –6

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

FLUENCY

5–7(½)

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter 7 Parabolas and other graphs

9–10(½)

7J 9 a

9–11(½)

10–11(½)

PROBLEM-SOLVING

532

3 Decide whether the point (1, 3) lies on the hyperbola y = . x

5 b Decide whether the point (1, –5) lies on the hyperbola y = – . x 2 c Decide whether the point (2, 1) lies on the hyperbola y = – . x 6 d Decide whether the point (–3, 6) lies on the hyperbola y = – . x 10 Find the coordinates of the points where y = a y=2 e y=x

b y=6 f y = 4x

1 intersects these lines. x c y = –1 g y = 2x

d y = –10 h y = 5x

2 11 Find the coordinates of the points where y = – intersects these lines. x 1 a y = –3 b y=4 c y=– 2 1 e y = –2x f y = –8x g y=– x 2 12

f

y

B

2

x

−2

y (−1, 1)

1 O

y = 3–x

C

2

−2

13–15

c y = 2x

e x2 + y 2 = 4

y

A

1 3

h y = –x 12, 13

12 Match equations a–f with graphs A–F. 4 1 a y= b y=– x x 1 d y= x+1 2

d y=

REASONING

Example 24

O

x

x

O (1, −1)

−2 y

D

y

E

F

y

(1, 4) (1, 2)

(−1, 3)

1 O

x

O

x

1 O

x

(−1, − 4)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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REASONING

Number and Algebra

1 13 Is it possible for a line on a number plane to not intersect the graph of y = ? If so, give x an example.

533

7J

14 Write the missing word (zero or infinity) for these sentences. 1 For y = , when x approaches infinity, y approaches x

.

1 b For y = , when x approaches negative infinity, y approaches x

.

1 For y = , when x approaches zero from the right, y approaches x 1 d For y = , when x approaches zero from the left, y approaches x c

15 Compare the graphs of y =

16 The graphs of y =

.

1 1 1 and y = . Describe the effect of the coefficient of x in y = . x 2x 2x

To intersect or not!

10A

.

—

—

16

1 and y = x + 1 intersect at two points. To find the x

points we set: 1 =x+1 x

ENRICHMENT

a

y 1 −1 O

1 = x(x + 1)

x

0 = x2 + x – 1

√ √ – 1± 5 1± 5 Using the quadratic formula, x = and y = . 2 2 a

Find the exact coordinates of the intersection of y = i

y=x–1

ii y = x – 2

1 and these lines. x iii y = x + 2

1 and y = –x + 1. What do you notice? x What part of the quadratic formula confirms this result?

b Try to find the coordinates of the intersection of y =

c

Write down the equations of the two straight lines (which have gradient –1) that intersect 1 y = only once. x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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534

Chapter 7 Parabolas and other graphs

7K Further transformations of graphs

10A

Earlier in this chapter we considered a wide range of transformations of parabolas but a limited number of transformations of circles, exponentials and hyperbolas. We will now look more closely at translations of these relations and the key features of their graphs.

Let’s start: Translations, translations, translations Use technology to assist in the discussion of these questions. • How does the graph of (x – 1)2 + (y + 2)2 = 9 compare with that of x2 + y2 = 9? •

What is the effect of h, k and r in (x – h)2 + (y – k)2 = r2 ?

•

How does the graph of y = 2x – 2 + 1 compare with that of y = 2x ?

• What is the effect of h and k in y = 2x – h + k? 1 1 – 1 compare with that of y = ? • How does the graph of y = x+2 x 1 + k? • What is the effect of h and k in y = x–h

Key ideas

The equation of a circle in standard form is (x – h)2 + (y – k)2 = r2 . • (h, k) is the centre. • r is the radius.

For the graph of the exponential equation y = ax – h + k the graph of y = ax is: • translated h units to the right • translated k units up. The equation of the asymptote is y = k.

y 2 1 x −2 −1−1O 1 2 3 −2 (1, −1) −3 −4 (x − 1)2 + (y + 1)2 = 9 y y = 2x − 1 + 1 3 2

asymptote O

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

(1, 2)

1

y=1 1

x

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Number and Algebra

1 +k For the graph of the hyperbola y = x–h 1 the graph of y = is: x • translated h units to the right • translated k units up. The asymptotes are x = h and y = k.

Key ideas

y y=

535

1 +1 x−2 1 1 2

O

y=1 1

x

x=2

Example 25 Sketching with transformations Sketch the graphs of the following relations. Label important features. a

(x – 2)2 + (y + 3)2 = 9

b y = 2x + 2 – 3

c

y=

1 +2 x+1

SOL UTI ON

EX P L A NA TI ON

a (x – 2)2 + (y + 3)2 = 9 Centre (2, –3) Radius = 3 x-intercept is 2. y-intercepts at x = 0: (0 – 2)2 + (y + 3)2 = 9

For (x – h)2 + (y – k)2 = r2 , (h, k) is the centre and r is the radius. Radius is 3 so, from centre (2, –3), point on circle is (2, 0). Find the y-intercepts by substituting x = 0 and solving for y.

4 + (y + 3)2 = 9 (y + 3)2 = 5

√ y+3= ± 5 √ y = –3 ± 5

Label centre and axes intercepts.

y

O −3 + √5

2

x

(2, −3) −3 − √5

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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536

Chapter 7 Parabolas and other graphs

b y = 2x + 2 – 3 y-intercept is 1.

For y = 2x – h + k, y = k is the equation of the asymptote.

y

Substitute x = 0 to find the y-intercept. (We will leave the x-intercept until we study logarithms later.)

1 x

O (−2, −2)

y = −3

1 +2 x+1

1 + k, h = –1 and k = 2, so the x–h asymptotes are x = –1 and y = 2.

y

Substitute to find the x- and y-intercepts.

y=2 O

x = −1 1 + 2 = 3. 1 y-intercept is 3. 3 x-intercept is – . 2

At x = 0, y =

Exercise 7K 1

Alternatively, substitute x = 1 to label a second point. For y =

3

3 − 2 −1

(0, 1) in y = 2x is translated 2 units to the left and 3 units down to (–2, –2).

x

x-intercepts (y = 0): 1 At y = 0, 0 = +2 x+1 1 –2 = x+1 1 x+1=– 2 3 x=– 2

1(½)

Choose the word left, right, up or down to complete each sentence. a The graph of y = 2x + 1 is the translation of the graph of y = 2x b The graph of y = 2x – 3 is the translation of the graph of y = 2x c The graph of y = 2x – 4 is the translation of the graph of y = 2x d The graph of y = 2x + 1 is the translation of the graph of y = 2x 1 1 is the translation of the graph of y = e The graph of y = x–3 x 1 1 is the translation of the graph of y = f The graph of y = x+2 x 1 1 g The graph of y = + 3 is the translation of the graph of y = x x 1 1 h The graph of y = – 6 is the translation of the graph of y = x x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

1(½)

by 1 unit. by 3 units. by 4 units. by 1 unit.

—

UNDERSTANDING

c y=

At x = 0, y = 22 – 3 = 1.

by 3 units. by 2 units. by 3 units. by 6 units.

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The graph of (x – 1)2 + y2 = 1 is the translation of the graph of x2 + y2 = 1 The graph of (x + 3)2 + y2 = 1 is the translation of the graph of x2 + y2 = 1 The graph of x2 + (y + 3)2 = 1 is the translation of the graph of x2 + y2 = 1 The graph of x2 + (y – 2)2 = 1 is the translation of the graph of x2 + y2 = 1 2–4(½)

2–4(½)

Example 25a

2 Sketch the graph of the following circles. Label the coordinates of the centre and find the x- and y-intercepts, if any. a (x – 3)2 + (y + 1)2 = 1 b (x + 2)2 + (y – 3)2 = 4 c (x – 1)2 + (y + 3)2 = 25 2 2 2 2 d (x + 3) + (y – 2) = 25 e (x + 2) + (y – 1) = 9 f x2 + (y – 4)2 = 36 g (x + 1)2 + y2 = 9 h (x – 2)2 + (y – 5)2 = 64 i (x + 3)2 + (y – 1)2 = 5

Example 25b

3 Sketch the graph of these exponentials. Label the asymptote and the y-intercept. a y = 2x – 2 b y = 2x + 1 c y = 2x – 5 d y = 2x – 1 e y = 2x + 3 f y = 2x + 1 x – 1 x + 2 g y=2 +1 h y=2 –3 i y = 2x – 3 – 4

Example 25c

4 Sketch the graph of these hyperbolas. Label the asymptotes and find the x- and y-intercepts. 1 1 1 a y= +2 b y= –1 c y= x x x+3 1 1 1 d y= e y= +1 f y= –3 x–2 x+1 x–1 1 1 1 g y= +2 h y= –1 i y= +6 x–3 x+4 x–5 5(½), 6

5(½), 6, 7

7, 8

5 The graphs of these exponentials involve a number of transformations. Sketch their graphs, labelling the y-intercept and the equation of the asymptote. a y = –2x + 1 b y = –2x – 3 c y = –2x + 3 x – 2 x + 1 d y = –2 e y = –2 –1 f y = –2x + 2 + 5 6 Sketch these hyperbolas, labelling asymptotes and intercepts. –2 –2 –1 a y= +2 b y= –1 c y= –2 x+1 x+2 x–3 1 7 The following hyperbolas are of the form y = + k. Write the rule for each graph. x–h a b c y y y

3 x

O y = −1

−1O

x=2 x = −1 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

7K

FLUENCY

2–4(½)

by 1 unit. by 3 units. by 3 units. by 2 units.

537

PROBLEM-SOLVING

i j k l

UNDERSTANDING

Number and Algebra

y=3 y = 32 x

x O

x=1

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Chapter 7 Parabolas and other graphs

PROBLEM-SOLVING

7K 8 Find the coordinates of the intersection of the graphs of these equations. 1 1 a y= and y = x + 2 b y= + 1 and y = x + 3 x+1 x–2 –1 c y= – 3 and y = –2x – 1 d (x – 1)2 + y2 = 4 and y = 2x x+2 1 f x2 + (y + 1)2 = 10 and y = x – 1 e (x + 2)2 + (y – 3)2 = 16 and y = –x – 3 3 9

9, 10

10, 11

9 A circle has equation (x – 3)2 + (y + 2)2 = 4. Without sketching a graph, state the minimum and maximum values for: a x b y 10 Find a rule for each graph. a y

y

b

(2, 3)

(2, 1)

−1

x

O

x

−2 O

y

c

REASONING

538

d

y

x

−2 O

y=1

(−5, −3)

O

x

x=1 y

e

O

y

f

− 1_ 2

x

O 1 −3

y = −1

x = −2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

x

x = −3

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REASONING

Number and Algebra

11 Explain why the graphs of the following pairs of relations do not intersect. 1 b (x – 1)2 + (y + 2)2 = 4 and y = 1 a y = and y = –x x 2 1 c y = 2x – 1 + 3 and y = x – 3 d y= – 1 and y = x+3 3x —

—

12

12 By expanding brackets of an equation in the standard form of a circle, we can write: (x – 1)2 + (y + 2)2 = 4

(1)

x2 – 2x + 1 + y2 + 4y + 4 = 4 x2 – 2x + y2 + 4y + 1 = 0

7K

ENRICHMENT

Circles by completing the square

539

(2)

Note that in equation (2) it is not obvious that the centre is (1, –2) and that the radius is 2. It is therefore preferable to write the equation of a circle in standard form (i.e. as in equation (1)). If given an equation such as (2), we can complete the square in both x and y to write the equation of the circle in standard form. x2 – 2x + y2 + 4y + 1 = 0 (x – 1)2 – 12 + (y + 2)2 – 22 + 1 = 0 (x – 1)2 – 1 + (y + 2)2 – 4 + 1 = 0 (x – 1)2 + (y + 2)2 = 4 The radius is 2 and centre (1, –2). a

Write these equations of circles in standard form. Then state the coordinates of the centre and the radius. i x2 + 4x + y2 – 2y + 1 = 0 ii x2 + 8x + y2 + 10y + 5 = 0 2 2 iii x – 6x + y – 4y – 3 = 0 iv x2 – 2x + y2 + 6y – 5 = 0 v x2 + 10x + y2 + 8y + 17 = 0 vi x2 + 6x + y2 + 6y = 0 2 2 vii x + 3x + y – 6y + 4 = 0 viii x2 + 5x + y2 – 4y – 2 = 0 ix x2 – x + y2 + 3y + 1 = 0 x x2 – 3x + y2 – 5y – 4 = 0

b Give reasons why x2 + 4x + y2 – 6y + 15 = 0 is not the equation of a circle.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

540

Chapter 7 Parabolas and other graphs

Investigation Painting bridges A bridge is 6 m high and has an overall width of 12 m and an archway underneath, as shown. We need to determine the actual surface area of the face of the arch so that we can order paint to refurbish it. The calculation for this area would be: Bridge area = 12 × 6 – area under arch = 72 – area under parabola 1 Consider an archway modelled by the formula h = – (d – 4)2 + 4, 4 where h metres is the height of the arch and d metres is the distance from the left. To estimate the area under the arch, divide the area into rectangular regions. If we draw rectangles above the arch and calculate their areas, we will have an estimate of the area under the arch even though it is slightly too large. Use the rule for h to obtain the height of each rectangle.

h 4 O

d

2 4 6 8

Area = (2 × 3) + (2 × 4) + (2 × 4) + (2 × 3) =6+8+8+6 = 28 m2 ∴ Area is approximately 28 m2 . We could obtain a more accurate answer by increasing the number of rectangles; i.e. by reducing the width of each rectangle (called the strip width). Overestimating the area under the arch a

Construct an accurate graph of the parabola and calculate the area under it using a strip width of 1.

b

Repeat your calculations using a strip width of 0.5.

c

Calculate the surface area of the face of the arch using your answer from part b.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

h 4

O

1 2 3 4 5 6 7 8

d

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Number and Algebra

541

Underestimating the area under the arch a

Estimate the area under the arch by drawing rectangles under the graph with a strip width of 1.

b

Repeat the process for a strip width of 0.5.

c

Calculate the surface area of the face of the arch using your answer from part b.

h 4

O

1 2 3 4 5 6 7 8

d

Improving accuracy a

Suggest how the results from parts 1 and 2 could be combined to achieve a more accurate result.

b

Explore how a graphics or CAS calculator can give accurate results for finding areas under curves.

Sketching regions in the plane A region in a plane is an infinite set of points defined on one side of a line or other curve. We describe the region with an inequality like y < 2x + 1, x + 2y ≥ 3 or x2 + y2 < 1. All points that satisfy the inequality form the region.

y

Half planes A half plane is the region on one side of a line. Here are two examples. 1 1 1 For y ≥ x + 1 we sketch y = x + 1, draw the full line (since y = x + 1 2 2 2 is included in the region) and then check the point (0, 0). As (0, 0) does 1 not satisfy y ≥ x + 1, the region must lie on the other side of the line. 2 a

For the given graph of x + y < 2, explain why the line is dashed.

b

Show that (0, 0) is inside the region.

c

Sketch these half planes. i y < 2x + 2 ii y ≥ – x – 3 iii 3x – y ≥ – 2

d

e

iv x + 2y < 1

Sketch the half planes that satisfy these inequalities. i x≤3 ii x < –1 iii y > –2

y≥

1 2x

+1 1 x

O

−2

y 2 O x+y 2(x + 3)(x + 1) v y < x2 – x – 6 vi y ≤ x2 + 3x – 40 b

Sketch the intersecting region defined by y ≥ x2 – 4 and y < x + 2. Find all significant points.

Circular regions The region shown satisfies x2 + y2 ≤ 4.

y 2

Sketch these regions. a x2 + y2 ≤ 9 b x2 + y2 < 16

−2

c x2 + y2 > 4

O

2

x

−2

d (x – 1)2 + y2 ≥ 9 e (x + 2)2 + (y – 1)2 < 3 f (x – 3)2 + (y – 4)2 ≥ 10

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Number and Algebra

Up for a challenge? If you get stuck on a question, check out the 'Working with unfamiliar problems' poster at the end of the book to help you.

Problems and challenges 1

Solve these inequalities for x. a 6x2 + x – 2 ≤ 0

b 12x2 + 5x – 3 > 0

2

Sketch the region defined by x2 – 4x + y2 – 6y – 3 ≤ 0.

3

Prove the following.

543

c x2 – 7x + 2 < 0

a The graphs of y = x – 4 and y = x2 – x – 2 do not intersect. b

The graphs of y = –3x + 2 and y = 4x2 – 7x + 3 touch at one point.

c

The graphs of y = 3x + 3 and y = x2 – 2x + 4 intersect at two points.

4

Prove that there are no points (x, y) that satisfy x2 – 4x + y2 + 6y + 15 = 0.

5

For what values of k does the graph of y = kx2 – 2x + 3 have: a one x-intercept? b two x-intercepts?

c no x-intercepts?

For what values of k does the graph of y = 5x2 + kx + 1 have: a one x-intercept? b two x-intercepts?

c no x-intercepts?

Find the rules for these parabolas. a b y

c

6

7

y

3

-1 O

O

3

x

y (4, 5)

x

x

O (−2, −3)

−3

(2, −3)

8

A graph of y = ax2 + bx + c passes through the points A(0, –8), B(–1, –3) and C(1, –9). Use your knowledge of simultaneous equations to find the values of a, b and c and, hence, find the turning point for this parabola, stating the answer using fractions.

9

Determine the maximum vertical distance between these two parabolas at any given x value between the points where they intersect: y = x2 + 3x – 2 and y = –x2 – 5x + 10

10 Two points, P and Q, are on the graph of y = x2 + x – 6. The origin (0, 0) is the midpoint of the line segment PQ. Determine the exact length of PQ. 11 A parabola, y1 = (x – 1)2 + 2, is reflected in the x-axis to become y2 . Now y1 and y2 are each translated 2 units horizontally but in opposite directions, forming y3 and y4 . If y5 = y3 + y4 , sketch the graphs of the possible equations for y5 .

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter summary

544

Chapter 7 Parabolas and other graphs

Transformations and turning point form x2 +

bx + c Sketching y = e.g. y = x 2 − 4x − 12 y-intercept: x = 0, y = −12 x-intercepts: y = 0, x 2 − 4x − 12 = 0 Factorise (x − 6)(x + 2) = 0 Null Factor Law x − 6 = 0 or x + 2 = 0 x = 6 or x = −2 Turning point: halfway between x-intercepts 6 + (−2) =2 x= 2 y Substitute x = 2. x y = 22 − 4(2) − 12 6 −2 O = −16 (2, −16) minimum −12

Sketching y = a (x − h )2 + k Any quadratic can be written in turning point form by completing the square. e.g. Sketch y = (x − 1)2 − 3. Turning point (1, −3) minimum y y-intercept: x = 0, y = 1 − 3 = −2 1 − √3 1 + √3 O −2

x

h > 0 move right h < 0 move left

Parabolas

Using substitution we can solve simultaneously to find 0, 1 or 2 intersection points.

y-intercept (x = 0) O

x

x-intercepts (y = 0)

x

minimum turning x=h point axis of symmetry

(h, k )

y

maximum turning point x

Hyperbolas

x

y=a x

y (1, a )

r2 = 4 r∴ =2

Exponential graphs y

y = a −x

(1, a) (1, 1 )

1 O

a

O x

(1, −a )

asymptote y=0

y = −a x, a > 1

y = ± r2 − x2 x = ± r2 − y2

y

Circle with equation (10A) (x − h)2 + (y − k )2 = r 2 is centred at (h, k ) with radius r.

O

y (−1, a )

y = a x, a > 1

y = a x − h + k (10A) y=k x

x

O

(−1, −a )

−1

−2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Parabola arises from a quadratic equation

Other graphs

x 2 + y 2 = r 2 is a circle centred at (0, 0) with radius r. y 2 x2 + y2 = 4 2

x y = x2 − 4

−4 Turning point form y = a (x − h)2 + k has turning point at (h, k ).

Circles

O

x

2

O

O

−2

y = x2 y = (x − 2)2

x

O y

Δ 0 move up y = x2 k < 0 move down

Quadratic formula (10A) For y = ax 2 + bx + c, the quadratic formula can be used to find x-intercepts. i.e. if ax 2 + bx + c = 0

two x-intercepts

x

−3 y = (x + 3)2

(2, −16)

2 x = −b ± b − 4ac 2a the discriminant Δ = b 2 − 4ac tells how many x-intercepts: y Δ>0

y = x2 y = 12 x 2

y

(1, −3)

x-intercepts: y = 0, 0 = (x − 1)2 − (√3)2 0 = (x − 1 − √3)(x − 1 + √3) ∴ x = 1 + √3, 1 − √3

y y = 2x 2

y = ax 2 dilates graph compared to y = x 2 0 < a < 1 graph is wider a > 1 graph is narrower

Asymptotes x = 0 and y = 0

y = − a (reflect in x- or y-axis) x x (1, −a ) y

a y= + k (10A) x− h find asymptotes and intercepts to sketch

y=k O

x

x=h

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Multiple-choice questions 38pt 7A

1

y

The equation of the axis of symmetry of the graph shown is: A B C D E

y = –4 x=3 x = –4 y=3 y = 3x

O

5

1 (3, −4)

38pt 7B

2 Compared to the graph of y = x2 , the graph of y = (x – 3)2 is: A B C D E

38pt 7B

translated 3 units down translated 3 units left dilated by a factor of 3 translated 3 units right translated 3 units up

3 The coordinates and type of turning point of y = –(x + 2)2 + 1 is: A B C D E

38pt 7B

a minimum at (–2, –1) a maximum at (2, 1) a minimum at (2, 1) a maximum at (2, –1) a maximum at (–2, 1)

4 The y-intercept of y = 3(x – 1)2 + 4 is: A 1

38pt 7C

38pt 7C

B 4

1 3

C

5 The x-intercept(s) of the graph of y = x2 + 3x – 10 are: A 2, –5 B –10 D –5, –2 E 5, 2

D 7

E 3

C 5, –2

6 A quadratic graph has x-intercepts at x = –7 and x = 2. The x-coordinate of the turning point is: 7 9 5 A x=– B x=– C x= 2 2 3 D x=

38pt 7D

x

545

Chapter review

Number and Algebra

5 2

E x=–

9 2

7 The quadratic rule y = x2 – 4x – 3, when written in turning point form, is: A y = (x – 2)2 – 3 B y = (x – 4)2 + 1 C y = (x – 2)2 – 7 D y = (x + 4)2 – 19 E y = (x + 2)2 – 1

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter review

546

Chapter 7 Parabolas and other graphs

38pt 10A 7E 10A

38pt 7F 10A

38pt 7H

38pt 7I

8 A quadratic graph y = ax2 + bx + c has two x-intercepts. This tells us that: A

The graph has a maximum turning point. b 0 E b2 – 4ac = 0 9 A toy rocket follows the path given by h = –t2 + 4t + 6, where h is the height above ground, in metres, t seconds after launch. The maximum height reached by the rocket is: A 10 metres B 2 metres C 6 metres D 8 metres E 9 metres 10 The equation of a circle centred at the origin with radius 4 units is: A y = 4x2 B x2 + y2 = 4 C x2 + y2 = 8 D y = 4x 11 The graph of y = 3x intersects the y-axis at: A (0, 3)

38pt 7I

B (3, 0)

12 The graph of y = 4–x is: A y

C (0, 1)

y

B

x

y

E

38pt 10A 7K 10A

38pt 10A 7K 10A

x

−1

(1, − 1) 4

y x

O

(1, 1) 4

O



x

O

1

1 0, 3

O

(1, −4)

D



y

C

−1

4

E

D (1, 3)

O

38pt 7I

E x2 + y2 = 16

x

(1, − 1) 4

1 13 The graphs of y = 3x and y = intersect at the point: 3   1 A (1, 3) –1, B C (–1, 3) 3 1 14 The graph of y = + 2 has asymptote(s) at: x–1 A x = 1, y = 2 B y = 2 C x = –1

D



1 1 , 9 3



D x = 2, y = 1

E



1,

1 3



E x = –1, y = 2

15 The circle with equation (x + 1)2 + (y – 3)2 = 16 has centre coordinates and radius: A (1, –3), r = 4 B (1, –3), r = 16 C (–1, 3), r = 4 D (–1, 3), r = 16 E (–1, –3), r = 4

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Short-answer questions 38pt 7A

1

y

State the following features of the quadratic graph shown. a b c d

turning point and whether it is a maximum or a minimum axis of symmetry x-intercepts y-intercept

38pt 7A/B

2

38pt 7C

3 Sketch the quadratics below by first finding: i the y-intercept ii the x-intercepts, using factorisation iii the turning point. a y = x2 – 4 b y = x2 + 8x + 16

O −1−1 −2 −3 −4

1 2 3

(1, −4)

State whether the graphs of the following quadratics have a maximum or a minimum turning point and give its coordinates. a y = (x – 2)2 b y = –x2 + 5 c y = –(x + 1)2 – 2 d y = 2(x – 3)2 + 4

c y = x2 – 2x – 8

38pt 7C

4 Complete the following for each quadratic below. i State the coordinates of the turning point and whether it is a maximum or a minimum. iii Find the y-intercept. iii Find the x-intercepts (if any). iv Sketch the graph, labelling the features above. a y = –(x – 1)2 – 3 b y = 2(x + 3)2 – 8

38pt 7D

5 Sketch the following quadratics by completing the square. Label all key features with exact coordinates. a y = x2 – 4x + 1 b y = x2 + 3x – 2

38pt 7E

38pt 7E 10A

x

547

Chapter review

Number and Algebra

6 State the number of x-intercepts of the following quadratics either by using the discriminant or by inspection where applicable. a y = (x + 4)2 b y = (x – 2)2 + 5 c y = x2 – 2x – 5 d y = 2x2 + 3x + 4 10A 10A 7 For the following quadratics: i

Find the y-intercept. b ii Use x = – to find the coordinates of the turning point. 2a iii Use the quadratic formula to find the x-intercepts, rounding to one decimal place. iv Sketch the graph. a y = 2x2 – 8x + 5

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b y = –x2 + 3x + 4

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter review

548

Chapter 7 Parabolas and other graphs

38pt 7F

8

200 metres of fencing is to be used to form a rectangular paddock. Let x metres be the width of the paddock. a b c d e f

38pt 7G 10A

Write an expression for the length of the paddock in terms of x. Write an equation for the area of the paddock (A m2 ) in terms of x. Decide on the suitable values of x. Sketch the graph of A versus x for suitable values of x. Use the graph to determine the maximum paddock area that can be formed. What will be the dimensions of the paddock to achieve its maximum area?

9 Solve these equations simultaneously. a y = x2 + 4x – 2 b y = 2x2 + 5x + 9 y = 10

38pt 7G 10A

y = –x + 4

c

y = x2 + 1 2x + 3y = 4

10 Use the discriminant to show that the line y = x + 4 intersects the parabola y = x2 – x + 5 in just one place.

38pt 7H

11 Sketch these circles. Label the centre and axes intercepts. a x2 + y2 = 25 b x2 + y2 = 7

38pt 7H

12 Find the exact coordinates of the points of intersection of the circle x2 + y2 = 9 and the line y = 2x. Sketch the graphs, showing the points of intersection.

38pt 7I

13 Sketch the following graphs, labelling the y-intercept and the point where x = 1. a y = 4x b y = –3x c y = 5–x

38pt 7J

14 Sketch these hyperbolas, labelling the points where x = 1 and x = –1. 2 3 a y= b y=– x x 4 15 Find the points of intersection of the hyperbola y = and these lines. x a y=3 b y = 2x

38pt 7J

38pt 7K 10A

16 Sketch the following graphs, labelling key features with exact coordinates. 1 a (x + 1)2 + (y – 2)2 = 4 b y = 2x – 1 + 3 c y= –3 x+2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Extended-response questions 1

1 (x – 200)2 + 30, where 800 h metres is the distance above the base of the bridge and x metres is the distance from the left side of the bridge. a Determine the turning point of the graph of the equation. b If the bridge is symmetrical, determine the suitable values of x. c Determine the range of values of h. d Sketch a graph of the equation for the suitable values of x. e What horizontal distance does the cable span? f What is the closest distance of the cable from the base of the bridge? g What is the greatest distance of the cable from the base of the bridge? The cable for a suspension bridge is modelled by the equation h =

549

Chapter review

Number and Algebra

2 The amount, A grams, of a radioactive element remaining after t years is given by the rule A = 450(3–t ). a What is the initial amount (i.e. at t = 0) of radioactive material? b How much material remains after: i 1 year? ii 4 years? (Round your answer to one decimal place.) c Determine when the amount will reach 50 grams. d Sketch a graph of A vs t for t ≥ 0. e Use trial and error to find, to one decimal place, when less than 0.01 grams will remain.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter

8

Probability

What you will learn

Australian curriculum

8A Review of probability (Consolidating) 8B Unions and intersections 8C The addition rule 8D Conditional probability 8E Two-step experiments using tables 8F Using tree diagrams 8G Independent events

S TAT I S T I C S A N D P R O B A B I L I T Y

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Chance Describe the results of two- and three-step chance experiments, both with and without replacement, assign probabilities to outcomes and determine probabilities of events. Investigate the concept of independence. Use the language of ‘if . . . then’, ‘given’, ‘of’, ‘knowing that’ to investigate conditional statements, and identify common mistakes when interpreting such language. (10A) Investigate reports of studies in digital media and elsewhere for information on the planning and implementation of such studies, and the reporting of variability.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Online resources • Chapter pre-test • Videos of all worked examples • Interactive widgets • Interactive walkthroughs • Downloadable HOTsheets • Access to HOTmaths Australian Curriculum courses

Disaster probability We have all heard about or experienced a disaster of some kind in Australia. Floods, bushfires or cyclones, for example, are very real possibilities in many parts of the country and, because of this, most people take out insurance to cover for any potential losses. Consequently, insurance companies try to predict the likelihood of such disasters and set their premiums

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

to suit. This is called risk analysis and includes calculating the experimental probability of certain disasters, depending on a number of factors. Some factors include: • geographic location • historical records of previous disasters • local facilities, such as hospitals and power supply • area topography.

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

4UBUJTUJDT BOE 1SPCBCJMJUZ



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Choice 1 Choice 2

R

R

B

Outcome Probability

R

(R, R)

R

(R, R)

B

(R, B)

R

(R, R)

R

(R, R)

B

(R, B)

R

(B, R)

R

(B, R)

B

(B, B)

1_ 9 1_ 9 1_ 9 1_ 9 1_ 9 1_ 9 1_ 9 1_ 9 1_ 9

Choice 1 Choice 2 Outcome Probability

2_ 3

1_ 3

2_ 3

R (R, R)

2_ × 2_ 3 3

=

4_ 9

1_ 3

B (R, B)

2_ × 1_ 3 3

=

2_ 9

2_ 3

R (B, R)

1_ × 2_ 3 3

=

2_ 9

1_ 3

B (B, B)

1_ × 1_ 3 3

=

1_ 9

R

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

4UBUJTUJDT BOE 1SPCBCJMJUZ

Box

D

1_ 2

Probability

(A, red)

1_ 2

×

1_ 2

= 1_4

1_ 2

red

1_ 2

green (A, green)

1_ 2

×

1_ 2

= 1_4

1_ 4

red

1_ 2

×

1_ 4

= 1_8

3_ 4

green (B, green)

1_ 2

×

3_ 4

= 3_8

(B, red)

B

1S # SFE  

F

Outcome

A

1_ 2

E

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B Selection 1 Selection 2

5_ 8

3_ 8

Outcome Probability

4_ 7

B

(B, B)

5_ 8

×

4_ 7

5 = __ 14

3_ 7

W

(B, W)

5_ 8

×

3_ 7

15 = __ 56

5_ 7

B

(W, B)

3_ 8

×

5_ 7

15 = __ 56

2_ 7

W

(W, W)

3_ 8

×

2_ 7

3 = __ 28

B

W

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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M F (M, F) ( _________ )

F

5_ 8

F

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M ( __, __ ) ( _________ ) F

F ( __, __ ) ( _________ )

5_ 8

o

( __, __ ) ( _________ )

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M (M, M) ( _________ )

M

M ( __, __ ) ( _________ )

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yellow

(A, yellow)

1_ 2

orange

__________

________

yellow

__________

________

______

__________

________

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9 M (M, M) ( 3_8 × 3_8 = __ 64 )

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Probability × 1_4 = 1_8

A

B

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Car

Colour white

1_ 2

1_ 2

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130#-&.40-7*/(

 

Probability

__________ ________

Falcon _____ __________ ________ _____ __________ ________ Commodore _____ __________ ________

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.



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Up for a challenge? If you get stuck on a question, check out the 'Working with unfamiliar problems' poster at the end of the book to help you.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Review • Sample space is the list of all possible

Unions and intersections • Union A ∪ B (A or B ) A B

outcomes.

• Pr(event) = number of favourable outcomes

• Intersection A ∩ B (A and B)

total number of outcomes

A

• Complement of A is A′′ (not A) Venn diagram A

Two-way table A A′ B 2 5 7 B′ 4 1 5 6 6 12

B

4 2 5 1

B

A

A′

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A

B

• Mutually exclusive events A ∩ B = ∅

A

B



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Addition rule Pr(A ∪ B) = Pr(A) + Pr(B) – Pr(A ∩ B)

Probability

Conditional probability Pr(A⎪B ) = Pr(A ∩ B ) or n(A ∩ B ) Pr(B)

A

B

4 2

5 1

If A and B are mutually exclusive: Pr(A ∩ B) = 0 and Pr(A ∪ B) = Pr(A) + Pr(B)

n(B)

A B 2 B′ 4 6

A′ 5 7 1 5 6 12

Tree diagrams 3 white 4 black

Pr (A⎪B ) = 2 7

With replacement Choice 1 Choice 2 Outcome Probability 3_ 7 3_ 7

Independent events •- Pr(A⎪B ) = Pr(A) •- Pr(A ∩ B) = Pr(A) × Pr(B)

4_ 7

W (W, W)

9 __ 49

4_ 7 3_ 7

B

(W, B)

12 __ 49

W (B, W)

12 __ 49

4_ 7

B

16 __ 49

W

B

(B, B)

Without replacement

3_ 7

W

Tables With replacement

Without replacement

A B C A (A, A) (B, A) (C, A) B (A, B) (B, B) (C, B) C (A, C) (B, C) (C, C)

A B C A × (B, A) (C, A) B (A, B) × (C, B) C (A, C) (B, C) ×

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B

2_ 6

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1 __ 7

4_ 6 3_ 6

B

(W, B)

2 __ 7

W (B, W)

2 __ 7

3_ 6

B

2 __ 7

(B, B)

Pr (1 white, 1 black) = 2_7

2 + _ 7

= 4_ 7

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter

9

Statistics

W h a t y o u w ill le a r n

Au s t r a l i a n c u r r ic u lu m

9 A Collecting and using data 9 B Review of statistical graphs (Consolidating) 9 C Summary statistics 9 D Box plots 9 E Standard deviation (10A) 9 F Time-series data 9 G Bivariate data and scatter plots 9 H Line of best fit by eye 9 I Linear regression with technology (10A)

S TAT I S T I C S A N D P R O B A B I L I T Y

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Data representation and interpretation Determine quartiles and interquartile range. Construct and interpret box plots and use them to compare data sets. Compare shapes of box plots with corresponding histograms and dot plots. Use scatter plots to investigate and comment on relationships between two continuous variables. Investigate and describe bivariate numerical data for which the independent variable is time. Evaluate statistical reports in the media and other places by linking claims to displays, statistics and representative data. (10A) Calculate and interpret the mean and standard deviation data and use these to compare data sets. (10A) Use information technologies to investigate bivariate 16x16line to numerical data sets. Where appropriate, use a straight describe the relationship, allowing for variation. Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

32

O n lin e re s o u rc e s • • • • • •

Chapter pre-test Videos of all worked examples Interactive widgets Interactive walkthroughs Downloadable HOTsheets Access to HOTmaths Australian Curriculum courses

C lo s in g th e g a p According to the Australian Bureau of Statistics the population of Indigenous Australians has increased from 351 000 in 1991 to 548 000 in 2011. This represents an annual increase of 2.3% compared with a total Australian annual population increase of 1.2%. The life expectancy for Indigenous Australians, however, is not as high as for non-Indigenous Australians. For males it is about 69 years (compared to about 80 years Australia-wide) and about

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

74 years for females (compared to about 83 years Australia-wide). This represents a life expectancy gap for Indigenous Australians of about 11 years for males and 9 years for females. The Australian government works with departments such as the Australian Bureau of Statistics to collect accurate data, like the examples above, so it can best determine how to allocate resources in assisting Indigenous communities.

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612

Chapter 9 Statistics

9A Collecting and using data There are many reports on television and radio that begin with the words ‘A recent study has found that …’. These are usually the result of a survey or investigation that a researcher has conducted to collect information about an important issue, such as unemployment, crime or obesity. Sometimes the results of these surveys are used to persuade people to change their behaviour. Sometimes they are used to pressure the government into changing the laws or to change the way the government spends public money. Results of surveys and other statistics can sometimes be misused or displayed in a way to present a certain point of view.

Let’s start: Improving survey questions Here is a short survey. It is not very well constructed. Question 1: How old are you? Question 2: How much time did you spend sitting in front of the television or a computer yesterday? Question 3: Some people say that teenagers like you are lazy and spend way too much time sitting around when you should be outside exercising. What do you think of that comment? Have a class discussion about the following. • What will the answers to Question 1 look like? How could they be displayed? • What will the answers to Question 2 look like? How could they be displayed? • Is Question 2 going to give a realistic picture of your normal daily activity? • Do you think Question 2 could be improved somehow? • What will the answers to Question 3 look like? How could they be displayed? • Do you think Question 3 could be improved somehow?

Key ideas

Surveys are used to collect statistical data. • Survey questions need to be constructed carefully so that the person knows exactly what sort of answer to give. Survey questions should use simple language and should not be ambiguous. • Survey questions should not be worded so that they deliberately try to provoke a certain kind of response. • If the question contains an option to be chosen from a list, the number of options should be an odd number, so that there is a ‘neutral’ choice. For example, the options could be: strongly agree

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

agree

unsure

disagree

strongly disagree

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Statistics and Probability

A population is a group of people, animals or objects with something in common. Some examples of populations are: • all the people in Australia on Census Night • all the students in your school • all the boys in your maths class • all the tigers in the wild in Sumatra • all the cars in Brisbane • all the wheat farms in NSW.

613

Key ideas

A sample is a group that has been chosen from a population. Sometimes information from a sample is used to describe the whole population, so it is important to choose the sample carefully. Statistical data can be divided into subgroups.

Data

Categorical

Numerical

Nominal Ordinal Discrete (Categories (Categories can (Data can take have no order; be ordered; on only a limited e.g. colours: e.g. low, medium, number of values; red, blue, yellow) high) e.g. number of children in a family)

Continuous (Data can take on any value in a given range; e.g. time taken to complete a race)

Example 1 Describing types of data What type of data would the following survey questions generate? a How many televisions do you have in your home? b To what type of music do you most like to listen? SOL UTI ON

EX P L A NA TI ON

a numerical and discrete

The answer to the question is a number with a limited number of values; in this case, a whole number.

b categorical and nominal

The answer is a type of music and these categories have no order.

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614

Chapter 9 Statistics

Example 2 Choosing a survey sample A survey is carried out on the internet to determine Australia’s favourite musical performer. Why will this sample not necessarily be representative of Australia’s views? SO L U T I O N

EX P L A N A T I O N

An internet survey is restricted to people with a computer and internet access, ruling out some sections of the community from participating in the survey.

The sample may not include some of the older members of the community or those in areas without access to the internet. Also, the survey would need to be set up so that people can do it only once so that ‘fake’ surveys are not completed.

1

1–4

3, 4

—

Match each word (a–e) with its definition (A–E). a population A a group chosen from a population b census B a tool used to collect statistical data c sample C all the people or objects in question d survey D statistics collected from every member of the population e data E the factual information collected from a survey or other source

UNDERSTANDING

Exercise 9A

2 Match each word (a–f) with its definition (A–F). a numerical A categorical data that has no order b continuous B data that are numbers c discrete C numerical data that take on a limited number of values d categorical D data that can be divided into categories e ordinal E numerical data that take any value in a given range f nominal F categorical data that can be ordered 3 Classify each set of data as categorical or numerical. a 4.7, 3.8, 1.6, 9.2, 4.8 b red, blue, yellow, green, blue, red c low, medium, high, low, low, medium d 3 g, 7 g, 8 g, 7 g, 4 g, 1 g, 10 g 4 Which one of the following survey questions would generate categorical data? A How many times do you eat at your favourite fast-food place in a typical week? B How much do you usually spend buying your favourite fast food? C How many items did you buy last time you went to your favourite fast-food place? D Which is your favourite fast-food?

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Statistics and Probability

5 Year 10 students were asked the following questions in a survey. Describe what type of data each question generates. Choices a How many people under the age of 18 years are there in your immediate family? categorical numerical b How many letters are there in your first name? c Which company is the carrier of your mobile telephone calls? Optus/Telstra/ ordinal nominal discrete continuous Vodafone/Virgin/Other (Please specify.) d What is your height? e How would you describe your level of application in Maths? (Choose from very high, high, medium or low.)

6, 7 Example 2

5

6–8

FLUENCY

Example 1

5

9A

7–9

6 The principal decides to survey Year 10 students to determine their opinion of Mathematics. a In order to increase the chance of choosing a representative sample, the principal should: A Give a survey form to the first 30 Year 10 students who arrive at school. B Give a survey form to all the students studying the most advanced Maths subject. C Give a survey form to five students in every Maths class. D Give a survey form to 20% of the students in every class.

PROBLEM-SOLVING

5

615

b Explain your choice of answer in part a. Describe what is wrong with the other three options. 7 Discuss some of the problems with the selection of a survey sample for each given topic. a A survey at the train station of how Australians get to work. b An email survey on people’s use of computers. c Phoning people on the electoral roll to determine Australia’s favourite sport. 8 Choose a topic in which you are especially interested, such as football, cricket, movies, music, cooking, food, computer games or social media. Make up a survey about your topic that you could give to the people in your class. It must have four questions. Question 1 must produce data that are categorical and ordinal. Question 2 must produce data that are categorical and nominal. Question 3 must produce data that are numerical and discrete. Question 4 must produce data that are numerical and continuous.

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9A 9 A television news reporter surveyed four companies and found that the profits of three of these companies had reduced over the past year. They report that this means the country is facing an economic downturn and that only one in four companies is making a profit. a What are some of the problems in this media report? b How could the news reporter improve their sampling methods? c Is it correct to say that only one in four companies is making a profit? Explain.

10

10, 11

11, 12

a b c d

8 ne

l1

7 Ch

an

l1

6

ne an

l1 Ch

ne an

Ch

Ch

an

ne

l1

5

Number of people

8 l1

7

30 25 20 15 10 5

ne an

Ch

Ch

an

ne

l1

6 l1 ne

an Ch

Ch

an

ne

l1

5

Number of people

10 Here are two column graphs, each showing the same results of a survey that asked people which TV channel they preferred. Graph B Graph A TV channel preference TV channel preference

27 26 25 24 23

PROBLEM-SOLVING

Chapter 9 Statistics

REASONING

616

Which graph could be titled ‘Channel 15 is clearly most popular’? Which graph could be titled ‘All TV channels have similar popularity’? What is the difference between the two graphs? Which graph is misleading and why?

11 Describe three ways that graphs or statistics could be used to mislead people and give a false impression about the data. 12 Search the internet or newspaper for ‘misleading graphs’ and ‘how to lie with statistics’. Explain why they are misleading.

—

—

13, 14

13 Research the 2011 Australian Census on the website of the Australian Bureau of Statistics. Find out something interesting from the results of the 2011 Australian Census and write a short news report.

ENRICHMENT

The 2011 Australian Census

14 It is often said that Australia has an ageing population. What does this mean? Search the internet for evidence showing that the ‘average’ Australian is getting older every year.

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Statistics and Probability

9B Review of statistical graphs

617

CONSOLIDATING

Statistical graphs are an essential element in the analysis and representation of data. Graphs can help to show the most frequent category, the range of values, the shape of the distribution and the centre of the data. By looking at statistical graphs the reader can quickly draw conclusions about the numbers or categories in the data set and interpret this within the context of the data.

Let’s start: Public transport analysis

Number of people

A survey was carried out to find out how many times people in the group had used public transport in the past month. The results are shown in this histogram.

8 6 4 2 0

10

20

30 40 Transport use

50

60

Discuss what the histogram tells you about this group of people and their use of public transport. You may wish to include these points: • How many people were surveyed? • Is the data symmetrical or skewed? • Is it possible to work out the exact mean? Why/why not? • Do you think these people were selected from a group in your own community? Give reasons. The types of statistical data that we saw in the previous section; i.e. categorical (nominal or ordinal) and numerical (discrete or continuous), can be displayed using different types of graphs to represent the different data. Graphs for a single set of categorical or discrete data • Dot plot

1 2 3 4

Key ideas

• Stem-and-leaf plot Stem 0 1 2 3

Leaf 13 259 1467 04

2 | 4 means 24

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618

Chapter 9 Statistics

Key ideas

lle nt

oo d

Ex ce

G

ed iu m

M

Lo w

Po or

Frequency

• Column graph 5 4 3 2 1 0

Score

Frequency 3 5 8 4

Percentage frequency 15 25 40 20

8 6 4 2 0

40 30 20 10 10

20 Score

30

% Frequency

Class interval 0– 10– 20– 30–40

Frequency

Histograms can be used for grouped discrete or continuous numerical data. The interval 10– includes all numbers from 10 (including 10) to fewer than 20.

40

Measures of centre include: sum of all data values • mean (x) x = number of data values • median the middle value when data are placed in order The mode of a data set is the most common value. Data can be symmetrical or skewed.

symmetrical

positively skewed tail

negatively skewed tail

Example 3 Presenting and analysing data Twenty people were surveyed to find out how many times they use the internet in a week. The raw data are listed. 21, 19, 5, 10, 15, 18, 31, 40, 32, 25 11, 28, 31, 29, 16, 2, 13, 33, 14, 24 a

Organise the data into a frequency table using class intervals of 10. Include a percentage frequency column.

b Construct a histogram for the data, showing both the frequency and percentage frequency on the one graph. c

Construct a stem-and-leaf plot for the data.

d Use your stem-and-leaf plot to find the median. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Statistics and Probability

SOL UTI ON

EX P L A NA TI ON

a

Calculate each percentage frequency by dividing the frequency by the total (i.e. 20) and multiplying by 100.

Class interval Frequency Percentage frequency 0– 2 10 10– 8 40 20– 5 25 30– 4 20 40–50 1 5 Total 20 100

c

40 30 20 10

8 6 4 2 0

10

Stem 0 1 2 3 4

20 30 Score

Leaf 25 01345689 14589 1123 0

40

50

Percentage frequency

Frequency

b Number of times the internet is accessed

619

Transfer the data from the frequency table to the histogram. Axis scales are evenly spaced and the histogram bar is placed across the boundaries of the class interval. There is no space between the bars.

Order the data in each leaf and also show a key (e.g. 3 | 1 means 31).

3 | 1 means 31

Median =

19 + 21 2

= 20

Exercise 9B 1

After counting the scores from the lowest value (i.e. 2), the two middle values are 19 and 21, so the median is the mean of these two numbers.

1, 2

2(a)

—

A number of families were surveyed to find the number of children in each. The results are shown in this dot plot. a How many families were surveyed? b Find the mean number of children in the families surveyed. c Find the median number of children in the families surveyed. d Find the mode for the number of children in the families 0 1 2 3 surveyed. e What percentage of the families have, at most, two children?

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UNDERSTANDING

d

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Chapter 9 Statistics

UNDERSTANDING

9B 2 Complete these frequency tables. a

Class interval 0– 10– 20– 30–40 Total

Frequency 2 1 5 2

Percentage frequency

b

Class interval 80– 85– 90– 95–100 Total

Frequency 8 23 13

Percentage frequency

50

3–6 Example 3

3–6

3, 5, 6

3 The number of wins scored this season is given for 20 hockey teams. Here are the raw data. 4, 8, 5, 12, 15, 9, 9, 7, 3, 7, 10, 11, 1, 9, 13, 0, 6, 4, 12, 5

FLUENCY

620

a Organise the data into a frequency table using class intervals of 5 and include a percentage frequency column. b Construct a histogram for the data, showing both the frequency and percentage frequency on the one graph. c Construct a stem-and-leaf plot for the data. d Use your stem-and-leaf plot to find the median.

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Statistics and Probability

Type of transport Car Train Tram Walking Bicycle Bus Total

Frequency 16 6 8 5 2 3 40

FLUENCY

4 This frequency table displays the way in which 40 people travel to and from work. Percentage frequency

621

9B

a Copy and complete the table. b Use the table to find: i

the frequency of people who travel by train

ii

the most popular form of transport

iii the percentage of people who travel by car iv the percentage of people who walk or cycle to work v

the percentage of people who travel by public transport, including trains, buses and trams.

5 Describe each graph as symmetrical, positively skewed or negatively skewed. a b

Very low

5 6 7 8 9 c

d

Frequency

9 6 3 0

20

40 Score

60

80

6 For the data in these stem-and-leaf plots, find: i the mean (rounded to one decimal place) ii the median iii the mode a b Stem Leaf 2 3 4

137 2899 46

3 | 2 means 32

Stem 4 5 6 7 8

Low Medium High

Very high

Leaf 16 0548 18999 278 38

4 | 6 means 46

Stem 0 1 2 3

Leaf 4 049 178 2

2 | 7 means 27

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Chapter 9 Statistics

7, 8

9B

8, 9

9, 10

7 Two football players, Nick and Jayden, compare their personal tallies of the number of goals scored for their team over a 12-match season. Their tallies are as follows. Game Nick Jayden

a b c d

1 0 0

2 2 0

3 2 4

4 0 1

5 3 0

6 1 5

7 2 0

8 1 3

9 2 1

10 3 0

11 0 4

12 1 0

PROBLEM-SOLVING

622

Draw a dot plot to display Nick’s goal-scoring achievement. Draw a dot plot to display Jayden’s goal-scoring achievement. How would you describe Nick’s scoring habits? How would you describe Jayden’s scoring habits?

8 Three different electric sensors, A, B and C, are used to detect movement in Harvey’s backyard over a period of 3 weeks. An in-built device counts the number of times the sensor detects movement each night. The results are as follows. Day Sensor A Sensor B Sensor C

1 2 3 0 0 1 0 15 1 4 6 8

4 5 6 7 0 0 1 1 2 18 20 2 3 5 5 5

8 0 1 4

9 10 11 12 13 14 15 16 17 18 19 20 21 0 2 0 0 0 0 0 1 1 0 0 1 0 3 25 0 0 1 15 8 9 0 0 2 23 2 8 2 3 3 1 2 2 1 5 4 0 4 9

a Using class intervals of 3 and starting at 0, draw up a frequency table for each sensor. b Draw histograms for each sensor. c Given that it is known that stray cats consistently wander into Harvey’s backyard, how would you describe the performance of: i sensor A? ii sensor B? iii sensor C? Possums could set off the sensors

9 This tally records the number of mice that were weighed and categorised into particular mass intervals for a scientific experiment. Mass (grams) Tally a Construct a table using these column headings: Mass, 10– ||| Frequency and Percentage frequency. 15– |||| |  b Find the total number of mice weighed in the 20– ||||  ||||  |||| |  experiment. 25– ||||  ||||  ||||  |||| |  c State the percentage of mice that were in the 20– gram 30–35 |||| interval. d Which was the most common weight interval? e What percentage of mice were in the most common mass interval? f What percentage of mice had a mass of 15 grams or more?

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10 A school symphony orchestra contains four musical sections: strings, woodwind, brass and percussion. The number of students playing in each section is summarised in this tally. a Construct and complete a percentage frequency table for the data. b What is the total number of students in the school orchestra?

Section String Woodwind Brass Percussion

Tally ||||  ||||  ||||  |||| |  |||| |||  |||| ||  ||||

PROBLEM-SOLVING

Statistics and Probability

623

9B

c What percentage of students play in the string section? d What percentage of students do not play in the string section? e If the number of students in the string section increases by 3, what will be the percentage of students who play in the percussion section? Round your answer to one decimal place. f What will be the percentage of students in the string section of the orchestra if the entire woodwind section is absent? Round your answer to one decimal place. 11, 12

12, 13

Frequency

11 This histogram shows the distribution of test scores for a class. Explain why the percentage of scores in the 20–30 range is 25%.

12 10 8 6 4 2 0

10

20 30 Score

40

REASONING

11

50

12 Explain why the exact value of the mean, median and mode cannot be determined directly from a histogram. 13 State the possible values of a, b and c in this ordered stem-and-leaf plot. Stem 3 4 5 6

Leaf 23a7 b4899 0149c 26

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9B

Chapter 9 Statistics

Cumulative frequency curves and percentiles

—

—

14

14 Cumulative frequency is obtained by adding a frequency to the total of its predecessors. It is sometimes referred to as a ‘running total’. cumulative frequency Percentage cumulative frequency = × 100 total number of data elements

ENRICHMENT

624

Percentage Number of Cumulative cumulative hours Frequency frequency frequency 0– 1 1 3.3 4– 4 5 16.7 8– 7 12 40.0 12– 12 24 80.0 16– 3 27 90.0 20– 2 29 96.7 24–28 1 30 100.0

Percentage cumulative frequency

A cumulative frequency graph is one in which the heights of the columns are proportional to the corresponding cumulative frequencies. The points in the upper right-hand corners of these rectangles join to form a smooth curve called the cumulative frequency curve. If a percentage scale is added to the vertical axis, the same graph can be used as a percentage cumulative frequency curve, which is convenient for the reading of percentiles.

Cumulative frequency 100 75 50 25

30 25 20 15 10 5 4

9.5 13

8 12 16 20 24 28 Hours

25th percentile 80th percentile 50th percentile

The following information relates to the amount, in dollars, of winter gas bills for houses in a suburban street. Amount ($) 0– 40– 80– 120– 160– 200–240

a b c d e f

Frequency 2 1 12 18 3 1

Cumulative frequency

Percentage cumulative frequency

Copy and complete the table. Round the percentage cumulative frequency to one decimal place. Find the number of houses that have gas bills of less than $120. Construct a cumulative frequency curve for the gas bills. Estimate the following percentiles. i 50th ii 20th iii 80th In this street, 95% of households pay less than what amount? What percentage of households pay less than $100?

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Statistics and Probability

625

Using calculators to graph grouped data 1 Enter the following data in a list called data and find the mean and median. 21, 34, 37, 24, 19, 11, 15, 26, 43, 38, 25, 16, 9, 41, 36, 31, 24, 21, 30, 39, 17 2 Construct a histogram using intervals of 3 and percentage frequency for the data above.

Using the TI-Nspire:

Using the ClassPad:

1 In a Lists and spreadsheets page type in the list name data and enter the values as shown. Use b>Statistics>Stat Calculations>One-Variable Statistics and press ·. Scroll to view the statistics.

1 In the Statistics application enter the data into list1. Tap Calc, One-Variable and then OK. Scroll to view the statistics.

2

2 Tap SetGraph, ensure StatGraph1 is ticked and then tap Setting. Change the Type to Histogram, set XList

Insert a Data and Statistics page and select the data variable for the horizontal axis. Use b>Plot Type> Histogram. Then use b>Plot Properties> Histogram Properties>Bin Settings>Equal Bin Width. Choose the Width to be 3 and Alignment to be 0. Use b>Window/Zoom>Zoom-Data to auto rescale.

to list1, Freq to 1 and then tap on Set. Tap HStart to 9 and HStep to 3.

and set

Use b>Plot Properties>Histogram Properties>Histogram Scale>Percent to show the percentage frequency.

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626

Chapter 9 Statistics

9C Summary statistics In addition to the median of a single set of data, there are two related statistics called the upper and lower quartiles. When data are placed in order, then the lower quartile is central to the lower half of the data and the upper quartile is central to the upper half of the data. These quartiles are used to calculate the interquartile range, which helps to describe the spread of the data, and determine whether or not any data points are outliers.

Let’s start: House prices A real estate agent tells you that the median house price for a suburb in 2015 was $753 000 and the mean was $948 000. • Is it possible for the median and the mean to differ by so much? • Under what circumstances could this occur? Discuss.

Key ideas

Five-figure summary • Minimum value (min): the minimum value • Lower quartile (Q1 ): the number above 25% of the ordered data • Median (Q2 ): the middle value above 50% of the ordered data • Upper quartile (Q3 ): the number above 75% of the ordered data • Maximum value (max): the maximum value Measures of spread • Range = max value – min value • Interquartile range (IQR) IQR = upper quartile – lower quartile = Q 3 – Q1 • The standard deviation is discussed in section 9E.

Odd number

↓

{

{

1 2 2 3) 5 (6 6 7 9

↓

↓

Even number 2 3 3 4 7 8 8 9 9 9

↓

↓

↓

Q1 (2) Q2 (5) Q3 (6.5)

Q1 (3) Q2 (7.5) Q3 (9)

IQR (4.5)

IQR (6)

Outliers are data elements outside the vicinity of the rest of the data. More formally, a data point is an outlier when it is below the lower fence (i.e. lower limit) or above the upper fence (i.e. upper limit). • Lower fence = Q1 – 1.5 × IQR • Upper fence = Q3 + 1.5 × IQR Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Statistics and Probability

627

Example 4 Finding the range and IQR Determine the range and IQR for these data sets by finding the five-figure summary. a 2, 2, 4, 5, 6, 8, 10, 13, 16, 20 b

1.6, 1.7, 1.9, 2.0, 2.1, 2.4, 2.4, 2.7, 2.9

SO L U T I O N

EX P L A N A T I O N

a Range = 20 – 2 = 18 2 2 4 5 6 8 10 13 16 20

Range = max – min

↑

↑

↑

Q1

Q2 (7)

Q3

Q2 = 7, so Q1 = 4 and Q3 = 13. IQR = 13 – 4 = 9 b Range = 2.9 – 1.6 = 1.3

First, split the ordered data in half to locate the 6+8 median, which is = 7. 2 Q1 is the median of the lower half and Q3 is the median of the upper half. IQR = Q3 – Q1 Max = 2.9, min = 1.6

1.6 1.7 1.9 2.0 2.1 2.4 2.4 2.7 2.9

1.6 1.7 1.9 2.0 2.1 2.4 2.4 2.7 2.9

↑

Q1 1.7 + 1.9 = 1.8 2 2.4 + 2.7 = 2.55 Q3 = 2 IQR = 2.55 – 1.8 = 0.75 Q1 =

↑

Q2

↑

Q3

Leave the median out of the upper and lower halves when locating Q1 and Q3 .

Example 5 Finding the five-figure summary and outliers The following data set represents the number of flying geese spotted on each day of a 13-day tour of England. 5, 1, 2, 6, 3, 3, 18, 4, 4, 1, 7, 2, 4 a

For the data, find: i the minimum and maximum number of geese spotted ii the median iii the upper and lower quartiles iv the IQR v

b

any outliers by determining the lower and upper fences.

Can you give a possible reason for why the outlier occurred?

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628

Chapter 9 Statistics

SOL UTI ON

EX P L A NA TI ON

a

Look for the largest and smallest numbers and order the data:

i

Min = 1, max = 18

ii 1, 1, 2, 2, 3, 3, 4, 4, 4, 5, 6, 7, 18 ∴ Median = 4 2+2 iii Lower quartile = 2

1 1 2 2 3 3) 4 (4 4 5 6 7 18

↑

=2 Upper quartile =

↑

Q1

5+6 2

Q2

↑

Q3

Since Q2 falls on a data value, it is not included in the lower or higher halves when Q1 and Q3 are calculated.

= 5.5 iv IQR = 5.5 – 2

IQR = Q3 – Q1

= 3.5 A data point is an outlier when it is less than Q1 – 1.5 × IQR or greater than Q3 + 1.5 × IQR.

v Lower fence = Q1 – 1.5 × IQR = 2 – 1.5 × 3.5 = –3.25 Upper fence = Q3 + 1.5 × IQR = 5.5 + 1.5 × 3.5 = 10.75 ∴ The outlier is 18.

There are no numbers less than –3.25 but 18 is greater than 10.75.

b Perhaps a flock of geese was spotted that day.

1

a b c d

1–3

3

State the types of values that must be calculated for a five-figure summary. Explain the difference between the range and the interquartile range. What is an outlier? How do you determine if a score in a single data set is an outlier?

—

UNDERSTANDING

Exercise 9C

2 This data set shows the number of cars in 13 families surveyed. 1, 4, 2, 2, 3, 8, 1, 2, 2, 0, 3, 1, 2 a Arrange the data in ascending order. b Find the median (i.e. the middle value). c By first removing the middle value, determine: i the lower quartile Q1 (middle of lower half) ii the upper quartile Q3 (middle of upper half). d e f

Determine the interquartile range (IQR). Calculate Q1 – 1.5 × IQR and Q3 + 1.5 × IQR. Are there any values that are outliers (numbers below Q1 – 1.5 × IQR or above Q3 + 1.5 × IQR)?

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UNDERSTANDING

Statistics and Probability

3 The number of ducks spotted in eight different flocks are given in this data set. 2, 7, 8, 10, 11, 11, 13, 15 a

i

Find the median (i.e. average of the middle two numbers).

ii

Find the lower quartile (i.e. middle of the smallest four numbers).

629

9C

iii Find the upper quartile (i.e. middle of the largest four numbers). b Determine the IQR. c Calculate Q1 – 1.5 × IQR and Q3 + 1.5 × IQR. d Are there any outliers (i.e. numbers below Q1 – 1.5 × IQR or above Q3 + 1.5 × IQR)?

Example 4

4–5(½), 6, 7

4–5(½), 6, 7

4 Determine the range and IQR for these data sets by finding the five-figure summary. a 3, 4, 6, 8, 8, 10, 13 b 10, 10, 11, 14, 14, 15, 16, 18 c 1.2, 1.8, 1.9, 2.3, 2.4, 2.5, 2.9, 3.2, 3.4 d 41, 49, 53, 58, 59, 62, 62, 65, 66, 68

FLUENCY

4–5(½), 6

5 Determine the median and mean of the following sets of data. Round to one decimal place where necessary. a 6, 7, 8, 9, 10 b 2, 3, 4, 5, 5, 6 c 4, 6, 3, 7, 3, 2, 5 Example 5

6 The following numbers of cars, travelling on a quiet suburban street, were counted on each day for 15 days. 10, 9, 15, 14, 10, 17, 15, 0, 12, 14, 8, 15, 15, 11, 13 For the given data, find: a the minimum and maximum number of cars counted b the median c the lower and upper quartiles d the IQR e any outliers by determining the lower and upper fences f a possible reason for the outlier.

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630

Chapter 9 Statistics

a 4, 5, 10, 7, 5, 14, 8, 5, 9, 9

ii the median (Q2 ) iv the IQR b 24, 21, 23, 18, 25, 29, 31, 16, 26, 25, 27

8, 9

9, 10

10, 11

PROBLEM-SOLVING

7 Summarise the data sets below by finding: i the minimum and maximum values iii the lower and upper quartiles (Q1 and Q3 ) v any outliers.

FLUENCY

9C

8 Twelve different calculators had the following numbers of buttons. 36, 48, 52, 43, 46, 53, 25, 60, 128, 32, 52, 40 a For the given data, find: i the minimum and maximum number of buttons on the calculators ii the median iii the lower and upper quartiles iv the IQR v any outliers vi the mean. b Which is a better measure of the centre of the data, the mean or the median? Explain. c Can you give a possible reason why the outlier has occurred? 9 Using the definition of an outlier, decide whether or not any outliers exist in the following sets of data. If so, list them. a b c d

3, 6, 1, 4, 2, 5, 9, 8, 6, 3, 6, 2, 1 8, 13, 12, 16, 17, 14, 12, 2, 13, 19, 18, 12, 13 123, 146, 132, 136, 139, 141, 103, 143, 182, 139, 127, 140 2, 5, 5, 6, 5, 4, 5, 6, 7, 5, 8, 5, 5, 4

10 For the data in this stem-and-leaf plot, find: a the IQR b any outliers c the median if the number 37 is added to the list d the median if the number 22 is added to the list instead of 37.

Stem 0 1 2 3

Leaf 1 68 046 23

2 | 4 means 24

11 Three different numbers have median 2 and range 2. Find the three numbers.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Statistics and Probability

12, 13

12 Explain what happens to the mean of a data set if all the values are: a increased by 5 b multiplied by 2 c divided by 10. 13 Explain what happens to the IQR of a data set if all values are: a increased by 5 b multiplied by 2

c divided by 10.

14 Give an example of a small data set that satisfies the following. a median = mean b median = upper quartile

c range = IQR

9C

13–15

REASONING

12

631

15 Explain why, in many situations, the median is preferred to the mean as a measure of centre.

—

—

16

16 Use the internet to search for data about a topic that interests you. Try to choose a single set of data that includes between 15 and 50 values. a Organise the data using: i a stem-and-leaf plot ii a frequency table and histogram. b Find the mean and the median. c Find the range and the interquartile range. d Write a brief report describing the centre and spread of the data, referring to parts a to c above. e Present your findings to your class or a partner.

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ENRICHMENT

Some research

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marises the average 632for a Chapter 9 Statistics r 100 km) nd European-made cars.

mmary (min, Q1 , Q2 , Q3 , max) can be represented in graphical form as a box plot. hs that summarise single data sets. They clearly display the minimum and he median, the quartiles and any outliers. Box plots also give a clear indication five-figure summary Q1 , Q2box. , Q3 , max) can be represented in graphical form as a box plot. ead, as the IQRThe is shown by the width of(min, the central Box plots are graphs that summarise single data sets. They clearly display the minimum and maximum values, the median, the quartiles and any outliers. Box plots also give a clear indication el consumption say about howofthe fuel consumption compares andcentral box. how data are spread, as the IQRbetween is shownAustralian-made by the width of the

9D Box plots

lot summarises the average the box plot represent? litres per 100 km) for astart: Fuel consumption Let’s cross (฀) represents on the European cars box plot? n-made and European-made cars. This parallel box plot summarises the average ed a box-and-whisker plot) can be used to summarise a data set. fuel consumption (litres per 100 km) for a ata values in each quarter (25%) are approximately equal. group of Australian-made and European-made cars.

Australian cars European cars

ox plots say about how the fuel consumption compares between Australian-made and 4 6 10 8 12 14 16 e cars? Fuel consumption L/100 km h part of the box plot represent? with a cross ( × ). ink the cross (฀)• represents European cars box how plot?the fuel consumption compares between Australian-made and What doon thethe box plots say about ter than Q3 + 1.5 × IQR or less than Q1 – 1.5 × IQR. European-made cars? etch to the lowest and highest data values that are not outliers. also called a box-and-whisker plot)part can of bethe used to plot summarise a data set. • What does each box represent? ber of data values eachdoquarter (25%) approximately equal. • in What you think theare cross ( ) represents on the European cars box plot?

Key A box plot (also called a box-and-whisker plot) can be used to summarise a data set. • The number datascale. valuesThey in each are approximately equal. ideas e two or more box plots drawn on theofsame are quarter used to (25%) compare

ame context.

25 %

25 %

25 %

25 %

marked with a cross ( × ). Q2 Q3 Maximum 1 1 – 1.5 × IQR. r is greater than Q3 + 1.5 ×Minimum IQR or less thanQQ value value kers stretch to the lowest and highest data values that are not outliers. Whisker Whisker Box An outlier is marked with a cross ( ). • An outlier is greater than Q3 + 1.5 × IQR or less than Q1 – 1.5 × IQR. • The whiskers stretch to the lowest and highest data values that are not outliers.

plots are two or more box plots drawn on the same scale. They are used to compare hin the same context. Minimum Q1 Q2 Q3 Outlier value Parallel box plots are two or more box plots drawn on the same scale. They are used to compare data sets within the same context.

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Statistics and Probability

633

Example 6 Constructing box plots Consider the given data set: 5, 9, 4, 3, 5, 6, 6, 5, 7, 12, 2, 3, 5 a Determine whether any outliers exist by first finding Q1 and Q3 . b

Draw a box plot to summarise the data, marking outliers if they exist.

SOL UTI ON

EX P L A NA TI ON

{

↑

Q1

Order the data to help find the quartiles.

{

2 3 3 4 5 5 ) 5 ( 5 6 6 7 9 12

a

↑

Q2

3 4 2 =35

Q1 =

↑

Q3

Locate the median Q2 then split the data in half above and below this value.

6 7 2 =65

Q1 is the middle value of the lower half and Q3 the middle value of the upper half.

Q1 =

∴ IQR = 6.5 – 3.5

Determine IQR = Q3 – Q1 .

=3 Q1 – 1.5 × IQR = 3.5 – 1.5 × 3 = –1

Check for any outliers; i.e. values below Q1 – 1.5 × IQR or above Q3 + 1.5 × IQR.

Q3 + 1.5 × IQR = 6.5 + 1.5 × 3 = 11 ∴ 12 is an outlier.

There are no data values below –1 but 12 > 11.

2 3 4 5 6 7 8 9 10 11 12

Exercise 9D 1

Draw a line and mark in a uniform scale reaching from 2 to 12. Sketch the box plot by marking the minimum 2 and the outlier 12 and Q1 , Q2 and Q3 . The end of the five-point summary is the nearest value below 11; i.e. 9.

1–3

2

For this simple box plot, find:

a c e g

5 10 15 20 25 the median (Q2 ) the maximum the lower quartile (Q1 ) the interquartile range (IQR).

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b the minimum d the range f the upper quartile (Q3 )

—

UNDERSTANDING

b

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634

Chapter 9 Statistics

UNDERSTANDING

9D 2 Complete the following for this box plot.

0 4 8 12 16 20 a Find the IQR. b Calculate Q1 – 1.5 × IQR. c Calculate Q3 + 1.5 × IQR. d State the value of the outlier. e Check that the outlier is greater than Q3 + 1.5 × IQR. 3 Construct a box plot that shows these features. a min = 1, Q1 = 3, Q2 = 4, Q3 = 7, max = 8 b outlier = 5, minimum above outlier = 10, Q1 = 12, Q2 = 14, Q3 = 15, max = 17

Example 6

4–5(½)

4–5(½)

FLUENCY

4–5(½)

4 Consider the data sets below. i ii a b c d

Determine whether any outliers exist by first finding Q1 and Q3 . Draw a box plot to summarise the data, marking outliers if they exist.

4, 6, 5, 2, 3, 4, 4, 13, 8, 7, 6 1.8, 1.7, 1.8, 1.9, 1.6, 1.8, 2.0, 1.1, 1.4, 1.9, 2.2 21, 23, 18, 11, 16, 19, 24, 21, 23, 22, 20, 31, 26, 22 0.04, 0.04, 0.03, 0.03, 0.05, 0.06, 0.07, 0.03, 0.05, 0.02

5 First, find Q1 , Q2 and Q3 and then draw box plots for the given data sets. Remember to find outliers and mark them on your box plot if they exist. 11, 15, 18, 17, 1, 2, 8, 12, 19, 15 37, 48, 52, 51, 51, 42, 48, 47, 39, 41, 65 0, 1, 5, 4, 4, 4, 2, 3, 3, 1, 4, 3 124, 118, 73, 119, 117, 120, 120, 121, 118, 122 6, 7

7, 8

7, 8

6 Consider these parallel box plots, A and B. A B

PROBLEM-SOLVING

a b c d

1 3 5 7 9 11 13 15 17 19 21 23 a What statistical measure do these box plots have in common? b Which data set (A or B) has a wider range of values? c Find the IQR for: i data set A ii data set B. d How would you describe the main difference between the two sets of data from which the parallel box plots have been drawn?

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PROBLEM-SOLVING

Statistics and Probability

7 The following masses, in kilograms, of 15 Madagascan lemurs are recorded as part of a conservation project. 14.4, 15.5, 17.3, 14.6, 14.7 15.0, 15.8, 16.2, 19.7, 15.3 13.8, 14.6, 15.4, 15.7, 14.9

635

9D

a Find Q1 , Q2 and Q3 . b Which masses, if any, would be considered outliers? c Draw a box plot to summarise the lemurs’ masses. 8 Two data sets can be compared using parallel box plots on the same scale, as shown below. A B

10 12 14 16 18 20 22 a What statistical measures do these box plots have in common? b Which data set (A or B) has a wider range of values? c Find the IQR for: i data set A ii data set B. d How would you describe the main difference between the two sets of data from which the parallel box plots have been drawn? 9, 10

10, 11

9 Select the box plot (A or B) that best matches the given dot plot or histogram. a B

REASONING

9

A 1 2 3 4 5 6 7

1 2 3 4 5 6 7 b

B A 4 5 6 7 8 9

Frequency

c

6 5 4 3 2 1 0

4 5 6 7 8 9

B A 10

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

20 30 Score

40

0

10 20 30 40

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Chapter 9 Statistics

9D 10 Fifteen essays are marked for spelling errors by a particular examiner and the following numbers of spelling errors are counted. 3, 2, 4, 6, 8, 4, 6, 7, 6, 1, 7, 12, 7, 3, 8 The same 15 essays are marked for spelling errors by a second examiner and the following numbers of spelling errors are counted. 12, 7, 9, 11, 15, 5, 14, 16, 9, 11, 8, 13, 14, 15, 13

REASONING

636

a Draw parallel box plots for the data. b Do you believe there is a major difference in the way the essays were marked by the two examiners? If yes, describe this difference. 11 The results for a Year 12 class are to be compared with the Year 12 results of the school and the State, using the parallel box plots shown. a

Describe the main differences between the performance of: i

the class against the school

ii

the class against the State

iii

the school against the State.

State School Class 0

20

40 60 Score %

80

100

b Why is an outlier shown on the class box plot but not shown on the school box plot?

12 a

—

—

12

Choose an area of study for which you can collect data easily, for example: • heights or weights of students • maximum temperatures over a weekly period • amount of pocket money received each week for a group of students. b Collect at least two sets of data for your chosen area of study – perhaps from two or three different sources, including the internet. Examples: • Measure student heights in your class and from a second class in the same year level. • Record maximum temperatures for 1 week and repeat for a second week to obtain a second data set. • Use the internet to obtain the football scores of two teams for each match in the previous season. c Draw parallel box plots for your data. d Write a report on the characteristics of each data set and the similarities and differences between the data sets collected.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

ENRICHMENT

Creating your own parallel box plots

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Statistics and Probability

637

Using calculators to draw box plots 1 Type these data into lists and define them as Test A and Test B. Test A: 4, 6, 3, 4, 1, 3, 6, 4, 5, 3, 4, 3 Test B: 7, 3, 5, 6, 9, 3, 6, 7, 4, 1, 4, 6 2 Draw parallel box plots for the data.

Using the TI-Nspire:

Using the ClassPad:

1 In a Lists and spreadsheets page type in the list names testa and testb and enter the values as shown.

1

In the Statistics application enter the data into the lists. Give each column a title.

2 Insert a Data and Statistics page and select the testa variable for the horizontal axis. Change to a box plot using b>Plot Type>Box Plot. Trace (or hover over) to reveal the statistical measures. To show the box plot for testb, use b>Plot Properties>Add X Variable and select testb.

Tap . For graph 1, set Draw to On, Type to MedBox, XList to main\TestA and Freq to 1. For graph 2, set Draw to On, Type to MedBox, XList to main\TestB and

2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Freq to 1. Tap Set. Tap

.

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638

Chapter 9 Statistics

9E Standard deviation

10A

For a single data set we have already discussed the range and interquartile range to describe the spread of the data. Another statistic commonly used to describe spread is standard deviation. The standard deviation is a number that describes how far data values are from the mean. A data set with a relatively small standard deviation will have data values concentrated about the mean, and if a data set has a relatively large standard deviation then the data values will be more spread out from the mean. The standard deviation can be calculated by hand but, given the tedious nature of the calculation, technology can be used for more complex data sets. In this section technology is not required but you will be able to find a function on your calculator (often denoted s) that can be used to find the standard deviation.

Let’s start: Which is the better team? These histograms show the number of points scored by the Eagles and the Monsters basketball teams in an 18-round competition. The mean and standard deviation are given for each team. • Which team has the higher mean? What does this say about the team’s performance? • Which team has the smaller standard deviation? What does this say about the team’s performance? Discuss.

Monsters

Frequency

Eagles

5 4 3 2 1

Mean = 52 Standard deviation = 21

10 20 30 40 50 60 70 80 90 100 Score

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7 6 5 4 3 2 1

Mean = 43 Standard deviation = 11

10 20 30 40 50 60 70 80 90 100 Score

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Statistics and Probability

The standard deviation is a number that describes the spread of data about the mean. • The symbol used for standard deviation is s (sigma). • The sample standard deviation is for a sample data set drawn from the population. • If every data value from a population is used, then we calculate the population standard deviation. To calculate the sample standard deviation, follow these steps. 1 Find the mean (¯x). 2

Find the difference between each value and the mean (called the deviation).

3

Square each deviation.

4

Sum the squares of each deviation.

5

Divide by the number of data values less 1 (i.e. n – 1).

639

Key ideas

 (x1 – ¯x)2 + (x2 – ¯x)2 + . . . + (xn – ¯x)2 s= (n – 1)

6 Take the square root. If the data represent the complete population, then divide by n instead of (n – 1). This would give the population standard deviation. Dividing by (n – 1) for the sample standard deviation gives a better estimate of the population standard deviation. If data are concentrated about the mean, then the standard deviation is relatively small. If data are spread out from the mean, then the standard deviation is relatively large. In many common situations we can expect 95% of the data to be within two standard deviations of the mean.

Example 7 Calculating the standard deviation Calculate the mean and sample standard deviation for this small data set, correct to one decimal place. 2, 4, 5, 8, 9 SOL UTI ON

¯x =

2+4+5+8+9 5

EX P L A NA TI ON

Sum all the data values and divide by the number of data values (i.e. 5) to find the mean.

= 5.6  (2 – 5.6)2 + (4 – 5.6)2 + (5 – 5.6)2 + (8 – 5.6)2 + (9 – 5.6)2 s= Deviation 1 is 2 – 5.6 (the 5–1 difference between the data value and the mean).  Sum the square of all the (–3.6)2 + (–1.6)2 + (–0.6)2 + (2.4)2 + (3.4)2 = deviations, divide by n – 1 (i.e. 4) 4 and then take the square root. = 2.9 (to 1 d.p.)

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640

Chapter 9 Statistics

Example 8 Interpreting the standard deviation This back-to-back stem-and-leaf plot shows the distribution of distances that 17 people in Darwin and Sydney travel to work. The means and standard deviations are given. Darwin Leaf 8742 99553 87430 522

Stem 0 1 2 3 4 5

Sydney Leaf 15 237 0556 2599 446 2

Sydney ¯x = 27.9 s = 15.1 Darwin ¯x = 19.0 s = 10.1

3 | 5 means 35 km

Consider the position and spread of the data and then answer the following. a By looking at the stem-and-leaf plot, suggest why Darwin’s mean is less than that of Sydney. b

Why is Sydney’s standard deviation larger than that of Darwin?

c

Give a practical reason for the difference in centre and spread for the data for Darwin and Sydney.

SO L U T I O N

EX P L A N A T I O N

a The maximum score for Darwin is 35. Sydney’s mean is affected by several values larger than 35.

The mean depends on every value in the data set.

b The data for Sydney are more spread out from the mean. Darwin’s scores are more closely clustered near its mean.

Sydney has more scores with a large distance from its mean. Darwin’s scores are closer to the Darwin mean.

c Sydney is a larger city and more spread out, so people have to travel farther to get to work.

Higher populations often lead to larger cities and longer travel distances.

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Statistics and Probability

1

1–4

3, 4

—

Insert the word smaller or larger into each sentence. a If data are more spread out from the mean, then the standard deviation is b If data are more concentrated about the mean, then the standard deviation is

. .

2 Here are two dot plots, A and B. A

UNDERSTANDING

Exercise 9E

641

B

1 2 3 4 5 1 2 3 4 5 a Which data set (A or B) would have the higher mean? b Which data set (A or B) would have the higher standard deviation? 3 These dot plots show the results for a class of 15 students who sat tests A and B. Both sets of results have the same mean and range. A

B

3 4 5 6 7 8 9 Mean

3 4 5 6 7 8 9 Mean a Which data set (A or B) would have the higher standard deviation? b Give a reason for your answer in part a.

4 This back-to-back stem-and-leaf plot compares the number of trees or shrubs in the backyards of homes in the suburbs of Gum Heights and Oak Valley. a

Which suburb has the smaller mean number of trees or shrubs? Do not calculate the actual means. b Without calculating the actual standard deviations, which suburb has the smaller standard deviation?

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Gum Heights Leaf 731 8640 9872 964

Stem 0 1 2 3 4

Oak Valley Leaf 0 0236889 4689 36

2 | 8 means 28

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Chapter 9 Statistics

5(½), 6

9E Example 7

5(½), 6

5(½), 6

5 Calculate the mean and sample standard deviation for these small data sets. Use the formula for the sample standard deviation. Round the standard deviation to one decimal place where necessary. a 3, 5, 6, 7, 9 b 1, 1, 4, 5, 7 c 2, 5, 6, 9, 10, 11, 13 d 28, 29, 32, 33, 36, 37

FLUENCY

642

6 Calculate the mean and sample standard deviation for the data in these graphs, correct to one decimal place. a b Stem Leaf 0 1 2

1 2 3 4

4 137 02

1 | 7 means 17

Example 8

7 This back-to-back stem-and-leaf plot shows the distribution of distances travelled by students at an inner-city and an outer-suburb school. The means and standard deviations are given.

Inner-city Leaf 964311 9420 71

7, 8

8, 9

Outer-suburb Leaf 349 2889 134 4 1

Stem 0 1 2 3 4

Inner-city ¯x = 10.6 s = 8.0 Outer-suburb ¯x = 18.8 s = 10.7

PROBLEM-SOLVING

7, 8

Consider the position and spread of the data and then answer the following. 2 | 4 means 24 a Why is the mean for the outer-suburb school larger than that for the inner-city school? b Why is the mean for the outer-suburb school larger than that for the inner-city school? c Why is the standard deviation for the inner-city school smaller than that for the outer-suburb school? d Give a practical reason for the difference in centre and spread for the two schools. 8 Consider these two histograms, and then state whether the following are true or false. A 5

B 5

4 3 2 1 0

4 3 2 1 0

10

20

30

40

50

10

20

30

40

50

a The mean for set A is greater than the mean for set B. b The range for set A is greater than the range for set B. c The standard deviation for set A is greater than the standard deviation for set B. 9 Find the mean and sample standard deviation for the scores in these frequency tables. Round the standard deviations to one decimal place. a b Score Frequency Score Frequency 1 2 3

3 1 3

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

4 5 6

1 4 3

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Statistics and Probability

10, 11

11, 12

10 Two simple data sets, A and B, are identical except for the maximum value, which is an outlier for set B. A: 4, 5, 7, 9, 10 B: 4, 5, 7, 9, 20 a b c d

REASONING

10

643

9E

Is the range for set A equal to the range for set B? Is the mean for each data set the same? Is the median for each data set the same? Would the standard deviation be affected by the outlier? Explain.

11 Data sets 1 and 2 have means ¯x1 and ¯x2 , and standard deviations s1 and s2 . a If ¯x1 > ¯x2 , does this necessarily mean that s1 > s2 ? Give a reason. b If s1 < s2 does this necessarily mean that ¯x1 < ¯x2 ? 12 Data sets A and B each have 20 data values and are very similar except for an outlier in set A. Explain why the interquartile range might be a better measure of spread than the range or the standard deviation. —

—

13

13 The Mathematics study scores (out of 100) for 50 students in a school are as listed. 71, 85, 62, 54, 37, 49, 92, 85, 67, 89 96, 44, 67, 62, 75, 84, 71, 63, 69, 81 57, 43, 64, 61, 52, 59, 83, 46, 90, 32 94, 84, 66, 70, 78, 45, 50, 64, 68, 73 79, 89, 80, 62, 57, 83, 86, 94, 81, 65

ENRICHMENT

Study scores

The mean (¯x) is 69.16 and the population standard deviation (s) is 16.0. a

Calculate: i ¯x + s iv ¯x – 2s

ii ¯x – s v ¯x + 3s

iii ¯x + 2s vi ¯x – 3s

b Use your answers from part a to find the percentage of students with a score within: i one standard deviation from the mean ii two standard deviations from the mean iii three standard deviations from the mean. c

i

Research what it means when we say that the data are ‘normally distributed’. Give a brief explanation.

ii For data that are normally distributed, find out what percentage of data are within one, two and three standard deviations from the mean. Compare this with your results for part b above.

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644

Chapter 9 Statistics

Progress quiz 38pt 9A

1

What type of data would these survey questions generate? a How many pets do you have? b What is your favourite ice-cream flavour?

38pt 9B

2

A Year 10 class records the length of time (in minutes) each student takes to travel from home to school. The results are listed here. 15 8

38pt 9C

3

32 16

6 33

14 20

44 19

28 27

15 23

9 12

25 38

18 15

a

Organise the data into a frequency table, using class intervals of 10. Include a percentage frequency column.

b

Construct a histogram for the data, showing both the frequency and percentage frequency on the one graph.

c

Construct a stem-and-leaf plot for the data.

d

Use your stem-and-leaf plot to find the median.

Determine the range and IQR for these data sets by finding the five-figure summary. a 4, 9, 12, 15, 16, 18, 20, 23, 28, 32 b 4.2, 4.3, 4.7, 5.1, 5.2, 5.6, 5.8, 6.4, 6.6

38pt 9C

4

The following numbers of parked cars were counted in the school car park and adjacent street each day at morning recess for 14 school days. 36, 38, 46, 30, 69, 31, 40, 37, 55, 34, 44, 33, 47, 42 a

b 38pt 9D

38pt 9E 10A

5

6

For the data, find: i

the minimum and maximum number of cars

ii

the median

iii

the upper and lower quartiles

iv

the IQR

v

any outliers.

Can you give a possible reason for why the outlier occurred?

The ages of a team of female gymnasts are given in this data set: 18, 23, 14, 28, 21, 19, 15, 32, 17, 18, 20, 13, 21 a

Determine whether any outliers exist by first finding Q1 and Q3 .

b

Draw a box plot to summarise the data, marking outliers if they exist.

Find the sample standard deviation for this small data set, correct to one decimal place. Use the sample standard deviation formula. 2, 3, 5, 6, 9

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Statistics and Probability

645

9F Time-series data A time series is a sequence of data values that are recorded at regular time intervals. Examples include temperature recorded on the hour, speed recorded every second, population recorded every year and profit recorded every month. A line graph can be used to represent time-series data and these can help to analyse the data, describe trends and make predictions about the future.

Let’s start: Share price trends A company’s share price is recorded at the end of each month of the financial year, as shown in this time-series graph. • Describe the trend in the data at different times of the year. • At what time of year do you think the company starts reporting bad profit results? • Does it look like the company’s share price will return to around $4 in the next year? Why?

$ 4 3 2 1 0

Share price

July

October

January Month

April

Key ideas

Time-series data are recorded at regular time intervals.

Variable

The graph or plot of a time series uses: • time on the horizontal axis • line segments connecting points on the graph.

Time If the time-series plot results in points being on or near a straight line, then we say that the trend is linear.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Downward linear trend

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646

Chapter 9 Statistics

Example 9 Plotting and interpreting a time-series plot The approximate population of an outback town is recorded from 1990 to 2005. Year 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 Population 1300 1250 1250 1150 1000 950 900 950 800 700 800 750 850 950 800 850

a

Plot the time series.

b Describe the trend in the data over the 16 years. E X P LA NA TI ON

a

Use time on the horizontal axis. Break the y-axis so as to not include 0–700. Join points with line segments.

1300 1200 1100 1000 900 800 700 0

1990 1992 1994 1996 1998 2000 2002 2004 2006 Time (year)

b The population declines steadily for the first 10 years. The population rises and falls in the last 6 years, resulting in a slight upwards trend.

Exercise 9F

2

Variable

Describe the following time-series plots as having a linear (i.e. straight-line trend), non-linear trend (i.e. a curve) or no trend. a b

Variable

1

1, 2

Interpret the overall rise and fall of the lines on the graph.

Time

Time d

Variable

Variable

c

—

UNDERSTANDING

Population

SOL UTI ON

Time Time Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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UNDERSTANDING

9F

.

. 4

p. m

p. m 2

12

p. m

.

. a.m

. 8

d Describe the general trend in the temperature for the 8-hour school day.

Time 3–5

3, 5

3 The approximate population of a small village is recorded from 2005 to 2015. Year 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 Population 550 500 550 600 700 650 750 750 850 950 900

FLUENCY

3, 4 Example 9

647

0

10

b What was the maximum temperature? c During what times did the temperature: i stay the same? ii decrease?

36 34 32 30 28

a.m

2 This time-series graph shows the temperature over the course of an 8-hour school day. a State the temperature at: i 8 a.m. ii 12 p.m. iii 1 p.m. iv 4 p.m.

Temperature (°C)

Statistics and Probability

a Plot the time-series graph. b Describe the general trend in the data over the 11 years. c For the 11 years, what was the: i minimum population? ii maximum population?

4 A company’s share price over 12 months is recorded in this table. Month J F M A M J J A S O N D Price ($) 1.30 1.32 1.35 1.34 1.40 1.43 1.40 1.38 1.30 1.25 1.22 1.23

a Plot the time-series graph. Break the y-axis to exclude values from $0 to $1.20. b Describe the way in which the share price has changed over the 12 months. c What is the difference between the maximum and minimum share price in the 12 months? 5 The pass rate (%) for a particular examination is given in a table over 10 years. Year 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 Pass rate (%) 74 71 73 79 85 84 87 81 84 83

a Plot the time-series graph for the 10 years. b Describe the way in which the pass rate for the examination has changed in the given time period. c In what year was the pass rate a maximum? d By how much had the pass rate improved from 1995 to 1999?

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Chapter 9 Statistics

6, 7

9F

6, 7

7, 8

700 600 500 400 300

20 0 20 8 0 20 9 1 20 0 1 20 1 1 20 2 1 20 3 1 20 4 1 20 5 1 20 6 17

Price ($’000)

6 This time-series plot shows the upwards trend of house prices in an Adelaide suburb over 7 years from 2008 to 2014.

PROBLEM-SOLVING

648

a Would you say that the general trend in house prices is linear or non-linear? b Assuming the trend in house prices continues for this suburb, what would you expect the house price to be in: i 2015? ii 2017? 7 The two top-selling book stores for a company list their sales figures for the first 6 months of the year. Sales amounts are in thousands of dollars. City Central ($’000) Southbank ($’000)

a

July 12 17

August 13 19

September 12 16

October 10 12

November 11 13

December 13 9

What was the difference in the sales volume for: i August? ii December?

b In how many months did the City Central store sell more books than the Southbank store? c Construct a time-series plot for both stores on the same set of axes. d Describe the trend of sales for the 6 months for: i City Central ii Southbank Based on the trend for the sales for the Southbank store, what would you expect the approximate sales volume to be in January?

b Describe the trend in the distance from the recording machine for: i Blue Crest ii Green Tail c Assuming that the given trends continue, predict the time when the pigeons will be the same distance from the recording machine.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

8 7 6 5 4 3 2 1

Blue Crest

Green Tail

1 p. 2 m. p. 3 m. p. 4 m. p. 5 m. p. 6 m. p. 7 m. p. 8 m. p. m .

8 Two pigeons (Green Tail and Blue Crest) each have a beacon that communicates with a recording machine. The distance of each pigeon from the machine is recorded every hour for 8 hours. a State the distance from the machine at 3 p.m. for: i Blue Crest ii Green Tail

Distance (km)

e

Time (hours)

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Statistics and Probability

9, 10

10, 11

30 25 20 15 10 5 0

9F

Jan Feb Mar April May June July Aug Sep Oct Nov Dec

Average temp. (°C)

9 The average monthly maximum temperature for a city is illustrated in this graph.

REASONING

9

649

Month

a

10 The balance of an investment account is shown in this time-series plot. a b

Describe the trend in the account balance over the 7 years. Give a practical reason for the shape of the curve that models the trend in the graph.

Balance ($’000)

Explain why the average maximum temperature for December is close to the average maximum temperature for January. b Do you think this graph is for an Australian city? c Do you think the data are for a city in the Northern Hemisphere or the Southern Hemisphere? Give a reason.

25 20 15 10 5 0

1 2 3 4 5 6 7 Year

11 A drink at room temperature is placed in a fridge that is at 4◦ C. a

Sketch a time-series plot that might show the temperature of the drink after it has been placed in the fridge. b Would the temperature of the drink ever get to 3◦ C? Why?

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9F

Chapter 9 Statistics

Moving run average

—

—

12

12 In this particular question, a moving average is determined by calculating the average of all data values up to a particular time or place in the data set. Consider a batsman in cricket with the following runs scored from 10 completed innings. Innings Score Moving average

1 26 26

2 38 32

3 5 23

4 10

5 52

6 103

7 75

8 21

9 33

ENRICHMENT

650

10 0

a

Complete the table by calculating the moving average for innings 4–10. Round to the nearest whole number where required. b Plot the score and moving averages for the batter on the same set of axes. c Describe the behaviour of the: i score graph ii moving average graph. d Describe the main difference in the behaviour of the two graphs. Give reasons.

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Statistics and Probability

651

9G Bivariate data and scatter plots

Height

When we collect information about two variables in a given context, we are collecting bivariate data. As there are two variables involved in bivariate data, we use a number plane to graph the data. These graphs are called scatter plots and are used to illustrate a relationship that may exist between the variables. Scatter plots make it very easy to see the strength of the association between the two variables.

Weight

Let’s start: A relationship or not? Consider the two variables in each part below. • Would you expect there to be some relationship between the two variables in each of these cases? • If you think a relationship exists, would you expect the second listed variable to increase or to decrease as the first variable increases? a Height of person and Weight of person b

Temperature and Life of milk

c

Length of hair and IQ

d

Depth of topsoil and Brand of motorcycle

e

Years of education and Income

f

Spring rainfall and Crop yield

g

Size of ship and Cargo capacity

h

Fuel economy and CD track number

i

Amount of traffic and Travel time

j

Cost of 2 litres of milk and Ability to swim

k

Background noise and Amount of work completed Bivariate data include data for two variables. • The two variables are usually related; for example, height and weight.

Key ideas

A scatter plot is a graph on a number plane in which the axes variables correspond to the two variables from the bivariate data. The words relationship, correlation and association are used to describe the way in which variables are related.

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652

Chapter 9 Statistics

Key ideas

Types of correlation: Examples Strong positive correlation

Weak negative correlation

No correlation

An outlier can clearly be identified as a data point that is isolated from the rest of the data.

y outlier

x

Example 10 Constructing and interpreting scatter plots Consider this simple bivariate data set. x y

a

13 2.1

9 4.0

2 6.2

17 1.3

3 5.5

6 0.9

8 3.5

15 1.6

Draw a scatter plot for the data.

b Describe the correlation between x and y as positive or negative. c

Describe the correlation between x and y as strong or weak.

d

Identify any outliers.

SOL UTI ON

E X P LA NA TI ON

a

Plot each point using a × symbol on graph paper.

y 7 6 5 4 3 2 1 O

5

10

15

20

x

b negative correlation

As x increases, y decreases.

c strong correlation

The downwards trend in the data is clearly defined.

d The outlier is (6, 0.9).

This point defies the trend.

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Statistics and Probability

1

1, 2a

2b

—

Decide if it is likely for there to be a strong correlation between these pairs of variables. a b c d e f

Height of door and Thickness of door handle Weight of car and Fuel consumption Temperature and Length of phone calls Size of textbook and Number of textbook Diameter of flower and Number of bees Amount of rain and Size of vegetables in the vegetable garden

UNDERSTANDING

Exercise 9G

653

2 For each of the following sets of bivariate data with variables x and y: i Draw a scatter plot by hand. ii Decide whether y generally increases or decreases as x increases.

b

x y x y

1 3 0.1 10

2 2 0.3 8

3 4 0.5 8

4 4 0.9 6

5 5 1.0 7

6 8 1.1 7

7 7 1.2 7

8 9 1.6 6

1.8 4

3, 4, 5a, 6 Example 10

9 11

10 12

2.0 3

3, 5b, 6

2.5 1 3, 5c, 6

3 Consider this simple bivariate data set. (Use technology to assist if desired. See page 658.) x y

a b c d

1 1.0

2 1.1

3 1.3

4 1.3

5 1.4

6 1.6

7 1.8

8 1.0

FLUENCY

a

Draw a scatter plot for the data. Describe the correlation between x and y as positive or negative. Describe the correlation between x and y as strong or weak. Identify any outliers.

4 Consider this simple bivariate data set. (Use technology to assist if desired. See page 658.) x y

a b c d

14 4

8 2.5

7 2.5

10 1.5

11 1.5

15 0.5

6 3

9 2

10 2

Draw a scatter plot for the data. Describe the correlation between x and y as positive or negative. Describe the correlation between x and y as strong or weak. Identify any outliers.

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Chapter 9 Statistics

9G 5 By completing scatter plots (by hand or using technology) for each of the following data sets, describe the correlation between x and y as positive, negative or none. a

x y

1.1 22

b

x y

4 115

3 105

1 105

7 135

8 145

10 145

6 125

9 140

5 120

5 130

c

x y

28 13

32 25

16 22

19 21

21 16

24 9

27 19

25 25

30 15

18 12

1.8 12

1.2 19

1.3 15

1.7 10

1.9 9

1.6 14

1.6 13

1.4 16

1.0 23

FLUENCY

654

1.5 16

6 For the following scatter plots, describe the correlation between x and y. a y b

c

x

y

d y

x x 8, 9

8, 10

7 For common motor vehicles, consider the two variables Engine size (cylinder volume) and Fuel economy (number of kilometres travelled for every litre of petrol). a Do you expect there to be some relationship between these two variables? b As the engine size increases, would you expect the fuel economy to increase or decrease?

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

PROBLEM-SOLVING

7, 8

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c The following data were collected for 10 vehicles. Car Engine size Fuel economy

A 1.1 21

B 1.2 18

C 1.2 19

D 1.5 18

E 1.5 17

F 1.8 16

G 2.4 15

H 3.3 20

I 4.2 14

J 5.0 11

PROBLEM-SOLVING

Statistics and Probability

655

9G

i Do the data generally support your answers to parts a and b? ii Which car gives a fuel economy reading that does not support the general trend? 8 A tomato grower experiments with a new organic fertiliser and sets up five separate garden beds: A, B, C, D and E. The grower applies different amounts of fertiliser to each bed and records the diameter of each tomato picked. The average diameter of a tomato from each garden bed and the corresponding amount of fertiliser are recorded below. Bed Fertiliser (grams per week) Average diameter (cm)

A 20 6.8

B 25 7.4

C 30 7.6

D 35 6.2

E 40 8.5

a

Draw a scatter plot for the data with ‘Diameter’ on the vertical axis and ‘Fertiliser’ on the horizontal axis. Label the points A, B, C, D and E. b Which garden bed appears to go against the trend? c According to the given results, would you be confident in saying that the amount of fertiliser fed to tomato plants does affect the size of the tomato produced?

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Chapter 9 Statistics

9G 9 In a newspaper, the number of photos and number of words were counted for 15 different pages. Here are the results. Page 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Number of photos 3 2 1 2 6 4 5 7 4 5 2 3 1 0 1 Number of words 852 1432 1897 1621 912 1023 817 436 1132 1201 1936 1628 1403 2174 1829

PROBLEM-SOLVING

656

a

Sketch a scatter plot using ‘Number of photos’ on the horizontal axis and ‘Number of words’ on the vertical axis. b From your scatter plot, describe the general relationship between the number of photos and the number of words per page. Use the words positive, negative, strong correlation or weak correlation. 10 On 14 consecutive days, a local council measures the volume of sound heard from a freeway at various points in a local suburb. The volume of sound, in decibels, is recorded against the distance (in metres) between the freeway and the point in the suburb.

Distance (m) 200 350 500 150 1000 850 200 450 750 250 300 1500 700 1250 Volume (dB) 4.3 3.7 2.9 4.5 2.1 2.3 4.4 3.3 2.8 4.1 3.6 1.7 3.0 2.2

a

Draw a scatter plot of Volume against Distance, plotting Volume on the vertical axis and Distance on the horizontal axis. b Describe the correlation between Distance and Volume as positive, negative or none. c Generally, as Distance increases does Volume increase or decrease? 11, 12

12, 13

11 A government department is interested in convincing the electorate that a larger number of police on patrol leads to a lower crime rate. Two separate surveys are completed over a one-week period and the results are listed in this table.

Survey 1 Survey 2

a

Area Number of police Incidence of crime Number of police Incidence of crime

A 15 28 12 26

B 21 16 18 25

C 8 36 9 20

D 14 24 12 24

E 19 24 14 22

F 31 19 26 23

REASONING

11

G 17 21 21 19

By using scatter plots, determine whether or not there is a relationship between the number of police on patrol and the incidence of crime, using the data in: i survey 1 ii survey 2

b Which survey results do you think the government will use to make its point? Why?

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12 A student collects some data and finds that there is a positive correlation between height and the ability to play tennis. Does that mean that if you are tall you will be better at tennis? Explain.

657

9G

y Exam marks (%)

13 A person presents you with this scatter plot and suggests a strong correlation between the amount of sleep and exam marks. What do you suggest is the problem with the person’s graph and conclusions?

REASONING

Statistics and Probability

100 50 x O 4 8 Amount of sleep (hours)

—

—

14

14 A university graduate is conducting a test to see whether a student’s general knowledge is in some way linked to the number of hours of television watched. Twenty Year 10 students sit a written general knowledge test marked out of 50. Each student also provides the graduate with details about the number of hours of television watched per week. The results are given in the table below.

ENRICHMENT

Does television provide a good general knowledge?

Student A B C D E F G H I J K L M N O P Q R S T Hours of TV 11 15 8 9 9 12 20 6 0 15 9 13 15 17 8 11 10 15 21 3 Test score 30 4 13 35 26 31 48 11 50 33 31 28 27 6 39 40 36 21 45 48

a

Which two students performed best on the general knowledge test, having watched TV for the following numbers of hours? i fewer than 10 ii more than 4

b Which two students performed worst on the general knowledge test, having watched TV for the following numbers of hours? i fewer than 10 ii more than 4 c

Which four students best support the argument that the more hours of TV watched, the better your general knowledge will be? d Which four students best support the argument that the more hours of TV watched, the worse your general knowledge will be? e From the given data, would you say that the graduate should conclude that a student’s general knowledge is definitely linked to the number of hours of TV watched per week?

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658

Chapter 9 Statistics

Using calculators to draw scatter plots Type the following data about car fuel economy into two lists and draw a scatter plot. Car Engine size Fuel economy

A 1.1 21

B 1.2 18

C 1.2 19

D 1.5 18

E 1.5 17

F 1.8 16

G 2.4 15

H 3.3 20

I 4.2 14

J 5.0 11

Using the TI-Nspire:

Using the ClassPad:

1 In a Lists and spreadsheets page type in the list names engine and fuel and enter the values as shown.

1 In the Statistics application, assign a title to each column then enter the data into the lists.

2

2 Tap . For graph 1 set Draw to On, Type to Scatter, XList to main\Engine, YList to main\Fuel, Freq

Insert a Data and Statistics page and select the engine variable for the horizontal axis and fuel for the vertical axis. Hover over points to reveal coordinates.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

to 1 and Mark to square. Tap Set. Tap Analysis, Trace to reveal coordinates.

. Tap

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Statistics and Probability

659

9H Line of best fit by eye When bivariate data have a strong linear correlation, we can model the data with a straight line. This line is called a trend line or line of best fit. When we fit the line ‘by eye’, we try to balance the number of data points above the line with the number of points below the line. This trend line and its equation can then be used to construct other data points within and outside the existing data points.

Let’s start: Size versus cost This scatter plot shows the estimated cost of building a house of a given size, as quoted by a building company. The given trend line passes through the points (200, 200) and (700, 700).

Cost of construction ($’000)

y 800 (700, 700) 600 400 200 O • • •

(200, 200) 200 400 600 800 Size of house (m2)

x

Do you think the trend line is a good fit to the points on the scatter plot? Why? How can you find the equation of the trend line? How can you predict the cost of a house of 1000 m2 with this building company? A line of best fit or trend line is positioned by eye by balancing the number of points above the line with the number of points below the line. • The distance of each point from the trend line also must be taken into account. The equation of the line of best fit can be found using two points that are on the line of best fit. For y = mx + c: y –y m = 2 1 and substitute a point to find the value of c. x2 – x1 • Alternatively, use y – y1 = m(x – x1 ). The line of best fit and its equation can be used for: • interpolation: constructing points within the given data range • extrapolation: constructing points outside the given data range.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Key ideas

y (x1, y1)

(x2, y2) x

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660

Chapter 9 Statistics

Example 11 Fitting a line of best fit Consider the variables x and y and the corresponding bivariate data. x y

a

1 1

2 2

3 2

4 3

5 4

6 4

7 5

Draw a scatter plot for the data.

b Is there positive, negative or no correlation between x and y? c

Fit a line of best fit by eye to the data on the scatter plot.

d

Use your line of best fit to estimate: i y when x = 3.5

ii y when x = 0

iii x when y = 1.5

iv x when y = 5.5

SOL UTI ON

EX P L A NA TI ON

a

Plot the points on graph paper.

y 6 5 4 3 2 1 O

1 2 3 4 5 6 7 8

x

b Positive correlation

As x increases, y increases.

c

Since a relationship exists, draw a line on the plot, keeping as many points above as below the line. (There are no outliers in this case.)

y 6 5 4 3 2 1 O

d

1 2 3 4 5 6 7 8

i y ≈ 2.7 ii y ≈ 0.4 iii x ≈ 1.7 iv x ≈ 7.8

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

x Extend vertical and horizontal lines from the values given and read off your solution. As they are approximations, we use the ≈ sign and not the = sign.

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Statistics and Probability

661

Example 12 Finding the equation of a line of best fit

b

c

Use your equation to estimate a Literature score if the English score is: i 50 ii 86 Use your equation to estimate the English score if the Literature score is: i 42 ii 87

SOL UTI ON a

y 100 Literature mark (%)

This scatter plot shows a linear relationship between English marks and Literature marks in a small class of students. A trend line passes through (40, 30) and (60, 60). a Find the equation of the trend line.

80 60

(60, 60)

40 (40, 30)

20 O

20

60 80 40 English mark (%)

100

x

EX P L A NA TI ON

y = mx + c 60 – 30 30 3 = = 60 – 40 20 2 3 ∴ y= x+c 2 m=

Use m =

3 (40, 30): 30 = (40) + c 2 30 = 60 + c c = –30

y2 – y1 for the two given points. x2 – x1

Substitute either (40, 30) or (60, 60) to find c.

3 ∴ y = x – 30 2 b

c

3 i y = (50) – 30 = 45 2 ∴ Literature score is 45. 3 ii y = (86) – 30 = 99 2 ∴ Literature score is 99. 3 42 = x – 30 2 3 72 = x, so x = 48. 2 ∴ English score is 48. 3 ii 87 = x – 30 2 3 117 = x, so x = 78. 2 ∴ English score is 78. i

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Substitute x = 50 and find the value of y.

Repeat for x = 86.

Substitute y = 42 and solve for x.

Repeat for y = 87.

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662

Chapter 9 Statistics

1

1–3

3

—

Practise fitting a line of best fit on these scatter plots by trying to balance the number of points above the line with the numbers of points below the line. (Using a pencil might help.) a b

c

UNDERSTANDING

Exercise 9H

d

2 For each graph find the equation of the line in the form y = mx + c. First, find the gradient y –y m = 2 1 and then subsititute a point. x2 – x1 a y b y

(1, 5)

(11, 9)

(4, 3)

(3, 5) x

x

5 3 Using y = x – 3, find: 4 a y when: i x = 16

ii x = 7

b x when: ii y =

y=4

1 2

4–6 Example 11

4–6

4 Consider the variables x and y and the corresponding bivariate data. x y

a b c d

1 2

2 2

3 3

4 4

5 4

6 5

7 5

4, 6

FLUENCY

i

Draw a scatter plot for the data. Is there positive, negative or no correlation between x and y? Fit a line of best fit by eye to the data on the scatter plot. Use your line of best fit to estimate: i y when x = 3.5 ii y when x = 0 iii x when y = 2 iv x when y = 5.5

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5 For the following scatter plots, pencil in a line of best fit by eye, and then use your line to estimate the value of y when x = 5. a b y y

10

20

5

10

O

5

x

10

O

y

c

9H

x

10

100 50

0.5

Example 12

663

y

d

1.0

O

5

FLUENCY

Statistics and Probability

O

x

2.5 5 7.5 10

5

10

x

6 This scatter plot shows a linear relationship between Mathematics marks and Physics marks in a small class of students. A trend line passes through (20, 30) and (70, 60). a Find the equation of the trend line. b Use your equation to find the Physics score if the Mathematics score is: i 40 ii 90 c Use your equation to find the Mathematics score if the Physics score is: i 36 ii 78

y 100

Physics (%)

80 60 40 20 O

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

(70, 60)

(20, 30)

20

60 80 40 Mathematics (%)

100

x

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Chapter 9 Statistics

7

9H

7, 8

7, 8

PROBLEM-SOLVING

664

7 Over eight consecutive years, a city nursery has measured the growth of an outdoor bamboo species for that year. The annual rainfall in the area where the bamboo is growing was also recorded. The data are listed in the table.

Rainfall (mm) Growth (cm)

450 25

620 45

560 25

830 85

680 50

650 55

720 50

540 20

a Draw a scatter plot for the data, showing growth on the vertical axis. b Fit a line of best fit by eye. c Use your line of best fit to estimate the growth expected for the following rainfall readings. You do not need to find the equation of the line. i 500 mm ii 900 mm d Use your line of best fit to estimate the rainfall for a given year if the growth of the bamboo was: i 30 cm ii 60 cm 8 A line of best fit for a scatter plot, relating the weight (kg) and length (cm) of a group of dogs, passes through the points (15, 70) and (25, 120). Assume weight is on the x-axis. a Find the equation of the trend line. b Use your equation to estimate the length of an 18 kg dog. c Use your equation to estimate the weight of a dog that has a length of 100 cm. 9, 10

9 Describe the problem when using each trend line below for interpolation. a b

10, 11

REASONING

9

10 Describe the problem when using each trend line below for extrapolation. a b

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REASONING

Statistics and Probability

11 A trend line relating the percentage scores for Music performance (y) and Music theory (x) is given by 4 y = x + 10. 5 a Find the value of x when: i y = 50 ii y = 98

665

9H

b What problem occurs in predicting Music theory scores when using high Music performance scores?

—

—

12

12 Two independent scientific experiments confirmed a correlation between Maximum heart rate (in beats per minute or b.p.m.) and Age (in years). The data for the two experiments are as follows. Experiment 1 Age (years) 15 18 22 25 30 34 35 40 40 52 60 65 71 Max. heart rate (b.p.m.) 190 200 195 195 180 185 170 165 165 150 125 128 105

ENRICHMENT

Heart rate and age

Experiment 2 Age (years) 20 20 21 26 27 32 35 41 43 49 50 58 82 Max. heart rate (b.p.m.) 205 195 180 185 175 160 160 145 150 150 135 140 90

a Sketch separate scatter plots for experiment 1 and experiment 2. b By fitting a line of best fit by eye to your scatter plots, estimate the maximum heart rate for a person aged 55 years, using the results from: i experiment 1 ii experiment 2 c

Estimate the age of a person who has a maximum heart rate of 190, using the results from: i experiment 1 ii experiment 2

d For a person aged 25 years, which experiment estimates a lower maximum heart rate? e Research the average maximum heart rate of people according to age and compare with the results given above.

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666

Chapter 9 Statistics

9I Linear regression with technology

10A

In section 9H we used a line of best fit by eye to describe a general linear (i.e. straight line) trend for bivariate data. In this section we look at the more formal methods for fitting straight lines to bivariate data. This is called linear regression. There are many different methods used by statisticians to model bivariate data. Two common methods are least squares regression and median–median regression. These methods are best handled with the use of technology.

Let’s start: What can my calculator or software do? Explore the menus of your chosen technology to see what kind of regression tools are available. For CAS calculator users, refer to page 667. • Can you find the least squares regression and median–median regression tools? • Use your technology to try Example 13 below.

Key ideas

Linear regression involves using a method to fit a straight line to bivariate data. • The result is a straight line equation that can be used for interpolation and extrapolation. The least squares regression line minimises the sum of the square of the deviations of each point from the line. • Outliers have an effect on the least squares regression line because all deviations are included in the calculation of the equation of the line. The median–median or 3 median regression line uses the three medians from the lower, middle and upper groups within the data set. • Outliers do not have an effect on the median–median line as they do not significantly alter the median values.

Example 13 Finding and using regression lines Consider the following data and use a graphics or CAS calculator or software to help answer the questions below. x y

a

1 1.8

2 2

2 1.5

4 1.6

5 1.7

5 1.3

6 0.8

7 1.1

9 0.8

11 0.7

Construct a scatter plot for the data.

b Find the equation of the least squares regression line. c

Find the equation of the median–median regression line.

d Sketch the graph of the regression line onto the scatter plot. e

Use the least squares regression line to estimate the value of y when x is: i 4.5 ii 15

f

Use the median–median regression line to estimate the value of y when x is: i 4.5 ii 15

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Statistics and Probability

667

Using calculators to find equations of regression Using the TI-Nspire:

Using the ClassPad:

a, b, d

a, b, d

In a Lists & Spreadsheet page enter the data in the lists named xvalue and yvalue. Insert a Data & Statistics page and select xvalue as the variable on the horizontal axis and yvalue as the variable on the vertical axis. To show the linear regression line and equation use b>Analyze>Regression>Show Linear (mx+b)

In the Statistics application enter the data into the lists. Tap Calc, Regression, Linear Reg and set XList to list1, YList to list2, Freq to 1, Copy Formula to y1 and Copy Residual to Off. Tap OK to view the regression equation. Tap on OK again to view the regression line. Tap Analysis, Trace and then scroll along the regression line.

Least squares: y = –0.126201x + 1.986245

c, d

To show the median-median regression line use b>Analyze>Regression>Show Median-Median

c, d

To produce the median-median regression line, tap Calc, Regression, MedMed Line and apply the settings stated previously.

Median-Median: y = –0.142857x + 2.111905 e

i y ≈ 1.42 ii y ≈ 0.09

f i y ≈ 1.47 ii y ≈ – 0.03

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668

Chapter 9 Statistics

1

1, 2

2

—

A regression line for a bivariate data set is given by y = 2.3x – 4.1. Use this equation to find: a the value of y when x is: i 7 ii 3.2 b the value of x when y is: i 12

UNDERSTANDING

Exercise 9I

ii 0.5

2 Give a brief reason why a linear regression line is not very useful in the following scatter plots. a y b y

x

x

Example 13

3(A, C), 4

3(B), 4

3 Consider the data in tables A–C and use a graphics or CAS calculator or software to help answer the following questions. A

x y

1 3.2

2 5

B

x y

3 3.8

6 3.7

7 3.9

C

x y

0.1 8.2

0.2 5.9

0.5 6.1

3 5.6

4 5.4

5 6.8

6 6.9

7 7.1

8 7.6

10 3.6

14 3.1

17 2.5

21 2.9

26 2.1

0.8 4.3

0.9 4.2

1.2 1.9

1.6 2.5

1.7 2.1

a b c d e

Construct a scatter plot for the data. Find the equation of the least squares regression line. Find the equation of the median–median regression line. Sketch the graph of the regression line onto the scatter plot. Use the least squares regression line to estimate the value of y when x is: i 7 ii 12

f

Use the median–median regression line to estimate the value of y when x is: i 7 ii 12

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FLUENCY

3(A, B), 4

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FLUENCY

Statistics and Probability

4 The values and ages of 14 cars are summarised in these tables. Age (years) Price ($’000)

5 20

2 35

4 28

9 14

10 11

8 12

7 15

Age (years) Price ($’000)

11 5

2 39

1 46

4 26

7 19

6 17

9 14

669

9I

a

Using Age for the x-axis, find: i the least squares regression line ii the median–median regression line. b Use your least squares regression line to estimate the value of a 3-year-old car. c Use your median–median regression line to estimate the value of a 12-year-old car. d Use your least squares regression line to estimate the age of a $15 000 car. e Use your median–median regression line to estimate the age of an $8000 car. 5, 6

6, 7

5 A factory that produces denim jackets does not have air-conditioning. It was suggested that high temperatures inside the factory were having an effect on the number of jackets able to be produced, so a study was completed and data collected on 14 consecutive days. Max. daily temp. inside factory (◦ C) 28 32 36 27 24 25 29 31 34 38 41 40 38 31 Number of jackets produced 155 136 120 135 142 148 147 141 136 118 112 127 136 132

PROBLEM-SOLVING

5, 6

Use a graphics or CAS calculator to complete the following. a Draw a scatter plot for the data. b Find the equation of the least squares regression line. c Graph the line onto the scatter plot. d Use the regression line to estimate how many jackets would be able to be produced if the maximum daily temperature in the factory was: i 30◦ C ii 35◦ C iii 45◦ C 6 A particular brand of electronic photocopier is considered for scrap once it has broken down more than 50 times or if it has produced more than 200 000 copies. A study of one particular copier gave the following results. Number of copies ( × 1000) 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 Total number of breakdowns 0 0 1 2 2 5 7 9 12 14 16 21 26 28 33

a b c d e

Sketch a scatter plot for the data. Find the equation of the median–median regression line. Graph the median–median regression line onto the scatter plot. Using your regression line, estimate the number of copies the photocopier will have produced at the point when you would expect 50 breakdowns. Would you expect this photocopier to be considered for scrap because of the number of breakdowns or the number of copies made?

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Chapter 9 Statistics

9I 7 At a suburban sports club, the distance record for the hammer throw has increased over time. The first recorded value was 72.3 m in 1967 and the most recent record was 118.2 m in 1996. Further details are as follows. Year New record (m)

1967 72.3

1968 73.4

1969 82.7

1976 94.2

1978 99.1

1983 101.2

1987 111.6

1996 118.2

PROBLEM-SOLVING

670

a Draw a scatter plot for the data. b Find the equation of the median–median regression line. c Use your regression equation to estimate the distance record for the hammer throw for: i 2000 ii 2020 d Would you say that it is realistic to use your regression equation to estimate distance records beyond 2020? Why? 8, 9

8, 9

8 a Briefly explain why the least squares regression line is affected by outliers. b Briefly explain why the median–median regression line is not strongly affected by outliers. 9 This scatter plot shows both the least squares regression line and the median–median regression line. Which line (i.e. A or B) do you think is the least squares line? Give a reason.

y

REASONING

8

B A

x

Correlation coefficient

—

10

ENRICHMENT

—

10 Use the internet to find out about the Pearson correlation coefficient and then answer these questions. a b

c

What is the coefficient used for? Do most calculators include the coefficient as part of their statistical functions? What does a relatively large or small correlation coefficient mean? Statisticians work in many fields of industry, business, finance, research, government and social services. As computers are used to process the data, they can spend more time on higher-level skills, such as designing statistical investigations, and data analysis.

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Statistics and Probability

671

Investigation Indigenous population comparison The following data was collected by the Australian Bureau of Statistics during the 2011 National Census. It shows the population of Indigenous and non-Indigenous people in Australia and uses class intervals of 5 years.

Age Total all ages 0–4 years 5–9 years 10–14 years 15–19 years 20–24 years 25–29 years 30–34 years 35–39 years 40–44 years 45–49 years 50–54 years 55–59 years 60–64 years 65 years and over

Total 21 507 719 1 421 048 1 351 921 1 371 055 1 405 798 1 460 675 1 513 237 1 453 777 1 520 138 1 542 879 1 504 142 1 447 403 1 297 247 1 206 117 3 012 282

Indigenous 548 371 67 416 64 935 64 734 59 200 46 455 38 803 33 003 34 072 33 606 28 818 24 326 18 640 13 592 20 771

Non-Indigenous 19 900 765 1 282 738 1 222 111 1 241 794 1 282 018 1 333 622 1 387 922 1 345 763 1 414 171 1 438 346 1 407 494 1 357 677 1 220 529 1 138 395 2 828 185

Indigenous status not stated 1 058 583 70 894 64 875 64 527 64 580 80 598 86 512 75 011 71 895 70 927 67 830 65 400 58 078 54 130 163 326

Indigenous histogram a

Use the given data to construct a histogram for the population of Indigenous people in Australia in 2011.

b

Which age group contained the most Indigenous people?

c

Describe the shape of the histogram. Is it symmetrical or skewed?

Non-Indigenous histogram a

Use the given data to construct a histogram for the population of non-Indigenous people in Australia in 2011. Try to construct this histogram so it is roughly the same width and height as the histogram for the Indigenous population. You will need to rescale the y-axis.

b

Which age group contains the most number of non-Indigenous people?

c

Describe the shape of the histogram. Is it symmetrical or skewed?

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672

Chapter 9 Statistics

Comparisons a

Explain the main differences in the shapes of the two histograms.

b

What do the histograms tell you about the age of Indigenous and non-Indigenous people in Australia in 2011?

c

What do the graphs tell you about the difference in life expectancy for Indigenous and non-Indigenous people?

Antarctic ice According to many different studies, the Antarctic ice mass is decreasing over time. The following data show the approximate change in ice mass in gigatonnes (Gt; 109 tonnes) from 2007 to 2014. Year Change (Gt)

2007 450

2008 300

2009 400

2010 200

2011 –100

2012 –400

2013 –200

2014 –700

A change of 300, for example, means that the ice mass has increased by 300 Gt in that year. Data interpretation a

By how much did the Antarctic ice mass increase in: i 2007? ii 2010?

b

By how much did the Antarctic ice mass decrease in: i 2011? ii 2014?

c

What was the overall change in ice mass from the beginning of 2010 to the end of 2012?

Time-series plot a

Construct a time-series plot for the given data.

b

Describe the general trend in the change in ice mass over the 8 years.

Line of best fit a

Fit a line of best fit by eye to your time-series plot.

b

Find an equation for your line of best fit.

c

Use your equation to estimate the change in ice mass for: i 2015 ii 2025

Regression a

Use technology to find either the least squares regression line or the median–median regression line for your time-series plot.

b

How does the equation of these lines compare to your line of best fit found above?

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Statistics and Probability

Problems and challenges

Up for a challenge? If you get stuck on a question, check out the 'Working with unfamiliar problems' poster at the end of the book to help you.

1

The mean mass of six boys is 71 kg, and the mean mass of five girls is 60 kg. Find the average mass of all 11 people put together.

2

Sean has a current four-topic average of 78% for Mathematics. What score does he need in the fifth topic to have an overall average of 80%?

3

A single-ordered data set includes the following data. 2, 4, 5, 6, 8, 10, x What is the largest possible value of x if it is not an outlier?

4

Find the interquartile range for a set of data if 75% of the data are above 2.6 and 25% of the data are above 3.7.

5

A single data set has 3 added to every value. Describe the change in: a the mean b the median c the range d the interquartile range

6

673

e the standard deviation.

Three key points on a scatter plot have coordinates (1, 3), (2, 3) and (4, 9). Find a quadratic equation that fits these three points exactly.

y 9 6 3 O

3

6

9

7

Six numbers are written in ascending order: 1.4, 3, 4.7, 5.8, a, 11. Find all possible values of a if the number 11 is considered to be an outlier.

8

The class mean, ¯x, and standard deviation, s, for some Year 10 term tests are: Maths (¯x = 70%, s = 9%); Physics (¯x = 70%, s = 6%); Biology (¯x = 80%, s = 6.5%). If Emily gained 80% in each of these subjects, which was her best and worst result? Give reasons for your answer.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter 9 Statistics

Collecting data A survey can be used to collect data from a population. The sample of the population chosen to be surveyed should be selected without bias and should be representative of the population.

Grouped data

Frequency

Percentage Class interval Frequency frequency 0– 2 20 10– 4 40 20– 3 30 30–40 1 10 Total 10 100

Very high

Low

Categorical Numerical • Nominal • Discrete (1, 2, 3, ...) (red, blue, ...) • Continuous • Ordinal (0.31, 0.481, ...) (low, medium, ...)

Graphs for single set of categorical or discrete data Dot plot Column graphs Stem-and-leaf plot

High

Data

Medium

Chapter summary

674

A

B C Category

• Median (Q2) = middle value (Mode = most common value)

10

20 30

Percentage frequency

Frequency

40 30 20 10 40

Statistics Quartiles Q1: above 25% of the data Q3: above 75% of the data 2 3 5 7 8 9 11 12 14 15 Q1

Q2 = 8.5

D

Measures of centre _ sum of all values • Mean (x ) = number of scores

Histogram 4 3 2 1 O

Measures of spread • Range = Max − Min • Interquartile range (IQR) = Q3 − Q1 • Sample standard deviation (σ ) for n data values (10A) _ _ _ = (x1 − x )2 + (x2 − x )2 + ... + (xn − x )2 n−1 – If σ is relatively small the data is concentrated about the mean. – If σ is relatively large the data is spread out from the mean.

Q3

1 4 7 ( 8 ( 12 16 21 Q1

Q2

Outliers • Single data set – less than Q1 − 1.5 × IQR or more than Q3 + 1.5 × IQR • Bivariate – not in the vicinity of the rest of the data

Q3

Box plots

Min

Q 1 Q2

Max Outlier

Q3

Time-series data Downwards linear trend

Bivariate data – Two related variables – Scatter plot y line of best fit (y = mx + c ) A strong, positive correlation x

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Stem Leaf 0 1 6 1 2 7 9 2 3 8 3 4 2 3 means 23

Weak, negative correlation

Time No correlation

Linear regression with technology (10A) – Least squares line – Median–median line

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Multiple-choice questions 38pt 9A

1

The type of data generated by the survey question What is your favourite food? is: A numerical and discrete B numerical and continuous C a sample D categorical and nominal E categorical and ordinal

Questions 2–4 refer to the dot plot shown at right. 38pt 9B

38pt 9B

38pt 9B

2 The mean of the scores in the data is: A 3.5 B 3.9 D 4 E 5 3 The mode for the data is: A 3.5 B 2 4 The dot plot is: A symmetrical D bimodal

C 3

C 3

2 3 4 5 Score D 4

B positively skewed E correlated

675

Chapter review

Statistics and Probability

E 5 C negatively skewed

Questions 5 and 6 refer to this box plot.

2 4 6 8 10 12 14 16 18 20 22 38pt 9D

38pt 9D

38pt 9G

5 The interquartile range is: A 8 B 5

C 3

D 20

E 14

6 The range is: A 5

C 20

D 14

E 8

B 3

7 The variables x and y in this scatter plot could be described as having: A no correlation B a strong, positive correlation C a strong, negative correlation D a weak, negative correlation E a weak, positive correlation

y

x

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Chapter review

676

Chapter 9 Statistics

38pt 9H

8 The equation of the line of best fit for a set of bivariate data is given by y = 2.5x – 3. An estimate for the value of x when y = 7 is: A –1.4 B 1.2 C 1.6 D 7 E 4

38pt 9E

9 The sample standard deviation for the small data set 1, 1, 2, 3, 3 is: A 0.8 B 2 C 1 D 0.9

10A

38pt 9H

E 2.5

10 The equation of the line of best fit connecting the points (1, 1) and (4, 6) is: 5 2 5 8 A y = 5x + 3 B y= x– C y=– x+ 3 3 3 3 5 8 3 2 D y= x– E y= x– 3 3 5 3

Short-answer questions 38pt 9B

1

38pt 9D

2 For each set of data below, complete the following tasks. i Find the range. ii Find the lower quartile (Q1 ) and the upper quartile (Q3 ). iii Find the interquartile range. iv Locate any outliers. v Draw a box plot.

A group of 16 people was surveyed to find the number of hours of television they watch in a week. The raw data are listed: 6, 5, 11, 13, 24, 8, 1, 12 7, 6, 14, 10, 9, 16, 8, 3 a Organise the data into a table with class intervals of 5 and include a percentage frequency column. b Construct a histogram for the data, showing both the frequency and percentage frequency on the graph. c Would you describe the data as symmetrical, positively skewed or negatively skewed? d Construct a stem-and-leaf plot for the data, using 10s as the stem. e Use your stem-and-leaf plot to find the median.

a b c

2, 2, 3, 3, 3, 4, 5, 6, 12 11, 12, 15, 15, 17, 18, 20, 21, 24, 27, 28 2.4, 0.7, 2.1, 2.8, 2.3, 2.6, 2.6, 1.9, 3.1, 2.2

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Statistics and Probability

3 Compare these parallel box plots, A and B, and answer the following as true or false. a b c d

38pt 9G

The range for A is greater than the range for B. The median for A is equal to the median for B. The interquartile range is smaller for B. 75% of the data for A sit below 80.

B A 0

20

40

60

80

100

4 Consider the simple bivariate data set. x y

a b c d 38pt 9H

Chapter review

38pt 9D

1 24

4 15

3 16

2 20

1 22

c

3 5

2 17

5 6

5 8

Draw a scatter plot for the data. Describe the correlation between x and y as positive or negative. Describe the correlation between x and y as strong or weak. Identify any outliers.

5 The line of best fit passes through the two points labelled on this graph. a b

4 11

677

y

Find the equation of the line of best fit. Use your equation to estimate the value of y when: i x=4 ii x = 10 Use your equation to estimate the value of x when: i y=3 ii y = 12

(6, 5)

(1, 2) x 38pt 9E 10A

38pt 9E

6 Calculate the mean and sample standard deviation for these small data sets. Round the standard deviation to one decimal place. a 4, 5, 7, 9, 10 b 1, 1, 3, 5, 5, 9 7 The Cats and The Vipers basketball teams compare their number of points per match for a season. The data are presented in this back-to-back stem-and-leaf plot. The Cats Leaf 2 83 74 97410 762

Stem 0 1 2 3 4 5

The Vipers Leaf 9 9 0489 24789 28 0

2 | 4 means 24

State which team has: a the higher range b the higher mean

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Chapter review

678

Chapter 9 Statistics

c the higher median 10A

38pt 9F

d the higher standard deviation.

8 Describe the trend in these time-series plots as linear, non-linear or no trend. a b

Time

Time 38pt 9I

9 For the simple bivariate data set in Question 4, use technology to find the equation of the: a b

10A

least squares regression line median–median regression line.

Extended-response questions 1

The number of flying foxes taking refuge in two different fig trees was recorded over a period of 14 days. The data collected are given here. Tree 1 Tree 2

a

b

c d

56 73

38 50

47 36

59 82

63 15

43 24

49 73

51 57

60 65

77 86

71 51

48 32

50 21

62 39

Find the IQR for: i tree 1 ii tree 2 Identify any outliers for: i tree 1 ii tree 2 Draw parallel box plots for the data. By comparing your box plots, describe the difference in the ways the flying foxes use the two fig trees for taking refuge.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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2 The approximate number of shoppers in an air-conditioned shopping plaza was recorded for 14 days, along with the corresponding maximum daily outside temperatures for those days. Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Max. daily temp. (T ) (°C) 27 26 28 33 38 36 28 30 32 25 25 27 29 33 No. of shoppers (N ) 1050 950 1200 1550 1750 1800 1200 1450 1350 900 850 700 950 1250

10A

a Draw a scatter plot for the number of shoppers versus the maximum daily temperatures, with the number of shoppers on the vertical axis, and describe the correlation between the variables as either positive, negative or none. b Find the equation of the median–median regression line for the data, using technology. c Use your regression equation to estimate: i

the number of shoppers on a day with a maximum daily temperature of 24◦ C

ii

the maximum daily temperature if the number of shoppers is 1500.

679

Chapter review

Statistics and Probability

d Use technology to determine the least squares regression line for the data. e Use your least squares regression equation to estimate: i

the number of shoppers on a day with a maximum daily temperature of 24◦ C

ii

the maximum daily temperature if the number of shoppers at the plaza is 1500.

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10

Chapter

Logarithms and polynomials

W h a t y o u w ill le a r n

Au s t r a l i a n c u r r ic u lu m

10A Logarithms (10A) 10B Logarithm laws (10A) 10C Exponential equations using logarithms (10A) 10D Polynomials (10A) 10E Expanding and Simplifying polynomials (10A) 10F Division of polynomials (10A) 10G The remainder and factor theorems (10A) 10H Solving polynomial equations (10A) 10I Graphs of polynomials (10A)

NUMBER AND ALGEBRA

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Patterns and algebra (10A) Investigate the concept of a polynomial and apply the factor and remainder theorems to solve problems. Real numbers (10A) Use the definition of a logarithm to establish and apply the laws of logarithms. (10A) Investigate the concept of a polynomial and apply the factor and remainder theorems to solve problems. Linear and non-linear relationships (10A) Apply understanding of polynomials to sketch a range 32x32 16x16 of curves and describe the features of these curves from their equation.

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O n lin e re s o u rc e s • Chapter pre-test • Videos of all worked examples • Interactive widgets • Interactive walkthroughs • Downloadable HOTsheets • Access to HOTmaths Australian Curriculum courses

M o d e llin g in fin a n c e Economists and financial analysts often deal with complex mathematical relationships linking variables such as cost, output, account balance and time. To model these relationships they use mathematical rules, including polynomials (linear, quadratic, cubic etc.) and exponentials, the graphs of which give a visual representation of how the variables are related.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Polynomials (sums of powers of x ) can be used when the x variable is the base, as in the quadratic y = 2x 2 – x + 1 or in the cubic equation P(x ) = 3x 3 – x 2 + 14. Exponentials combine with logarithms to model exponential growth and decay, and to solve complex problems involving compound interest, investment returns and insurance.

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682

Chapter 10 Logarithms and polynomials

10A Logarithms

10A

Logarithms (‘logical arithmetic’) are an important idea in mathematics and were invented by John Napier in the 17th century to simplify arithmetic calculations. Logarithms are linked directly to exponentials and can be used to solve a range of exponential equations. Recall that 23 = 8 (2 to the power 3 equals 8). We can also say that the logarithm of 8 to the base 2 equals 3 and we write log2 8 = 3. So for exponential equations such as y = 2x , a logarithm finds x for a given value of y. A logarithm can often be evaluated by hand but calculators can also be used.

The magnitude of an earthquake is expressed using logarithms.

Logarithms can also be used to create logarithmic scales, which are commonly used in science, economics and engineering. For example, the Richter scale, and the moment magnitude scale that replaced it, are logarithmic scales that illustrate the strength of an earthquake.

Let’s start: Can you work out logarithms? We know that 32 = 9, so log3 9 = 2. This means that log3 9 is equal to the index that makes 3 to the power of that index equal 9. Similarly, 103 = 1000 so log10 1000 = 3. Now find the value of the following. • log10 100 • log10 10 000 • log2 16 • log2 64

Key ideas

• log3 27

• log4 64

A logarithm of a number to a given base is the power (or index) to which the base is raised to give the number. • For example: log2 16 = 4 since 24 = 16. • The base a is written as a subscript to the operator word ‘log’; i.e. loga . In general, if ax = y then loga y = x with a > 0 and y > 0. • We say ‘the logarithm of y to the base a is x’.

Example 1 Writing equivalent statements involving logarithms Write an equivalent statement to the following. a log10 1000 = 3 b 25 = 32 SOL UTI ON

EX P L A NA TI ON

a 103 = 1000

loga y = x is equivalent to ax = y.

b log2 32 = 5

ax = y is equivalent to loga y = x.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

683

Example 2 Evaluating logarithms a

Evaluate the following logarithms. i log2 8 ii log5 625

b

Evaluate the following. 1 i log3 ii log10 0.001 9 c Evaluate, correct to three decimal places, using a calculator. i log10 7 ii log10 0.5 SO L U T I O N

EX P L A N A T I O N

a i log2 8 = 3

Ask the question ‘2 to what power gives 8?’ Note: 23 = 8 54 = 5 × 5 × 5 × 5 = 625

ii log5 625 = 4 1 b i log3 = –2 9

3–2 =

1 1 = 32 9

10–3 =

ii log10 0.001 = –3 c i log10 7 = 0.845 (to 3 d.p.)

ii log10 0.5 = –0.301 (to 3 d.p.)

1 1 = = 0.001 103 1000

Use the log button on a calculator and use base 10. (Some calculators will give log base 10 by pressing the log button.) Use the log button on a calculator.

Example 3 Solving simple logarithmic equations Find the value of x in these equations. a log4 64 = x b log2 x = 6 SO L U T I O N

EX P L A N A T I O N

a

loga y = x then ax = y.

log4 64 = x 4x = 64 x=3

b

log2 x = 6 6

2 =x

43 = 64 Write in index form: 26 = 2 × 2 × 2 × 2 × 2 × 2 = 64

x = 64

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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684

Chapter 10 Logarithms and polynomials

1

1–5

4, 5

—

Complete this table. x 2x 3x 4x 5x 10x

0

1

2

3

4

5 243

UNDERSTANDING

Exercise 10A

256 5 100

2 State the value of the unknown number for each statement. a 2 to the power of what number gives 16? b 3 to the power of what number gives 81? c 7 to the power of what number gives 343? d 10 to the power of what number gives 10 000?

Example 1a

Example 1b

c 2–2 g 4–3

4 Write the following in index form. a log2 16 = 4 b log10 100 = 2 1 d log2 = –2 e log10 0.1 = –1 4 5 Write the following in logarithmic form. a 23 = 8 b 34 = 81 d 42 = 16

e 10–1 =

d 2–5 h 6–2 c log3 27 = 3 1 f log3 = –2 9 c 25 = 32

1 10

f 6–7(½)

Example 2a

Example 2b

6 Evaluate the following logarithms. a log2 16 b log2 4 e log3 3 f log4 16 i log7 49 j log11 121 m log2 1 n log5 1 7 Evaluate the following. 1 1 a log2 b log2 8 4 1 1 e log7 f log3 49 81 i log10 0.1 m log2 0.125

Example 2c

j log10 0.001 n log5 0.2

c g k o

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

e log10 0.17

6–8(½)

log2 64 log5 125 log10 100 000 log37 1

1 9 1 g log5 625

6–8(½)

d h l p

log3 27 log10 1000 log9 729 log1 1

1 1000 1 h log8 8

c log3

d log10

k log10 0.00001 o log5 0.04

l log2 0.5 p log3 0.1˙

8 Evaluate, correct to three decimal places, using a calculator. a log10 5 b log10 47 d log10 0.8

1 125

5–3 =

FLUENCY

3 Write these numbers as fractions. a 0.0001 b 0.5 –3 e 3 f 5–2

c log10 162 f

log10

1 27

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Number and Algebra

Example 3

9 Find the value of x in these equations. a log3 27 = x b log2 32 = x e log10 1000 = x f log6 36 = x i log10 x = 3 j log3 x = –2 m logx 27 = 3 n logx 32 = 5 q logx 81 = 4 r logx 10 000 = 4

c g k o s

9(½), 10

d h l p t

log2 64 = x log2 x = 4 log4 x = –1 logx 64 = 3 logx 0.5 = –1

10 A single bacterium cell divides into two every minute. a Complete this cell population table. b Write a rule for the population, P, after t minutes. c Use your rule to find the population after 8 minutes.

9(½), 10

Time (minutes) Population

0 1

PROBLEM-SOLVING

9(½), 10

log5 625 = x log3 x = 4 log7 x = –3 logx 64 = 2 log4 0.25 = x

1 2

2

3

4

685

10A

5

d Use trial and error to find the time (correct to the nearest minute) for the population to rise to 10 000. e Write the exact answer to part d as a logarithm.

11, 12

12–14

11 Is it possible for a logarithm (of the form loga b) to give a negative result? If so, give an example and reasons. 12 If loga y = x and a > 0, is it possible for y ≤ 0 for any value of x? Give examples and reasons.

REASONING

11

13 Evaluate: a log2 4 × log3 9 × log4 16 × log5 25 b 2 × log3 27 – 5 × log8 64 + 10 × log10 1000 4 × log5 125 2 × log3 9 + log2 64 log10 10

14 Explain why

1 does not exist. loga 1

Fractional logarithms

—

—

15

1 √ √ √ 1 1 15 We know that we can write 2 = 2 2 , so log2 2 = and log2 3 2 = . Now evaluate the following 2 3 without the use of a calculator. √ √ √ √ a log2 4 2 b log2 5 2 c log3 3 d log3 3 3 √ √ √ √ e log7 7 f log10 3 10 g log10 3 100 h log2 3 16 √ √ √ √ i log3 4 9 j log5 4 25 k log2 5 64 l log3 7 81

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

ENRICHMENT

c

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686

Chapter 10 Logarithms and polynomials

10B Logarithm laws

10A

From the study of indices you will recall a number of index laws that can be used to manipulate expressions involving powers. Similarly, we have laws for logarithms and these can be derived using the index laws. Recall index law 1: am × an = am + n Now let x = am and y = an so m = loga x and n = loga y

(1) (2)

From equation (1) xy = am × an = am + n m + n = loga (xy)

So: From (2) So:

(using index law 1)

m + n = loga x + loga y loga (xy) = loga x + loga y

This is a proof for one of the logarithm laws and we will develop the others later in this section.

Let’s start: Proving a logarithm law In the introduction above there is a proof of the first logarithm law, which is considered in this section.   It uses the first index law. x • Now complete a similar proof for the second logarithm law, loga = loga x – loga y, using the second y index law.

Key ideas

Law 1: loga x + loga y = loga (xy) • This relates to index law 1: am × an = am + n .   x Law 2: loga x – loga y = loga y • This relates to index law 2: am ÷ an = am – n . Law 3: loga (xn ) = n loga x • This relates to index law 3: (am )n = am × n . Other properties of logarithms. • loga 1 = 0, (a π 1) using a0 = 1 • loga a = 1, using a1 = a 1 • loga = loga x–1 = – loga x from law 3. x

Before the invention of the electronic calculator, multiplication and division of numbers with many digits was done with tables of logarithms or slide rules with logarithmic scales. Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

687

Example 4 Simplifying logarithmic expressions Simplify the following. a loga 4 + loga 5

b loga 22 – loga 11

c

3 loga 2

SO L U T I O N

E X P LA N A T I O N

a loga 4 + loga 5 = loga 20

This is logarithm law 1: loga x + loga y = loga (xy)

b loga 22 – loga 11 = loga 2

This uses logarithm law 2: ( ) x loga x – loga y = loga y Note: loga

c

3 loga 2 = loga 23

22 = loga 2 11

Using logarithm law 3: n loga x = loga xn

= loga 8

Example 5 Evaluating logarithmic expressions Simplify and evaluate the following expressions. a log2 1 b log5 5 c

log6

1 36

d log2 6 – log2 3

SO L U T I O N

EX P L A N A T I O N

a log2 1 = 0

20 = 1

b log5 5 = 1

51 = 5

c

log6

1 = log6 6–2 36 = –2 × log6 6 = –2 × 1

1 Alternatively, use the rule loga = – loga x. x So log6

1 = – log6 36 36 = –2

= –2 d

log2 6 – log2 3 = log2 2 =1

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

( ) 6 log2 = log2 2 3 and 21 = 2

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688

Chapter 10 Logarithms and polynomials

1

1, 2–3(½)

3

—

UNDERSTANDING

Exercise 10B

Copy and complete the rules for logarithms using the given pronumerals. a logb (xy) = logb x + ( ) x b logb = y

–

c loga bm = m × d loga a = e logc 1 = 1 f loga = b

d loga 2 + loga g loga 34 =

= loga 8 loga 3

3 Evaluate: a log10 100 d –2 log5 25

b log3

c log

=0

e loga 36 – loga 1 h loga = loga 7 7

= loga 3

f i

b log2 32 e 4 log10 1000

Example 4b

Example 4c

Example 5a,b

loga 5 = loga 53 1 loga 3 = loga 3

c log3 27 f –6 log5 1

4–6(½) Example 4a

32 = 5

4–7(½)

4–7(½)

4 Simplify using the first logarithm law. a loga 3 + loga 2 b loga 5 + loga 3 d logb 6 + logb 3 e logb 15 + logb 1

c loga 7 + loga 4 f logb 1 + logb 17

5 Simplify using the second logarithm law. a loga 10 – loga 5 b loga 36 – loga 12 d logb 28 – logb 14 e logb 3 – logb 2

c loga 100 – loga 10 f logb 7 – logb 5

6 Simplify using the third logarithm law. a 2 loga 3 b 2 loga 5 d 4 loga 2 e 5 loga 2

c 3 loga 3 f 3 loga 10

7 Evaluate: a log3 1 d log4 4 g 5 log2 1 j

2 log 10 3 10

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b log7 1 e log18 18 h 3 log4 4 k

log15 225 2

FLUENCY

2 Write the missing numbers. a log2 =1

c logx 1 f loga a 1 i log 7 3 7 l

log3 243 10

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Number and Algebra

Example 5c

8 Simplify and evaluate. 1 a log2 4

b log3

1 5

e log10

d log5 Example 5d

9 Simplify and evaluate. a log2 10 – log2 5 d log4 8 + log4 2

1 27

8–9(½)

c log4

1 100

f

b log3 30 – log3 10 e log8 16 + log8 4

8–10(½)

PROBLEM-SOLVING

8–9(½)

1 64

log10

1 100 000

689

10B

c log4 128 – log4 2 f log10 50 + log10 2

10 Simplify using a combination of log law number 3 with law 1 and 2. a 2 log3 2 + log3 5 b 4 log10 2 + log10 3 d 5 log7 2 – log7 16

e

1 log 4 + 2 log3 2 2 3

f

1 log5 3 – log5 9 2

g

1 1 log 27 – log2 64 3 2 3

h

1 1 log 16 + log5 243 4 5 5

11

11, 12

11–13

1 1 1 √ √ √ 11 Recall that x = x 2 and 3 x = x 3 and in general n x = x n . Use this to simplify the following. √ √ √ a log2 8 b log2 32 c log2 3 16 √ √ √ d log10 1000 e log7 3 7 f log5 5 625

REASONING

c 3 log10 2 – log10 4

12 Prove that: 1 a loga = – loga x using logarithm law 2 x 1 b loga = – loga x using logarithm law 3 x √ loga x 13 Prove that loga n x = using logarithm law 3. n

—

—

14

14 Read the proof for logarithm law 1 in the introduction and then complete the following tasks. a

Complete a proof giving all reasons for logarithm law 1: loga (xy) = loga x + loga y. ( ) x b Complete a proof for logarithm law 2: loga = loga x – loga y. y c

ENRICHMENT

Proving the laws for logarithms

Complete a proof for logarithm law 3: loga xn = n loga x.

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690

Chapter 10 Logarithms and polynomials

10C Exponential equations using logarithms

10A

When solving a simple exponential equation like 2x = 16 we know that the solution is x = 4 because 24 = 16. Solving 2x = 10, however, is trickier and requires the use of logarithms. Depending on what calculator functions you have, one of two different methods can be chosen. These methods can be used to solve many types of problems in science and finance.

Let’s start: Trial and error versus logarithms Consider the equation 10x = 20. • First, use a calculator and trial and error to find a value of x (correct to three decimal places) that satisfies the equation. • Now write 10x = 20 in logarithmic form and use the log function on your calculator to find the value of x. • Check the accuracy of your value of x obtained by trial and error. The growth of bacteria can be modelled by exponential equations.

Key ideas

Solving for x if ax = y • Using the given base: x = loga y • Using base 10: ax = y log10 ax = log10 y

(taking log10 of both sides)

x log10 a = log10 y log10 y x= log10 a

(using law 3) (dividing by log10 a)

Most calculators can evaluate using log base 10, but only some calculators can work with any base.

Example 6 Solving using the given base Solve the following using the given base. Round your answer to three decimal places. a 2x = 7 b 50 × 1.1x = 100 SOL UTI ON a

EX P L A NA TI ON

2x = 7

If ax = y then x = loga y.

x = log2 7

This method can be used on calculators that have a log function loga y, where both a and y can be entered.

= 2.807 (to 3 d.p.) b

50 × 1.1x = 100 1.1x = 2 x = log1.1 2

Divide both sides by 50.

= 7.273 (to 3 d.p.) Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Write in logarithmic form, then use a calculator for the approximation.

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Number and Algebra

691

Example 7 Solving using base 10 Solve using base 10 and evaluate, correct to three decimal places. a 3x = 5 b 1000 × 0.93x = 100 SO L U T I O N

EX P L A N A T I O N

3x = 5

a

log10 3x = log10 5

Take log10 of both sides.

x log10 3 = log10 5

Use law 3: loga xn = n loga x.

x=

log10 5 log10 3

Divide by log10 3. Use the log function on a calculator.

= 1.465 (to 3 d.p.)

1000 × 0.93x = 100 0.93x = 0.1

Divide both sides by 1000.

x

log10 0.93 = log10 0.1

Take log10 of both sides.

x log10 0.93 = log10 0.1

Use law 3 and solve for x by dividing both sides by log10 0.93.

log10 0.1 x= log10 0.93

Use the log function on a calculator.

= 31.729 (to 3 d.p.)

Exercise 10C 1

1–3

2–3(½)

Write the logarithmic form of these equations. a 23 = 8 d 3x = 10

2 State the missing number. a 5 = 125 d log2

=3

1

b 52 = 25 e 7x = 2

c 42 = 2 f 1.1x = 7

b 10

c 8

= 10 000 1 e log4 = 2

—

UNDERSTANDING

b

f

= 64

log10

= –1

3 Use a calculator to evaluate the following, correct to three decimal places. a log10 7 d

log10 12 log10 7

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b log10 0.6 e

log10 1.3 log10 1.4

3 4 log10 37 log10 56

c log10 f

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Chapter 10 Logarithms and polynomials

4–6(½)

4–6(½)

Example 6a

4 Solve the following using the given base and round to three decimal places where necessary. a 3x = 5 b 2x = 11 c 5x = 13 x x d 1.2 = 3.5 e 2.9 = 3.5 f 0.2x = 0.04

Example 6b

5 Solve the following using the given base and round to three decimal places where necessary. a 10 × 2x = 20 b 25 × 3x = 75 c 4 × 1.5x = 20 d 3.8 × 1.7x = 9.5 e 300 × 0.9x = 150 f 7.3 × 0.4x = 1.8

Example 7

6 Solve using base 10 and evaluate, correct to three decimal places. a 2x = 6 b 3x = 8 x d 11 = 15 e 1.8x = 2.5 g 10 × 2x = 100 h 7 × 3x = 28 x j 4 × 1.5 = 20 k 100 × 0.8x = 50

7, 8

c f i l

FLUENCY

4–5(½)

10C

5x = 7 0.9x = 0.5 130 × 7x = 260 30 × 0.7x = 20

7, 8

8, 9

7 The rule modelling a population (P) of mosquitoes is given by P = 8t , where t is measured in days. Find the number of days, correct to three decimal places where necessary, required for the population to reach: a 64 b 200 c 1000

PROBLEM-SOLVING

692

8 An investment of $10 000 is expected to grow by 5% p.a., so the balance $A is given by the rule A = 10 000 × 1.05n , where n is the number of years. Find the time (to two decimal places) for the investment to grow to: a $20 000 b $32 000 c $100 000 9 50 kg of a radioactive isotope in a set of spent nuclear fuel rods is decaying at a rate of 1% per year. The mass of the isotope (m kg) is therefore given by m = 50 × 0.99n , where n is the number of years. Find the time (to two decimal places) when the mass of the isotope reduces to: a 45 kg b 40 kg c 20 kg

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Number and Algebra

10, 11

11, 12

10 The value of a bank balance increases by 10% per year. The initial amount is $2000. a Write a rule connecting the balance $A with the time (n years). b Find the time, correct to the nearest year, when the balance is double the original amount.

REASONING

10

693

10C

11 The value of a Ferrari is expected to reduce by 8% per year. The original cost is $300 000. a Find a rule linking the value of the Ferrari ($F ) and the time (n years). b Find the time it takes for the value of the Ferrari to reduce to $150 000. Round your answer to one decimal place. 12 The half-life of a substance is the time it takes for the substance to reduce to half its original mass. Round answers to the nearest year. a Find the half-life of a 10 kg rock if its mass reduces by 1% per year. b Find the half-life of a 20 g crystal if its mass reduces by 0.05% per year.

—

—

13

13 If ax = y then we can write x = loga y. Alternatively, if ax = y we can find the logarithm of both sides, as shown here. ax = y

ENRICHMENT

Change of base formula

logb ax = logb y x logb a = logb y logb y logb a logb y ∴ loga y = logb a x=

This is the change of base formula. a Use the change of base formula to write the following with base 10. i log2 7 ii log3 16 iii log5 1.3 b Change to log base 10 and simplify. i log5 10 ii log2 1000 c

iii log3 0.1

Make x the subject and then change to base 10. Round your answer to three decimal places. i 3x = 6 ii 9x = 13 iii 2 × 1.3x = 1.9

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694

Chapter 10 Logarithms and polynomials

10D Polynomials

10A

x We are familiar with linear expressions such as 3x – 1 and 4 + and with quadratic expressions such as 2 x2 – 3 and – 4x2 + 2x – 4. These expressions are in fact part of a larger group called polynomials, which are sums of powers of a variable using whole number powers {0, 1, 2, . . . }. For example, 2x3 – 3x2 + 4 is a cubic polynomial and 1 – 4x3 + 3x7 is a polynomial of degree 7. The study of polynomials opens up many ideas in the analysis of functions and graphing that are studied in many senior mathematics courses.

Let’s start: Is it a polynomial? A polynomial is an expression that includes sums of powers of x with whole number powers {0, 1, 2, . . . }. Decide, with reasons, whether the following are polynomials. √ • 5 + 2x + x2 • x + x2 •

2 +3 x

• 4x4 – x2 – 6

1

• 4x 3 + 2x2 + 1

Key ideas

• 5

A polynomial is an expression of the form an xn + an – 1 xn – 1 + an – 2 xn – 2 + · · · + a0 x0 , where: • n is a whole number {0, 1, 2, . . . } • an , an–1 , . . . a0 are coefficients which can be any real number. • a0 x0 = a0 is the constant term • an xn is the leading term Naming polynomials Polynomials are named by the highest power of x. This is called the degree of the polynomial. • constant For example: 2 • linear For example: 3x – 7 • quadratic For example: 2x2 – 4x + 11 • cubic For example: – 4x3 + 6x2 – x + 3 1 • quartic For example: x4 – x2 – 2 2 • of degree 8 For example: 3x8 – 4x5 + x – 3 Function notation • A polynomial in x can be called P(x). For example: P(x) = 2x3 – x is a cubic polynomial. • P(k) is the value of the polynomial at x = k. For example: If P(x) = 2x3 – x, then: P(3) = 2(3)3 – (3) and P(–1) = 2(–1)3 – (–1) = 51

= –2 + 1 = –1

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Number and Algebra

695

Example 8 Deciding if an expression is a polynomial Decide if the following expressions are polynomials. a

√ 2 b 2x2 – x + x

4x2 – 1 + 7x4

SO L U T I O N

EX P L A N A T I O N

a yes

Powers of x are whole numbers {0, 1, 2, . . . }. 1 √ 2 2x2 – x + = 2x2 – x 2 + 2x–1 x 1 Powers include and –1, which are not 2 allowed in the polynomial family.

b no

Example 9 Evaluating polynomials If P(x) = x3 – 3x2 – x + 2, find: a P(2) SO L U T I O N a

b P(–3) EX P L A N A T I O N

P(2) = (2)3 – 3(2)2 – (2) + 2

Substitute x = 2 and evaluate.

= 8 – 12 – 2 + 2 = –4 b

P(–3) = (–3)3 – 3(–3)2 – (–3) + 2 = –27 – 27 + 3 + 2

Substitute x = –3 and note (–3)3 = –27 and (–3)2 = 9.

= – 49

1

1–3

A polynomial expression is given by 3x4 – 2x3 + x2 – x + 2. a How many terms does the polynomial have? b State the coefficient of: i x4 ii x3 iii x2

2, 3(½)

—

iv x

UNDERSTANDING

Exercise 10D

c What is the value of the constant term?

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter 10 Logarithms and polynomials

UNDERSTANDING

10D 2 Decide if these polynomials are constant, linear, quadratic, cubic or quartic. a 2x – 5 b x2 – 3 c x4 + 2x3 + 1 2 d 1 + x + 3x2 e 6 f 3– x 5 i

3 State the coefficient of x2 in each of the following. a 2x3 + 4x2 – 2x + 1 b 5x4 + 3x2 – 1 d –7x5 + x3 – 2x2 + 8 e –3x6 + 2x4 – 9x2 + 1

c x4 – 2x2 – 2 f –3x5 + 2x2 – 3x3 + 10

4–6(½) Example 8

Example 9

4–7(½)

4 Decide if the following are polynomials. a 3x3 + x2 – x + 3 7 1 d – +2 x2 x √ g x + 2 – x2

b 2x4 – x2 – 4 2 e x4 – x3 + 3 x √ √ √ 4 3 h x+ x+ x

4x – x3 + x2

c f i

4–7(½)

2 3 – +2 x x 4 – 7x8 1 x3 + √ x

5 Evaluate the quadratic polynomial x2 – x + 2, using: a x=4 b x = 10 c x = –2

d x = –1

6 If P(x) = x3 – 3x2 – 2x + 3, find: a P(2) b P(4)

c P(–1)

d P(–3)

7 If P(x) = 2x4 – 3x3 + 5x – 4, find: a P(1) b P(3)

c P(–1)

d P(–2)

8, 9

8 If P(x) = x3 – x2 and Q(x) = 4 – 3x, find: a P(1) + Q(2) b P(3) + Q(–1) ( )2 ( )2 d Q(1) – P(3) e P(2) + Q(1)

9 Find the coefficient of x2 in these polynomials. 4 – 2x2 x3 + 7x2 + x – 3 a P(x) = b P(x) = 4 –7 10 Evaluate P(–2) for these polynomials. a P(x) = (x + 2)2 b P(x) = (x – 2)(x + 3)(x + 1)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

FLUENCY

h 1 – x4

g –2

8–10

9–11

c P(–2) – Q(–2) ( )3 ( )3 f P(–1) – Q(–1) c P(x) =

PROBLEM-SOLVING

696

x3 – 4x2 –8

c P(x) = x2 (x + 5)(x – 7)

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PROBLEM-SOLVING

Number and Algebra

11 The height (P metres) of a roller coaster track above a platform is given by the equation P(x) = x3 – 12x2 + 35x, where x metres is the horizontal distance from the beginning of the platform. a Find the height of the track using: i x = 2 ii x = 3 iii x = 7

10D

Polynomials are used in the design of roller coasters. 12

12, 13(½)

13–14(½)

REASONING

b Does the track height ever fall below the level of the platform? If so, find a value of x for which this occurs.

697

12 a What is the maximum number of terms in a polynomial of degree 7? b What is the maximum number of terms in a polynomial of degree n? c What is the minimum number of terms in a polynomial of degree 5? d What is the minimum number of terms in a polynomial of degree n? 3 2 13 If P(x) simplify these without  the use of a calculator.   =x – x – 2x, evaluateand 1 1 1 1 a P b P c P – d P – 2 3 2 4             4 1 1 3 3 2 e P – f P g P – +P h P – +P 3 5 2 2 4 4

14 If P(x) = 2x3 – x2 – 5x – 1, find the following and simplify where possible. a P(k) b P(b) c P(2a) d P(–a) e P(–2a) f P(–3k) g P(ab) h P(–ab) —

—

15 If P(x) = x3 – 2x2 + bx + 3 and P(1) = 4, we can find the value of b as follows. P(1) = 4 (1)3 – 2(1)2 + b(1) + 3 = 4

15

ENRICHMENT

Finding unknown coefficients

2+b=4 b=2 a Use this method to find the value of b if P(x) = x3 – 4x2 + bx – 2 and if: i P(1) = 5 ii P(2) = –6 iii P(–1) = –8 iv P(–2) = 0 v P(–1) = 2 vi P(–3) = –11 b If P(x) = x4 – 3x3 + kx2 – x + 2, find k if: i P(1) = 2 ii P(–2) = 0 c If P(x) =

x3

+ ax2

iii P(–1) = –15

+ bx – 3 and P(1) = –1 and P(–2) = –1, find the values of a and b.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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698

Chapter 10 Logarithms and polynomials

Using calculators to work with polynomials 1 Define the polynomial P(x) = x3 – 2x2 + 5x + 4 and evaluate at x = –2. 2 Expand and simplify (x2 – x + 1)(x2 + 2).

Using the TI-Nspire:

Using the ClassPad:

1 In a Calculator page define the polynomial using b>Actions>Define as shown. Evaluate for x = 2 p(–2).

1 In the Main application, type and highlight the polynomial expression. Tap Interactive, Define, OK. Evaluate by typing p(–2).

2

Use b>Algebra>Expand, then type in the expression and press ·.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

2 In the Main application, type and highlight expression. Tap Interactive, Transformation, expand. OK. EXE.

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Number and Algebra

10E Expanding and simplifying polynomials

699

10A

From your work on quadratics, you will remember how to use the distributive law to expand brackets. For example, (2x – 1)(x + 5) expands to 2x2 + 10x – x – 5, and after collecting like terms this simplifies to 2x2 + 9x – 5. In a similar way we can expand the product of two or more polynomials of any degree. To do this we also multiply every term in one polynomial with every term in the next polynomial.

Let’s start: The product of two quadratics The equation (x2 – x + 3)(2x2 + x – 1) = 2x4 – x3 + 4x2 + 4x – 3 is written on the board. • Is the equation true for x = 1? • Is the equation true for x = –2? • How can you prove the equation to be true for all values of x? Expand products of polynomials by multiplying each term in one polynomial by each term in the next polynomial. Simplify by collecting like terms.

Key ideas

Example 10 Expanding polynomials Expand and simplify. a x3 (x – 4x2 )

b (x2 + 1)(x3 – x + 1)

SOL UTI ON

EX P L A NA TI ON

a x3 (x – 4x2 ) = x4 – 4x5

x3 × x1 = x4 and x3 × (– 4x2 ) = – 4x5 using index law 1.

b

(x2 + 1)(x3 – x + 1) = x2 (x3 – x + 1) + 1(x3 – x + 1) = x5 – x3 + x2 + x3 – x + 1

(x2 + 1)(x3 − x + 1) –x3 cancels with x3 .

= x5 + x2 – x + 1

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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700

Chapter 10 Logarithms and polynomials

Example 11 Expanding P(x) × Q(x) If P(x) = x2 + x – 1 and Q(x) = x3 + 2x + 3, expand and simplify the following. a P(x) × Q(x) b (Q(x))2 SO L U T I O N a

EX P L A N A T I O N

P(x) × Q(x) 2

3

= (x + x – 1)(x + 2x + 3) = x2 (x3 + 2x + 3) + x(x3 + 2x + 3) – 1(x3 + 2x + 3)

Each term in the first polynomial is multiplied by each term in the second polynomial.

= x5 + 2x3 + 3x2 + x4 + 2x2 + 3x – x3 – 2x – 3 = x5 + x4 + x3 + 5x2 + x – 3 b

(Q(x))2

(Q(x))2 = Q(x) × Q(x)

= (x3 + 2x + 3)2 3

Expand to gain 9 terms then collect and simplify.

3

= (x + 2x + 3)(x + 2x + 3) = x3 (x3 + 2x + 3) + 2x(x3 + 2x + 3) + 3(x3 + 2x + 3) = x6 + 2x4 + 3x3 + 2x4 + 4x2 + 6x + 3x3 + 6x + 9 = x6 + 4x4 + 6x3 + 4x2 + 12x + 9

1

1–3(½)

Expand and simplify these quadratics. a x(x + 2) b x(1 – 3x) d (x – 5)(x + 11) e (2x – 5)(3x + 1)

3

—

UNDERSTANDING

Exercise 10E

c (x + 1)(x – 1) f (4x – 3)(2x – 5)

2 Collect like terms to simplify. a 2x4 – 3x3 + x2 – 1 – x4 – 2x3 + 3x2 – 2 b 5x6 + 2x4 – x2 + 5 – 5x4 + x3 + 8 – 6x6 c 7x8 – 1 + 2x6 + x – 10x8 – 3x6 + 4 – 7x 3 Use substitution to confirm that this equation is true for the given x values. (x3 – x + 3)(x2 + 2x – 1) = x5 + 2x4 – 2x3 + x2 + 7x – 3 a x=1 b x=0 c x = –2

Example 10a

4 Expand and simplify. a x2 (x – 3) d x3 (1 – x) g –2x3 (x2 + x)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b x2 (x2 – 1) e x3 (x2 + 3x) h –x2 (x5 – x2 )

4–5(½), 6

c 2x2 (1 + 3x) f –3x2 (x4 – x) i – 4x3 (x4 – 2x7 )

4–5(½), 6, 7

FLUENCY

4–5(½), 6

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Example 10b

Example 11

5 Expand and simplify. a (x2 + 1)(x3 + 2) d (x2 – 2)(x3 + x – 2) g (x3 – x2 – 1)(x3 + x – 2)

b (x2 – 1)(x3 + x) e (x3 – x)(x2 + 2x + 3) h (x3 – 5x2 + 2)(x3 – x + 1)

6 If P(x) = x2 – 2x + 1 and Q(x) = x3 + x – 1, expand and simplify: a P(x) × Q(x) b (Q(x))2

c (x2 – x)(x3 – 3x) f (x3 – x2 )(x2 – x + 4) i (x4 – x2 + 1)(x4 + x – 3)

FLUENCY

Number and Algebra

701

10E

c (P(x))2

7 If P(x) = x3 + 2x2 – x – 4 and Q(x) = x2 + x – 2, expand and simplify: a P(x) × Q(x) b (Q(x))2 c (P(x))2

8 If P(x) = x2 – 5x + 1 and Q(x) = x3 + x, expand and simplify: a P(x) + Q(x) b Q(x) – P(x) d 1 – P(x)Q(x) e 4 – (Q(x))2

8(½), 9

8(½), 9, 10

PROBLEM-SOLVING

8(½)

c 5P(x) + 2Q(x) f (P(x))2 – (Q(x))2

9 Find the square of P(x) when P(x) = (x2 + x – 1)2 . 10 Show that (x2 – x – 1)2 – (x2 – x + 1)2 = 4x – 4x2 . 11, 12

12, 13

REASONING

11

11 If P(x) and Q(x) are polynomials, does P(x)Q(x) = Q(x)P(x) for all values of x? 12 Give the degree of the polynomial P(x) × Q(x) when: a b c d

P(x) is quadratic and Q(x) is linear P(x) is quadratic and Q(x) is cubic P(x) is cubic and Q(x) is quartic P(x) is of degree 7 and Q(x) is of degree 5.

13 If P(x) is of degree m and Q(x) is of degree n and m > n, what is the highest possible degree of the following polynomials? a P(x) + Q(x) b P(x) – Q(x) c P(x) × Q(x) d (P(x))2 e (P(x))2 – Q(x) f (Q(x))3

14 Expand and simplify. a x(x2 + 1)(x – 1) c (x + 2)(x – 1)(x + 3) e (5x – 2)(x – 2)(3x + 5)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

—

—

b x3 (x + 3)(x – 1) d (x + 4)(2x – 1)(3x + 1) f (x2 + 1)(x2 – 2)(x + 3)

14

ENRICHMENT

Triple expansions

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702

Chapter 10 Logarithms and polynomials

10F Division of polynomials

10A

Division of polynomials requires the use of the long division algorithm. You may have used this algorithm for the division of whole numbers in primary school. Recall that 7 divided into 405 can be calculated in the following way.

5×7

 57 7 405

7×7

35

7 into 4 does not go. 7 into 40 gives 5 and 5 × 7 = 35.

55

Then subtract 405 – 350.

49 6

7 into 55 gives 7 and 7 × 7 = 49. Subtract to give remainder 6. Division of polynomials occurs often in the field of signal processing.

So 405 ÷ 7 = 57 and 6 remainder. The 57 is called the quotient. Another way to write this is 405 = 7 × 57 + 6. We use this technique to divide polynomials.

Let’s start: Recall long division Use long division to find the quotient and remainder for the following. • 832 ÷ 3 • 2178 ÷ 7

Key ideas

We use the long division algorithm to divide polynomials. The result is not necessarily a polynomial. Example:

dividend

x3

x2 + x − 1 15 = x 2 − 3x + 7 − x+2 x+2

divisor

remainder

quotient

We can write this as: x3 − x2 + x − 1 = (x + 2)(x2 − 3x + 7) − 15 dividend

divisor

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

quotient

remainder

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Number and Algebra

703

Example 12 Dividing polynomials a

Divide P(x) = x3 + 2x2 – x + 3 by (x – 2) and write in the form P(x) = (x – 2)Q(x) + R, where R is the remainder.

b

Divide P(x) = 2x3 – x2 + 3x – 1 by (x + 3) and write in the form P(x) = (x + 3)Q(x) + R.

SO L U T I O N a

EX P L A N A T I O N 2 ) x + 4x + 7

x – 2 x3 + 2x2 – x + 3 x2 (x – 2) x3 – 2x2

4x2 – x + 3 4x2 – 8x

4x(x – 2)

7x + 3 7(x – 2)

7x – 14 17

First, divide x from (x – 2) into the leading term (i.e. x3 ). So divide x into x3 to give x2 . x2 (x – 2) gives x3 – 2x2 and subtract from x3 + 2x2 – x + 3. After subtraction, divide x into 4x2 to give 4x and repeat the process above. After subtraction, divide x into 7x to give 7. Subtract to give the remainder 17.

∴ x3 + 2x2 – x + 3 = (x – 2)(x2 + 4x + 7) + 17 b

2 ) 2x – 7x + 24

x + 3 2x3 – x2 + 3x – 1 2x2 (x + 3)

2x3 + 6x2 –7x2 + 3x – 1

–7x(x + 3)

–7x2 – 21x 24x – 1

24(x + 3)

24x + 72 –73

First, divide x from (x + 3) into the leading term. So divide x into 2x3 to give 2x2 . After subtraction, divide x into –7x2 to give –7x. After subtraction, divide x into 24x to give 24. Subtract to give the remainder –73.

∴ 2x3 – x2 + 3x – 1 = (x + 3)(2x2 – 7x + 24) – 73

1

1, 2

Use long division to find the remainder. a 208 ÷ 9 b 143 ÷ 7

2 Copy and complete the equation with the missing numbers. a If 182 ÷ 3 = 60 remainder 2, then 182 = × 60 + . b If 2184 ÷ 5 = 436 remainder 4, then 2184 = × 436 + . c If 617 ÷ 7 = 88 remainder 1, then 617 = 7 × + .

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

2

c 2184 ÷ 3

—

UNDERSTANDING

Exercise 10F

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Chapter 10 Logarithms and polynomials

3, 4

10F

3, 4, 5(½)

3, 4, 5(½)

Example 12a

3 Divide P(x) = x3 + x2 – 2x + 3 by (x – 1) and write in the form P(x) = (x – 1)Q(x) + R, where R is the remainder.

Example 12b

4 Divide P(x) = 3x3 – x2 + x + 2 by (x + 1) and write in the form P(x) = (x + 1)Q(x) + R, where R is the remainder.

FLUENCY

704

5 For each of the following, express in this form: Dividend = divisor × quotient + remainder (as in the examples) a (2x3 – x2 + 3x – 2) ÷ (x – 2) b (2x3 + 2x2 – x – 3) ÷ (x + 2) 3 2 c (5x – 2x + 7x – 1) ÷ (x + 3) d (–x3 + x2 – 10x + 4) ÷ (x – 4) e (–2x3 – 2x2 – 5x + 7) ÷ (x + 4) f (–5x3 + 11x2 – 2x – 20) ÷ (x – 3)

6, 7(½)

b (8x5 – 2x4 + 3x3 – x2 – 4x – 6) ÷ (x + 1)

7 Divide the following and express in the usual form. a (x3 – x + 1) ÷ (x + 2) b (x3 + x2 – 3) ÷ (x – 1) 4 c (x – 2) ÷ (x + 3) d (x4 – x2 ) ÷ (x – 4)

8

8, 9

PROBLEM-SOLVING

6 Divide and write in this form: Dividend = divisor × quotient + remainder a (6x4 – x3 + 2x2 – x + 2) ÷ (x – 3)

6, 7(½)

9–11

8 There are three values of k for which P(x) = x3 – 2x2 – x + 2 divided by (x – k) gives a remainder of zero. Find the three values of k.

REASONING

6

9 Prove that (6x3 – 37x2 + 32x + 15) ÷ (x – 5) leaves remainder 0. 10 Find the remainder when P(x) is divided by (2x – 1) given that: a P(x) = 2x3 – x2 + 4x + 2 b P(x) = –3x3 + 2x2 – 7x + 5 11 Find the remainder when P(x) = –3x4 – x3 – 2x2 – x – 1 is divided by these expressions. a x–1 b 2x + 3 c –3x – 2 —

—

12

12 Divide the following and express in the form P(x) = divisor × Q(x) + R, where R is a function of x. a (x3 – x2 + 3x + 2) ÷ (x2 – 1) b (2x3 + x2 – 5x – 1) ÷ (x2 + 3) c (5x4 – x2 + 2) ÷ (x3 – 2)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

ENRICHMENT

When the remainder is not a constant

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Number and Algebra

Progress quiz 38pt 10A

38pt 10A

38pt 10A

38pt 10B

1

2

3

4

705

10A

Express as a logarithm. a 25 = 32 b 103 = 1000

c a1 = a

Rewrite using exponential notation. a log10 100 = 2 b log2 8 = 3

c log7 1 = 0

Find the value of x. a log2 16 = x c log12 1 = x e log10 100 000 = x

b log3 81 = x d log5 x = 4 f log10 0.000001 = x

Simplify the following, using logarithm laws. 1 b log2 + log2 12 3

a log10 20 + log10 50 c log3 18 – log3 2 √ e log10 10

1 9 log4 24 – (log4 2 + log4 3)

d log3 f

38pt 10C

5 Solve and round each answer to three decimal places. a 3x = 7 b 1.2x = 200 c 500(1.09)x = 1000

38pt 10D

6

38pt 10D

7

Consider the polynomial P(x) = 3x4 – 2x3 + x2 + 7x + 8. Find: a P(0) b P(–1) c P(k)

38pt 10D

8

For the polynomial P(x) = 4x3 + 3x2 + 2x + 1, state: a the degree of the polynomial b the constant term 2 c the coefficent of x d the leading term.

38pt 10E

9 Expand and simplify the following. a x4 (x3 – 2x + 1)

√ Explain why 4x3 – x + 6x – 1 is not a polynomial.

b (x2 – 1)(x2 + 2x + 6)

38pt 10E

10 If P(x) = x3 + x + 2 and Q(x) = x4 + 2x, find the following in their simplest forms. a P(x) + Q(x) b P(x) – Q(x) 2 c (Q(x)) d P(x) × Q(x)

38pt 10F

11 Divide P(x) = 2x3 – x2 + 3x – 1 by (x + 2) and write in the form P(x) = (x + 2)Q(x) + R, where R is the remainder.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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706

Chapter 10 Logarithms and polynomials

10G The remainder and factor theorems

10A

Using long division we can show, after dividing (x – 2) into P(x) = x3 – x2 + 2x – 3, that P(x) = x3 – x2 + 2x – 3 = (x – 2)(x2 + x + 4) + 5, where 5 is the remainder. Using the right-hand side to evaluate P(2), we have: P(2) = (2 – 2)(22 + 2 + 4) + 5 = 0 × (22 + 2 + 4) + 5 =0+5 =5 This shows that the remainder when P(x) is divided by (x – 2) is P(2). More generally, when P(x) is divided by (x – a) we obtain: P(x) = (x – a)Q(x) + R So P(a) = 0 × Q(a) + R =R So the remainder is P(a) and this result is called the remainder theorem. This means that we can find the remainder when dividing P(x) by (x – a) simply by evaluating P(a). We also know that when factors are divided into a number there is zero remainder. So if P(x) is divided by (x – a) and the remainder P(a) is zero, then (x – a) is a factor of P(x). This result is called the factor theorem.

Let’s start: Which way is quicker? A polynomial P(x) = x3 – 3x2 + 6x – 4 is divided by (x – 2). • Show, using long division, that the remainder is 4. • Find P(2). What do you notice? • Explain how you can find the remainder when P(x) is divided by: a x–3 b x–5 • Show that when P(x) is divided by (x + 1) the remainder is –14. • What would be the remainder when P(x) is divided by (x – 1)? What do you notice and what does this say about (x – 1) in relation to P(x)?

Key ideas

Remainder theorem: When a polynomial P(x) is divided by (x – a) the remainder is P(a). • When dividing by (x – 3) the remainder is P(3). • When dividing by (x + 2) the remainder is P(–2).

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

Factor theorem: When P(x) is divided by (x – a) and the remainder is zero (i.e. P(a) = 0), then (x – a) is a factor of P(x). P(2) = 0 P(x) = x3 − 3x2 − 3x + 10

= (x − 2)(x2 − x − 5) factor

707

Key ideas

(x – 2) is a factor with zero remainder.

quotient

Example 13 Using the remainder theorem Find the remainder when P(x) = x3 – 5x2 – x + 4 is divided by: a x–2 b x+1 SOL UTI ON a

EX P L A NA TI ON

P(x) = x3 – 5x2 – x + 4 3

For (x – 2) substitute x = 2.

2

P(2) = (2) – 5(2) – 2 + 4 = 8 – 20 – 2 + 4

Using the remainder theorem, P(2) gives the remainder.

= –10 The remainder is –10. b

P(x) = x3 – 5x2 – x + 4

For (x + 1) substitute x = –1. Note: (–1)3 = –1, (–1)2 = 1 and –(–1) = 1.

P(–1) = –1 – 5 + 1 + 4 = –1 The remainder is –1.

Example 14 Finding a linear factor Decide whether each of the following are factors of P(x) = x3 + x2 – 3x – 6. a x+1 b x–2 SOL UTI ON a

P(x) = x3 + x2 – 3x – 6 P(–1) = –1 + 1 + 3 – 6

EX P L A NA TI ON If (x + 1) is a factor of P(x), then P(–1) = 0. But this is not true as the remainder is –3.

= –3 ∴ (x + 1) is not a factor. b

P(x) = x3 + x2 – 3x – 6

Substitute x = 2 to evaluate P(2).

P(2) = 8 + 4 – 6 – 6

Since P(2) = 0, (x – 2) is a factor of P(x).

=0 ∴ (x – 2) is a factor.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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708

Chapter 10 Logarithms and polynomials

Example 15 Applying the remainder theorem Find the value of k such that (x3 – x2 + 2x + k) ÷ (x – 1) has a remainder of 5. SO L U T I O N

EX P L A N A T I O N

Let P(x) = x3 – x2 + 2x + k. The remainder is P(1), which is 5.

P(1) = 5 3

2

(1) – (1) + 2(1) + k = 5

Substitute x = 1 and solve for k.

2+k=5 k=3

1

1–3

3

If P(x) = 2x3 – x2 – x – 1, find the value of the following. a P(1) b P(3) c P(–2)

—

d P(– 4)

2 What value of x do you substitute into P(x) to find the remainder when a polynomial P(x) is divided by: a x – 3? b x + 2?

UNDERSTANDING

Exercise 10G

3 What is the remainder when an expression is divided by one of its factors?

Example 13

Example 14

4–7(½)

4–7(½)

4 Find the remainder when P(x) = x3 – 2x2 + 7x – 3 is divided by: a x–1 b x–2 c x–3 e x+4 f x+2 g x+1

d x–4 h x+3

5 Find the remainder when P(x) = x4 – x3 + 3x2 is divided by: a x–1 b x–2 c x+2

d x+1

6 Decide which of the following are factors of P(x) = x3 – 4x2 + x + 6. a x–1 b x+1 c x–2 e x–3 f x+3 g x–4

d x+2 h x+4

FLUENCY

4–6(½)

7 Decide which of the following are factors of P(x) = x4 – 2x3 – 25x2 + 26x + 120. a x–2 b x+2 c x+3 d x–3 e x–4 f x+4 g x–5 h x+5

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

8–9(½)

8–9(½)

PROBLEM-SOLVING

8(½)

8 Use the factor theorem and trial and error to find a linear factor of these polynomials. a P(x) = x3 + 2x2 + 7x + 6 b P(x) = x3 + 2x2 – x – 2 c P(x) = x3 + x2 + x + 6 d P(x) = x3 – 2x – 4

709

10G

9 Use the factor theorem to find all three linear factors of these polynomials. a P(x) = x3 – 2x2 – x + 2 b P(x) = x3 – 2x2 – 5x + 6 c P(x) = x3 – 4x2 + x + 6 d P(x) = x3 – 2x2 – 19x + 20

Example 15

10–12

11–13

REASONING

10

10 For what value of k will (x3 – 2x2 + 5x + k) ÷ (x – 1) have the following remainders? a 0 b 2 c –10 d 100 11 For what value of k will (x4 – 2x3 + x2 – x + k) ÷ (x + 2) have zero remainder? 12 Find the value of k in these polynomials. P(x) = x3 + 2x2 + kx – 4 and when divided by (x – 1) the remainder is 4. P(x) = x3 – x2 + kx – 3 and when divided by (x + 1) the remainder is –6. P(x) = 2x3 + kx2 + 3x – 4 and when divided by (x + 2) the remainder is –6. P(x) = kx3 + 7x2 – x – 4 and when divided by (x – 2) the remainder is –2.

13 Find the value of k when: a (x + 2) is a factor of x3 – kx2 – 2x – 4

Simultaneous coefficients

b (x – 3) is a factor of 2x3 + 2x2 – kx – 3

—

—

14

14 Use simultaneous equations and the given information to find the value of a and b in these cubics. a P(x) = x3 + ax2 + bx – 3 and P(1) = –1 and P(2) = 5 b P(x) = 2x3 – ax2 – bx – 1 and P(–1) = –10 and P(–2) = –37

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

ENRICHMENT

a b c d

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710

Chapter 10 Logarithms and polynomials

10H Solving polynomial equations

10A

We know from our work with quadratics that the Null Factor Law can be used to solve a quadratic equation that has been factorised. For example: x2 – 3x – 40 = 0 (x – 8)(x + 5) = 0 Using the Null Factor Law: x–8=0

or

x=8

or

x+5=0 x = –5

We can also apply this method to polynomials. If a polynomial is not in a factorised form, we need to use the remainder and factor theorems to help find its factors. Long division can also be used.

Let’s start: Solving a cubic Consider the cubic equation P(x) = 0, where P(x) = x3 + 6x2 + 5x – 12. • Explain why (x – 1) is a factor of P(x). • Use long division to find P(x) ÷ (x – 1). • Write P(x) in the form (x – 1)Q(x). • Now complete the factorisation of P(x). • Show how the Null Factor Law can be used to solve P(x) = 0. Why are there three solutions?

Key ideas

A polynomial equation of the form P(x) = 0 can be solved by: • factorising P(x) • using the Null Factor law: If a × b × c = 0 then a = 0, b = 0 or c = 0. To factorise a polynomial follow these steps. • Find one factor using the remainder and factor theorems. Start with (x – 1) using P(1) or (x + 1) using P(–1). If required, move to (x – 2) or (x + 2) etc. • A good idea is to first consider factors of the constant term of the polynomial to reduce the number of trials. • Use long division to find the quotient after dividing by the factor. • Factorise the quotient (if possible). • Continue until P(x) is fully factorised.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

711

Example 16 Using the Null Factor Law Solve for x. a (x – 1)(x + 2)(x + 5) = 0

b (2x – 3)(x + 7)(3x + 1) = 0

SO L U T I O N

EX P L A N A T I O N

a (x – 1)(x + 2)(x + 5) = 0 x – 1 = 0 or x + 2 = 0 or x=1 x = –2

Using the Null Factor Law, if a × b × c = 0 then a = 0 or b = 0 or c = 0.

b (2x – 3)(x + 7)(3x + 1) = 0 2x – 3 = 0 or x + 7 = 0 or 2x = 3 x = –7 x=

3 2

x+5=0 x = –5 3x + 1 = 0 3x = –1

x = –7

x=–

Equate each factor to 0 and solve for the three values of x.

1 3

Example 17 Factorising and solving Solve x3 + 2x2 – 5x – 6 = 0. SO L U T I O N

EX P L A N A T I O N

Let P(x) = x3 + 2x2 – 5x – 6.

Try to find a factor using the remainder and factor theorems. Start with (x – 1) using P(1) or (x + 1) using P(–1). If required, move to (x – 2) or (x + 2) of other factors of 6.

P(1) = 1 + 2 – 5 – 6 π 0 P(–1) = –1 + 2 + 5 – 6 = 0 ∴ x + 1 is a factor. 2 )x + x – 6 3 x + 1 x + 2x2 – 5x – 6 2

3

P(–1) = 0 so (x + 1) is a factor. Divide (x + 1) into P(x) to find the quotient using long division.

2

x (x + 1) x + x x2 – 5x – 6 x(x + 1)

x2 + x –6x – 6

–6(x + 1)

–6x – 6

Note: The remainder is 0, as expected (P(–1) = 0).

0 ∴ P(x) = (x + 1)(x + x – 6) 2

P(x) = (x + 1)Q(x) + R but R = 0.

= (x + 1)(x + 3)(x – 2) Solve P(x) = 0:

x2 + x – 6 factorises to (x + 3)(x – 2). Use the Null Factor Law to now solve for x.

(x + 1)(x + 3)(x – 2) = 0 x + 1 = 0 or x = –1

x + 3 = 0 or x = –3

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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712

Chapter 10 Logarithms and polynomials

1

1, 2

2

—

UNDERSTANDING

Exercise 10H

Give a reason why (x + 1) is a factor of P(x) = x3 – 7x – 6. Hint: Find P(–1).

2 Use the Null Factor Law to solve these quadratic equations. a (x – 1)(x + 3) = 0 b (x + 2)(x – 2) = 0 c (2x + 1)(x – 4) = 0 d x2 – x – 12 = 0 2 e x + 6x + 9 = 0 f x2 + 7x + 12 = 0

3 Solve for x using the Null Factor Law. a (x + 3)(x – 2)(x – 1) = 0 c (x – 4)(x + 4)(x – 3) = 0 e (2x + 1)(x – 3)(3x + 2) = 0 ( ) 1 x + (3x + 11)(11x + 12) = 0 g 2

Example 17

3–4(½)

b (x + 2)(x + 7)(x – 1) = 0 ( ) ( ) 1 1 d x + (x – 3) x + =0 2 3 f (4x – 1)(5x – 2)(7x + 2) = 0 ( ) 1 h (5x + 3)(19x + 2) x – =0 2

4 For each of the following cubic equations, follow these steps as in Example 17. • Use the factor theorem to find a factor. • Use long division to find the quotient. • Factorise the quotient. • Write the polynomial in a fully factorised form. • Use the Null Factor Law to solve for x. a x3 – 4x2 + x + 6 = 0 c x3 – 6x2 + 11x – 6 = 0 e x3 – 3x2 – 16x – 12 = 0

b x3 + 6x2 + 11x + 6 = 0 d x3 – 8x2 + 19x – 12 = 0 f x3 + 6x2 – x – 30 = 0

5

5, 6

5 Use the quadratic formula to solve for x, expressing your answers in exact form. a (x – 1)(x2 – 2x – 4) = 0 b (x + 2)(x2 + 6x + 10) = 0 6 Solve by first taking out a common factor. a 2x3 – 14x2 + 14x + 30 = 0

b 3x3 + 12x2 + 3x – 18 = 0

7 Solve for x. a x3 – 13x + 12 = 0

b x3 – 7x – 6 = 0

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

6, 7

PROBLEM-SOLVING

Example 16

3–4(½)

FLUENCY

3–4(½)

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Number and Algebra

8, 9

9–11

REASONING

8

8 What is the maximum number of solutions to P(x) = 0 when P(x) is of degree: a 3? b 4? c n?

713

10H

9 Show that the following equations can be factorised easily without the use of long division, and then give the solutions. a x3 – x2 = 0 b x3 + x2 = 0 c x3 – x2 – 12x = 0 d 2x5 + 4x4 + 2x3 = 0 10 Explain why x4 + x2 = 0 has only one solution. 11 Explain why (x – 2)(x2 – 3x + 3) = 0 has only one solution.

—

—

12

12 Factorising a quartic may require two applications of the factor theorem and long division. Solve these quartics by factorising the left-hand side first. a x4 + 8x3 + 17x2 – 2x – 24 = 0 b x4 – 2x3 – 11x2 + 12x + 36 = 0 4 3 2 c x + x – 11x – 9x + 18 = 0 d 2x4 – 3x3 – 7x2 + 12x – 4 = 0

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

ENRICHMENT

Quartics with four factors

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714

Chapter 10 Logarithms and polynomials

Using calculators to factorise and solve polynomials 1 Define the polynomial P(x) = x3 – 2x2 – 5x + 6 and factorise. 2 Solve x3 – 2x2 – 5x + 6 = 0.

Using the TI-Nspire:

Using the ClassPad:

1 In a Calculator page define the polynomial using b>Actions>Define. Factor the polynomial using b>Algebra>Factor as shown.

1 In the Main application, type and highlight the polynomial. Tap Interactive, Define. Type p for function name and tap OK. Type p(x )) in the next entry line. Highlight and tap Interactive, Transformation, factor, factor.

2

Solve using b>Algebra>Solve. Then type p(x ) = 0, x ) as shown. Alternatively use Solve using b>Algebra>Zeros for solving equations equalling zero.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

2 Type and highlight p(x ) = 0, Tap Interactive, Advanced, Solve.

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715

Number and Algebra

10I Graphs of polynomials

10A

So far in Year 10 we have studied graphs of linear equations (straight lines) and graphs of quadratic equations (parabolas). We have also looked at graphs of circles, exponentials and hyperbolas. In this section we introduce the graphs of polynomials by focusing on those of degree 3 and 4. We start by considering the basic cubic y = x3 and quartic y = x4 , and then explore other cubics and quartics in factorised form.

Let’s start: Plotting y = x 3 and y = x 4 y

Complete the table before plotting points and considering the discussion points below. x y=x

–2

–1

–

1 2

0

1 2

1

16 14 12 10 8 6 4 2

2

2

y = x3 y = x4

• •

Describe the features and shape of each graph. Describe the differences between the graphs of y = x2 and y = x4 . Where do they intersect?

-2 -1-2O -4 -6 -8 -10 -12 -14 -16

• y=

• y=

x4

y

y y = x4 y = x2

(1, 1) (-1, -1)

x

Key ideas

Graphs of basic polynomials x3

1 2

x

O gradient = 0 at (0, 0)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

(-1, 1)

(1, 1) O

x

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716

Key ideas

Chapter 10 Logarithms and polynomials

To sketch cubic graphs in factorised form with three different factors: • Find the three x-intercepts using the Null Factor Law. • Find the y-intercept. • Connect points to sketch a positive or negative cubic graph. Positive cubic

Negative cubic

(The coefficient of

x3

(The coefficient of x3 is negative.)

is positive.)

y = (x – 1)(x + 2)(x + 3)

y = –(x + 4)(x – 2)(x + 1)

y

y

-3 -2

O

8

x

1

-4

-6

-1 O

2

x

Further consideration is needed to find turning points of cubics, as they are not located symmetrically between x-intercepts. This will be studied at more senior levels of mathematics.

Example 18 Sketching cubic graphs Sketch the graphs of the following by finding the x- and y-intercepts. a y = (x + 2)(x – 1)(x – 3) b y = –x(x + 3)(x – 2) SOL UTI ON

EX P L A NA TI ON

a y = (x + 2)(x – 1)(x – 3) y-intercept at x = 0: y = (2)(–1)(–3)

Substitute x = 0 to find the y-intercept.

=6 x-intercepts at y = 0: 0 = (x + 2)(x – 1)(x – 3) ∴ x + 2 = 0 or x – 1 = 0 or x = –2 x=1 y

O

Use the Null Factor Law. x–3=0 x=3 Mark the four intercepts and connect to form a positive cubic graph. The coefficient of x3 in the expansion of y is positive, so the graph points upwards to the right.

6

-2

Substitute y = 0 to find the x-intercepts.

1

3

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

x

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Number and Algebra

b y = –x(x + 3)(x – 2) y-intercept at x = 0: y = –0(3)(–2) = 0 x-intercepts at y = 0: 0 = –x(x + 3)(x – 2) ∴ –x = 0 or x + 3 = 0 or x=0 x = –3 y

Find the y-intercept using x = 0.

The three factors are –x, x + 3 and x – 2. x–2=0 x=2 The coefficient of x3 in the expansion of y is negative, so the graph points downwards at the right.

x

2

Exercise 10I 1

1, 3

2

—

Join the given x- and y-intercepts to form a smooth cubic curve. Each graph has been started for you on the right-hand side. a b y y

3

-3

-1 O

UNDERSTANDING

O

-3

717

3

2

-5

x

O

2

5

x

2 Find the x- and y-intercepts of the graphs of these cubics. a y = (x + 1)(x – 3)(x – 4) b y = (x + 3)(x – 7)(x + 1) c y = –x(x + 2)(x – 4) d y = –2x(x + 7)(x – 5) 3 On the same set of scaled axes plot graphs of the following. Use this table to help you. x y=x

–2

–1

–

1 2

0

1 2

1

2

2

y = x3 y = x4

a y = x2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

–

–

b y = x3

c y = x4

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Chapter 10 Logarithms and polynomials

4(½)

Example 18

4(½)

4(½)

FLUENCY

10I

4 Sketch the graphs of the following by finding x- and y-intercepts. a y = (x + 2)(x – 1)(x – 3) b y = (x – 3)(x – 4)(x + 1) c y = (x – 5)(x – 1)(x + 2)

d y=

e y = x(x – 2)(x + 3)

f

y = x(x – 5)(x + 1) 1 h y = – x(x + 1)(x – 3) 3   1 j y = –(x + 3) x – (x + 1) 2

g y = –2x(x – 1)(x + 3) i

1 (x + 3)(x – 2)(x – 1) 2

y = –(x + 2)(x + 4)(x – 1)

5, 6

5 Sketch the graph of: a y = –x3

b y = –x4

6 Find a cubic rule for these graphs. a y

b

5, 6

5–7

PROBLEM-SOLVING

718

y

9 -1 O

-4

2

x -3

-8

y

c

O

3

x

3

2

x -3

7 Sketch these quartics. a y = (x – 5)(x – 3)(x + 1)(x + 2)

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

1

y

d

(-1, 3) -3

O

-1O

2

x

b y = –x(x + 4)(x + 1)(x – 4)

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Number and Algebra

8–9(½)

REASONING

8–9(½)

8 We know that the graph of y = (x – 2)2 – 1 is the graph y = x2 translated 2 units to the right and 1 unit down. Use this idea to sketch graphs of the following. a y = (x – 2)3 – 1 b y = (x + 2)3 3 c y=x –2 d y = x4 – 1 e y = (x + 3)4 f y = (x – 2)4 – 3 9 If a polynomial has a repeated factor (x – a), then the point at x = a is an x-intercept and also a turning point; e.g. y = x(x – 2)2 . Now sketch these polynomials. a y = x(x – 3)2 b y = –2x(x + 1)2 2 c y = –(x + 2) (x – 3) d y = (x + 4)(x + 1)2 e y = (2 – x)(x + 1)2 f y = –x2 (x + 2)(x – 2)

Polynomial with the lot

—

—

y

O

x

2

10

10 To sketch a graph of a polynomial that is not in factorised form you must factorise the polynomial to help find the x-intercepts. Complete the following for each polynomial. i Find the y-intercept. ii Factorise the polynomial using the factor theorem and long division. iii Find the x-intercepts. iv Sketch the graph. a y = x3 + 4x2 + x – 6 c y = x4 + 2x3 – 9x2 – 2x + 8

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

10I

ENRICHMENT

8

719

b y = x3 – 7x2 + 7x + 15 d y = x4 – 34x2 + 225

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720

Chapter 10 Logarithms and polynomials

Using calculators to sketch polynomials Sketch the graphs of y = x2 and y = x4 on the same set of axes and find the intersection points.

1

2 Sketch P(x) = x3 + 2x2 – 11x – 12 and find the x-intercepts.

Using the TI-Nspire:

Using the ClassPad:

1 In a Graphs page, enter the rules y = x 2 and y = x 4 . Adjust the scale using Window Settings and find their intersection points using b>Analyze >Intersection. Alternatively use b>Geometry>Points & Lines >Intersection Point(s) to display all three intersections simultaneously as shown.

1 In the Graph&Table application, enter the rules y 1 = x 2

2

Enter the rule P (x ) = x 3 + 2x 2 – 11x – 12. Adjust the scale using Window Settings. Find the x -intercepts using Trace>Graph Trace or using Analyze Graph>Zero and set the lower and upper bounds by scrolling left and right.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

and y 2 = x 4 . Tap to adjust the scale. Tap see the graph. Tap Analysis, G-Solve, Intersect.

2 Enter the rule y 1 = x 3 + 2x 2 – 11x – 12. Tap see the graph. Adjust the scale by tapping on Analysis, G-Solve, root to find x -intercepts.

to

to . Tap

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Number and Algebra

721

Investigation The logarithmic graph y

You will be familiar with the shape of an exponential curve, which in its basic form is a graph of the rule y = ax , where a > 0 and a π 1. This is shown here. The graph of a logarithmic equation can also be plotted to form a curve that holds a special relationship with the exponential curve.

y = ax 1 O

x

Forming the logarithmic curve a Complete the table of values for y = 2x and y = log2 x. x y = 2x x

–3

–2

–1

0

1

2

3

1 8

1 4

1 2

1

2

4

8

y = log2 x

b On a set of axes with –8 ≤ x ≤ 8 and –8 ≤ y ≤ 8, plot the graphs of: i y = 2x ii y = log2 x c On the same set of axes, now sketch the graph of y = x. Features of the logarithmic graph a What is the equation of the asymptote of: i y = 2x ?

ii y = log2 x?

b What special position is held by the graph y = x between the two graphs of y = 2x and y = log2 x? c Describe the general shape of a logarithmic curve. Different bases a Use technology or plot the following equations by hand to compare these logarithmic curves with different bases. i y = log2 x ii y = log3 x iii y = log5 x b Describe at least two common features of these three logarithmic curves. c Describe the effect of changing the base when graphing these logarithmic curves.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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722

Chapter 10 Logarithms and polynomials

Logarithmic scales – the moment magnitude scale The moment magnitude scale, like the Richter scale it replaced, is a logarithmic scale with base 10. It is used to measure the strength of an earthquake. The moment magnitude is obtained by multiplying a calculation of the area and the amount of displacement in the slip. Because it is a logarithmic scale, an earthquake of magnitude 4.0, for example, is 10 times more powerful than one of magnitude 3.0. An earthquake of magnitude 4.0 is considered ‘light’, and the shaking of the ground is noticeable but unlikely to cause damage. An earthquake of magnitude 6.0 would be 100 times more powerful than that of magnitude 4.0 and is likely to cause significant damage in populated areas. The earthquake in Christchurch, New Zealand, in 2011 was of magnitude 6.3. The Great East Japan Earthquake in 2011 was of magnitude 9.0, nearly 1000 times as strong.

a

Explain why an earthquake of magnitude 4.0 is 10 times more powerful than one of magnitude 3.0.

b

How much stronger is an earthquake of magnitude 7.0 than an earthquake of magnitude: i 6.0? ii 5.0? iii 2.0?

c

Investigate some of the most powerful earthquakes ever recorded. Describe their strength, using the moment magnitude scale, and the damage they caused.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Number and Algebra

723

Up for a challenge? If you get stuck on a question, check out the 'Working with unfamiliar problems' poster at the end of the book to help you.

Problems and challenges 1

Simplify the following without the use of a calculator. 16 1 b – log2 + 3 log2 4 a 2 log3 4 – log3 9 4 √ 1 d 2 log2 27 ÷ log2 9 c log5 125 + log3 3

2

Solve these equations using log base 10. Round your answers to two decimal places. a 5x – 1 = 2 b 0.2x = 10 c 2x = 3x + 1

3

Solve for x: 2 log10 x = log10 (5x + 6)

4

Given that loga 3 = p and loga 2 = q, find an expression for loga 4.5a2 .

5

Solve these inequalities using log base 10. Round your answers to two decimal places. a 3x > 10 b 0.5x ≤ 7

6

If y = a × 2bx and the graph of y passes through (–1, 2) and (3, 6), find the exact values of a and b.

7

An amount of money is invested at 10% p.a., compound interest. How long will it take for the money to double? Give an exact value.

8

Find the remainder when x4 – 3x3 + 6x2 – 6x + 6 is divided by (x2 + 2).

9

x3 + ax2 + bx – 24 is divisible by (x + 3) and (x – 2). Find the values of a and b.

10 Prove the following, using division. a x3 – a3 = (x – a)(x2 + ax + a2 )

b x3 + a3 = (x + a)(x2 – ax + a2 )

11 Solve for x. a (x + 1)(x – 2)(x – 5) ≤ 0

b x3 – x2 – 16x + 16 > 0

y

12 A cubic graph has a y-intercept at 2, a turning point at (3, 0) and another x-intercept at –2. Find the rule for the graph.

2

-2

O

3

x

13 Given that x2 – 5x + 1 = 0, find the value of x4 – 2x3 – 16x2 + 13x + 14 without solving the first equation. 14 A quartic graph has a turning point at (0, 0) and two x-intercepts at 3 and –3. Find the rule for the graph if it also passes through (2, 2). Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Chapter summary

724

Chapter 10 Logarithms and polynomials

Definition

Other properties

loga y = x is equivalent to a x = y , a > 0

• loga 1 = 0 (a 0 = 1) • loga a = 1 (a 1 = a) • loga x1 = −loga x ( x1 = x −1)

• log2 8 = 3 since 23 = 8 -3 • log10100× =102 since 102 = 100 • log3 1 = -3 since 3-3 = 1 27 27 • log100.001 = -3 since 10-3 = 0.001

Logarithms (10A)

Log law 3 loga x n = n loga x

Log law 1 loga x + loga y = loga (xy ) e.g. loga 5 + loga 6 = loga 30

1

Log law 2 x loga x − loga y = loga (y )

e.g. log2 8 2 = 12 log2 8 = 12 ´ 3

e.g. log318 − log3 2 = log3 9 =2

Solving exponential equations 2x

=5 x = log2 5

or

2x = 5 log10 2 x = log10 5 x log10 2 = log10 5 log10 5 x= log10 2

= 32

Polynomial of degree n

Expanding polynomials Example: (x 2 − 1)(x 2 + 2x − 3) = x 2(x 2 + 2x − 3) − 1(x 2 + 2x − 3) = x 4 + 2x 3 − 3x 2 − x 2 − 2x + 3 = x 4 + 2x 3 − 4x 2 − 2x + 3

an + an −1x n −1 + .... + a 0x 0 e.g. cubic x 3 − 2x 2 + 3x − 4 xn

Powers of x are whole numbers.

Remainder theorem

Division

If P (x ) is divided by (x − a) then the remainder is P(a).

(x 3 − 2x 2 + 5x + 1) ¸ (x − 1)

Polynomials (10A)

x2 − x + 4 x − 1 x 3 − 2x 2 + 5x + 1

Factor theorem

x 2(x − 1) x 3 − x 2 −x 2 + 5x + 1

If P(x ) is divided by (x − a) and P(a) = 0 then (x − a) is a factor of P(x ).

−x (x − 1) 4(x − 1)

∴ x 3 − 2x 2 + 5x + 1 = (x − 1)(x 2 − x + 4) + 5

Null Factor Law (x + 2)(x − 3)(2x + 1) = 0 x + 2 = 0 or x − 3 = 0 or 2x + 1 = 0 x = −2

x =3

−x 2 + x 4x + 1 4x − 4 5

dividend

divisor quotient remainder

x = − 12 Graphs y

y

y

Factorising • Find one factor using the factor theorem. • Use long division to find the quotient. • Factorise the quotient.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

0

y = −(x + 2)(x − 1)(x − 2)

y = x4

y = x3 x 0

x

−2 −1 O

1 2

x

−4

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

10A

38pt 10A

38pt 10A

38pt 10A

38pt 10B

Multiple-choice questions 1

Which of the following is equivalent to 53 = 125? A log3 125 = 5 B log3 5 = 125 C log5 125 = 3

38pt 10D

38pt 10D

38pt 10G

38pt 10H

38pt 10I

10A

38pt 10A

C 32

D 128

E 642

3 If 5x = 7, then x is equal to: 7 B logx 5 A log10 5

C log10 7

D log7 5

E log5 7

C 36

D 6

E 0

5 x6 – 2x2 + 1 is a polynomial of degree: A 1 B –2

C 2

D 6

E 0

6 Which of these is a polynomial? 1 √ A x 2 + x2 B 3 x + x2

C x3 + x2 +

D 4x5 – 2x3 – 1

E

4 log6

1 simplifies to: 6 B 1

1 x2

1 –x x

7 If P(x) = x3 – x and Q(x) = –2x2 + 1, then P(–1) – Q(1) is equal to: A 1 B –1 C 2 D 3

E –3

8 The remainder when P(x) = 2x3 + 4x2 – x – 5 is divided by (x + 1) is: A –4 B –2 C –10 D 0

E –1

9 The three solutions to (x – 3)(x + 5)(2x – 1) = 0 are x equals: 1 1 1 1 A , – and 2 B 5, –3 and – C –5, 3 and – D –5, 3 and 1 3 5 2 2 10 The equation of this graph could be: A y = x(x + 5)(x – 1) B y = (x + 5)(x + 1)(x + 3) C y = (x – 5)(x – 1)(x + 3) D y = (x – 5)(x + 1)(x – 3) E y = (x + 5)(x + 1)(x – 3)

E –5, 3 and

1 2

y

-5

x

-1 O 3

Short-answer questions 1

Write the following in logarithmic form. a 24 = 16

38pt 10A

E log125 3 = 5

2 If log2 64 = x, then x is equal to: A 5 B 6

A –1 38pt 10D

D 1253 = 5

725

Chapter review

Number and Algebra

b 103 = 1000

c 3–2 =

1 9

2 Write the following in index form. a log3 81 = 4

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

b log4

1 = –2 16

c log10 0.1 = –1

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Chapter review

726

Chapter 10 Logarithms and polynomials

38pt 10A

38pt 10B

3 Evaluate the following. a log10 1000

b log3 81

c log2 16

d log7 1

1 e log3 27

f

g log4 0.25

h log10 0.0001

i

4 Simplify using the laws for logarithms. a loga 4 + loga 2 c logb 24 + logb 6 e 2 loga 2 g log10 25 + log10 4 √ i log2 8

b d f h j

1 125 log3 0.1˙ log5

logb 7 + logb 3 loga 1000 – loga 100 3 loga 10 log3 60 – log3 20 3 loga 2 + loga 2

38pt 10C

5 Solve these equations using logarithms with the given base. a 3x = 6 b 20 × 1.2x = 40

38pt 10C

6 Solve for x, in exact form, using base 10. a 2x = 13

b 100 × 0.8x = 200

7 If P(x) = x3 – x2 – x – 1, find: a P(0) b P(2)

c P(–1)

8 Expand and simplify. a (x2 + 2)(x2 + 1) c (x2 + x – 3)(x3 – 1)

b x3 (x2 – x – 3) d (x3 + x – 3)(x3 + x – 1)

38pt 10D

38pt 10E

d P(–3)

38pt 10F

9 Use long division to express each of the following in this form: Dividend = divisor × quotient + remainder a (x3 + x2 + 2x + 3) ÷ (x – 1) b (x3 – 3x2 – x + 1) ÷ (x + 1) 3 2 c (2x – x + 4x – 7) ÷ (x + 2) d (–2x3 – x2 – 3x – 4) ÷ (x – 3)

38pt 10G

10 Use the remainder theorem to find the remainder when P(x) = 2x3 – 2x2 + 4x – 7 is divided by: a x–1 b x+2 c x+3 d x–3

38pt 10G

11 Using the factor theorem, decide if the following are factors of P(x) = x3 – 2x2 – 11x + 12. a x+1 b x–1 c x–4 d x+3

38pt 10H

12 Solve these cubic equations. a (x – 3)(x – 1)(x + 2) = 0

b (x – 5)(2x – 3)(3x + 1) = 0

13 Factorise and solve these cubic equations. a x3 + 4x2 + x – 6 = 0

b x3 – 9x2 + 8x + 60 = 0

14 Sketch the graphs of these polynomials. a y = x3 c y = –x(x – 3)(x + 2)

b y = (x + 1)(x – 1)(x – 4) d y = x4

38pt 10H

38pt 10I

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

10A

Extended-response questions 1

A share portfolio initially valued at $100 000 is invested, compounding continuously at a rate equivalent to 10% per annum. a If $A is the value of the investment and n is the number of years, which of the following is the correct rule linking A and n? 1.1n A A = 100 000 × 0.1n B A = 100 000 × 1.1n C A= 100 000 b Find the value of the investment, correct to the nearest dollar, after: i 2 years ii 18 months iii 10.5 years c Find the time, correct to two decimal places, when the investment is expected to have increased in value to: i $200 000 ii $180 000 iii $0.5 million

727

Chapter review

Number and Algebra

2 A cubic polynomial has the rule P(x) = x3 – 5x2 – 17x + 21. a Find: i P(–1) ii P(1) b Explain why (x – 1) is a factor of P(x). c Divide P(x) by (x – 1) to find the quotient. d Factorise P(x) completely. e Solve P(x) = 0. f Find P(0). g Sketch a graph of P(x), labelling x- and y-intercepts.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

728

Semester review 2

Semester review 2 Measurement Multiple-choice questions 1

Correct to two decimal places, the perimeter and area, respectively, for this shape are: A 12.71 cm, 16.77 cm2 5 cm B 20.71 cm, 22.07 cm2 C 25.42 cm, 29.14 cm2 3 cm D 18.36 cm, 43.27 cm2 E 17.71 cm, 17.25 cm2

2 0.04 m2 is equivalent to: A 4 cm2 B 40 mm2 10A

C 0.0004 cm2

E 0.00004 km2

D 400 cm2

3 A square-based pyramid has base area 30 m2 and height 7 m. Its volume is: A 105 m3 B 70 m3 C 210p m3 D 210 m3 E 140 m3 4 The curved surface area of this half cylinder, in exact form, is: A 80p cm2 B 105p cm2 C 92.5p cm2 2 2 D 120p cm E 160p cm

10 cm 10A

5

The volume of a sphere of diameter 30 cm is closest to: A 113 097 cm3 B 2827 cm3 C 11 310 cm3 D 14 137 cm3

16 cm E 7069 cm3

Short-answer questions 1

Find the perimeter and area of these shapes. Give answers correct to one decimal place where necessary. You will need to use Pythagoras’ theorem for part c. a 3 cm b c 50 mm 0.2 m

30 mm 4 cm

90 mm

10 cm 2 Find the surface area and volume for these solids. Give your answers to one decimal place. 10A b 10A c a 2 cm

7m

6 cm

6m 1m

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Semester review 2

729

3 A rectangular prism has length 5 cm, width 3 cm and volume 27 cm3 . a Find the height of the prism. b Find the total surface area of the prism. 10A

4 A cone has volume 90 cm3 and height 10 cm. Find the exact radius of the cone. Extended-response question

10A

A cylindrical glass vase is packaged inside a box that is a rectangular prism, so that the vase touches the box on all four sides and is the same height as the box. The vase has a diameter of 8 cm and height 15 cm. Round your answers to two decimal places where necessary. a Find the volume of the vase. b Find the volume of space inside the box but outside the vase. c A glass stirring rod is included in the vase. Find the length of the longest rod that can be packaged inside the vase. d Find the difference in the length of rod in part c and the longest rod that can fit inside the empty box. Round your answer to two decimal places.

Parabolas and other graphs Multiple-choice questions 1

The graph of y = (x – 2)2 + 3 could be: A B y

y

( 3, −2)

(−2, 3)

(2, −3)

O

y

x

O

x

O

D

1

7

3

y

C

x

y

E

7 4 (2, 3) O

x

O

(2, 3)

x

2 The graph of y = x2 – 4x has a turning point with coordinates: A (2, –4) B (0, –4) C (4, 0) D (–2, 12) 10A

E (1, –3)

3 For the quadratic y = ax2 + bx + c, b2 – 4ac < 0, we know that the graph has: A a maximum turning point B two x-intercepts C no y-intercept D no x-intercepts E a minimum turning point

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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730

Semester review 2

4 The graph with equation x2 + y2 = 9 is: A a circle with radius 9 B a parabola with turning point (0, 9) C a circle with radius 3 D a hyperbola with asymptote at x = 3 E an exponential curve with y-intercept (0, 3) 10A

5 The equation of the asymptote of y = 2x + 3 is: A x=3 B y=3 C x=2

D y=2

E y = 2x

Short-answer questions 1

Sketch the following parabolas and state the transformations from y = x2 . a y = 3x2 b y = –(x + 2)2 c y = x2 + 5

2 Consider the quadratic y = x2 + 4x – 5. a Find the y-intercept. b Find the x-intercepts by factorising. c Use symmetry to find the turning point. d Sketch the graph. 3 Consider the quadratic y = –2(x – 3)2 + 8. a State the coordinates of the turning point and whether it is a maximum or minimum. b Find the y-intercept. c Find the x-intercepts by factorising. d Sketch the graph. 4 Sketch the following quadratics by first completing the square. a y = x2 + 6x + 2 b y = x2 – 5x + 8 5 Consider the quadratic y = 2x2 – 4x – 7. a Use the discriminant to determine the number of x-intercepts of the graph. b Sketch its graph using the quadratic formula. Round x-intercepts to one decimal place. c Find the points of intersection of the quadratic and the line y = –6x – 3 by solving simultaneously.

10A

6 Sketch the following graphs, labelling key features. a x2 + y2 = 4 b x2 + y2 = 10 d y = 5–x

e y=

2 x

c y = 3x f

y=–

6 x

7 Find the coordinates of the points of intersection of the graphs of the following. a x2 + y2 = 15 and y = 2x b y = 2x and y = 16 2 c y = and y = 8x x 10A

8 Sketch the graphs of the following relations. Label important features. 1 a (x – 2)2 + (y + 1)2 = 16 b y = 2x + 3 + 1 c y= –3 x+2

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Semester review 2

731

Extended-response question 1 2 (x – 120x + 1100), where h is 40 the height above the ground and x is the horizontal distance from the start of the section. All distances are measured in metres and x can take all values between 0 and 200 metres. a Sketch the graph of h vs x for 0 ≤ x ≤ 200, labelling the endpoints. b What is the height above ground at the start of the section? c The rollercoaster travels through an underground tunnel. At what positions from the start will it enter and leave the tunnel? d What is the maximum height the rollercoaster reaches? e What is the maximum depth the rollercoaster reaches? A rollercoaster has a section modelled by the equation h =

Probability Multiple-choice questions 1

The number of tails obtained from 100 tosses of two fair coins is shown in the table. Number of tails Frequency

0 23

1 57

2 20

From this table, the experimental probability of obtaining two tails is: A 0.23 B 0.25 C 0.2 D 0.5 E 0.77 2 From the given two-way table Pr(A ∩ B ) is: B B

A 2

A 5 12

6

A

1 2

B

2 3

C

1 4

D

4 5

E

1 3

3 Two events, A and B, are such that Pr(A) = 0.7, Pr(B) = 0.4 and Pr(A ∩ B) = 0.3. Pr(A ∪ B) is equal to: A 1.4 B 0.8 C 0.6 D 0 E 0.58  4 From the information in the Venn diagram, Pr(A  B) is: A B 5 4 4 A B C 3 4 5 12 5 7 D

4 9

E

1 3

3

5 A bag of 5 marbles contains 2 green ones. Two marbles are randomly selected without replacement. The probability of selecting the 2 green marbles is: 9 2 1 2 4 A B C D E 20 25 10 5 25

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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732

Semester review 2

Short-answer questions 1

Consider events A and B. Event A is the set of letters in the word ‘grape’ and event B is the set of letters in the word ‘apricot’: A = {g, r, a, p, e} B = {a, p, r, i, c, o, t} a Represent the two events A and B in a Venn diagram. b If a letter is randomly selected from the alphabet, find: i Pr(A) ii Pr(A ∩ B) iii Pr(A ∪ B)

iv Pr(B )

c Are the events A and B mutually exclusive? Why or why not? 2 The Venn diagram shows the distribution of elements in two sets, A and B. a Transfer the information in the Venn diagram to a two-way table. b Find: i n(A ∩ B) ii n(A ∩ B) iii n(B ) iv n(A ∪ B) c Find: i Pr(A ∩ B) iii Pr(B)

A 1 3

B 4 4

ii Pr(A ∩ B )  iv Pr(B  A)

3 Two events, A and B, are such that Pr(A) = 0.24, Pr(B) = 0.57 and Pr(A ∪ B) = 0.63. Find: a Pr(A ∩ B) b Pr(A ∩ B ) 4 Two fair 4-sided dice numbered 1 to 4 are rolled and the total is noted. a List the sample space as a table. b State the total number of outcomes. c Find the probability of obtaining: i

a sum of 4

ii

a sum of at least 5

iii a sum of 7, given the sum is at least 5. 5 In a group of 12 friends, 8 study German, 4 study German only and 2 study neither German nor Mandarin. Let A be the event ‘studies German’. Let B be the event ‘studies Mandarin’. a Summarise the information in a Venn diagram. b Find:  i Pr(A) ii Pr(A  B) c State whether or not the events A and B are independent.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Semester review 2

733

Extended-response question Lindiana Jones selects two weights from her pocket to sit on a weight-sensitive trigger device after removing the goblet of fire. Her pocket contains three weights, each weighing 200 g, and five weights, each weighing 250 g. The two weights are selected randomly without replacement. Use a tree diagram to help answer the following. a Find the probability that Lindiana selects two weights totalling: i 400 g ii 450 g iii 500 g b c

If the total weight selected is less than 480 g, a poison dart will shoot from the wall. Find the probability that Lindiana is at risk from the poison dart. By feeling the weight of her selection, Lindiana knows that the total weight is more than 420 g. Given this information, what is the probability that the poison dart will be fired from the wall?

Statistics Multiple-choice questions 1

For the given stem-and-leaf plot, the range and median, respectively, of the data are: Stem 0 1 2

Leaf 2 2 0 1 3 3

6 2 5

7 3 7

 1  5 means 15

A 20, 12.5

5 9

B 7, 12

8

C 27, 12.5

D 29, 3

E 27, 13

2 The interquartile range (IQR) for the data set 2, 3, 3, 7, 8, 8, 10, 13, 15 is: A 5 B 8.5 C 7 D 13 E 8 3 The best description of the correlation between the variables for the scatter plot shown is: A weak, negative B strong, positive C strong, negative D weak, positive E no correlation

y

x

4 A line of best fit passes through the points (10, 8) and (15, 18). The equation of the line is: 2 1 A y= x+8 B y = 2x – 12 C y = – x + 13 3 2 D y = 2x + 6 10A

1 E y= x+3 2

5 The mean and sample standard deviation of the small data set 2, 6, 7, 10 and 12, correct to one decimal place, are: A ¯x = 7.4 and s = 3.8 B ¯x = 7 and s = 3.7 C ¯x = 7.4 and s = 3.4 D ¯x = 7 and s = 7.7 E ¯x = 27.1 and s = 9.9

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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734

Semester review 2

Short-answer questions 1

Twenty people are surveyed to find out how many days in the past completed month they used public transport. The results are as follows. 7, 16, 22, 23, 28, 12, 18, 4, 0, 5 8, 19, 20, 22, 14, 9, 21, 24, 11, 10 a Organise the data into a frequency table with class intervals of 5 and include a percentage frequency column. b Construct a histogram for the data, showing both the frequency and the percentage frequency on the one graph. c i State the frequency of people who used public transport on 10 or more days. ii

State the percentage of people who used public transport on fewer than 15 days.

iii State the most common interval of days for which public transport was used. Can you think of a reason for this?

2 By first finding quartiles and checking for outliers, draw box plots for the following data sets. a 8, 10, 2, 17, 6, 25, 12, 7, 12, 15, 4 b 5.7, 4.8, 5.3, 5.6, 6.2, 5.7, 5.8, 5.1, 2.6, 4.8, 5.7, 8.3, 7.1, 6.8 3 Farsan’s bank balance over 12 months is recorded below. Month J F M A M J J A S O N D Balance ($) 1500 2100 2300 2500 2200 1500 1200 1600 2000 2200 1700 2000

a b c d

Plot the time-series for the 12 months. Describe the way in which the bank balance has changed over the 12 months. Between which consecutive months did the biggest change in the bank balance occur? What is the overall change in the bank balance over the year?

4 Consider the variables x and y and the corresponding bivariate data below. x 4 5 6 7 8 9 10 y 2.1 2.5 3.1 2.8 4 3.6 4.9

a b c d

Draw a scatter plot for the data. Describe the correlation between x and y as either positive, negative or none. Fit a line of good fit by eye to the data on the scatter plot. Use your line of good fit to estimate: i y when x = 7.5 ii x when y = 5.5

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Semester review 2

10A

735

5 The back-to-back stem-and-leaf plot below shows the number of DVDs owned by people in two different age groups. Over 40 Under 40 Leaf Stem Leaf 9 7 6 6 4 3 2 0 7 8 6 4 3 2 2 0 1 2 5 5 7 8 2 4 4 6 8 3 3 2 6 9 4 1 8 2 4 means 24

10A

a

By considering the centre and spread of the data, state with reasons: i Which data set will have the higher mean? ii Which data set will have the smaller standard deviation?

b

Calculate the mean and sample standard deviation for each data set. Round your answers to one decimal place where necessary.

Extended-response question After a month of watering with a set number of millilitres of water per day, the height of a group of the same species of plant is recorded below. Water (mL) Height (cm)

a

b c

8 25

5 27

10 34

14 40

12 35

15 38

18 45

Using ‘Water’ for the x-axis, find the equation of: i

the least squares regression line

ii

the median–median regression line.

Use your least squares regression line to estimate the height (to the nearest cm) of a plant watered with 16 mL of water per day. Use your median–median regression line to estimate the daily amount of water (to the nearest mL) given to a plant of height 50 cm.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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736

Semester review 2

Logarithms and polynomials 10A

Multiple-choice questions 1

The value of log2 16 is: A 8

C 216

B 256

D

1 4

E 4

2 An equivalent statement to 3x = 20 is: A x = log3 20

B 20 = log3 x

C x = log20 3

  20 E x = log10 3

D 3 = logx 20

3 The expression that is not a polynomial is: A 3x2 + 1

B 2 – 5x5 + x

D 4x3 + 2x –

C 7x – 5

1 E 5x6 + 2x4 – 3x2 – x x

4 When P(x) = x4 – 3x3 + 2x + 1 is divided by (x – 2), the remainder is: A 37 B 2 C –3 D 13 E 5 5 A possible graph of y = –x(x + 3)(x – 2) is: A B y

y

−2 −3

O

x

2

y

C

y

D

−3 –3

O

x

3

O

2

O

2

x

x

y

E

−3 O

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

2

x

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Semester review 2

10A

737

Short-answer questions 1

Simplify where necessary and evaluate. 1 a log4 64 b log5 25 e log4 2 + log4 8 f log3 54 – log3 6

2 Solve for x. a log6 216 = x 3 a b

c log10 1000

d log7 1

g log8 8

h loga a3

b logx 27 = 3

Solve for x using the given base. i 3x = 30

c log3 x = 4 ii 15 × 2.4x = 60

Solve for x using base 10 and evaluate, correct to three decimal places. i 7x = 120 ii 2000 × 0.87x = 500

4 Consider the polynomials P(x) = x3 + 3x2 – 4x – 6 and Q(x) = 2x3 – 3x – 4. a Find: i P(2) ii P(–1) iii Q(–3) iv Q(1) b

Expand and simplify. i P(x) × Q(x)

ii (Q(x))2

5 Divide P(x) = x3 – 4x2 + 2x + 7 by (x – 3) and write in the form P(x) = (x – 3)Q(x) + R, where Q(x) is the quotient and R is the remainder. 6 Find the remainder when P(x) = x3 – 2x2 – 13x – 10 is divided by each of the following and, hence, state if it is a factor. a x–1 b x+2 c x–3 7 Solve for x. a (x + 1)(x – 3)(x + 6) = 0 c x3 + 5x2 + 2x – 8 = 0 10A

b –x(2x – 5)(3x + 2) = 0 d 2x3 – 3x2 – 3x + 2 = 0

Extended-response question A section of a train track that heads through a valley and then over a mountain is modelled by the equation P(x) = –2x3 + 3x2 + 23x – 12 for –5 ≤ x ≤ 6. a Show that (x + 3) is a factor of P(x). b Hence, factorise P(x) using division. c Sketch a graph of this section of the track, labelling axes, intercepts and endpoints.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers Working with unfamiliar problems: Part 1 x6

1 + 2 99

6x5 y

+

15x4 y2

+

20x3 y3

+

15x2 y4

+

6xy5

+

y6

3 i x = 15 ii x = 450 iii x = 6 4 1, $56 8 5 24 cm 6 a i 11.8 seconds ii 6.5 seconds b 1, 1, 1 4 2 4 7 XY = 5.6 cm 8 35 9 72

4 22◦

5 y = 31 4 6 x = 0.9, y = 3.3 √ 7 16 10 cm 8 20 students; 6 with 100%, 7 with 75%, 7 with 76%, mean = 82.85%. 9 9 1 units 8

10 Charlie 23 years, Bob 68 years 11 D 2 12 b = 1 1 3 13 k = 11 14 V = 27

 n b The perimeter increases indefinitely as 3x 4 →∞ 3 as n → ∞. The area approaches a finite value as area  n √ → 0 as n → ∞. change 3 x2 × 3 4 4 4 9 2 77 cm, 181 cm √ 3 2k(2 3 − 3)

10 a 11

b 28

d 3 2

c 20

11 640 cm3 ,

TSA = 54

cm2

15 4 cm < third side < 20 cm. Its length is between the addition and subtraction of the other two sides. 16 785

12 P to R : 145◦ , 1606 m; R to S : 295◦ , 789 m; S to Q : 51◦ , 1542 m; Q to P : 270◦ , 1400 m 13 y = 1 x2 − x + 1 or y = 1 (x − 4)2 − 1 8 8 y

17 n + 1 18 10

x= 4

19 3 : 5

(4, 1)

1

Working with unfamiliar problems: Part 2 1

a i

ii

 5 P = 3 × 45 × x5 or P = 3x 4 ; 3 3

O

1.2

(4, −1)

x

6.8 y = −3

P = 3 × 4n × xn 3 √ √  2 A = 3 x2 + 3 × 3 x + 4 4 3 √  2 √  2 3 × 4 × 3 x2 + 3 × 42 × 3 x3 4 3 4 3 √   Area change = 3 × 4n−1 × 3 xn 2 4 3

738 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Hours

10 a 3(x − 3)

1st and 6th 2nd and 5th 3rd and 4th

% change from equation

6.7%

18.3%

25%

% change from ‘rule of thumb’

8.3%

16.7%

25%

b 1

c −4

g 2 7

h −7 3

4 a yes

b yes

c no

5 a 9 6 a 10a

b −8 b 15d

c −8 c 0

d −9

e 1 2

8 a 5x + 5 d −20 − 5b g 6m + 18

c (−2)3 = (−2)(−2)(−2) = −8 15 a True b False, 1 − 2 π 2 − 1 1 2 d False, π e True 2 1 f False, 3 − (2 − 1) π (3 − 2) − 1

d −9 d 5xy

d −10hm h −21b2 d5

c ‘Half of the sum of a and b’ or ‘a plus b all divided by 2’.     17 a P = 4 + p x + 2, A = 1 + p x2 + x 2 4     b P = 6 + p x − 6, A = 3 − p x2 − 3x 2 4   c P = 2px, A = 1 + p x2 2

c −6ad g 12s2 t

j 24p3 q o −a 3

k −18h5 i5 l 63m2 pr p − ab q 2b 4

s −y 2

t −a 2

1

b 2x + 8

c 3x − 15

e −2y + 6 f −7a − 7c h 4m − 12n + 20

q −6a3 + 3a2 + 3a

j 2x2 + 10x l −12x2 + 16xy

n 36g − 18g2 − 45gh p 14y2 − 14y3 − 28y r −5t4 − 6t3 − 2t

s 6m4 − 2m3 + 10m2 t x4 − x 9 a 5x + 23 b 10a + 26 c 21y + 3 d 15m + 6 f 11t − 1 i −11d3

g True

h False, 8 ÷ (4 ÷ 2) π (8 ÷ 4) ÷ 2

b 25ab f 30bl

m 15y2 + 3yz − 24y o −8ab + 14a2 − 20a

c True

b It could refer to either of the above, depending on interpretation.

f 9t g 9b h −st2 j −0.7a2 b k 2gh + 5 l 12xy − 3y

i −2p + 6q + 4 k 6a2 − 24a

e 10 h 15z − 7

b x2 − 5x

16 a x + y or x + y 2 2

m 3a + 7b n 8jk − 7j o ab2 + 10a2 b p 2mn − m2 n q 5st − s2 t r 3x3 y4 + 2xy2

r −3x

12 a 2x2 + 6x

c P = 4x + 14, A = 7x + 12

1 C 2 D

n 3ab

p −3a(ab + 2b + 1) c 61 d 12 7 g − h 1 5

14 a (−2)(−2) = 4, negative signs cancel b a2 > 0 ∴ −a2 < 0

Exercise 1A

i 8a2 b4

o −y(y + 8z) 11 a −32 b 7 1 e − f 13 2 5

13 a P = 4x − 4, A = − 2x − 4 b P = 4x + 2, A = 3x − 1

Chapter 1

7 a 12ab e 30ht

l ab(7a + 1) n −6mn(1 + 3n)

x2

b The proportion of tide height change = 1 [cos(30t1 ) − cos(30t2 )] 2

e 4ab i −3m2 n

h 9y(y − 7) j x2 y(1 − 4y)

k 8a2 (b + 5) m −5t(t + 1)

in the 2nd and 5th hours and the same in the middle two hours. Overall, this is quite an accurate ‘rule of thumb’.

f −1 5

d 6(y + 5) f 2a(a + 4)

g 5x(x − 1) i xy(1 − y)

The percentage change per hour for the ‘rule of thumb’ is 1.6 points higher for the 1st and 6th hours, 1.6 points lower

3 a 7

b 4(x − 2)

c 10(y + 2) e x(x + 7)

Answers

14 a

g 3x2 + 15x j 9q4 − 9q3

mx

Exercise 1B a 1 2

b 2

c 5 6

d 37 30

2 a 2 3

b 3 7a

c − 7t 4xy

d −

3 a 12

b 6

c 14

d 2x

4 a 5x

b 4x

c a 4 g −9b k − 3x y

d 1 3a h −2y l 6b 7

e 5x i − 1 2p 5 a x+2

f −2x j − 4 9st b a−5

c 3x − 9

e 1 + 6b

f 1 − 3x

g 3−t

i x+2

j 3 − 2x

k a−1

b2 c 8x2 a

d 1 − 3y

h x−4 l 1 + 2a 3

739 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

6 a x−1 2x f 5a 2 7 a 3 f

1 25

b x+4 5x

c −4

d 4 9

g 2

h 15

i −1 2

b 3

c 18 5

d 3 4

g −5 3

h 2 5

i −1 3

b 4a + 3 8

c 3 − 15b 10

d 4x + 6 15

e 1 − 6a 9

f 3a + 8 4a

g 7a − 27 9a

h 16 − 3b 4b

i 4b − 21 14b

j 27 − 14y 18y

k − 12 − 2x 3x

l

b 7x + 11 12

c −x−7 4

d 2x + 15 9

e 4x + 7 6

f 8x + 3 10

g 7x + 2 24

h 5x − 1 5

i 5 − 3x 14

8 a 3a + 14 21

9 a 9x + 23 20

e 5

3 a false

b true

e 4 3

− 27 − 2x 6x

c −3 g −1 3

d 4

i −4

j 3 2

k −9 2

l −4 3

m −2

n 7

o −2

p 11 9

b 9

c 23 2

d −5 6

f 2 3 j 7 n 23

g 1

h 2

k −9 o 1

l 5

5 a 1 e − 9 11 i 4 m 19 6 a 10 e −5 i −9 m 20

c −22 g 16

d 4 h 4

j 8 n 15

k 6 o −9

l −7 p 5

7 a x + 3 = 7, x = 4 c x − 4 = 5, x = 9 e 2x + 5 = 13, x = 4

c e g i

3x − 1 (x − 3)(x + 5)

d

−21 (2x − 1)(x − 4)

f

41 − 7x (2x − 1)(x + 7)

h

14 − 17x (3x − 2)(1 − x)

11 a i a2 2a − 3 b i a2

x − 18 (x + 3)(x − 4)

14x − 26 (x − 5)(3x − 4) 3x + 17 (x − 3)(3x + 4)

d

3x + 14 iii 4x2

vi 7 2

5a + 2 a2 21x − 9x2 e 14(x − 3)2 b

3x − x2 (x − 2)2

15 a 2

3x + 5 (x + 1)2 f yz − xz − xy xyz c

b 1

Exercise 1C 1 a no 2 a true

d 15 − x = 22, x = −7

f 2(x − 5) = −15, x = − 5 2 h 2x − 5 = x − 3, x = 2

b 0

c −17

d 7 2

f 28 5

g 13 14

h 2 5

b 6

c 2

d 25

i 94 11 9 a 1 10 17 cm 12 24 km

a2 + a − 4 ii a2

14 a −1

e 27 23

b x + 8 = 5, x = −3

11 17 and 18

ii x2

12 The 2 in the second numerator needs to be subtracted, x − 2. 6 13 a −(3 − 2x) = −3 + 2x(−1 × −2x = 2x) 2 b i ii 2x iii x + 3 iv − 7 x−1 3−x 7−x 3 v − 12 x

8 a 1

h − 11 6

b 13 f 6

g 3x + 8 = 23, x = 5 7x − 13 b (x − 7)(x + 2)

7x + 22 10 a (x + 1)(x + 4)

c true

b 8 f 11 4

4 a 5 e 5 2

b no b false

c yes c false

13 a $214 b $582 c i 1 14 a 41 L

ii 10.5 iii 21 b 90 s = 1 min 30 s

c 250 s = 4 min 10 s 15 a 6 b 4

c −15

d 20

e 3

f 6 g 1 h −26 i −10 16 x = 9. Method 2 is better, expanding the brackets is unnecessary, given 2 is a factor of 8. 17 a 5 − a b a c 5 6 a d 2a + 1 a

e 3a + 1 a

f c−b a

18 a a =

c b+1

b a=

b b+1

c a=

1 c−b

d a=

b b−1

d yes

e a = −b 19 a 6a

b

ab a+b

f a = bc b−c c

abc b−a

740 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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1

b x≤2

7 a x>6

a 3, 6, 10 (Answers may vary.)

e x≤ 1 16

b −4, −3, −2 (Answers may vary.) d −8.5, −8.4, −8.3 (Answers may vary.) 2 a B b C 3 11, 12 or 13 rabbits

c A

4 a x≥1 b x −9 f 8 < x ≤ 11

g −9 < x < −7 i −1 ≤ x < 1

h 1.5 ≤ x ≤ 2.5

9 a i C < $1.30 b i less than 9 min

e x ≥ 27 29

x

4

b 4 − x ≥ −2, x ≤ 12 2

c 3(x + 1) ≥ 2, x ≥ − 1 3 d x + (x + 2) ≤ 24, x ≤ 10 since x must be even e (x − 6) + (x − 4) + (x − 2) + x ≤ 148, x ≤ 40

10 a x < −5 b x ≥ 11 4

5 a x 32

32

x

x

i x < −18

−18

b −9.5 < x ≤ −7

x

f x>3

3

13 a −4 ≤ x < 5 c x = 10 14 a 3 ≤ x ≤ 9

x

2

b x < 2 − 4a

c x < 1 − 7 or x < a − 7 if a > 0. a a Reverse the inequality if a < 0.

d x ≤ 10

k x ≤ 10 9 10 9

x

x

x

x

x

l x < −3 8

−0.375 6 a x ≥ −2 b x < 2 5 e x < −8 f x ≥ 4

b 1, 3 is the only whole number.

x

c x ≤ −5 d x ≤ −7 g x ≥ −10 h x < −21

1 15 a x ≥ 23

b x < 19 5

15 4

x

c x≤1

741 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

Exercise 1E 1

d m = −1, c = 2

y

a y = −2x + 5, m = −2, c = 5 b y = 2x − 3, m = 2, c = −3

y = −x + 2

c y = x − 7, m = 1, c = −7 d y = − 2x − 3 , m = − 2 , c = − 3 5 5 5 5

2 a i 3

ii 6

iii 21 2

b i 2

ii 6

iii 8 3

3 a i d iii

b iv e v

c ii f vi

4 a yes d no

b yes e yes

c no f no

2

(1, 1) x

O

e m = 1, c = −4

y y=x– 4

5 a m = 5, c = −3

y

x

O y = 5x −3

(1, 3)

–4

(1, 2) x

O

f m = −3,c = 1 2

−3

y y = − 32 x + 1

b m = 2, c = 3

y (1, 5)

1 O

y = 2x + 3

x (2,−2)

3 x

O

g m = 4 , c = −2 3

y c m = −2, c = −1

y y = −2x −1

−1 O

O −2

y = 43 x −2 (3, 2) x

x (1, −3)

742 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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6 a m = −3, c = 12

y 6

y = − 72 x + 6

O

12 (1, 9)

y = −3x + 12 x

(2, –1)

y

x

O

b m = −5, c = 5 2

i m = 0.5, c = −0.5

y

Answers

h m = −7,c = 6 2

y

y = 0.5x − 0.5 2.5 (2, 0.5)

O

x

O

x

0.5

(1, −2.5) 5 2

y = −5x + c m = 1, c = −7

j m = −1, c = 1

y

y y=1−x

x

O 1 O

1

x

(1, −6)

–7 d m = 1, c = −2

k m = 2,c = 3 3

y y=x–2

y y=

2 3x

+3

(3 , 5)

3

y = x –7

O x

O

e m = 4 , c = −3 3

l m = −0.2, c = 0.4

(1, –1) −2

x

y

y

y=

4 3x

-3

y = 0.4 − 0.2x (0.4)

(3, 1) (1, 0.2)

O

x

O -3

x

743 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

f m = −1, −c = − 1 3 y

c x = −2.5, y = 10

y = −x –

y

1 3

y = 4x + 10 10

– 13

O

x

(1, – ) 4 3

−2.5

x

O

g m = −4, c = −8

y d x = 4 , y = −4 3

y

x

O

y = 3x − 4 y = -4x - 8 -8

O

x

4 3

(1, -12) –4 h m = −1,c = 1 2 4

y

e x = 3.5, y = 7

y y = −2x + 7

0.25 O

(2, -0.75)

x

7 O

y = - 12 x + 14 7 a x = 2, y = −6

x

3.5

y f x = 8, y = 4

2

O

x

y y = − 12 x + 4

y = 3x − 6

−6 4

b x = −2, y = 4

O

y

8

x

y = 2x + 4 4 −2

O

x

744 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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y

y y = – 32 x + 6 6

2

y = 5x +

2.4 4

O

x

O

−6

x

12 5

Answers

l x = −6, y = 12 5

g x = 4, y = 6

h x = 5, y = 2

y

8 a

y=

– 25 x

y

+2 y = −4

2 O

5

x

O

x

−4 i x = −8, y = 6

y

b

y

y = 34 x + 6

6

y=1 1

O

−8

j x = 5, y = 2.5

x

O

x

c

y

y

x=2 y=

− 12

x+

5 2

2 O

x

2.5 O

x

5

d k x = 7,y = 7 3 4

x = − 52

y

−2.5

y = − 34 x + 74 1.75 O

y

7 3

O

x

x

745 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

e

k

y

y=0

l

y

4

y

9 a C = 2n + 10 b y

(1, 4)

35 30 25 20 15 10 5

y = 4x x

O

y

C = 2n + 10

1 2 3 4 5 6 7 8 9 10 c i $28 ii 23.5 kg 10 a V = 90 − 1.5t b y 90 80 70 60 50 40 30 20 10

x

O (1, −3)

y y = −13 x

0

x

O

c 11 a

(3, −1)

j

(10, 30)

0

y = −3x

i

x

O

x

O

h

(1, −1)

y

x=0

g

x

O

x

O

f

y

12 a c

y

13 a c

(2, 5) y = −52 x O

x

14 a b c d

x

V = 90 − 1.5t

5 10 15 20 25 30 35 40 45 50 55 60 i 82.5 L ii 60 hours $7 per hour b P = 7t $0.05/km b C = 0.05k C = 1200 + 0.05k m = 25 b The cyclist started 30 km from home. (0, 30) y = x + 1, gradient = 1 2 y = 0.5x + 1.5, y-intercept = 1.5 y = −3x + 7, gradient = −3 y = 1 x − 2, gradient = 1 2 2

x

746 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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9 a i V = 4t

b gradient = a, y-intercept = −b c gradient = − a, y-intercept = 3 b b 16 a d a

b d b

c −a b

17 a 12 sq. units c 121 sq. units 4

b 9 sq. units d 121 sq. units 5

e 32 sq. units 3

a y = 4x − 10

b y = −x + 3 d y = 1 x + 11 2 2

c y = −x − 7

c 0

d −4

b 2

c 5 2

d 3

f 0

g −1

h 5 2

j Infinite

k 3 l −3 2 2 b y=x−2

2 a 2

b 3

e −3 3 a 1 4

f Infinite

e 0 i −1 4 a y=x+3

c y = 3x + 6 d y = −3x + 4 e y=4 f y = −7x − 10

5 a y = 2x + 4 c y=x−4

e y = −3x − 4

b y = 4x − 5 d y = −2x + 12 f y = −3x − 2

6 a y = 3x + 5 b y = −2x + 4 1 3 c y= x− d y = −2x − 2 2 2 7 a A = 500t + 15 000 b $15 000 c 4 years more, i.e 10 years from investment d $21 250 8 a y

90 80 70 60 50 40 30 20 10

iii V = t + 1

c Initially the flask contains b litres and it is losing 1 litre per minute. 10 a m = − 5 = −1 b m = 5 = −1 5 −5 c It doesn’t matter in which pair of points is (x1 , y1 ) and which is (x2 , y2 ). 11 a − 4 b y = − 4x + 13 3 3 3 c y = − 4x + 13 3 3 d The results from parts b and c are the same (when

Exercise 1F 1

ii V = 3t

iv V = 1.5t + 2 b 1 L, 2 L

Answers

15 a gradient = 3, y-intercept = 7 a a

simplified). So it doesn’t matter which point on the line is used in the formula y − y1 = m(x − x1 ). 12 a i 1 = 0.02 ii 2 = 0.04 50 50 b i y = 0.02x + 1.5 ii y = 0.04x + 1.5 c The archer needs m to be between 0.02 and 0.04 to hit the target.

Exercise 1G 1

a 4

b 5

√ 41 d √ √ c 32 = 4 2

c



 3, 9 2

2 a 4 b 4 d (0, −3) √ √ 3 a d = 20 = 2 5, M = (2, 5) √ b d = 97, M = (2, 3.5) √ c d = 41, M = (−1, 1.5) √ d d = 37, M = (−1, −1.5) √ √ √ √ b 58 c 37 d 65 4 a 29 √ √ √ e 37 f 15 g 101 h 193 √ 37 i 5 a (1, 6.5) b (1.5, 2.5) c (−0.5, 1) e (1, −1.5) g (−3, −0.5)

d (−1, 4.5) f (−3.5, 3)

h (2, 2.5) i (−7, 10.5) 6 B and C are both 5 units away from (2, 3). 7 a a = 3, b = 5 c a = −2, b = 2

C = 10t + 20

8 a 3, 7 9 a 1478 m

b −1, 3 b 739 m

b a = −4, b = 5 d a = 11, b = 2 c −1, 9

d −6, 0

10 a (−0.5, 1) b (−0.5, 1) c These are the same. The order of the points doesn’t

x 0 1 2 3 4 5 6 7 8 9 b C = 10t + 20 c i $10 per hour ii $20 up-front fee

matter since addition is commutative (x1 + x2 . . .) (x1 + x2 = x2 + x1 ). d 5 e 5 f The order of the points doesn’t matter (x − y)2 = (y − x)2 , as (−3)2 = (3)2 .

747 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

11 a = −4, 0

7 a gradient = − 3, y-intercept = 1 2

y

y

(−4, 3)

4 3 2 1

y=3

3 d = √20

x

O

O −4 −3 −2 −1−1 −2 −3 −4

(−2, −1)

12 a d 13 a b c



   1,2 b −1,4 2 3 3     2, 16 e −3,1 5 4  2 2 (x − 7) + y  (x − 7)2 + (x + 3)2 i 721 m ii 707 m iii 721 m

 4,8 3 3   f 0, 8 5 c



y

4 3 2 1

iv 762 m

O −4 −3 −2 −1−1 −2 −3 −4

joining Sarah to the fence is perpendicular to the fence (when it has gradient −1). The closest point is (2, 5).

Progress quiz a 9a2 b + 2ab + 8b

2 a 4k e 9a 2 3 a 6+m 8 d

b a−4

b −12x3 y c 7−x 2

c 13m + 14 d m+3 3m

O −4 −3 −2 −1−1 −2 d x = −2 −3 −4

3m − 13 (m − 1)(m − 3)

4 a x=5

y 4 3 2 1

c 14 − 3a 24

b k = − 3 c m = 30 2

5 a a>3

0 1 2 3 4 5 b x ≥ −4 −5 −4 −3 −2 −1 0 c m 2x − 8

y

8 There are 36 ducks and 6 sheep. 9 43

y > 2x − 8

10 $6.15 (mangoes cost $1.10, apples cost $0.65) 11 70 12 1 hour and 40 minutes 13 1 of an hour 7 14 200 m

x 2

O

4

x

−8

751 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

d y ≤ 3x − 5

h y > −3x + 6

y y ≤ 3x − 5

O

y

6 x

5 3

O

−5 e y < −4x + 2

x

2 y > 6 − 3x

y i y ≤ −x

y < − 4x + 2

y

2 O

x

1 2

x

O y ≤ −x

f y ≤ 2x + 7

y j x>3

7

y

y ≤ 2x + 7 −72

x>3 x

O

3 O

g y < 4x

x

y k x < −2

y

4 O

x

1 y < 4x

x < −2

−2 O

x

752 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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d

y

y

y≥2 2

9

x

O

O

x

Answers

l y≥2

−6 5 a

e

y

y

(0, 5) 3 O

b

−2.5

x

9

O

f

y

x

y

1.5 x

O −3

c

O

−3

g

y

x

y

4 O

2

x

−5 O

x

−2

753 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

h

d

y

x

O

−9

y

O 1 (0, −3)

−4

6 a yes 7 a no

b no b yes

8 a y≤x+3 c y < −3x − 3 2 9 a

x

5

c no c no

d yes d no

e

y

b y ≥ −2x + 2 d y > 2x − 2 5

4 2 O

y

3

x

4

(6, −2) 4 2 O

4

x

f

y 10 (5, 5)

b

y

2.5 −5 (0, 3) 4

O 1

O

g

x

10 y

x 9 (4, 3) 2

c

y

O

−8 h

O

2

3

x

6

y

x

6

(0, −2)

2.5 −5

O

(1, 3) 2

x

754 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

f

y

y

Answers

10 a

y≥0 y ≥ 12 x + 3

(0, 1) (0, 0) O (2, 0) – 12 x

y≤

x≥0

(−5.5, 14.5)

x

y ≤ −x + 9

(4, 5)

(−1.43, 2.29)

+1

O b

y y ≥ 2x − 4 (0, 0) O

(2, 0)

x≥0

x

(0, −4) y≤0

1 4x

1

(4, 5)

(0, 4)

x≥0 y≤

x + 15

O

x

(2, 3) (2, 1.2)

x 0, y < 2

y

y≥

y ≥ −3x − 2

11 a y ≥ 0, y < 2x + 4, y ≤ −x + 7 b y > − 1 x + 6, y ≤ x + 3, x < 8 2 12 a 1

c

x

−3 −2O

x 2x + 3y ≥ −6

◦ corners  are not 90 . In particular, AB is not perpendicular to BC mAB π −1 . mBC 8 x = 2, y = −3, z = −1

9 24 units2 10 24, 15 years

Multiple-choice questions 1 E

2 D

3 B

4 C

5 D

6 B 11 E

7 C 12 B

8 A 13 A

9 C 14 A

10 D 15 D

16 C

755 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

Short-answer questions 1

c 3x 2 f x+2

a 5xy + 6x b 12a2 b e −2m2 + 12m

2 a 6 − 7a 14 d

e

y

d 3b + 21

y = 2x

b 5a + 18 c 7x + 26 6a 30

3 a 3x − 1

b

(1, 2)

(0, 0)

11 − x (x + 1)(x − 3)

x

O

2 x+2

c 3 4

d x=2 4 a x = −3 b x = − 3 c x = 1 4 5 5 a x5 b x ≥ 10 c x > −3 d x ≤ 2 7 7 a V = 2 − 0.4t b 1.4 L c 5 minutes

y

f

y = −5x

(0, 0) O

d ≤ 3.5 minutes 8 a y

x (1, 5)

(3, 0)

x

O

g

y

y = 3x − 9 4

y = 4 − –x 2

O

8

(0, 9)

b

y h

y

(0, 5) 3

(2.5, 0) O

c

x

x

O

y = 5 − 2x

3x y = 3 − –– 8 8

x

y 9 a y = 1x + 3 2

y=3

3

x

O

d

y

O

x=5

x

b y = − 3 x + 15 2 2 c y = 2x − 3 10 a m = − 3 b y = − 3 x + 34 5 5 5 √ √ 11 a m = (4, 8), d = 52 = 2 13   √ b M = 11 , 1 , d = 61 2   √ √ c M = 1 , − 5 , d = 18 = 3 2 2 2 12 a y = 3x − 2

b y = −1 c y = − 1 x + 5 2

d y = 3x − 1

13 a a = 7 b b = −8 c c = 0 or 4 14 a (−3, −1) b (−8, −21) 15 a (−3, −1)

b (0, 2)

756 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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16 A regular popcorn costs $4 and a small drink costs $2.50.

2 a A(0, 0), B(8, 6), C(20, 0)

y

y

Answers

17 a

B (8, 6) O

x

4 3

D (14, 3) A (0, 0) O

−4

b

x

b 43.4 km c The drink station is at (14, 3).

y

d i y = 3x 4

8 3

ii y = − 1 x + 10 2

iii y = 0

e y ≥ 0, y ≤ 3 x, y ≤ − 1 x + 10 4 2

x

O

−4

C (20, 0)

f y = − 4 x + 80 3 3 18 The point of intersection is (4, 0)

Chapter 2

y

Exercise 2A 2 O

1

x

4

a i h = 4t + 25 b 16 cm

ii h = 6t + 16

c Shrub B because its gradient is greater. d y

(12, 88) (12, 73)

0

2 4 6 8 10 12 e after 4.5 months

f false

g true

d true h false

3 a a = 110 (angles on a line), b = 70 (vertically opposite) b a = 140 (angles in a revolution)

Extended-response questions

90 80 70 60 50 40 30 20 10

octagon, nonagon, decagon 2 a false b true c true e false i true

6

1

triangle, quadrilateral, pentagon, hexagon, heptagon,

c a = 19 (complementary) d a = 113 (cointerior angles in  lines), b = 67 (alternate

angles in  lines), c = 67 (vertically opposite to b) e a = 81 (isosceles triangle), b = 18 (angles in a triangle) f a = 17 (angles in a triangle), b = 102 (angles at a line) g a = 106 (cointerior angles in  lines), b = 74 (opposite angles in a parallelogram) h a = 17 (angles in a triangle), b = 17 (complementary)

i a = 90 (vertically opposite), b = 60 (angles in a triangle) 4 a 72 b 60 c 56 5 a 60 (equilateral triangle) b 60 (exterior angle theorem) c 110 (isosceles, angles in a triangle)

x

d 80 (angles in a triangle) e 10 (exterior angle theorem)

f i 1.24 m ii 26.25 months iii between 8.75 and 11.25 months

f 20 (isosceles, angles in a triangle) g 109 (angles on a line) h 28 (diagonals meet at a right angle in a rhombus) i 23 (angles in a triangle)

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b ∠FED = ∠CBA = 90◦ (given) R

Answers

j 121 (vertically opposite to cointerior angle in  lines) k 71 (isosceles, cointerior angles in  lines) l 60 (isosceles, cointerior angles in  lines)

FD = CA (given) H FE = CB (given) S

6 a 50 (angle sum in a quadrilateral) b 95 (angle sum in a quadrilateral)

∴ FED ≡ CBA (RHS)

c AC = DF (given) S

c 125 (angle sum in a pentagon) d 30 (angle sum in a pentagon)

BC = EF (given) S

e 45 (angle sum in a hexagon) f 15 (angle sum in a quadrilateral)

AB = DE (given) S

b 135◦ c 144◦ 7 a 108◦ 8 a 95 (alternate + cointerior) b 113 (2 × alternate)

∴ ACB ≡ DFE (SSS)

d ∠EDF = ∠BAC (given) A ∠DFE = ∠ACB (given) A

c 85 (alternate + cointerior) d 106 (cointerior) e 147, (cointerior, angles in a revolution)

EF = BC (given) S

f 292, (angles in a revolution, alternate + cointerior) 9 a 176.4◦ b 3.6◦

∴ EDF ≡ BAC (AAS) 4 a x = 7.3, y = 5.2 b x = 12, y = 11

10 a 12 b 20 11 x = 36, y = 144

c a = 2.6, b = 2.4 5 a AD = CB (given) S

c 48

12 115, equilateral and isocles triangle 60 + 55 13 a Expand the brackets. b n = S + 360 180

DC = BA (given) S AC is common; S

360 c I = S = 180(n − 2) d E = 180 − I = n n n 14 a ∠BCA = 180◦ − a◦ − b◦ (angles in a triangle) b c◦ = 180◦ − ∠BCA = a◦ + b◦ (angles at a line)

∴ ADC ≡ CBA (AAS)

b ∠ADB = ∠CBD (given) A ∠ABD = ∠CDB (given) A

15 a alternate angles (BA  CD) b ∠ABC + ∠BCD = 180◦ (cointerior), so

BD is common; S ∴ ADB ≡ CBD (AAS)

a + b + c = 180. c Angle sum of a triangle is 180◦ .

c ∠BAC = ∠DEC (alternate, AB  DE) A ∠CBA = ∠CDE (alternate, AB  DE) A

16 ∠ACB = ∠DCE (vertically opposite), so ∠CAB = ∠CBA = ∠CDE = ∠CED (isosceles) since ∠CAB = ∠CED (alternate) AB  DE. 17 Answers may vary.

BC = DC (given) S ∴ BAC ≡ DEC (AAS)

d DA = DC (given) S

18 a 15 (alternate angles in parallel lines) b 315 (angle sum in an octagon)

∠ADB = ∠CDB (given) A

19 Let M be the midpoint of AC. Then ∠AMB = is equilateral). ∠BMC = 120◦ (supplementary).

60◦

(ABM

Therefore, ∠MBC = 30◦ (MBC is isosceles). So ∠ABC = ∠ABM + ∠MBC = 60◦ + 30◦ = 90◦ .

DB is common; S ∴ ADB ≡ CDB (SAS)

e OA = OC (radii) S

OB = OD (radii) S

20 Let ∠AOB = x and ∠COD = y. 2x + 2y = 180◦ (angles at a line). So ∠BOD = x + y = 90◦ .

Exercise 2B 1

a SAS e RHS

2 a 5 f 2

d x = 16, y = 9

AB = CD (given) S f

b SSS f RHS

c AAS

d SAS

b 4

c 3

d 5

∴ OAB ≡ OCD (SSS)

∠ADC = ∠ABC = 90◦ (given) R AC is common; H DC = BC (given) S

e 2

∴ ADC ≡ ABC(RHS)

6 a OA = OC (radii) S

3 a AB = DE (given) S

∠AOB = ∠COB (given) A

∠ABC = ∠DEF (given) A

OB is common; S

BC = EF (given) S ∴ ABC ≡ DEF (SAS)

∴ AOB ≡ COB (SAS) b AB = BC (corresponding sides in congruent triangles) c 10 mm

758 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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7 a BC = DC (given) S ∠BCA = ∠DCE (vertically opposite) A

∠AOD = ∠BOC (vertically opposite) A

AC = EC (given) S

OA = OB (given) S

∴ ABC ≡ EDC (SAS) b AB = DE (corresponding sides in congruent triangles) c ∠ABC = ∠CDE (corresponding sides in congruent

∴ AOD ≡ BOC (SAS)

∠OAD = ∠OBC (corresponding angles in congruent triangles)

triangles). ∠ABC and ∠CDE are alternate angles. ∴ AB  DE.

f

AD = AB (given) S

Answers

e OD = OC (given) S

∠DAC = ∠BAC (given) A

d 5 cm 8 a AB = CD (given) S

AC is common; S

AD = CB (given) S

∴ ADC ≡ ABC (SAS)

BD is common; S

∠ACD = ∠ACB (corresponding angles in congruent

∴ ABD ≡ CDB (SSS) b ∠DBC = ∠BDA (corresponding angles in congruent

triangles) ∠ACD = ∠ACB are supplementary.

triangles) c ∠DBC and ∠BDA are alternate angles (and equal). ∴ AD  BC. 9 a CB = CD (given) S

∴ ∠ACD = ∠ACB = 90◦ ∴ AC ⊥ BD

10 a OA = OB (radii) S

∠BCA = ∠DCE (vertically opposite) A

OM is common; S

CA = CE (given) S

AM = BM (M is midpoint) S

∴ BCA ≡ DCE (SAS)

∴ OAM ≡ OBM (SSS)

∠BAC = ∠DEC (corresponding angles in congruent

∠OMA = ∠OMB (corresponding angles in congruent

triangles)

triangles)

∴ Alternate angles are equal, so AB || DE.

∠OMA and ∠ OMB are supplementary.

b ∠OBC = ∠OBA = 90◦ (given) R

∴ ∠OMA = ∠OMB = 90◦

OA = OC (radii) H

∴ OM ⊥ AB

OB is common; S

b OA = OB (radii of same circle) S

∴ OAB ≡ OCB (RHS)

CA = CB (radii of same circle) S

AB = BC (corresponding sides in congruent triangles)

OC is common; S

∴ OB bisects AC.

∴ OAC ≡ OBC (SSS)

c AB = CD (given) S

∠AOC = ∠BOC (corresponding angles in congruent

AC is common; S

triangles)

AD = CB (given) S

c ∠CAB = ∠CBA = x (ABC is isosceles)

∴ ACD ≡ CAB (SSS)

∠EAB = ∠DBA = x 2

∠DAC = ∠BCA (corresponding angles in congruent

∴ AFB is isosceles, so AF = BF.

triangles) ∴ Alternate angles are equal, so AD || BC.

d AB = AE (given) S

Exercise 2C

∠ABC = ∠AED (ABE is isosceles) A

1

ED = BC (given) S

2 a rectangle, square b rectangle, square c parallelogram, rhombus, rectangle, square

∴ ABC ≡ AED (SAS) AD = AC (corresponding sides in congruent triangles)

a rectangle d rhombus

b parallelogram

c square

d rhombus, square e rhombus, square 3 a A trapezium does not have both pairs of opposite sides parallel. b A kite does not have two pairs of opposite sides parallel.

759 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

4 a ∠BAC = ∠DCA (alternate angles)

So ∠ECB = ∠ECD since ABC and ADC are

isosceles. DC = BC (given)

∠BCA = ∠DAC (alternate angles) AC is common.

EC is common. ∴ CDE ≡ CBE (SAS)

∴ ABC ≡ CDA (AAS) b As ABC ≡ CDA, AD = CB, AB = CD (corresponding sides). 5 a ∠ABE = ∠CDE (alternate angles)

So ∠CED = ∠CEB = 90◦ . b From part a, ∠ECD = ∠ECB. 11

∠BAE = ∠DCE (alternate angles)

D

C

AB = CD (opposite sides of parallelogram) ∴ ABE ≡ CDE (AAS) b AE = CE (corresponding sides), BE = DE

A B As ABCD is a parallelogram, ∠BDC = ∠DBA (alternate angles) and ∠DBC = ∠BDA (alternate angles).

(corresponding sides). 6 a AB = CB (given)

BD is common. So CBD ≡ ADB (AAS).

AD = CD (given)

So ∠BAD = ∠DCB = 90◦ .

BD is common.

Similarly, ∠ADC = ∠180◦ − ∠BAD (cointerior angles) = 90◦ and similarly for ∠ABC.

∴ ABD ≡ CDB (SSS) b ∠ABD = ∠ADB = ∠CBD = ∠CDB (equal angles in congruent isosceles triangles). Therefore, BD bisects ∠ABC and ∠CDA.

12 D

C E

7 a AE = CE (given) BE = DE (given)

A B First, prove AED ≡ BEC (SAS).

∠AEB = ∠CED (vertically opposite angles) ∴ ABE ≡ CDE (SAS) b ∠ABE = ∠CDE (corresponding angles), ∠BAE =

Hence, corresponding angles in the isosceles triangles are equal and CED ≡ BEA (SAS).

∠DCE (corresponding angles). Therefore, AB  DC (alternate angles are equal). ∠ADE = ∠CBE

Hence, corresponding angles in the isosceles triangles are equal.

(corresponding angles), ∠DAE = ∠BCE (corresponding angles). Therefore, AD  BC (alternate angles are equal).

So ∠ADC = ∠DCB = ∠CBA = ∠BAC, which sum to 360◦ .

8 a AD = CB (given)

∠DAC = ∠BCA (alternate angles)

Therefore, all angles are 90◦ and ABCD is a rectangle. 13

G

C

F

AC is common. ∴ ABC ≡ CDA (SAS) b ∠BAC = ∠DCA (corresponding angles), therefore AB  DC (alternate angles are equal).

D

9 a ABE ≡ CBE ≡ ADE ≡ CDE (SAS) b ∠ABE = ∠CDE (corresponding angles), ∠BAE =

H

∠DCE (corresponding angles), therefore AB  CD. ∠ADE = ∠CBE (corresponding angles), ∠DAE = ∠BCE (corresponding angles), therefore AD  CB. Also, AB = AD = CB = CD (corresponding sides). Therefore, ABCD is a rhombus. 10 a

C

D

E

Exercise 2D a Yes, both squares have all angles 90◦ and all sides of equal length. b 3

B A ∠CAB = ∠ACD and ∠CAD = ∠ACB (alternate angles).

A

First, prove all four corner triangles are congruent (SAS). So EF = FG = GH = HE, so EFGH is a rhombus.

1

E

B

c 15 cm

2 a 2

b 8 5

c 4 3

3 a A

b ∠C

c FD

d 3 2

d ABC ||| EFD

760 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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d 3 cm 2

b 12.5 f 14.5

7 1.7 m 8 a 1.6

b 62.5 cm

d 4.5 11 a true e false i true

b EF = GH AB CD

c 4 3

e 10.5 m

6 a 1.2 e 11.5

9 a 2 10 a BC

c 3 2

e 4 cm 3

5 a ABCD ||| EFGH d 12 m

b AB = DE FG IJ

c 4.8

d 3.75

b true

c false

d false

f false j true

g false

h false

Cube

Length

Area

Volume

Small Large

2 3 3 2

4 9 9 4

8 27 27 8

Scale factor (fraction)

e Scale factor for area = (scale factor for length)2 ; Scale factor for volume = (scale factor for length)3 . b3 b2 ii 3 a2 a 14 Answer will vary

5 a 3 b 19.5 c 2.2 d a = 4, b = 15 2 e x = 0.16, y = 0.325 f a = 43.2, b = 18 6 a ∠ABC = ∠EDC (alternate angles)

∠AEB = ∠ADC (corresponding angles) ∠BAE = ∠CAD (common) ∴ ABE ||| ACD (AAA). c ∠DBC = ∠AEC (given) ∠BCD = ∠ECA (common) ∴ BCD ||| ECA (AAA). d AB = 3 = 0.4 CB 7.5 EB = 2 = 0.4 (ratio of corresponding sides) DB 5 ∠ABE = ∠CBD (vertically opposite angles)

f i

∴ AEB ||| CDB (SAS). 7 a ∠EDC = ∠ADB (common)

Exercise 2E

∠CED = ∠BAD = 90◦ ∴ EDC ||| ADB (AAA).

c AB

b 4 cm 3 8 a ∠ACB = ∠DCE (common)

2 a ∠D (alternate angles) b ∠A (alternate angles)

∠BAC = ∠EDC = 90◦ ∴ BAC ||| EDC (AAA).

c ∠ECD d CA e ABC ||| EDC 3 a SAS b AAA

AC = 32 = 4 (ratio of corresponding sides) DF 8 ∴ ABC ||| DEF (SSS).

∴ ABC ||| EDC (AAA). b ∠ABE = ∠ACD (corresponding angles)

ii 9 ii 27

a E b ∠C d ABC ||| DEF

DE = 8 = 2 (ratio of corresponding sides) CB 4 ∠ABC = ∠FED = 90◦ ∴ ABC ||| FED (RHS).

∠BAC = ∠DEC (alternate angles) ∠ACB = ∠ECD (vertically opposite angles)

angles are equal. 13 a 3 2

1

c DF = 10 = 2 CA 5

BC = 16 = 4 EF 4

d 4.3 c 1

12 Yes, the missing angle in the first triangle is 20◦ and the missing angle in the second triangle is 75◦ , so all three

d

∴ ABC ||| DEF (SAS).

d AB = 28 = 4 DE 7

b 1 c 1.875 b ABC ||| EDC

b i 4 c i 8

∠ABC = ∠DEF = 120◦

Answers

4 a ABCDE ||| FGHIJ

c SAS

d SSS

4 a ∠ABC = ∠DEF = 65◦ ∠BAC = ∠EDF = 70◦

b 1.25 m 9 1.90 m

10 4.5 m 11 a Yes, AAA for both.

∴ ABC ||| DEF (AAA).

b 20 m c 20 m

b DE = 2 = 2 AB 1 EF = 6 = 2 (ratio of corresponding sides) BC 3

d Less working required for May’s triangles. 12 The missing angle in the smaller triangle is 47◦ , and the missing angle in the larger triangle is 91◦ . Therefore the two triangles are similar (AAA).

761 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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13 a ∠AOD = ∠BOC (common)

∠ABP = ∠CBP (diagonals of a rhombus bisect the interior

Answers

∠OAD = ∠OBC (corresponding angles) ∠ODA = ∠OCB (corresponding angles)

angles through which they cross) ∴ ABP ≡ CBP (SAS)

and ∠APB = ∠BPC (corresponding angles of congruent triangles).

So OAD ||| OBC (AAA). OC = 3 = 3 (ratio of corresponding sides), therefore OD 1

And ∠APB + ∠BPC = 180◦ (straight line) ∴ diagonal AC ⊥ diagonal DB.

OB = 3 OA

A

B

OB = 3OA b ∠ABC = ∠EDC (alternate angles) ∠BAC = ∠DEC (alternate angles)

P

∠ACB = ∠ECD (vertically opposite) So ABC ||| EDC (AAA). CE = CD = 2, therefore AC + CE = 1 + 2 = 7. AC BC 5 AC 5 5 But AC + CE = AE, so AE = 7 and AE = 7 AC. AC 5 5 14 a ∠BAD = ∠BCA = 90◦ ∠ABD = ∠CBA (common) So ABD ||| CBA (AAA). Therefore, AB = BD. CB AB AB2 = CB × BD

B

D

C

∠BAC = ∠ACD (alternate angles AB  CD) ∠BCA = ∠DAC (alternate angles AD  CB)

∴ ABC ≡ CDA (AAS) and AB = DC as well as AD = BC (corresponding sides in

So ABD ||| CAD (AAA). Therefore, AD = BD. CD AD AD2 = CD × BD c Adding the two equations:

congruent triangles). 5 a ABE ||| ACD (all angles equal)

2

AB + AD = CB × BD + CD × BD = BD(CB + CD)

b 2.5 c x = 7.5

6 a ∠CAB = ∠FDE (given) AC = AB = 1 DF DE 3

= BD × BD = BD2

CAB ||| FDE (SAS) b ∠BAO = ∠CDO (alternate angles AB  DC)

Progress quiz 1

A

Let ABCD be any parallelogram with opposite sides parallel. AC is common.

b ∠BAD = ∠ACD = 90◦ ∠ADB = ∠CDA (common)

2

C

D 4

a x = 78 (exterior angle of a triangle) b w = 89 (angle sum of a quadrilateral) c x = 120 (interior angle of a regular hexagon) d x = 35 (alternate angles in parallel lines) e x = 97 (cointerior angles in parallel lines, vertically opposite angles equal) f w = 47 (angle sum of an isosceles triangle)

2 a AB = QB (given) ∠ABC = ∠QBP (vertically opposite) ∠CAB = ∠PQB (alternate angles AC  PQ) ∴ ABC ≡ QBP (AAS)

b CB = PB corresponding sides of congruent triangles and B is the midpoint of CP.

∠AOB = ∠DOC (vertically opposite) ∴ ABO ||| DCO (AAA)

7 a ∠D is common ∠ABD = ∠ECD (corresponding angles equal since AB  EC) ABD ||| ECD (AAA)

b 3 cm 8 ∠A is common, as Q and P are both midpoints. AP = 1 AB 2

and

AQ AC

=1 2

∴ AQP ||| ACB (SAS) and QP = 1 (corresponding sides in the same ratio). CB 2

3 Let ABCD be any rhombus with diagonals intersecting at P. AB = BC (sides of a rhombus equal)

762 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Exercise 2F

AC = BC (radii of same circle) CD is common.

a–e

∴ ACD ≡ BCD (SSS). b AC = BC

chord minor sector

centre

∠ACE = ∠BCE (corresponding angles in congruent triangles)

radius

major sector 2 a 55◦ b 90◦

CE is common. ∴ ACE ≡ BCE (SAS).

c 75◦

Exercise 2G

3 a each b ∠AOB = ∠COD (chord theorem 1)

4 a 1 cm each b 52◦ each

b ∠ADC

c ∠ADC

d ∠AFC

e ∠AEC 2 a ∠AOB

f ∠AEC b ∠ACB

c 80◦

d 61◦

180◦

c AM = BM and ∠AOM = ∠BOM (chord theorem 3) 5 a ∠DOC = 70◦ (chord theorem 1) b OE = 7.2 cm (chord theorem 2) c XZ = 4 cm and ∠XOZ = 51◦ (chord theorem 3) 6 The perpendicular bisectors of two different chords of a circle

8 a 140 e 30

b 40 f 54

c 19

d 90◦ d 72

9 6m √ √ 10 3 + 128 mm = 3 + 8 2 mm 11 a Triangles are congruent (SSS), so angles at the centre of the circle are corresponding, and therefore equal. b Triangles are congruent (SAS), so chords are corresponding sides, and therefore equal. 12 a Triangles are congruent (SSS), so the angles formed by the chord and radius are corresponding, and therefore equal. Since these angles are also supplementary, they must be 90◦ . b Triangles are congruent (SAS), so the angles formed by the chord and radius are corresponding, and therefore equal. Since these angles are also supplementary, they must be 90◦ .

A

3 a 4 a 50

e 250 i 18

90◦

60◦

b b 40

c c 80

d 7◦ d 60

f 112.5

g 38

h 120

5 a 70 b 25 c 10 6 a ∠ABC = 72◦ , ∠ABD = 22◦ b ∠ABC = 70◦ , ∠ABD = 45◦ c ∠ABC = 72◦ , ∠ABD = 35◦ 7 a ∠ADC = 75◦ , ∠ABC = 75◦ b ∠ABC = 57.5◦ , ∠ADC = 57.5◦ c ∠AOD = 170◦ , ∠ABD = 85◦ 8 a 100◦ e 70◦

b 94.5◦ f 66◦

c 100◦

d 119◦

9 a 58◦ e 19◦

b 53◦ f 21◦

c 51◦

d 45◦

10 a 70◦ b 90◦ c The angle in a semicircle is 90◦ . d Theorem 2 is the specific case of theorem 1 when the angle at the centre is 180◦ . 11 a i false b i false

ii true ii true

iii true iii true

iv false iv false

b 360 − 2x 12 a 2x◦ 13 a ∠AOC = 180◦ − 2x◦ (AOC is isosceles)

b ∠BOC = 180◦ − 2y◦ (BOC is isosceles) c ∠AOB = 360◦ − ∠AOC − ∠BOC = 2x◦ + 2y◦

d ∠AOB = 2(x◦ + y◦ ) = 2∠ACB 14 a ∠BOC = 180◦ − 2x◦ (BOC is isosceles).

∠AOB = 180◦ − ∠BOC = 180◦ − (180◦ − 2x◦ ) = 2x◦

O B

a ∠ADC

1

c 0.9 cm each d OE = OF (chord theorem 2)

13

c Using the converse of chord theorem 3 since ∠ACE = ∠BCE, CD ⊥ AB.

d 140◦

85◦

intersect at the centre of the circle. 7 a 3.5 m b 9m c 90◦

Answers

1

14 a AD = BD (radii of same circle)

b ∠AOC = 180◦ − 2x◦ (AOC is isosceles) ∠BOC = 180◦ − 2y◦ (BOC is isosceles)

C

First, prove OAB ≡ OAC (AAS), which are isosceles. So AB = AC, corresponding sides in congruent triangles.

Reflex ∠AOB = 360◦ − ∠AOC − ∠BOC = 360◦ − (180◦ − 2x◦ ) − (180◦ − 2y◦ )

= 2x◦ + 2y◦ = 2(x + y)◦ = 2∠ACB.

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Answers

c ∠OBC = x◦ + y◦ (OCB is isosceles) ∠COB = 180◦ − 2(x + y)◦ ∠AOB = 180◦ − 2x◦ − (180◦ − 2(x + y)◦ )

= 2y◦ 15 ∠AOB = 180◦ − 2x◦ (AOB is isosceles)

∠BOC = 180◦ − 2y◦ (BOC is isosceles) ∠AOB + ∠BOC = 180◦ (supplementary angles), therefore (180 − 2x) + (180 − 2y) = 180 360 − 2x − 2y = 180 2x + 2y = 180 2(x + y) = 180 x + y = 90

Exercise 2H 1

a ∠ACD

b ∠ACD

2 a ∠ABD and ∠ACD c ∠BAC and ∠BDC

c ∠ACD b 85◦ d 17◦

4 a x = 37 b x = 20 e x = 22.5 f x = 55

c x = 110 d x = 40

5 a x = 60 b x = 90 e x = 72, y = 108

c x = 30 d x = 88 f x = 123

6 a 72 e 52

c 69 g 30

i 108 7 a a = 30, b = 100

1 a Once 2 a ∠BAP

b 90◦ b ∠BPX

3 a 180◦ 4 a a = 19

b 360◦ b a = 62

5 a a = 50 6 a 50◦

b a = 28 b 59◦

d 57 h 47

b a = 54, b = 90

c a = 105, b = 105, c = 75 d a = 55, b = 70 e a = 118, b = 21

c 5 cm c ∠ABP

d ∠APY

c a = 70 c a = 25

d a = 63

7 a a = 73, b = 42, c = 65 b a = 26, b = 83, c = 71 c a = 69, b = 65, c = 46 8 a 5 cm b 11.2 cm 9 a a = 115 b a = 163 c a = 33 e a = 26 f a = 26 g a = 36

d a = 28 h a = 26

i a = 30 10 a a = 70

d a = 40

e a = 19 11 4 cm

3 a Supplementary angles sum to 180◦ . b 117◦ c 109◦ d Yes, 117◦ + 109◦ + 63◦ + 71◦ = 360◦

b 43 f 48

Exercise 2I

b a = 50

c a = 73

f a = 54

12 a OA and OB are radii of the circle. b ∠OAP = ∠OBP = 90◦ c ∠OAP = ∠OBP = 90◦ OP is common OA = OB ∴ OAP ≡ OBP (RHS)

d AP and BP are corresponding sides in congruent triangles. 13 a ∠OPB = 90◦ − x◦ , tangent meets radii at right angles b ∠BOP = 2x◦ , using angle sum in an isocles triangle

c ∠BAP = x◦ , circle theorem 1 14 ∠BAP = ∠BPY (alternate segment theorem) ∠BPY = ∠DPX (vertically opposite angles) ∠DPX = ∠DCP (alternate segment theorem) ∴ ∠BAP = ∠DCP, so AB  DC (alternate angles are equal).

f a = 45, b = 35 8 a 80◦ b 71◦ c ∠CBE + ∠ABE = 180◦ (supplementary angles) ∠CBE + ∠CDE = 180◦ (circle theorem 4) ∴ ∠CBE + ∠ABE = ∠CBE + ∠CDE. ∴ ∠ABE = ∠CDE 9 a ∠ACD = ∠ABD = x◦ and ∠DAC = ∠DBC = y◦ (circle theorem 3) b Using angle sum of ACD, ∠ADC = 180◦ − (x◦ + y◦ ).

c ∠ABC and ∠ADC are supplementary. 10 a i 80◦ ii 100◦ iii 80◦

b ∠BAF + ∠DCB = 180◦ , therefore AF  CD (cointerior angles are supplementary). 11 a ∠PCB = 90◦ (circle theorem 2) b ∠A = ∠P (circle theorem 3) c sin P = a 2r d As ∠A = ∠P, sin A = a , therefore 2r = a . 2r sin A

15 AP = TP and TP = BP, hence AP = BP. 16 a Let ∠ACB = x◦ , therefore ∠ABC = 90◦ − x◦ . Construct OP. OP ⊥ PM (tangent). (OPC is isosceles). Construct OM.

∠OPC = x◦

OAM ≡ OPM (RHS), therefore AM = PM. ∠BPM = 180◦ − 90◦ − x◦ = 90◦ − x◦ . Therefore,

BPM is isosceles with PM = BM. Therefore, AM = BM. b Answers may vary

Exercise 2J 1 a 3 b 6 c 7 2 a 21 b 5 c 33 2 2 7 3 a AP × CP = BP × DP b AP × BP = DP × CP c AP × BP = CP2

4 a 5

b 10

d 8 d 27 7

c 112 15

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b 178 9

6 a 32 3 √ 7 a 65 8 a 64 7

b 16 c 35 3 2 √ b 77 b 209 c 81 10 7 √ f 65 − 1

c 161 9

Multiple-choice questions 1 C 6 A d 74 7

e 153 20 9 a x(x + 5) = 7 × 8, x2 + 5x = 56, x2 + 5x − 56 = 0 b x(x + 11) = 10 × 22, x2 + 11x = 220, x2 + 11x − 220 = 0

c x(x + 23) = 312 , x2 + 23x = 961, x2 + 23x − 961 = 0 10 For this diagram, the third secant rule states: AP2 = DP × CP and BP2 = DP × CP, so BP = AP. 11 AP × BP = DP × CP AP × BP = AP × CP since AP = DP. BP = CP

12 a ∠A = ∠D and ∠B = ∠C (circle theorem 3) b ∠P is the same for both triangles (vertically opposite), so

2 B 7 E

3 B 8 C

4 C 9 D

5 B 10 B

Short-answer questions 1

a 65

d 46 2 a 148

b 120

c x = 62, y = 118

b 112

Answers

5 a 143 8

3 a AB = DE (given) ∠ABC = ∠DEF (given) ∠BAC = ∠EDF (given) ∴ ABC ≡ DEF (AAS).

b AB = AD (given)

∠BAC = ∠DAC (given) AC is common. ∴ ABC ≡ ADC (SAS).

c AB = CD (given)

ABP ||| DCP (AAA).

AD = CB (given)

c AP = BP DP CP

BD is common.

d AP = BP , cross-multiplying gives AP × CP = BP × DP. DP CP 13 a ∠B = ∠C (circle theorem 3) b PBD ||| PCA (AAA) c AP = CP , so AP × BP = DP × CP. DP BP 14 a yes b alternate segment theorem c BPC ||| CPA (AAA) d BP = CP , so CP2 = AP × BP. CP AP  √ 15 d = (4r1 r2 ) = 2 r1 r2

∴ ABD ≡ CDB (SSS). 4 a AB = CD (given) ∠BAC = ∠DCA (alternate angles) AC is common. ∴ ABC ≡ CDA (SAS).

b ∠BCA = ∠DAC (alternate angles), therefore AD  BC (alternate angles are equal). 5 a DE = 10.5 = 1.5 AB 7 EF = 14.7 = 1.5 (ratio of corresponding sides) BC 9.8 ∠ABC = ∠DEF (given)

Problems and challenges

∴ ABC ||| DEF (SAS)

1 21 units2 2 BD = 5 cm, CE = 19 cm

b ∠EAB = ∠DAC (common)

x = 19.5

3 ∠ADE = ∠ABE, ∠EFD = ∠BFA, ∠DEB = ∠DAB, ∠DFB = ∠EFA, ∠CDB = ∠CAE, ∠DAE = ∠DBE, ∠ADB = ∠AEB, ∠ABD = ∠AED = ∠CBD = ∠CEA 4 42.5% 5 Check with your teacher. 6 a ∠FDE = ∠DFC = ∠ABC (alternate and corresponding angles in parallel lines) ∠FED = ∠EFB = ∠ACB (alternate and corresponding

∠EBA = ∠DCA (corresponding with EB  DC) ∴ ABE ||| ACD (AAA)

x = 6.25 c ∠BAC = ∠EDC (given) ∠ACB = ∠DCE (vertically opposite) ∴ ABC ||| DEC (AAA) x = 8.82

angles in parallel lines) ∠DFE = ∠BAC (angle sum of a triangle) ABC ||| FDE (AAA) b i 4:1 ii 16 : 1 c 4n−1 : 1

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Answers

d ∠ABD = ∠DBC (given) ∠DAB = ∠CDB = 35◦ (angle sum of triangle) ∴ ABD ||| DBC (AAA) x = 100 7 6 a 65 (chord theorem 1) b 7 (chord theorem 2) c 6 (chord theorem 3) 7 a a = 25 b a = 50, b = 40

c a = 70

d b = 54 e a = 115 f a = 30, b = 120 8 a x = 26, y = 58, z = 64 b a = 65, b = 130, c = 50, d = 8 c t = 63 9 a 5 b 6 c 40 3

Extended-response questions 1

a ∠BAC = ∠BDE = 90◦ ∠B is common. ∴ ABC ||| DBE (AAA). b 1.2 km AC = 3 c i DE 2 ∴ AB = 3 (ratio of corresponding sides in similar DB 2 triangles) x+1=3 x 2 ∴ 2(x + 1) = 3x ii 2

d 44.4% 2 a i 106.26◦

√ √ √ √ 5 a 2 3 b 3 5 c 2 6 d 4 3 √ √ √ √ f 10 5 g 7 2 h 3 10 e 5 3 √ √ √ √ i 8 2 j 6 10 k 9 2 l 4 5 √ √ √ √ b 6 5 c 16 3 d 6 7 6 a 6 2 √ √ √ √ e 21 2 f 20 5 g 5 h 7 √ √ √ √ 6 6 5 11 i j k l 2 4 5 6 √ √ √ √ n 2 2 o 2 2 p 2 17 m 3 11 7 √ √ √ √ 2 3 3 3 3 3 s q r 4 6 t 2 3 2 √ √ √ √ 7 a 2 2 b 2 3 c 3 2 d 11 3 7 5 5 √ √ √ √ e 10 f 21 g 13 h 14 3 12 4 5 √ √ √ √ 5 3 5 3 l √14 i j k √ 3 2 2 2 19 √ √ √ √ b 32 c 50 d 27 8 a 12 √ √ √ √ e 45 f 108 g 128 h 700 √ √ √ √ 810 j 125 k 245 l 363 i √ √ √ √ 9 a 15 3 b 13 7 c 19 5 d 31 3 √ √ √ 10 a 4 2 m b 2 30 cm c 4 15 mm √ √ 11 a radius = 2 6 cm, diameter = 4 6 cm √ √ b radius = 3 6 m, diameter = 6 6 m √ √ c radius = 8 2 m, diameter = 16 2 m √ √ √ 12 a 2 5 cm b 3 5 m c 145 mm √ √ √ d 11 m e 11 mm f 2 21 cm √ √ 13 72 = 36 × 2 (i.e. 36 is highest square factor of 72) √ =6 2 √ 14 a 9, 25, 225 b 15 2 15 a Draw triangle with shorter sides length 1 cm and 3 cm. b Draw triangle with shorter sides length 2 cm and 5 cm.

ii 73.74◦

c

b 12 cm c 25 cm

d

1

2 √22

√6

d 70 cm

1

1

Chapter 3 Exercise 3B

a irrational d rational

b root

2 a 4

b 9

e 16 f 16 3 a irrational b rational e rational

−1

e no

c 25

d 4

g 36 c rational

h 36 d rational

√5

0.18 2

0

1 57

2 5

−√2 4 a yes

c non recurring

f irrational g irrational h irrational

−1.24 −2

4

16 Check with your teacher

Exercise 3A 1

1

2

2√3 3 π

b yes

c no

d no

f yes

g yes

h no

4

1

a yes e yes

2 a 6x e 17a √ 3 a 4 3 √ b i 5 3 √ 4 a 10 2 √ b i 5 2 √ 5 a 6 5 √ e 11 5 √ i −2 21

b no f no

c no g yes

d yes h yes

b 7y f −5x

c −5x g t

d −2b h −2y

√ ii −3 3 √ ii −16 2 √ b 3 3 c √ f 3 g √ j −6 11 k

√ iii 17 3 iii 0 √ 4 2 √ 6 10 √ − 13

√ d 3 2 √ h 5 2 √ l −7 30

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c e h 7 a e i 8 a d f h j 9 a e i 10 a c e 11 a b 12 a b c d e f 13 a b c d e f 14 a e j

√ √ √ √ 3+5 2 b 3 6 + 7 11 √ √ √ √ d −2 2 + 5 4 5−7 2 √ √ √ 4 3 f 0 g −3 2 − 3 10 √ √ −2 5 + 3 15 √ √ √ √ 2 b 5 2 c 4 3 d 5 √ √ √ √ f 12 3 g 8 11 h 3 2 7 2 √ √ √ √ 5 6 j 5 k 32 2 l 20 2 √ √ √ √ b 9 6 c 2 5− 7 13 2 √ √ √ √ 5 5+6 7 e 6−3 2 √ √ √ √ √ 2 3 + 11 5 − 5 2 g 9 3 + 2 2 √ √ 11 − 9 3 i 9 + 18 2 √ √ −9 2 + 15 5 √ √ √ √ 5 3 b 7 5 c 2 d 7 6 12 30 6 √ √ √ √ 3 5 − 2 13 13 − 7 3 f g h 10 14 18 30 √ − 11 10 24 √ √ √ 4 3 + 2 5 cm b 14 2 cm √ √ √ √ 10 + 3 2 cm d 2 10 + 4 5 cm √ √ √ 4 3 + 30 cm f 12 3 cm √ √ 20 = 2 5 √ √ √ √ 3 72 = 18 2, 338 = 13 2 √ √ √ 5 3−6 3+ 3=0 √ √ √ 6+2 6−3 6=0 √ √ √ 6 2−8 2+2 2=0 √ √ √ 2 2−3 2+ 2=0 √ √ √ 4 5−7 5+3 5=0 √ √ √ √ √ 3 2−6 3−5 2+6 3+2 2=0 √ √ 6 3 − 3 2, unlike surds √ √ 8 2 + 2 5, unlike surds √ √ 5 2 − 6 5, unlike surds √ √ 10 10 + 10 3, unlike surds √ √ 20 2 + 30 3, unlike surds √ √ 4√ 5 − 6 6, unlike √ √ surds √ 7 2 b 2 3 c 5 d −3 2 15 3 12 4 √ √ √ √ 3 h 29 6 f −7 7 g − 2 2 15 28 √ √ √ 6 6 5 29 k 8 3 l 35 42

Exercise 3C 1

a

c 2 a e i 3 a e i

 15 = √5 3 √ √ 6 × 5 = 30 √ √ 15 b 21 √ √ f − 30 − 30 √ 70 √ √ 10 b 6 √ √ 3 f 10 √ − 5

 42 = √6 7 √ √ d 11 × 2 = 22 √ √ c 26 d 35 √ √ g 66 h 6

b

√ c − 3 √ g 5

√ d 5 √ h − 13

4 a f k 5 a e i 6 a e 7 a c e g i k 8 a e h 9 a 10 a 11 a b c 12 a

i 0

√ 21

√ √ 10 c 30 d 3 e 5 √ √ √ √ h 2 11 i 3 6 j 5 2 9 g 7 2 √ 4 6 l 10 √ √ √ √ b 21 2 c 12 14 d −50 3 10 3 √ √ √ √ −18 3 f 15 5 g 42 6 h −60 10 √ √ √ √ k 24 30 l 216 7 −20 10 j 42 2 √ √ √ −4 2 2 d √ b 3 6 c 5 2 13 √ 2 5 − 1 √ f 3 3 7 √ √ √ √ 6 + 15 b 14 − 10 √ √ √ √ − 55 − 65 d −2 15 − 2 21 √ √ √ f 20 − 20 2 6 26 − 3 22 √ √ √ √ 30 2 + 15 30 h −12 3 + 12 2 √ √ √ j 90 3 − 24 10 42 + 63 2 √ √ −16 + 24 10 l 42 2 + 30 √ √ 28 b 18 c −75 d 2− 6 √ √ √ 3 3 + 4 f − 10 + 5 g 2 √ √ i 2−6 8 2 √ √ 2 6 b 30 c 6 √ 3 cm2 b 2 6 cm 4 √ √ √ √ 6 × 6 = 6 × 6 = 36 = 6 √ √ √ √ − 8 × 8 = − 8 × 8 = − 64 = −8 √ √ √ √ − 5 × − 5 = + 5 × 5 = 25 = 5 Simplify each surd before multiplying. b

Answers

6 a

b Allows for the multiplication of smaller surds, which is simpler. √ √ √ c i 3 2×3 3=9 6 √ √ √ ii 2 6 × 2 5 = 4 30 √ √ √ iii 5 2 × 3 5 = 15 10 √ √ √ iv 3 6 × 5 3 = 45 2 √ √ √ v 6 2 × 4 3 = 24 6 √ √ √ vi 6 3 × −10 5 = −60 15 √ √ √ vii −12 3 × −2 7 = 24 21 √ √ √ viii 7 2 × 10 3 = 70 6 √ √ √ ix 12 2 × 12 5 = 144 10 e 2 13 a 3 b 2 c −9 d −1 5 5 f 3 √ √ √ 14 a 54 2 b 375 3 c 162 3 d 25 e 9 √ √ √ f 128 2 g −120 5 h −108 2 i 720 √ √ √ j 14 7 k 27 2 l 81 m 100 3 2 √ √ 5 n 144 o −96 15 p 81 3 q √ 25 3 3 √ r 9 6 2

Exercise 3D 1

a

√ 21

e 13 i 162

√ b − 10

f 12

√ c 6 6

d 11

g 125

h 147

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Answers

2 a 0 b 0 √ √ f 6 3 e 2 3 3 a x2 + 5x + 6 c x2 + x − 12 e 6x2 − 11x − 10 h 4x2 − 9 j x2 + 4x + 4

c 0

d 2x2 − 9x − 5 f 6x2 − 17x − 28 i 25x2 − 36 k 4x2 − 4x + 1

7 a 7 f −6 c e g i

b 19 g 3 √ 118 + 28 10 √ 195 + 30 30 √ 207 − 36 33 √ 66 + 36 2 √ 107 − 40 6

9 a 97 e 26

1

b x2 − 4x − 5

l 9x2 − 42x + 49 √ 4 a − 2−4 √ √ d 7+3 3 e 5+ 7 √ g 6 5 − 13 √ i 23 − 8 7 √ 5 a 27 + 12 2 √ c 25 + 32 5 √ e 35 − 13 10 √ g 23 3 − 46 √ i 24 5 − 89 √ 6 a 14 − 6 5 √ c 23 + 8 7 √ e 28 + 10 3 √ g 9 + 2 14 √ i 13 − 2 30 √ k 40 + 2 391

8 a

Exercise 3E

d 0

b 17 f 10

√ √ b 2 5−3 c 4+ 6 √ f −13 − 2 2 √ h 30 − 9 10 b d f h

√ b 10 − 4 6 √ d 15 + 4 11 √ f 54 − 14 5 √ h 16 − 2 55 √ j 32 + 2 247 √ l 60 − 2 899

c 13 h 6 b d f h

d 6 i −4 √ 139 + 24 21 √ 176 − 64 6 √ 87 − 12 42 √ 140 + 60 5

c 41 g 0

e −2 √ 2 a 3 √ e 3

3 4

√ 21 + 2 3 √ 2 6 − 118 √ 18 7 − 65 √ 43 − 19 2

i −40 j −90 √ 10 a 7 + 4 3 cm2 b 2 m2 √ √ √ c 15√ 6 − √ 5 2 −√18 + 2 3 mm2 e 7 6 − 7 2 − 3 + 1 cm2 2 √ f 81 − 30 2 mm2 11 a −5 b 7 √ √ 12 a 11 21 − 26 3 √ √ c 5 35 + 31 7 √ √ e 2−2 6

g x2 − 16

5

6

7 e 4 8

d 163 h −33 √ d 5 + 6 5 m2

c 128 √ √ b 2 5 + 30 √ √ d 19 7 + 2 √ √ f 10 + 3 5 √ √ 13 Yes. Possible example a = 12, b = 3 √ √ 14 a 19 − 2 6 b 16 c 2 15 − 85 √ √ e 30 − 10 2 f 0 d 10 3 − 37 √ √ √ g 4 3 − 14 h 47 2 − 10 30 + 11

a 1

9

10

11

b 1 f −9 √ b 5 √ f 7

c 1 2 g 6 c √ 10 g √3 3

d −1 2 h −1 √ d √5 h √7 7

√ i √13 13 a 0.377… b 2.886… c 16.31… All√ pairs of numbers √ are equal. √ √ 2 a b 7 c 3 11 d 4 5 2 7 11 5 √ √ √ √ 3 15 5 g e f 4 2 h 14 3 3 7 √ √ √ √ 6 35 66 a b c d 10 3 7 11 5 √ √ √ √ 21 42 30 e f g h 34 3 7 3 2 √ √ √ √ a 4 14 b 5 6 c 3 10 d 3 42 7 3 2 7 √ √ 30 105 7 2 e f 10 15 √ √ √ √ √ 21 6 4 a b c 35 d 2 2 e 2 5 15 3 3 5 15 √ √ f 10 g 9 2 h 3 7 9 2 2 √ √ √ √ 3 + 6 a b 3 7 + 35 3 7 √ √ √ √ 5 − 15 6 − 10 2 c d 5 2 √ √ √ √ f 30 − 21 e 35 + 14 7 3 √ √ √ √ 2 5 3 + 42 2 +2 5 g h 6 10 √ √ √ √ 8 30 − 5 2 3 − 15 2 i j 5 6 √ √ √ √ 3 6 2 + 2 5 5 + 5 6 k l 2 2 √ 3 2 5 2 2 b m a cm 3 3 √ √ 10 + 15 c mm2 10 √ √ √ √ a 2 3+3 2 b 6 5+5 2 6 10 √ √ √ √ 7 − 14 3 3 −2 2 9 5 c d 21 6 √ √ √ √ e 2 2+5 3 f 9 5+4 3 12 30 √ √ √ 14 30 +4 6 −2 6 g h 15 9 √ √ 10 − 2 42 3 i 9 √ x As √ is equal to 1. x

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d g 13 a b

√ √ √ √ √ √ 21 + 7a b 30 + 5a c 2 3 + 6a 7 5 6 √ √ √ 1 − 5a f 1 − 1√ − 3a √ e √ √ √ 7a √ 4 10a + 5 2 h 6a + 2 i 2 14a + 7 2 10 2 14 i 14 ii 2 iii 47 Each question is a difference of perfect squares, and each

answer √ is an integer. 4 + √2 c 4+ 2 √ d i 12 + 3 2 14 √ √ iii 2 2 + 6 √ 14 a 5 3 − 5 2 √ d −4 − 4 2 √ g −12 − 4 10 √ √ j 2 5−2 2

√ ii −3 3−3 2 √ iv −(6 + 2 30) 7

√ b 2 3+2 √ e −3−3 3 2 √ h −14 − 7 5 k

√

√ c 3 5+6 √ f 42 + 7 7 29 √ √ 2 11 +2 2 i 9 √ √ 14 − 2 l 6 √ o 10 − 4 5

√ √ 7− 3

√ √ n 14 + 2 2 m 6+ 6 5 √ √ √ √ b a−b b a a+a b p q a−b √ a−b a + b − 2 ab r b √ a −√ a b+b a t a−b

s

√ a − ab a−b

a 34

b 76

e 15y4

f 8a3 b2

c 83

4

3

d 6x3

2

1

= 25

d 5y15

y2 2

b t6

64t6

f

4u4

g

27r9

h 81p16 16 8 l u 8w v 256p8 q12 p 81r 4 d 1 e 5

4 2 n 9a2 b6 4p q b 3

3 9 o a t12 27g c 1

f 3 8 a x8

g −5 b x2 y 2

e m

f r4 s7

h 3 c x6 n 8 9x8 y2 g 2

i 2a2 b2

j 27m7 n14 k −45a8 b5 l 16 f 3 3 n 21y3 z2 o 1 p −6m2 n7

m 2m6 n3 9 a −27

b −27

x12

b

11 a 13 g 9

b 18 h 8

10 a

a105

c 81 a30 c 15 b c 81 d 64

d xy2 h 2y4

d −81 e 1

f 1

12 He has not included the minus sign inside the brackets, i.e. has only applied it afterwards. Need (−2)4 not −24 .

13 a 3 f 1

b 4

c 1

d 3

e 4

14 a 9

b 2

c 162

15 a ±2

b 5

c 2

b x = 8, y = 2 or x = 4, y = 3 or x = 64, y = 1, or x = 2, y = 6 c x = 9, y = 2, or x = 3, y = 4 or x = 81, y = 1 d x = 1, y = any positive integer

Exercise 3G

=x×x×x×x×x x×x×x x3

1

= x2 c (a ) = a × a × a × a × a × a = a6

0

d (2x) × 2x0 = 1 × 2 c b6 3 g p 5

d 14m5 7 h c 6 l 18v9 w2

e 6s7

f 2t16

9 s2 25 m 150x5 y6

j 6x3 y3

k 15a3 b6

n 12r7 s6

o 20m8 n10

a 2−2 , 2−3 , 2−4

c 3x−1 , 3x−2 , 3x−3 2 a 12 b 12 3 5 1 3 a b b ab a

2 3

i

c 4a6

r −

x 4 y6 k z8

0

x5

=2 b x5

n 4st

9 j x12 y

3

4 a a9

m 5b3

h m6 l 2m4 x p 1a 2

4 i a6 b 27f 6 m 125g3 7 a 8

3 a 2 ×2 =2×2×2×2×2 b

d b4

g j k 2p2 o 1 v2 4

e

2x 24 = 16 23 = 8 22 = 4 21 = 2 20 = 1 2

c q3

f d5 j 3r2 s

q −x 3 6 a x10

2 x

b a

e y5 i 2xy3

d −18 d 7 2 16 a x = 2, y = 4 or x = 4, y = 2 or x = 16, y = 1

Exercise 3F 1

5 a x3

Answers

12 a

4 a 15 x 2 e 3a3 b q3 r i 3p2

5 a x2 e 2b4 d3

b 14 a 3 f 4m3 n d2 f5 j 5e4

b 2y3 f 3m2 n4

b x−1 , x−2 , x−3 c 52 or 54 4 2

d −2 33

c 24 m 10y5 z g x2

d 37 y

2 7 k 3u w 8v6 c 4m7 4 3 g 4b a 3

2b3 5c5 d2 d 3b5 3 3 h 5h g 2

3 h 3z x4 y2

l

769 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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b a3

c 24 b

d 33 y

Exercise 3H

e 62 a

f 12 x

g −10 m6

h −18 a10

1

i 2x 3

2 j 7d 10

k 1 2

2 l 3b 4

m 53 3s

n

4 3f 2

o

1 2d2

p 52 6t

7 a 164 x

b

1 64m6

c 221 x

d 46 d

g 81 x20

h −8 x15

k 7j8

l 2t6

Answers

6 a x

e 98 t y4 i 16 8 a 1 xy e a3 b p i 2 q r m

f5 g5

f 54 x 12 j h 81 4y2 b 3 a q f 5 p

c

6 a5 b2

4 d 18b2 a

2 g a2 b

2 h m n

j

x 2y

k 4m3 7n

3 7 l 4r s 3

n

1 r6 s2

5 o wx 2

5 4 p 5c d 4

16p4 9q2 11 e 324r s

9 a a7 b2

c 54x7 y10

b

d 4a8 b3

f

2y14 x3

14 h m8 n

i 27x 2y

b 1 64

c 2 49

d −5 81

e 1 9

f 1

g 98

h −48

i 9 4

k 1 16

l 100

g a2 b18 10 a 1 25

j −64 125 11 0.0041 cm 12 a i 3 2

ii 7 5

iii y 2x

b b a 13 The negative index should only be applied to x not to 2: 2 2x−2 = 2 x 14 a 5 b 5 c 1 d − 7 6 18 3 12 e 71 48

f 106 9

 x 15 Proof: 1 = (2−1 )x = 2−1×x = 2−x 2 16 a −2 b −5 c −3 d −1 e −2 f −3 g −3 h −4 i 0

j 0

k 1

l 2

m −2

n 1

o −2

p 2

a 3 f 2

b 4 g 1

c 3 h 3

2 a 103 b 107 4 3 a 4.3 × 10 c 9.012 × 105 e 7.8 × 10−4

g 3 × 10−5 4 a 3120 d 8 213 000 g −10 120

j −55 000 5 a 0.0045

d 0.00783 g 0.0001002

d 2

e 3

c 10−6 d 10−3 5 b 7.12 × 10 d 1.001 × 104 f 1.01 × 10−3

h 3.00401 × 10−2 b 54 293 c 710 500 e 59 500 h 9 990 000

f −800 200 i 210 500 000

k 2 350 000 000 l 1 237 000 000 000 b 0.0272 c 0.0003085 e −0.000092 f 0.265 h −0.000006235 i 0.98

i −0.000000000545 k 0.000000000003285 l 0.000000875 6 a 6.24 × 103 b −5.73 × 105 c 3.02 × 104 e −1.01 × 104 g 7.25 × 104 i 1.10 × 108

k −4.56 × 106 7 a 2.42 × 10−3

c 1.25 × 10−4 e −7.08 × 10−4 g 6.40 × 10−6 i 1.30 × 10−4 k 9.89 × 10−9

8 a −2.4 × 104 c 7.0 × 108 e 1.9 × 10−3 g 9.8 × 10−6

i 5.00 × 10−5 9 a 7.7 × 106 km2

d 4.24 × 105 f 3.50 × 107 h 3.56 × 105 j 9.09 × 105

l 9.83 × 109 b −1.88 × 10−2 d 7.87 × 10−3 f 1.14 × 10−1 h 7.89 × 10−5

j 7.01 × 10−7 l −5.00 × 10−4

b 5.71 × 106 d 4.88 × 103

f −7.05 × 10−4 h −3.571 × 10−1

b 2.5 × 106

c 7.4 × 109 km e 1.675 × 10−27 kg

d 1 × 10−2 cm f 9.5 × 10−13 g

e 8.00 × 107 g 1.80 × 10−3

f 3.63 × 108 h 3.42 × 1015

10 a 2.85 × 10−3 c 4.41 × 10−8

i 8.31 × 10−2 11 328 minutes 12 38 is larger than 10. 13 a 2.1 × 104

b 1.55 × 10−3 d 6.38 × 10−3

b 3.94 × 109

c 6.004 × d 1.79 × 10−4 3 −1 e 2 × 10 f 7 × 10 g 1 × 107 101

h 6 × 106 i 4 × 10−3 j 3.1 × 10−14 −4 k 2.103 × 10 l 9.164 × 10−21

14 a 9 × 104 b 8 × 109 d 1.44 × 10−8 g 2.25 × 10−6

c 6.4 × 109 e 4 × 104 f 6.25 × 10−12 h 1.25 × 107

770 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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k 1.8 × 10−1 n 4 × 10−14

15 3 × 10−4 = 3 ÷ 10000 16 a i 9 × 1017 J

5 a 7x 2

l 2 × 102 m 8 × 10−1 o 2.5 × 104

4

iii 9.69 × 10−13 kg c 5.4 × 1041 J

iv 1.89 × 10−19 kg

b 3.142857 rational

22 7 π

2 3 4 5 6 7 8

9 a

a5

b

e a6

f 9m10

10 a 13 x

4 b 2b a2 c3

c

d 1 m 6 n3 2

g

4p8 q6 49r2 t4

h 6

b 8

c 216

g 1 16

h

1 125

1 16

k 1 81

l

1 100

9 a a2

b m3

c x

d b2

6 s7

f

1 y9

g 1

a2 h b

b

3t2

c 2t2

d 5t4

g

t3

e 10 a f

c 3, 5

d 5, 5

f 5, 5 b 5

c 11

d 25

e 2

f 3 k 2

g 5 l 10

h 4

i 2

j 3

e

f

h

m2

l 7r3 t2

iv −3

c $52.50

d $55.13

ii −10

e x

i 4ab4 m5 7

viii 81 10000

a $50

b $1050

2 , 0.98 c 4.52 kg 100 b growth c decay b

d decay

f decay

4 a A = amount of money at any time, n = number of years of investment A = 200 000 × 1.17n b A = house value at any time, n = number of years since initial valuation A = 530 000 × 0.95n

b 1.58, 1.58

1 7 43 t3

1

iii −2

2 a 4.9 kg

b 2, 3

b

1

e $1276.28

e 4, 4 2 a 3

1 1 22 a2

i 125

13 a i −3

1

a 3, 2

4 a

e 1 8

iv 81

e growth

1 35 3

d 32

1 10

2 p 10x n 2x o 2 3 x 11 a method B b i 32 ii 216 iii 128 v 625 vi 1 vii 32 27 3125 12 It equals 2 since 26 = 64.

3 a growth

3 a 1.91, 1.91 c 1.43, 1.43

k 2x2 y3

e 2

Exercise 3J

Exercise 3I

1 29 2

b4

d 7 i 1 3

b i no ii yes iii yes iv no c y is a real number when n is odd, for x < 0.

6 5 g 12m e 168 f a h 4d2 k 8 a5 3c 11 a i 7 012 000 ii 0.009206 ii 2.10 × 10−4 b i 3.52 × 108 4 10 d9 a 9x 12 a b 5b5 8c13

1

5s2

j 6m2 n

d 42 5d

c 7m2

1

l

1 10

f 1 9

h4

4

k 1 20

j

12x3 y4

1

c 4 h 1 2

8 a 4

3.14 3.15 3.16 3.17 √ √ √ √ 5 2 a 7 2 b 10 3 c d 5 5 2 4 √ 192 √ √ √ √ b 6 2+2 5 c 16 3 a 3 3 √ √ d 17 5 − 6 3 √ √ 2 b 35 6 c √ a − 15 3 5 √ 66 2 √ √ a 26√− 18 3 √ b 5√ − 2 6√c −4 a 3 7 b 2 15 c 2 3 − 3 10 7 5 2

3

b 3 g 1 3

√10

315%

1

1

g 5 2 or 125 2

7 a 6

j

a 3.16227766 irrational

5

i 4 3 or 256 3 √ √ 10 3 c 6 d 11 √ √ 5 7 g 8 h 81

f 5

c 3.141592654 irrational d 3.15 rational

3

f 2g 4 h 4

2

d 5p 3 r 3

h 7 2 or 343 2 √ √ 5 7 6 a 2 b 8 √ √ 3 3 e 9 f 49

Progress quiz 1

2

3

iv 9 × 1011 J ii 4.22 × 10−1 kg

c 3y3

b 6n 3

e 2a 3 b 3

ii 2.34 × 1021 J

iii 2.7 × 1015 J b i 1.11 × 108 kg

7

5

j 1.275 × 10−4

Answers

i 1 × 10−5

c A = car value at any time, n = number of years since purchase

c

2 x5

d

g

1 2 10 5 t 5

h 88 m2

3 b4 1

1

A = 14 200 × 0.97n

771 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

d A = population at any time, n = number of years since

Answers

initial census A = 172 500 × 1.15n

e A = litres in tank at any time, n = number of hours elapsed A = 1200 × 0.9n f A = cell area at any time, n = number of minutes elapsed A = 0.01 × 2n g A = Size of oil spill at any time, n = number of minutes elapsed A = 2 × 1.05n

h A = mass of substance at any time, n = number of hours elapsed

5

2 3

4200 4410

210 220.50

4 4630.50 231.53 5 4862.03 243.10 6 a $5105.13 b $11 946.33 c $13 652.22 d $9550.63

8 a $2254.32

b $87 960.39

d $789.84 9 $11 651.92

e $591.63

iv $3461.88 b $226.54

c P = 1000, n = 65, r = 0.036%, R = 0.936%, t = 2.5 years iii $42 672.53

d P = 3500, n = 30, r = 0.0053%, R = 1.9345%, t = 30 days

7 a V = 15 000 × 0.94t b i 12 459 L ii 9727 L

e P = 10 000, n = 10, r = 7.8%, R = 7.8%, t = 10 years

c 769.53 L d 55.0 hours

13 5.3% compounded bi-annually 14 a i approx. 6 years ii approx. 12 years

8 a V = 50 000 × 1.11n b i $75 903.52 c 6.64 years

9 a 3000 b i 7800

iii approx. 9 years v approx. 7 years

ii $403 115.58

iv approx. 5 years vi approx. 4 years

b Same answer as part a c yes

ii 20 280

iii 243 220

c 10 hours 11 minutes 10 a D = 10 × 0.875t , where t = number of 10 000 km travelled b 90 000 c yes

Exercise 3L 1

B

2 P = 750, r = 7.5, n = 5 3 I = 225P = 300 r = 3

11 a T = 90 × 0.92t b i 79.4◦ C ii 76.2◦ C c 3.2 minutes = 3 minutes 12 seconds 12 a i $1610.51 ii $2143.59 iii $4177.25

4 a i $7146.10 iv $7260

t = 25

ii $6955.64 v $7916.37

b i $1645.31 ii $2218.18 iii $4453.92 13 a $2805.10 b $2835.25 c $2837.47 14 a i 90 g b 66 years

b 6000 at 5.7% p.a., for 5 years 6 a i I $240, $240

ii 72.9 g

iii 53.1 g

b $550 c $55

d $605 e $605

2 a $1102.50 b $1102.50 c $1157.63 d $1157.63 b 1000(1.15)6 c 850(1.06)4 3 a 700(1.08)2 4 a 6, 3% d 14, 2.625%

b 60, 1% e 32, 3.75%

ii $7080

iii $7428

iv $7200

II $480, $494.40

ii

III $1200, $1352.90 I $240, $243.60

IV $2400, $3163.39 II $480, $502.04

iii

III $1200, $1375.67 I $240, $246.71

IV $2400, $3224.44 II $480, $508.64

III $1200, $1395.40 b compound interest

IV $2400, $3277.59

b 22.8 minutes

Exercise 3K

iii $6858.57

b $6000 at 5.7% p.a., for 5 years 5 a i $7080 v $7710

a $50

iii $3446.15

v $3465.96

12 a P = 300, n = 12, r = 7%, R = 14%, t = 6 years b P = 5000, n = 24, r = 2.5%, R = 30%, t = 2 years

b i $216 750 ii $96 173.13 c 3.1 years

1

c $1461.53

10 a $5075 b $5228.39 c $5386.42 11 a i $3239.42 ii $3348.15

c 7.3 years 6 a 300 000

15 a 60 L 16 0.7%

4862.03 5105.13

7 a $106 000 b $112 360 c $119 101.60 d $133 822.56 e $179 084.77 f $239 655.82

A = 30 × 0.92n 5 a 1.1 b i $665 500 ii $1 296 871.23 iii $3 363 749.97

4410 4630.50

c compound interest

c 52, 0.173% f 120, 0.8%

772 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

9 1 cent doubled every second for 30 seconds. Receive 230

Principal Rate Overall time Interest Amount $7000 5% 5 years $1750 $8750 $7000 10% $3300 10%

5 years 3 years

$3500 $990

$10 500 $4290

$8000 10% $9000 8%

3 years 2 years

$2400 $1440

$10 400 $10 440

2 years

$2880

$20 880

$18 000

8%

b i interest is doubled

cents, which is more than 1 million dollars. 10 a i 1 ii 1 iii 1 b No solutions. If a = 1, then ax = 1 for all values of x. d 3 e 5 11 a 2 b 1 c 2 3 4 4

ii no change

iii interest is doubled 8

Overall Principal Rate Period time $7000 4.56% annually 5 years

Interest $1750

Amount $8750

$7000 8.45% annually 5 years $3500 $9000 8% fornightly 2 years $1559.00

$10 500 $10 559.00

$18 000 8% fornightly 2 years 9 a 8.45%

b 8.19%

$2880

f 1 3 12 a i 0.25 b i 5−2

g 3 h −1 10 2 ii 0.125 iii 0.001 ii 2−4 iii 2−1

13 a −4

b −6

14 a 1 e −5 2 j 2

c 8.12%

The more than interest is calculated, the lower the required rate.

f −3 4

e −1.5

$21 118.01

c −5

b −1

c 8

d −3 2

f −2

g 3

h 2 3

k 0

l −2

1

Exercise 3M

3 1 5 4 a −8 b 22−a √ 5 length = 10 2 cm, width = 10 cm √ √ 6 −3 − 2 + 7 3 7

a i 4 b i 3

ii 8 ii 5

iv 32

2 a 16, 32, 64, 128, 256, 512 b 81, 243, 729, 2187, 6561 c 64, 256, 1024, 4096 e 216, 1296 32

53

d 125, 625, 3125

b

4 a 3 f 3

b 3 g 4

c 2 h 3

d 2 i 4

k 4 5 a −2

l 3 b −2

c −2

d −4

f 3 b 4 3

g 2 c 3 2

f 1 3

g 1 5

h 1 4

j −4

k −3 2

l −3 2

7 a 1 b i 2

ii 32

iii 260

d

h 6 d 3 2

f 2 j − 11 2

g 9

h 6 7

k 4

l −3 2

e 3 j 5

i 3 e 1 2

√

xy(x − y)

9 x = 3.5

Multiple-choice questions 1 C 6 D

2 D 7 D

11 C

12 D

1 iv 21440

i 15 4

b

 x 4 9

3 B 8 C

4 E 9 B

5 A 10 D

Short-answer questions

i −2

c i 3 min ii 8 min iii 10 min 1 8 a b 1 c 3 d 1 2

b

7 a x−y xy √ 8 12 + 8 2

27

3 a e 36

e −5 6 a 3 2

c

35

3n

2 a 5

iii 16 iii 6

i 1 5

Problems and challenges

10 a i 4.2% ii 8.7% b it increases by more than the factor.

1

iv 0.0016 iv 5−4 1 d 2

Answers

7 a

e 3 4

√ c 30 2 d √ √ h 2 5 g 5 7 5 √ √ √ 4+7 3 b 2 5+2 7 c √ √ √ 4 3 + 2 2√ e 2 30 f √ h 7 2 5 i 0 3 √ √ √ b 12 5 − 6 2 6+4 2 √ √ d 6 5 e 12 3 − 4 √ −9 g 16 + 6 7 h

√ a 2 6 √ f 2 2 3

2 a d g 3 a c f

√ b 6 2

√ 12 6

e 2 7

√ 5 2 √ −12 5

6 √ 56 − 16 6

773 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

√

√ √ c 3 6 d 2 14 √ √ g 10 + 2 2 2

√ 6 b 5 2 6 √ √ e 3 2 f 5 2 4 8 √ 6 − 3 4 h 3

4 a

5 a 25y6 e

3y4 2x3 1

6 a 21 2 e

1 3 10 2 x 2

7 a 5

c 20x7 y10

d

3x2 y4

27 4b8 1

5

2

c m3

d a5

f

g

h

1 1 2 3 a3 b 3

3 72

c 1 2

b 4

4 43

d 1 7

ii 4 024 000 iv 0.0000981

10−4

b i 3.08 × iii 5.68 × 106

9 22.3 m 10 7.54 m

13 a 4.5 cm b 8.5 mm 14 The student rounded tan 65◦ too early. 15 a 3.7 b 6.5 16 a i a = c sin q iii tan q = a b v Answers may vary

b 2

e −2

f −3

ii 7.18 × iv 1.20 × 108

i 4

j 3

c 1 g 3 2 k −4

d 6 h 2 3 l 0

Extended-response questions 1

√ √ √ √ a 36 15 + 3 45 = 36 15 + 9 5 cm2 √ √ √ b 360 3 + 144 15 + 90 + 36 5 cm3 √ c 4 3+1

d i 10 000 cm2 2 a V = 10 000 × 1.065n b i $11 342.25 c 11.0 years

ii 1.6% ii $13 700.87

c 7.7 ii b = c cos q iv tan q = c sin q = sin q c cos q cos q

b i a = c sin q ii b = c cos q iii c2 = a2 + b2 iv c2 = (c sin q)2 + (c cos q)2

10−6

c2 = c2 (sin q)2 + c2 (cos q)2 ∴ 1 = (sin q)2 + (cos q)2

9 a V = 800 × 1.07t b V = 3000 × 0.82t 10 a $1215.51 b $3519.60 c $5673.46 11 a 3

b 327 m

11 28.5 m 12 26.4 cm

b x3

e 1 f 1 10 5 8 a i 3210 iii 0.00759

b x = 12.26 cm, y = 6.11 cm c x = 0.20 m, y = 0.11 m 7 a 125 m 8 1.85 m

b 6 f

6 a x = 2.5 cm, y = 4.33 cm

Exercise 4B 23.58◦

b 1 2 b 60◦

c 11.31◦

25.84◦

e 3 a tangent

45◦

f b cosine

14.48◦

g c sine

h 31.79◦

4 a 60◦ e 52.12◦

b 45◦ f 32.74◦

c 48.59◦

d 30◦

1

a 60◦

2 a

c 0.75 d 5.74◦

5 a a = 60◦ , q = 30◦ b a = 45◦ , q = 45◦ ◦ ◦ c a = 53.1 , q = 36.9 d a = 22.6◦ , q = 67.4◦ e a = 28.1◦ , q = 61.9◦ f a = 53.1◦ , q = 36.9◦

d i $14 591 ii V = 14 591 × 0.97t iii $12 917; profit of $2917

6 a 44.4◦ , 45.6◦ c 58.3◦ , 31.7

b 74.7◦ , 15.3◦ d 23.9◦ , 66.1◦

Chapter 4

e 82.9◦ , 7.1◦ 7 70.02◦

f 42.4◦ , 47.6◦

Exercise 4A

8 31.1◦ 9 47.1◦

1

a 0.799

b 0.951

c 1.192

d 0.931

e 0.274 2 a sin q

f 11.664 b cos q

g 0.201 c tan q

h 0.999

3 a 1.80 e 22.33

b 2.94 f 12.47

c 3.42

d 2.38

4 a 1.15 e 2.25

b 3.86 f 2.79

c 13.74 g 1.97

d 5.07 h 13.52

i 37.02 5 a 8.55

j 9.30 b 4.26

k 10.17 c 13.06

l 13.11 d 10.04

e 5.55

f 1.52

g 22.38

h 6.28

i 0.06

j 12.12

k 9.80

l 15.20

10 a 66.4◦ 11 a i 45◦

b 114.1◦ c 32.0◦ ◦ ii 33.7

b 11.3◦ 12 a Once one angle is known, the other can be determined by subtracting the known angle from 90◦ . b a = 63.4◦ , b = 26.6◦ b tan 45◦ = x = 1 x

13 a

45°

c

√

45° 2x

774 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

b a = 60◦ √ c 3 d i 1 2 √ v 3 3

√

ii 1 iii 3 2 2 √ vi 3 √  √ 3+1 x e AB = 1 x + 3 x = 2 2 2

√ iv 3 2

14 a i 2.5 km b i 4.33 km c i 45.2◦

1

b 2.07 km

a

C

A

2

B D

2√3

c 35.3◦

2

b 1.50 m

A b 3.44◦

c 1.99◦

14 a i 8.7 cm b i 17.3 cm

2 61.4◦ 3 a 37.609 m b 45.47◦ 4 a 57.409 m b 57.91◦ 5 a i 26.57◦ ii 11.18 cm

ii 5 cm ii 20 cm

b 10.14◦ 6 a 7.31 m

c Answers may vary. 15 321.1 km/h ii 72◦ ii 2.38 m

C

2√2

d 45◦

12 Yes 13 89.12 m

iii 36◦ iii 2.02 m

iv 54◦ iv 1.47 m

c 3.85 m d 4.05 m

b 6.87 m

7 138.56 m 8 a i 2.25 m

ii 2.59 m

b 40.97◦ c 3.43 m 9 a i 1.331 km ii 1.677 km b 0.346 km

e Proof

Exercise 4D a 0◦

b 45◦

c 90◦

d 135◦

e 180◦

f 225◦

g 270◦

h 315◦

2 a 050◦ e 227◦

b 060◦ f 289◦

c 139◦

d 162◦

3 a 200◦ 4 a 220◦ T

b 082◦ b 305◦ T

c 335◦ c 075◦ T

d 164◦ d 150◦ T

5 a 1.7 km 6 a 121◦

b 3.6 km b 301◦

ii 236.3◦ ii 116.6◦

45◦

12 a 13 a i 1.55

35.26◦

b 1.41 units c ii 1.27 iii 2.82

d 0.2◦ d 1.73 units

b 34.34◦ 14 22◦

Exercise 4F b sin(135◦ ) ≈ 0.7, cos(135◦ ) ≈ −0.7

c sin(160◦ ) ≈ 0.34, cos(160◦ ) ≈ −0.94

2 a 149◦ e 41◦

9 a 217◦ b 37◦ 10 a 1.414 km b 1.414 km c 2.914 km b 5.92 km

10 a camera C b 609.07 m 11 a 5.5 m b 34.5◦ c 34.7◦

1

7 a 3.83 km b 6.21 km 8 a 14.77 cm b 2.6 cm

13 a i 045◦ b i 296.6◦

2

b

6 1509.53 m 7 32◦

11 a 1.62 km 12 10.032 km

c 4.81 km

2√2

5 320 m

1

b 55.1◦ , 12.3 km

Exercise 4E

3 28.31 m 4 4.3◦

16 a i 18◦ b i 0.77 m

iii 5.36 km

b i 99.32 m ii 69.20 m iii 30.11 m 17 a 38.30 km b 57.86 km c 33.50◦

1 1866.03 m 2 39 m

10 299 m 11 a 1.45◦

iii 5.32 km

ii 1.03 km ii 7.6 km

15 a 229.7◦ , 18.2 km 16 a 212.98 m

18 a 4.34 km

Exercise 4C

8 a 1.17 m 9 8.69 cm

ii 2.82 km

Answers

d sin 45◦ = √x = √1 cos 45◦ also equals √1 . 2x 2 2 14 a q = 30◦

3 a both 0.34 c both 0.63

c 2.16 km iii 26.6◦ iii 101.3◦

b 128◦ f 56◦

iv 315◦ iv 246.8◦

e 0.91, −0.91 g 1.19, −1.19

i 0.21, −0.21

c 135◦ g 29◦

d 93◦ h 69◦

b both 0.98 d 0.77, −0.77

f 0.42, −0.42 h 2.75, −2.75

775 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

4 a √1 2 f √1 3 5 a 140◦ e 138◦

b √1 2 √ g 3

c 1

b 160◦ f 99◦

c 115◦ g 143◦

√ 3 2 √ 3 i 2 d 155◦ h 124◦

b 55◦

c 86◦

d 70◦

21◦

h 78◦

d

h 1 2

172◦

i 6 a 30◦ 45◦

e i 37◦

f

7 a positive e negative √ 8 a 3 2 √ e − 3 2 √ 3 i 2 m −1

9◦

g

b positive f positive √ b 2 2 √ f − 3 √ j − 3 3 n 1

9 a 30◦ , 150◦ c 60◦ , 120◦ 120◦

c positive g negative √ c 3 √ g 2 2

d negative h negative √ d 2 2 √ h − 2 2

k −1 2 o 0

l 1 2 p undefined

b 45◦ , 135◦ 135◦

c

150◦

10 a e 135◦ √ 11 a 3 2 √ e 5 3

b f 150◦ √ b 3 2

12 a 45◦ √ 13 a 3

b 30◦ √ b 2

c 60◦

ii 12 13 √ b 2 7 7 √ b − 5 2

iii 5 12 √ c 2 21 21

√ c 20 3 3

d

120◦

d 14

f 3

5 13

15 a 1 3 16 a − 5 3 17 a i 0.17

iii 0.59

c i

d i 70◦

− q ii

ii 5◦ e i 90◦ − q ii b c √ f 2 5 5

−q

4 63◦

c 1 2

d 1

Exercise 4G 1

a = b = c sin A sin B sin C

2 a 1.9 3 a 50.3◦

b 3.6 b 39.5◦

c 2.5 c 29.2◦

4 a 7.9 e 8.4

b 16.5 f 22.7

c 19.1

d 9.2

5 a 38.0◦ e 47.5◦

b 51.5◦ f 48.1◦

c 28.8◦

d 44.3◦

6 a 1.367 km b 74◦ 7 27.0◦

c 2.089 km

8 131.0 m 9 a ∠ABC = 80◦ , ∠ACB = 40◦ 10 a ∠ABC = 80◦ b 61.3 km

b 122 km c 53.9 km

11 a 147.5◦ e 123.9◦

d 100.5◦

b 102.8◦ f 137.7◦

c 126.1◦

14 a 59.4◦ or 120.6◦ b B

3

35°

120.6°

59.4° C A 2

c 31.3◦ e

viii 0.37

B

35°

A

iv 0.59

2

C

d A triangle can only have one obtuse angle.

C 31.3°

iii 19◦ iii b c

iv 38◦

10 120° A

6

B

Exercise 4H

Progress quiz 1 a 12.58 b 38.14 2 a 39◦ b 58◦ √ √ 3 a i 80 or 4 5 √ 4 iii √ or 5 5 80 b 26.6◦

C 288◦

b 61◦ b 1 2 √ f − 3 √ b 32 3

8 a 36.77 m 9 a 1 2 1 e −√ √2 10 a 8 3

c −1 7

ii 0.17

90◦

B 150◦ 6 A 060◦ 7 a 13.65 km b 048.4◦

3

v 0.99 vi 0.99 vii 0.37 b sin a = cos b when a + b = 90◦ . 90◦

5 23.84 m

12 Impossible to find q as such a triangle does not exist. 13 37.6◦ or 142.4◦

14 a 13 b i

e 1 2

1 c 15.44 d 6.59 c 52◦ √ 8 ii √ or 2 5 5 80

a c2 = 32 + 42 − 2 × 3 × 4 × cos 105◦

b 72 = 52 + 92 − 2 × 5 × 9 × cos q 2 a 9.6 b 1.5 c 100.3◦ d 36.2◦ 3 a 16.07 cm b 8.85 m e 2.86 km f 8.14 m

c 14.78 cm d 4.56 m

4 a 81.79◦ e 92.20◦

c 64.62◦

b 104.48◦ f 46.83◦

d 61.20◦

5 310 m

776 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

7 a 145.9◦ 8 383 km 9 7.76 m 10 a cosine rule

d c2 = a2 − x2 + (b − x)2 = a2 + b2 − 2bx e cos C = x a f x = a cos C substitute into part d.

Exercise 4I 1 a 3.7 2 a a

b 28.8 b q

b 45.58◦ or 134.42◦

4 a 4.4 cm2 b 26.4 m2 c 0.9 km2 e 318.4 m2 f 76.2 cm2

d 13.7 m2

5 a 11.9 cm2 b 105.6 m2 c 1.6 km2 6 a x = 5.7 b x = 7.9 c x = 9.1

d x = 18.2

e x = 10.6 f x = 1.3 7 a 59.09 cm2 b 1.56 mm2

c 361.25 km2

cm2

m2

8 a 35.03 9 a 965.88 m2

c 6.37 km2 c 0.72 km2

b 51.68 b 214.66 m2

10 a 17.3 m2 b 47.2 cm2 c 151.4 km2 11 a Area = ab sin q √ 3 2 a b Area = 1 a2 sin 60◦ = 2 4 c Area = 1 a2 sin(180◦ − 2q) = 1 a2 sin 2q 2 2 12 a i 129.9 cm2 ii 129.9 cm2 b They are equal because sin 60◦ and sin 120◦ are equal. c Same side lengths with included angle 140◦ . 13 a 65.4◦ , 114.6◦

B

B 11 m 65.4° 8m

14 a i 540◦

11 m C ii 108◦

114.6° A

C 8m iii 11.89 cm2

iv 8.09 cm v 72◦ , 36◦ vi 19.24 cm2 vii 43.0 cm2 b 65.0 cm2 c Answers may vary.

Exercise 4J 1

d quadrants 1 and 4 f quadrants 3 and 4

a quadrant 1 c quadrant 4

q

0◦

90◦

180◦

270◦

360◦

sin q cos q

0 1

1 0

−1 0

0 1

tan q

0

undefined

0 −1

4 a 0.139 e −0.799 i 0.900

b 0.995 f −0.259 j −1.036

0

undefined

c −0.530

g 0.777 k 0.900

0

d −0.574

h −0.087 l −0.424

5 a quadrant 2, sin q positive, cos q negative, tan q negative b quadrant 4, sin q negative, cos q positive, tan q negative c quadrant 3, sin q negative, cos q negative, tan q positive d quadrant 1, sin q positive, cos q positive, tan q positive e quadrant 4, sin q negative, cos q positive, tan q negative f quadrant 2, sin q positive, cos q negative, tan q negative g quadrant 3, sin q negative, cos q negative, tan q positive

c 48.0 c b

3 a 56.44◦ or 123.56◦ c 58.05◦ or 121.95◦

A

b quadrants 2 and 4

c quadrants 2 and 3 e quadrants 1 and 3 3 b sine rule c sine rule

d cosine rule e sine rule f cosine rule 11 Obtuse, as cos of an obtuse angle gives a negative result. a2 + b2 − c2 b 121.9◦ 12 a cos c = 2ab 13 a AP = b − x b a2 = x2 + h2 2 2 2 c c = h + (b − x)

b

2 a quadrants 1 and 2

b 208.2◦

Answers

6 32.2◦ , 49.6◦ , 98.2◦

b quadrant 3 d quadrant 2

h quadrant 3, sin q negative, cos q negative, tan q positive i quadrant 3, sin q negative, cos q negative, tan q positive j quadrant 1, sin q positive, cos q positive, tan q positive k quadrant 4, sin q negative, cos q positive, tan q negative l quadrant 2, sin q positive, cos q negative, tan q negative 6 a −sin 80◦ b cos 60◦ c tan 40◦ d sin 40◦ e −cos 55◦ h −cos 58◦

k cos 66◦ 7 a 30◦ 71◦

e i 82◦ 8 a 42◦ e 33◦

f −tan 45◦ g −sin 15◦ i tan 47◦ j sin 68◦

l −tan 57◦ b 60◦ c 24◦ f

76◦

b 47◦ f 62◦

9 a 0 < q < 90◦ c 270◦ < q < 360◦

d 40◦

50◦

h 25◦

c 34◦ g 14◦

d 9◦ h 58◦

g

b 90◦ < q < 180◦ d 180◦ < q < 270◦

10 q 150◦ 315◦ 350◦ 195◦ 235◦ 140◦ 100◦ 35◦ 55◦ 2 11 a quadrant 4

b quadrant 1

c quadrant 2 e quadrant 1

d quadrant 2 f quadrant 3

12 a 45◦ √ b i − 2 2 c 30◦ d i −1 2 ◦ e 60√ 3 f i 2 √ 2 13 a b 2 e −1 √ i − 2 2 m0

√ ii − 2 2 ii

√ 3 2

iii 1 √ iii − 3 3

√ iii − 3 √ √ 0 c − 3 d − 3 2 2 √ √ f − 3 g −1 h − 3 2 2 √ j −1 k − 3 l 1 3 2 n undefined o 1 p −1 ii − 1 2

777 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

14 As tan q = sin q and both sin q and cos q are negative over cos q this range, tan q is positive in the third quadrant. 15 As tan q = sin q and cos q = 0 at 90◦ and 270◦ , the value cos q of sin q and, hence, tan q is undefined at these values. cos q 16 a true

b true

c false

d true

e true

f false

g true

h false

iii 17◦ , 163◦ v 204◦ , 336◦ vii 224◦ , 316◦ 6 a true b false

sin 110◦ = cos(−20◦ ) = 0.940, sin 195◦ = cos(−105◦ ) = −0.259 ii Their values are the same. iii They add to 90◦ . iv sin q = cos(90◦ − q) v True for these values. c Answers may vary

Exercise 4K q sin q

b

0◦ 0

30◦ 0.5

60◦ 0.87

90◦ 1

120◦ 0.87

sin θ

90°

180°

270°

360°

θ

−1

q b

0◦ 30◦ 60◦ 90◦ 120◦ 1

0.87 0.5

0

150◦

−0.5 −0.87

180◦ 210◦ 240◦ 270◦ 300◦ 330◦ 360◦

cos q −1 −0.87 −0.5

0

0.5

0.87

1

cos θ 1

O

iv 84◦ , 276◦ vi 102◦ , 258◦ viii127◦ , 233◦ iii −0.64 iv −0.77 vii −0.64 viii 0.94 ii 12◦ , 168◦ iv 64◦ , 116◦ vi 233◦ , 307◦ viii 186◦ , 354◦ c false d true g true k true

h true l true

7 a 110◦ e 27◦

b 60◦ f 326◦

c 350◦ g 233◦

d 260◦ h 357◦

8 a 280◦ e 136◦

b 350◦ f 213◦

c 195◦ g 24◦

d 75◦ h 161◦

9 a 30◦ e 55◦

b 60◦ f 80◦

c 15◦ g 55◦

d 70◦ h 25◦

k 63◦ l 14◦ ◦ b 44.4 , 135.6◦

c 53.1◦ , 306.9◦ e 191.5◦ , 348.5◦

d 36.9◦ , 323.1◦ f 233.1◦ , 306.9◦

g 113.6◦ , 246.4◦

h 49.5◦ , 310.5◦

151.3◦

i 11 a 0, the maximum value of sin q is 1.

O

cos q

vii −0.42 viii 0.57 ii 53◦ , 307◦

f true j false

28.7◦ ,

1

q

iv −0.77

iii 0.87

e false i true

i 36◦ j 72◦ ◦ 10 a 17.5 , 162.5◦

150◦ 0.5

q 180◦ 210◦ 240◦ 270◦ 300◦ 330◦ 360◦ sin q 0 −0.5 −0.87 −1 −0.87 −0.5 0

2 a

iii 73◦ , 287◦ v 114◦ , 246◦

v 0.34 vi −0.82 b i 37◦ , 143◦

j true k true l false ii True for these values. b i sin 60◦ = cos 30◦ = 0.866, sin 80◦ = cos 10◦ = 0.985,

a

ii −0.98

v −0.17 vi 0.26 b i 37◦ , 323◦

vii 143◦ , 217◦ 5 a i 0.42 ii 0.91

i false 17 a i Proof

1

4 a i 0.82

90°

180°

270°

360°

θ

−1 3 a i maximum = 1, minimum = −1 ii 0◦ , 180◦ , 360◦ b i maximum = 1, minimum = −1 ii 90◦ , 270◦ ◦ ◦ ◦ c i 90 < q < 270 ii 180 < q < 360◦

b 0, the minimum value of cos q is −1. 12 a q 0◦ 30◦ 45◦ 60◦ √ √ 1 2 3 sin q 0 2 2 2 √ √ 3 2 1 cos q 1 2 2 2 √ 1 1 2 b i ii − iii − 2 2 2 √ v 1 vi − 3 vii 0 2 2 √ √ ix − 3 x −1 xi − 2 2 2 2 √ √ √ 2 3 xiii − xiv xv − 3 2 2 2 b 60◦ , 120◦ 13 a 45◦ , 315◦ c 30◦ , 150◦ d 210◦ , 330◦

90◦ 1 0 iv 0 √ viii 2 2 xii − 1 2 √ xvi 3 2

e 120◦ , 240◦ f 150◦ , 210◦ 14 a Graph is reflected in the x-axis. b Graph is reflected in the x-axis. c Graph is dilated and constricted from the x-axis. d Graph is dilated and constricted from the y-axis. e Graph is translated up and down from the x-axis. f Graph is translated left and right from the y-axis.

778 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Problems and Challenges

ii 38.0◦

d i 21.9 m 2 a 33.646◦

b 8.7 cm

b 3177.54 m2 c 41.00 m

3 Use the cosine rule. 4 514 m

d 61.60 m e i 65.66◦ , 114.34◦

b 308◦

5 a 2 hours 9 minutes 6 17.93◦

ii 80.1 m, 43.1 m

7 4.33 cm

Chapter 5

Multiple-choice questions 1 D 6 B

2 B 7 C

3 E 8 A

4 D 9 D

5 A 10 C

3 a 4

4 6.1 m 5 a A = 115◦ , B = 315◦ , C = 250◦ , D = 030◦ 295◦

b i 6 a 98.3 km

070◦

ii b 228.8 km c 336.8◦

10 a 52.6◦ 11 a 12.5

p −4x +

−cos 30◦

12 a i ii iv √ −sin 45◦ √ 3 b i ii − 3 2 2 c i negative ii positive

iii iii −1 iii negative

13 a i 0.77 ii −0.97 b i 53◦ , 127 ii 197◦ , 343◦ ii true

a

c −5x − 15 f 12x + 4

n −18x2 + 6x q 4x + 8 5

o −10x + 10x2 r 6x − 15 4

w 9 x2 + 6x 4 b 6x2 − 3x e 2x2 − 2x

x 14 x − 6 x2 5 5 c 2x2 + 7x f 25x − 12x2

iii no value

e x2 + x − 30 g x2 − 4x − 21

i x2 − 13x + 40 7 a x2 + 10x + 25 x2

+ 12x + 36 c e x2 − 16x + 64

N

3 km

b 3x − 12 e 6x − 3

u −9x + 3 8

s −2x − 1 3

6 a x2 + 10x + 16 c x2 + 12x + 35

waterfall

g x2 − 16 4x2

−9 i k 16x2 − 25

8 a 6x2 + 13x + 5 c 10x2 + 41x + 21

5 km

entrance 325° b 2.9 km west

l −13x

t −2x + 3 2

16x2

v −2x − 14 9 5 a 2x2 + 3x d 8x2 + 7x

iii false

h −3x

i 2x2 + 5x l 6x − 3x2

√ iv − 2 2 iv positive

Extended-response questions 1

−tan 45◦

d −4x2

h −20x − 15 k 2x − 2x2

m −6x2 − 4x

b 105.4◦ b 42.8◦

c i true

3 a 6x e x 2 i −18x

c x2 + 8x + 16 c a 2 − b2

b 2x − 5 e a−b 2 b −20x c 2x2 f x g −4x 3 j 7x k 5x

g −10x + 6 j 3x2 − x

ii 111.3 cm2

sin 60◦

d (y + x)2

b x2 + 4x + 3

4 a 2x + 10 d −4x + 8

7 a 15.43 m b 52◦ 8 a i 15.5 cm ii 135.0 cm2 b i 14.9 cm 9 28.1 m

a x2 + 2x

2 a a2 + ab c x = 11.55, y = 5.42

b 13.17 b 64.8◦ √ b 5 3

Exercise 5A 1

Short-answer questions 1 a 14.74 2 a 45.6◦

Answers

1 a 120◦ , 60◦ 2 225◦

c 7.7 km

b x2 + 7x + 12 d x2 + 5x − 24

f x2 + x − 6 h x2 − 10x + 24 b x2 + 14x + 49 d x2 − 6x + 9 f x2 − 20x + 100

h x2 − 81

j 9x2 − 16 l 64x2 − 49

b 12x2 + 23x + 10 d 9x2 − 9x − 10

e 20x2 + 2x − 6 g 16x2 − 25

f 6x2 + 5x − 25 h 4x2 − 81 i 25x2 − 49

n 25x2 + 60x + 36

o 49x2 − 14x + 1

j 14x2 − 34x + 12 l 56x2 − 30x + 4

k 25x2 − 45x + 18 m 4x2 + 20x + 25

779 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

9 a 3

b 3

e 1 f 2 10 a 2x2 + 14x + 24 c −2x2 − 20x − 32 e 5x2 + 5x − 60

c 3

d −4x2 − 44x − 72 f 3x2 + 6x − 45

3 a (x − 1)(5 − a) c (x + 5)(a − 4)

l −6y2 + 42y − 72 n 18x2 + 12x − 48

i (x − 6)(1 − x) 4 a (x + 3)(x − 3)

h −5a2 + 30a + 80 j 3y2 − 27y + 60

o −6x2 − 10x + 56 q 4m2 + 40m + 100

p 2x2 + 12x + 18 r 2a2 − 28a + 98

s −3y2 + 30y − 75 u −12y2 + 72y − 108

11 a 2x2 + 10x + 11 c 2y2 − 4y + 5 e −24a − 45 g x2 + 10x + 18

i −4x2 + 36x − 78 12 a x2 − 12x + 36 cm2

q ab(b − a) s −12mn(m + n)

b 3x2 + 27x + 42

g −3a2 + 15a + 42 i 4a2 − 36a + 72 k −2y2 + 22y − 48 m 12x2 + 48x + 45

o −x(2 + x)

d 8

e (x − 4)(x − 2) g (x + 3)(a + 1)

c (y + 7)(y − 7) e (2x − 3)(2x + 3)

t 12b2 − 12b + 3

g (1 + 9y)(1 − 9y) i (5x − 2y)(5x + 2y)

b 2x2 + 20x + 44 d 2y2 − y − 43

k (3a + 7b)(3a − 7b) 5 a 2(x + 4)(x − 4)

f b2 + 54b + 5 h x2 − 14x + 40

j −25x2 − 30x + 5 b x2 + 10x − 200 cm2

13 a (a + b)(a − b) = a2 − ab + ba − b2 = a2 − b2 b (a + b)2 = (a + b)(a + b) = a2 + ab + ba + b2 = a2 + 2ab + b2 2

2

c (a − b) = (a − b)(a − b) = a − ab − ba + b 2

= a2 − 2ab + b2

d (a + b) − (a − b)2 =

a2 + ab + ba + b2 − (a2 − ab − ba + b2 ) =

2ab + 2ab = 4ab 14 a 618 b 220 e 1386 f 891

c 567 g 3960

15 a −x2 + 7x c 4x2 + 12x + 9

b 10a − 28 d 4x + 8

d 1664 h 3480

16 a x3 + 6x2 + 11x + 6 b x3 + 11x2 + 38x + 40 c x3 + 2x2 − 15x − 36 d 2x3 − 13x2 + 17x + 12

e 2x3 − x2 − 63x + 90 f 6x3 − 35x2 − 47x − 12 17 a 2ab b (a + b)2 − c2 c (a + b)2 − c2 = 2ab

c2 = a2 + 2ab + b2 − 2ab

Exercise 5B a 7

b 6

e 2a f 3a 2 a 3(x − 6)

c 8

d −5

g −5a h −3xy b 4(x + 5)

c 7(a + b) e −5(x + 6)

d 3(3a − 5) f −2(2y + 1)

k 6b(b − 3) m 5a(2 − a)

l 7a(2a − 3) n 6x(2 − 5x)

g −3(4a + 1) i x(4x + 1)

h −b(2a + c) j x(5x − 2)

r 2xy(xz − 2) t 3z2 (2xy − 1)

b (x + 2)(b + 3) d (x + 2)(x + 5) f (x + 1)(3 − x) h (x − 2)(x − 1) b (x + 5)(x − 5)

d (y + 1)(y − 1) f (6a − 5)(6a + 5)

h (10 − 3x)(10 + 3x) j (8x − 5y)(8x + 5y)

l (12a − 7b)(12a + 7b) b 5(x + 3)(x − 3)

c 6(y + 2)(y − 2) e 3(x + 5y)(x − 5y)

d 3(y + 4)(y − 4) f 3(a + 10b)(a − 10b)

k (a + 5)(a − 11) m (4x + 5)(2x + 5)

l (a − 8)(a − 6) n (y + 7)(3y + 7)

g 3(2x + 3y)(2x − 3y) h 7(3a + 4b)(3a − 4b) i (x + 9)(x + 1) j (x − 7)(x − 1)

o (3x + 11)(7x + 11) p 3x(3x − 10y) √ √ √ √ 6 a (x + 7)(x − 7) b (x + 5)(x − 5) √ √ √ √ c (x + 19)(x − 19) d (x + 21)(x − 21) √ √ √ √ e (x + 14)(x − 14) f (x + 30)(x − 30) √ √ √ √ g (x + 15)(x − 15) h (x + 11)(x − 11) √ √ √ √ i (x + 2 2)(x − 2 2) j (x + 3 2)(x − 3 2) √ √ √ √ k (x + 3 5)(x − 3 5) l (x + 2 5)(x − 2 5) √ √ √ √ m (x + 4 2)(x − 4 2) n (x + 4 3)(x − 4 3) √ √ √ √ o (x + 5 2)(x − 5 2) p (x + 10 2)(x − 10 2) √ √ q (x + 2 + 6)(x + 2 − 6) √ √ r (x + 5 + 10)(x + 5 − 10) √ √ s (x − 3 + 11)(x − 3 − 11) √ √ t (x − 1 + 7)(x − 1 − 7) √ √ u (x − 6 + 15)(x − 6 − 15) √ √ v (x + 4 + 21)(x + 4 − 21) √ √ w (x + 1 + 19)(x + 1 − 19) √ √ x (x − 7 + 26)(x − 7 − 26) 7 a (x + 4)(x + a) c (x − 3)(x + a)

c2 = a2 + b2

1

2

p −4y(1 + 2y)

e (x + 5)(x − b) g (x − a)(x − 4)

b (x + 7)(x + b) d (x + 2)(x − a)

f (x + 3)(x − 4a) h (x − 2b)(x − 5)

i (x − 2a)(3x − 7)   √  √  √  √  8 a x+ 2 x− 2 b x+ 3 x− 3 3 3 2 2   √  √  √  √  7 7 5 c x+ x− d x+ x− 5 4 4 6 6 √ √ e (x − 2 + 2 5)(x − 2 − 2 5) √ √ f (x + 4 + 3 3)(x + 4 − 3 3) √ √ g (x + 1 + 5 3)(x + 1 − 5 3) √ √ h (x − 7 + 2 10)(x − 7 − 2 10) √ √ √ √ i ( 3x + 2)( 3x − 2) j ( 5x + 3)( 5x − 3)

780 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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9 a (x + 2)(y − 3) c (a + 5)(x − 2)

b (a − 4)(x + 3) d (y − 4)(x − 3)

e (a − 3)(2x − 1) f (2a − 5)(x + 4) √ √ 10 a 5(x + 2 6)(x − 2 6) √ √ b 3(x + 3 6)(x − 3 6) √ √ c 7(x + 3 2)(x − 3 2) √ √ d 2(x + 4 3)(x − 4 3) √ √ e 2(x + 3 + 5)(x + 3 − 5) √ √ f 3(x − 1 + 7)(x − 1 − 7) √ √ g 4(x − 4 + 2 3)(x − 4 − 2 3) √ √ h 5(x + 6 + 3 2)(x + 6 − 3 2) 11 a 60 b 35 c 69 d 104 e 64 f 40 g 153 12 a 4 − (x + 2)2 = (2 − (x + 2))

h 1260

(2 + (x + 2)) = −x(x + 4) b i −x(x + 6) ii −x(x + 8) iii x(10 − x) iv (3 − x)(7 + x)

v (8 − x)(6 + x) vi (6 − x)(14 + x) 13 a (x + a)2 = x2 + 2ax + a2 π x2 + a2 b If x = 0, then (x + a)2 = x2 + a2 . Or if a = 0, then (x + a)2 = x2 + a2 is true for all real values of x. 14 x2 − 4 = 1 (9x2 − 4) = 1 (3x + 2)(3x − 2) 9 9 9    or: x2 − 4 = x + 2 x − 2 9 3 3 = 1 (3x + 2) 1 (3x − 2) 3 3 = 1 (3x + 2)(3x − 2) 9 15 a −(2x + 5) b −11(2y − 3) c 16(a − 1) d 20b

e −12s f −28y g (5w + 7x)(−w − x) h (4d + 3e)(−2d + 7e)

i 6f(2f + 6j) 16 a x2 + 5y − y2 + 5x

j 0

Exercise 5C b 10, 2 c 5, −3 f −10, 3 g −2, −5 2 a Possible answer: x − 10 = 1 x − 10

1

a 9, 2 e −8, 3

b Possible answer: 3(x − 7) = 3 x−7

c Possible answer: −5(x + 3) = −5 x+3 d Possible answer: x + 4 = 1 3(x + 4) 3 3 a x 2

b x 3

c 3

d 1 5

e 1 3

f 1 4

g 5

h 2 3

i x+1

j x−2

k x−3 l 1 − 2x 2 b (x + 3)(x + 2)

4 a (x + 6)(x + 1) c (x + 3)2 e (x + 4)(x + 3)

d (x + 5)(x + 2) f (x + 9)(x + 2)

g (x − 1)(x + 6) i (x + 4)(x − 2)

h (x + 3)(x − 2) j (x − 1)(x + 4)

o (x − 4)(x − 3) q (x − 6)(x − 3)

p (x − 1)2 r (x − 2)(x − 9)

k (x + 10)(x − 3) m (x − 2)(x − 5)

s (x − 6)(x + 2) u (x − 7)(x + 2)

c 2(x + 9)(x + 2) e 4(x − 5)(x + 1)

d 5(x − 2)(x + 1) f 3(x − 5)(x + 2)

g −2(x + 4)(x + 3) i −2(x − 7)(x + 2)

h −3(x − 2)(x − 1) j −4(x − 2)(x + 1)

k −5(x + 3)(x + 1)

l −7(x − 6)(x − 1)

g 2(x + 11)2 j −3(x − 6)2

i 5(x − 5)2 l −4(x + 9)2 c x−3 d 1 x+7

6 a (x − 2)2 d (x − 7)2

7 a x+6

b (x + 3)2 e (x − 9)2

h 3(x − 4)2 k −2(x − 7)2

b x−3

1 x−5

f

8 a

5 x+6

b x−3 3

e

4 x+7

f

= (x − y)(x + y) + 5(x + y)

i x−7 5

ii (x + y)(x − y − 2)

t (x − 5)(x + 4) v (x − 4)(x + 3)

x (x − 5)(x + 2) b 3(x + 4)(x + 3)

e

iii (2x + 3y)(2x − 3y + 2) iv (5y + 2x)(5y − 2x + 3)

l (x + 11)(x − 2) n (x − 4)(x − 2)

w (x + 8)(x − 4) 5 a 2(x + 5)(x + 2)

= x2 − y2 + 5x + 5y

= (x + y)(x − y + 5) b i (x + y)(x − y + 7)

d 4, −3 h −12, −3

Answers

√ √ √ √ k ( 7x + 5)( 7x − 5) √ √ √ √ l ( 6x + 11)( 6x − 11) √ √ √ √ m ( 2x + 3)( 2x − 3) n ( 5x + 4)( 5x − 4) √ √ √ √ o ( 3x + 10)( 3x − 10) √ √ √ √ p ( 13x + 7)( 13x − 7)

1 x−6

6 x−2

g

2 x−8

c (x + 6)2 f (x − 10)2

h x+4 3

c 2(x − 1) d 4 x+5 x+5 g x+2 x−1

h x−4 x+6

781 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

√ 9 a x− 7 d√ 1 5x − 3 √ g x+1− 2 10 a 2(x + 3) 3(x − 5)

√ b x + 10 e √ 1 3x + 4 √ h x−3+ 5 b x−3 4

√ c x−2 3 √ √ f 7x − 5

√ i x−6− 6 c 3 x−3

d 3 e x−2 2 x+3 t2 − 5t − 24 t2 − 49 × 2 = 11 5t − 40 2t − 8t − 42

f x+3 x−1

t − 7 )(t + 7) × ( t − 8 )( t + 3) = t + 7 ( 5( t − 8) 2( t − 7 )( t + 3) 10 12 a x − 3 b x+1 c x−8 e 4 f x−7 d 6 x−2 x+5 5 a2 − ab a2 + 2ab + b2 × 2 13 a a2 + ab a − b2  2 (a  + b)  (a  − b) a = ×  (a + b)  (a  − b) a (a  + b)  =1 b Answer will vary. 14 a a − b b 1 a

3x − 8 (x + 3)(x − 4)

b

x − 12 (x + 4)(x − 4)

c

d

x − 14 (x − 3)(x + 2)(x − 6)

e

7x − 36 (x + 2)(x − 9)

3x − 23 (x + 3)(x − 3)(x − 5)

h

g (3x − 4)(2x + 1) i (2x − 1)(4x + 3)

k (5x + 6)(2x − 3) 3 a (3x + 1)(x + 3) c (3x + 2)(x + 2) e (2x − 1)(x − 5)

g (3x + 1)(x − 4) i (7x − 5)(x + 1)

k (3x − 4)(x + 2) m (2x + 1)(x − 5)

o (5x − 2)(x − 4) q (3x − 4)(2x + 3)

s (3x + 2)(2x + 3) u (4x − 5)(2x − 1)

d (3x − 2)(x − 1) f (5x − 3)(x + 1)

h (3x + 1)(x − 1) j (2x − 7)(x − 1)

l (2x − 3)(x + 4) n (13x + 6)(x − 1)

p (4x − 5)(2x − 1) r (5x − 2)(2x + 3)

t (4x − 1)(x − 1) v (2x − 5)(4x − 3)

c (7x − 2)(3x + 4) e (8x + 3)(5x − 2)

d (5x − 2)(6x + 5) f (7x + 2)(4x − 3)

g (6x − 5)(4x − 3) i (5x − 2)(5x − 8)

5 a 2(3x + 4)(x + 5) c 3(8x + 1)(2x − 1) e 8(2x − 1)(x − 1)

g −5(5x + 4)(2x + 3) i 5(4x − 1)(x − 1) 2 7x − 2

i 5x + 4 7x − 2

h (9x − 2)(5x − 4)

b 3(2x + 3)(x − 4) d 4(4x − 5)(2x − 3)

f 10(3x − 2)(3x + 5)

h 3(2x − 3)2

2 3x + 2

b 4x − 1

c 3x − 2

d

4 2x − 3

g x+4 3x + 1

h 3x − 1 2x + 3

k 2x + 3 7x + 1

l 2x − 3 4x − 5

f

j 3x − 2 5x − 2

7 a (3x − 4)(x − 5) b −10 m; the cable is 10 m below the water. c x = 4 or x = 5 3

Two numbers which multiply ax 2 + bx + c a × c to give a × c and add to give b 6x2 + 13x + 6 36 9 and 4 8x2 + 18x + 4 12x2 + x − 6

32 −72

15x2 − 13x + 2

30

10x2 − 11x − 6 21x2 − 20x + 4

l (2x − 1)(6x − 5) b (2x + 1)(x + 1)

x (3x − 2)(3x + 5) b (4x + 3)(5x + 6)

9 − 3x (x + 5)(x − 5)(x − 1) 4x + 11 (x − 1)2 (x + 4)

h (x − 4)(3x + 2) j (x + 4)(5x − 2)

w (3x − 2)(2x − 3) 4 a (6x + 5)(3x + 2)

e

Exercise 5D 1

d (x − 7)(x − 2) f (x − 5)(x + 3)

6 a 2x − 5

14x + 9 f (x + 3)(x + 4)(x − 8) g

b (x + 4)(x + 6)

c (x + 3)(x + 7) e (x − 3)(x − 4)

2 c (a + b)2 (a − b)

− b) d (a + b)(a a2 15 a

2 a (x + 2)(x + 5)

−60 84

16 and 2 −8 and 9

−15 and 4 −6 and −14 −3 and −10

8 a 3x + 4 x−3 e 125 9

b 3x + 2 4

c 1−x 3

f x+2 5

g 1

−12x2 − 5x + 3

d 4x − 3 5x + 1 2 h (4x − 5)2 (x − 3)

= −(12x2 + 5x − 3) = −(3x − 1)(4x + 3)

= (1 − 3x)(4x + 3) a (3 − 2x)(4x + 5)

c (4 − 3x)(4x + 1) e (2 − 7x)(2x − 5)

10 Answers will vary. 9x + 2 11 a (2x − 3)(4x + 1)

b (5 − 2x)(3x + 2)

d (3 − 4x)(2x − 3) f (3 − 5x)(3x + 2) b

5x + 15 (3x − 1)(2x + 5)

782 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

e f g

16x2 + 5x (2x − 5)(4x + 1)

d

7x − 12x2 (3x − 2)(4x − 1)

8x − 5 (2x + 1)(2x − 1)(3x − 2)

b

11 − 3x (3x + 5)(3x − 5)(3x − 2) 2 (2x − 5)(3x − 2)

h

c

12x + 3 (5x − 2)(2x − 3)(2x + 7)

a 9

b 36

e 16

f 25

c 1 g 25 4

f

d 4 h 9 4

g

i 81 4

h 2)2

2 a (x + d (x − 6)2 3 a b

c d e f

d e

Exercise 5E 1

7 a

4)2

b (x + e (x − 3)2 √ √ (x + 1 + 5)(x + 1 − 5) √ √ (x + 2 + 7)(x + 2 − 7) √ √ (x + 4 + 10)(x + 4 − 10) √ √ (x − 3 + 11)(x − 3 − 11) √ √ (x − 6 + 22)(x − 6 − 22) √ √ (x − 5 + 3)(x − 5 − 3)

5)2

c (x + f (x − 9)2

8 a b c d e f g h

4 a 9, (x + 3)2

b 36, (x + 6)2

i

2)2

d 16, (x + 4)2 f 1, (x − 1)2

9 a

c 4, (x + e 25, (x − 5)2

g 16, (x − 4)2 h 36, (x − 6)2  2  2 i 25, x + 5 j 81, x + 9 4 2 4 2  2  2 k 49, x + 7 l 121, x + 11 4 2 4 2  2  2 m 9, x − 3 n 49, x − 7 4 2 4 2  2  2 o 1, x − 1 p 81, x − 9 4 2 4 2 √ √ 5 a (x + 2 + 3)(x + 2 − 3) √ √ b (x + 3 + 7)(x + 3 − 7) √ √ c (x + 1 + 5)(x + 1 − 5) √ √ d (x + 5 + 29)(x + 5 − 29) √ √ e (x − 4 + 3)(x − 4 − 3) √ √ f (x − 6 + 26)(x − 6 − 26) √ √ g (x − 2 + 7)(x − 2 − 7) √ √ h (x − 4 + 21)(x − 4 − 21) 6 a not possible b not possible √ √ c (x + 4 + 15)(x + 4 − 15) √ √ d (x + 2 + 2)(x + 2 − 2) √ √ e (x + 5 + 22)(x + 5 − 22) √ √ g not possible f (x + 2 + 10)(x + 2 − 10) √ √ h (x − 3 + 3)(x − 3 − 3) √ √ j not possible i (x − 6 + 34)(x − 6 − 34) √ √ k (x − 4 + 17)(x − 4 − 17) l not possible

b c d e f g h i

 √  √  x+3+ 5 x+3− 5 2 2  √  √  41 41 7 + 7 − x+ x+ 2 2  √  √  x + 5 + 33 x + 5 − 33 2 2  √  √  93 93 9 + 9 − x+ x+ 2 2  √  √  x−3+ 7 x−3− 7 2 2  √  √  x − 5 + 23 x − 5 − 23 2 2  √  √  31 31 5 + 5 − x− x− 2 2    √ √  x − 9 + 91 x − 9 − 91 2 2 √ √ 2(x + 3 + 5)(x + 3 − 5) √ √ 3(x + 2 + 5)(x + 2 − 5) √ √ 4(x − 1 + 5)(x − 1 − 5) √ √ 3(x − 4 + 14)(x − 4 − 14) √ √ −2(x + 1 + 6)(x + 1 − 6) √ √ −3(x + 5 + 2 6)(x + 5 − 2 6) √ √ −4(x + 2 + 7)(x + 2 − 7) √ √ −2(x − 4 + 3 2)(x − 4 − 3 2) √ √ −3(x − 4 + 11)(x − 4 − 11)  √  √  x+3− 5 3 x+3+ 5 2 2  √  √  3 + 3 − 37 37 5 x+ x+ 2 2  √  √  2 x − 5 + 17 x − 5 − 17 2 2  √  √  4 x − 7 + 37 x − 7 − 37 2 2  √  √  7 + 7 57 −3 x + x + − 57 2 2  √  √  65 65 7 + 7 − −2 x + x+ 2 2  √  √  −4 x − 3 + 29 x − 3 − 29 2 2  √  √  17 17 3 + 3 − −3 x − x− 2 2  √  √  −2 x − 5 + 41 x − 5 − 41 2 2

Answers

c

10 a x2 − 2x − 24

= x2 − 2x + (−1)2 − (−1)2 − 24

= (x − 1)2 − 25

= (x − 1 + 5)(x − 1 − 5) = (x + 4)(x − 6) b Using a quadratic trinomial and finding two numbers that multiply to −24 and add to −2.

11 a If the difference of perfect squares is taken, it involves the square root of a negative number.

783 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

b i yes

ii yes

iii no

6 a x = − 3 , −4 2

iv no

v no c i m≤4

vi yes vii yes viii no ii m ≤ 9 iii m ≤ 25   12 a 2(x + 4) x − 3 2  √  √  x + 2 − 13 b 3 x + 2 + 13 3 3  √  √  x − 7 − 305 c 4 x − 7 + 305 8 8

b x = −1,−7 2 2

c x = 5, 7 2

d x = 1 , 11 2

e x = −5,3 3

f x = −3,2 5

g x = 4,−5 3 2 7 a x = −2, −6

h x = 3 , −4 7 b x = −1, 11 e x = 3 , −2 2 b x = 8, −4

d x=2 8 a x = 6, −4

d x = −2, −5 g x = 4, −4

d Unable to be factorised.  √  √  x + 3 − 41 e −2 x + 3 + 41 4 4  √  √  7 + 13 13 7 − f −3 x + x+ 6 6

j x = −5

m x = 3, −1 9 a x = 12, −7 d x = 5 , −4 2

g Unable to be factorised.  √  √  x − 3 − 41 h −2 x − 3 + 41 4 4   7 i 2(x − 1) x + 2  √  √  2 + 19 j 3 x+ x + 2 − 19 3 3   k −2 x + 5 (x − 1) 2   l −3 x + 4 (x + 1) 3

g x = −3, 1 10 a i x = 1, −2

a 0, −1

b 0, 5

e −5, 4 f 1, −1 √ √ i 2 2, −2 2

l −3,−3 8 4 2 a x2 + 2x − 3 = 0 c x2 − 5x + 6 = 0

i −x2 − 4x + 12 = 0 k 3x2 + 2x + 4 = 0 3 a 2 f 2

b 2 g 1

4 a x = 0, 4 d x = 0, 4 √ √ g x = 7, − 7 √ √ i x = 5, − 5 l x = 0, − 1 7 o x = 6, −6 5 a x = −2, −1 d x = 5, 2 g x = 5, −4 j x = −2 mx=7

when the product does not equal zero. Correct solution is: x2 − 2x − 8 = 7

c 0, 4 d 3, −2 √ √ √ √ 3, − 3 h 5, − 5 j 1,−7 k 5,−2 2 3 4 5

x2 − 2x − 15 = 0 (x − 5)(x + 3) = 0 x = 5 or x = −3

b x2 − 3x − 10 = 0 d 5x2 − 2x − 7 = 0

f 4x2 + 4x − 3 = 0 h 2x2 − 6x − 5 = 0 j x2 − 3x − 2 = 0 l x2 + 3x − 6 = 0 d 1 i 1

b x=1

c x = 1,5 2

d x = 8, −6

e x = −6, −2

g x = 8, −3 j x = 4, −3

h x = 5, −3 k x = 5, −2

f x = 3 , −4 2 i x=2 l x = −5, 3

Progress quiz e 2

b x = 0, 3 c x = 0, −2 e x = 0, 5 f x = 0, −2 √ √ h x = 11, − 11 j x = 0, 2 k x = 0, −5 m x = 2, −2

n x = 3, −3

b x = −3, −2 e x = −6, 2

c x = 2, 4 f x = −5, 3

h x = 8, −3 k x = −5 n x = 12

h x = 1, 1 i x = 3, −2 2 ii x = 1, −2

i.e. x = −8. 12 The student has applied the null factor law incorrectly; i.e.

g

c 1 h 1

l x = 8, −8 o x = −1,−3 4 2 c x = −9, 2 f x = 2, − 5 6

a, the coefficient of 3 makes no difference when solving. 11 This is a perfect square (x + 8)2 , which only has 1 solution;

13 a x = −2, −1

e 3x2 − 14x + 8 = 0 g x2 + x − 4 = 0

k x=8

n x = − 2 , −4 3 b x = −5, 14 e x = −4,2 5

f x = 3, −3 i x = 5, −1

b no difference c 3x2 − 15x − 18 = 3(x2 − 5x − 6) and, as seen in part

Exercise 5F 1

e x = 5, 3 h x = −1, −9

i x = 5,2 4 5 c x=3 f x = 2,5 3 2 c x=3

i x = 4, 8 l x=4 o x = −9

1

a −8x2 + 10x 3 c m2 + 7m + 10 9m2

−4 e g 5x2 − 35x + 60

b 4a2 − 7a d k2 − 6k + 9

f 8h2 − 6h − 35 h 19p + 4

2 a 4(a − 5) c (x + 5)(4 − x)

b −6m(2m − 3) d (a − 9)(a + 9)

k (x + 5)(x + a) 3 a (x − 4)(x + 5)

l (x − 2m)(4x − 5) b (a − 3)(a − 7)

e (4a − 11b)(4a + 11b) f 5(m − 5)(m + 5) g (k − 5)(k + 9) √ √ h (x − 3)(x + 1) i (x − 2 5)(x + 2 5) √ √ j (h + 3 − 7)(h + 3 + 7)

c 3(k − 9)(k + 2)

d (m − 6)2

784 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

5 a b d 6 a

c x = 4 or x = −4 e x = −3

g x = 4 or x = −8 7 a (3a + 2)(2a + 5) c (3x − 2)(5x − 4) 2x +5 8 2x − 3

9 a x = 5 or x = − 7 2 3

d x = 3 or x = 4 f x = −5 or x = 11

h x = 2 or x = 5 b (2m − 3)(4m + 3) d (2k − 7)(3k + 5)

b x = −5 or x = 7 2

c x = 4 or x = − 1 5 6

16 150 m × 200 m

Exercise 5H 1

a 1

b width = 9 m, length = 7 m c width = 14 mm, length = 11 mm 3 height = 8 cm, base = 6 cm 4 height = 2 m, base = 7 m

2 a b c d

i j

c 14

l n o 5 a

11 5 cm 12 a 55 b i 7 ii 13 iii 23 13 a 3.75 m b t = 1 second, 3 seconds c The ball will reach this height both on the way up and on the e t = 2 seconds

f The ball reaches a maximum height of 4 m. g No, 4 metres is the maximum height. When h = 5, there is no solution.

d e g h

9 1m 10 father 64, son 8

way down. d t = 0 seconds, 4 seconds

b c

f

5 8 and 9 or −9 and −8 6 12 and 14

b 9 c 100 d 625 e 4 g 25 h 9 4 4 √ √ (x + 3)(x − 3) = 0 √ √ (x + 7)(x − 7) = 0 √ √ (x + 10)(x − 10) = 0 √ √ (x + 1 + 5)(x + 1 − 5) = 0 √ √ (x + 3 + 11)(x + 3 − 11) = 0 √ √ (x − 1 + 2)(x − 1 − 2) = 0 √ √ √ √ b x = 7, − 7 x = 2, − 2 √ √ √ √ x = 10, − 10 d x = 3 − 5, 3 + 5 √ √ √ √ x = 4 − 6, 4 + 6 f x = −5 − 14, −5 + 14 √ √ x = −3 − 6, −3 + 6 √ √ x = −2 − 2, −2 + 2 √ √ x = −5 − 10, −5 + 10 √ √ x = −2 − 6, −2 + 6 √ √ x = −4 − 19, −4 + 19 √ √ x = −3 − 14, −3 + 14 √ √ x = 4 − 17, 4 + 17 √ √ x = 6 − 39, 6 + 39 √ √ x = 1 − 17, 1 + 17 √ √ √ √ x = 5 − 7, 5 + 7 k x = 3 − 5, 3 + 5 √ √ √ √ x = 4 − 7, 4 + 7 m x = −3 − 13, −3 + 13 √ √ x = −10 − 87, −10 + 87 √ √ x = 7 − 55, 7 + 55 √ √ x = −4 − 2 3, −4 + 2 3 √ √ x = −3 − 2 2, −3 + 2 2 √ √ x = 5 − 2 5, 5 + 2 5 √ √ x = 2 − 3 2, 2 + 3 2 √ √ x = 5 − 2 7, 5 + 2 7 √ √ x = −4 − 2 6, −4 + 2 6 √ √ x = 1 − 4 2, 1 + 4 2 √ √ x = −6 − 3 6, −6 + 3 6 √ √ x = −3 − 5 2, −3 + 5 2

f 25

e 4 a

e width = 3 m, length = 8 m 2 a width = 6 m, length = 10 m

b 13

c x = 2 m or 98 m 15 5 m × 45 m

3 a c

b x+5 c x(x + 5) = 24 d x2 + 5x − 24 = 0, x = −8, 3

7 15 m 8 a 6

b The ball starts at the tee (i.e. at ground level) and hits the ground again 100 metres from the tee.

e f

Exercise 5G 1

14 a x = 0, 100

Answers

b x+5 2 √ √ (x + 4 + 13)(x + 4 − 13) √ √ c not possible (x − 6 + 10)(x − 6 − 10)  √  √  5 − 3 5 + 3 3 3 x+ x+ 2 2 √ √ x = 0 or x = 5 b x = 10 or x = − 10

4 a x−3

b c d e f g h i

6 a 2 f 2

b 2 g 2

c 0 h 0

d 0 i 0

e 0 j 2

k 2

l 0 √ √ 17 , −5 − 17 −5 + 7 a x= 2 2 √ √ b x = −3 + 5 , −3 − 5 2 2 √ √ 29 −7 + −7 − 29 c x= , 2 2

785 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

√ √ d x = 3 + 17 , 3 − 17 2 2 √ √ 13 13 1 + 1 − e x= , 2 2 √ √ f x = −5 + 33 , −5 − 33 2 2 √ √ 41 41 7 + 7 − g x= , 2 2 √ √ 61 61 9 + 9 − h x= , 2 2 √ √ i x = −1 + 17 , −1 − 17 2 2 √ √ 5 −9 + 3 −9 − 3 5 j x= , 2 2 √ 3 √ √ √ 3 k x = + 3, − 3 l x = − 5 + 5, − 5 − 5 2 2 2 2 √ 8 a No real solution. b x = −5 ± 17 2 √ √ −9 ± 17 69 5 ± d x= c x= 2 2 √ √ e x = −5 ± 21 f x=3± 5 2 √ √ 9 a x = −3 ± 29 b x = −5 ± 61 2 2 √ c No real solutions. d x=4± 5 √ f No real solutions. e x = −5 ± 2 5 √ √ −3 + 89 10 width = cm, length = 3 + 89 cm 2 2 11 a i 1.5 km b i 0 km or 400 km √ c 200 ± 100 2 km 12 a x2 + 4x + 5 = 0

1

13 Factorise by quadratic trinomial; i.e. (x + 6)(x − 5) = 0, 6 × (−5) = −30, and 6 + (−5) = −1.

2 a −8 e 41 3 a 1 4 a 2

b −31 f −44 b 0 b 0

f 2 k 0

g 0 l√ 2 5 a x = −3 ± 17 2 √ 29 7 ± c x= 2 f x = −1, −7 √ h x = −5 ± 37 6 √ 65 5 ± j x= 4 √ l x = −3 ± 19 5 √ 6 a x = −2 ± 3 √ c x = −3 ± 11

7 8

√ b x = −4 ± 10 √ d x=1± 6 √ f x = −5 ± 2 6

15 a Use the dimensions of rectangle BCDE and ACDF and the corresponding side lengths in similar rectangles. √ b a=1+ 5 2 √ √ 16 a x = −1 ± 6 b x = −1 ± 5 2 √ √ c x = 4 ± 11 d x=3± 5 2 √ √ 17 −5 ± −1 ± 13 e x= f x= 2 2

a a = 3, b = 2, c = 1 b a = 2, b = 1, c = 4 c a = 5, b = 3, c = −2 d a = 4, b = −3, c = 2

e a = 2, b = −1, c = −5 f a = −3, b = 4, c = −5

ii 1.5 km ii 200 km

(x + 2)2 + 1 = 0, no real solutions  2 b x − 3 + 3 = 0, no real solutions 2 4

Therefore, x = −6, 5. √ 14 a x = 3 ± 7 √ c x = −2 ± 11 √ e x=4±2 3

Exercise 5I

9 10

√ e x=2±2 2 √ g x=1± 7 2 √ 4 ± 31 i x= 5 √ −5 + 105 2 √ a x=3±2 3 3 √ c x = −5 ± 57 8 √ 13 −2 ± e x= 3 √ g x = 1 ± 11 5 √ i x = 5 ± 19 6 √ √ 3 + 53 , −3 + 53 2 2 √ 6 2 + 10 units

c 49

d −23

c 2 c 1

d 2

e 2

h 0

i 2

j 1

√

b x = −7 ± 65 2 e x = −1, −4 √ g x = −7 ± 65 8 √ 22 2 ± i x= 3 d x=4

k x = −4,1 3 √ b x = 3 ± 5√ d x = −3 ± 3 5 2 √ 4 ± 10 f x= 3 √ h x=3±2 3 3

√ b x = −2 ± 10 2 √ d x = 5 ± 17 4 √ f x=1± 6 √ h x = 3 ± 41 4

11 63 cm 12 When b2 − 4ac = 0, the solution reduces to x = −b; i.e. a 2a single solution. 13 Answers will vary. 14 k = 6 or −6 15 a i k > 4 b i k>9 8

ii k = 4 ii k = 9 8

iii k < 4 iii k < 9 8

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ii ±2

iii k > 2, k < −2 d i no values ii no values

8 a x = 0, −4 iii All values of k.

Problems and challenges 1

b = −4, c = 1

2 47 3 a ±2, ±1 b ±3

4 a x = 0, 1 b x = 1, −2 5 144 cm2 6 25 km/h 7 1.6 8 x2 − 2x + 2 = (x − 1)2 + 1, as (x − 1)2 ≥ 0, (x − 1)2 + 1 > 0 (x − y)2 9 Square area − rectangle area = > 0 for all 4 x and y; hence, square area is greater than rectangle area.

b x = 0, 3

c x = 5, −5 e x=4 g x = −2, 1 2

d x = 3, 7 f x = −9, 4 h x = 2,−5 3 2

Answers

c i −2 < k < 2

i x = 1,−3 9 2 9 a x = 3, −3 c x = 4, −7

b x = 5, −1 d x = −3, 6

d 2 solutions √ 13 a x = −3 ± 33 2 √ 2 ± 14 c x= 2

√ b x=1± 5 √ d x = 1 ± 37 6

10 length = 8 m, width = 6 m √ √ 11 a x = −2 ± b x=3±2 √ 2 √ 7 17 5 3 ± −5 ± 3 c x= d x= 2 2 12 a 1 solution b 2 solutions c 0 solutions

10 w : p = 1 : 3; t : q = 1 : 9

Extended-response questions Multiple-choice questions 1 D

2 B

3 C

4 A

5 B

a i 15 + 2x m ii 12 + 2x m b area = 4x2 + 54x + 180 m2

6 D 11 A

7 C 12 B

8 C

9 E

10 C

c trench = 4x2 + 54x m2

1

d Minimum width is 1 m. 2 a S = 63p m2

Short-answer questions 1

a −2x + 26 c 25x2 − 4

b 3x2 + 11x − 20 d x2 − 12x + 36

e 7x + 22 2 a x2 + 4x + 4

f 12x2 − 23x + 10 b 4x2 + 18x

c x2 + 3x + 21 3 a (x + 7)(x − 7)

b (3x + 4)(3x − 4)

c (2x + 1)(2x − 1) d 3(x + 5)(x − 5) √ √ e 2(x + 3)(x − 3) f (x + 11)(x − 11) √ √ g −2(x + 2 5)(x − 2 5) h (x + 5)(x − 3) √ √ i (x − 3 + 10)(x − 3 − 10)

4 a (x − 6)(x − 2) c −3(x − 6)(x − 1)

b 0.46 m c i 420 = 3pr2 + 12pr

b (x + 12)(x − 2)

5 a (3x + 2)(x + 5) b (2x − 3)(2x + 5) c (6x + 1)(2x − 3) d (3x − 2)(4x − 5) 6 a 2x b x−4 x+3 4  √ 7 a (x + 4 + 6)(x + 4 − 6) √ √ b (x + 5 + 29)(x + 5 − 29) √ √ c (x − 3 + 2 3)(x − 3 − 2 3)  √  √  d x + 3 + 17 x + 3 − 17 2 2  √  √  e x + 5 + 13 x + 5 − 13 2 2  √  √  31 31 7 + 7 − f x+ x+ 2 2

iii r = 4.97 m; i.e.

pr2

ii 3pr2 + 12pr − 420 = 0

+ 4pr − 140 = 0.

Semester review 1 Linear relations Multiple-choice questions 1 A

2 C

3 D

4 E

5 C

Short-answer questions 1

a 3 − 2x

b 20

c 3a − 8 4a

9x − 2 (x + 2)(x − 3) 2 a i x = −4 ii x = 2 iii x = 13 b i x ≤ 6, x 3 4 5 6 7 ii x < 3, x 0 1 2 3 4 d

iv x = 2

787 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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ii m = − 4, c = 2 3

Answers

3 a i m = 3, c = −2

y

Extended-response question a x = 3, y = 4 b, c y

y (1, 1)

1 O

2

x

1

−2

8.5

O

x

3

(3, 4)

(3,−2)

−2

O

b

y

i

intersecting region y = 4x − 8

2

x

17 3

3x + 2y = 17

ii

y O

−8

x

3

3 −6

d 167 500 m2

O

x

5

Geometry Multiple-choice questions

y

iii

iv

1 D

y

O

x

3

O

x

1

1

b y = 8x − 9 5 5 5 a a = −3 b a = −4 c a = 1 or a = 7

AC is common. ∠ABC = 90◦ = ∠ADC (given)

d x = 3, y = −5

O

∴ △ABC ≡ △ADC (RHS). x = 3 (corresponding sides in congruent triangles)

O

3 2

− 4.5

x

O

A

D

x B C ∠DBC = ∠BDA (alternate angles in parallel lines) ∠BDC = ∠DBA (alternate angles in parallel lines) BD is common. ∴ △BAD ≡ △DCB (AAS).

y

c

2

3

a AB = DE (given)

a = 35 (corresponding angles in congruent triangles) b BC = DC (given)

7 A hot dog costs $3.50 and a soft drink $2. 8 a b y y

3

5 C

∴ △ABC ≡ △DEF (SAS).

b x = −2, y = −4

c x = −1, y = 4

4 E

AC = DF (given) ∠BAC = 60◦ = ∠EDF (given).

4 a y = −x + 3

d a = −4 6 a x = −3, y = −7

3 C

Short-answer questions

(1, −2)

−2

2 B

x

−3

Using congruence, BC = AD and AB = DC, corresponding sides in congruent triangles. 3 a x = 6.75 b x = 2 4 a x=8 b x=5

c a = 32, b = 65

d x = 40 5 a x = 20 6 a x = 47 5

f a = 90, b = 60, c = 70 c a = 63, b = 55 c x = 39 5

e a = 55 b x=8 b x = 29 3

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3 a 32.174 m b 52.2◦ 4 a x = 9.8 5 95.1◦

a CD = 6 cm, chord theorem 2 b OA = OD (radii of circle)

6 a tan q ◦ b i q = 155◦ ii √ q = 35◦ iii q = √ 42 2 3 1 c i ii iii − 2 2 3 7 a ≈ 0.34 b q ≈ 233◦ , 307 c yes

OB = OC (radii of circle) AB = DC (chord theorem 2) c

b q = 125.3

∴ OAB ≡ OCD (SSS). OM = 4 cm, area = 12 cm2

d 30.6% e ∠BOD = 106.2◦

Extended-response question 1

Indices and surds

a 104.3 m

b

8 km 20°

Multiple-choice questions 1 B

2 D

108°

3 E

4 E

5 C

13 km 52°

Short-answer questions 1

√ a 3 6

√ b 20 3 √ f 48 3

√ c 3 6 √ g 3

e 21 √ i 10 2 7 √ √ 2 a 7 5− 7 √ √ 3 a 2 15 − 4 3 √ d 59√+ 24 6 √ 4 a 3 2 b 2 2 5

√ d √10 h 5 3

1

√ c − 2−4 √ b 11 5 − 62 c 4 b 0

1

ii 7 2 x3 , when x > 0

iv b i 8 a 1 5

3 15 2

√

√ 5 ii 20 b 1 16

6

9 a x=3

Start

iii c 3

√ 4

Multiple-choice questions 1 C

1 3

2 a (2x − y)(2x + y) c 3(x − 4)(x + 4)

iii 4x 5

5 a 6 a

b i $86 400 ii $108 839 c 11.91 years d 6% per year

Trigonometry Multiple-choice questions 4 D

Short-answer questions 2 a i 150◦ T b i 310◦ T

b q = 43.8, y = 9.4 ii 330◦ T ii 130◦ T

√ √ b (x + 2 + 7)(x + 2 − 7) d (x − 2)(x + 7)

b x = −4, 1 2 √ √ x = 4, −4 e x = 7, − 7 x = 8, −3 h x = −2, 1 3 x = −8, 5 b x = 3, 7 √ √ i (x − 3 + 5)(x − 3 − 5)

4 a x = 0, 3

a V = 80 000(1.08)n

a x = 19.5

5 D

c (5x − 4)(2x − 3)

d −1 2

Extended-response question

1

4 B

e (x − 5)2 f 2(x − 6)(x − 2) 3 a (3x + 4)(x − 2) b (3x − 1)(2x + 3)

d

3 A

3 D

a 9x2 − 1 b 4x2 − 20x + 25 c −x2 + 30x − 5

g

2 B

2 B

Short-answer questions

√ 4 73 or ( 7)3 d 1 2

c x=3 2

b x=2

Quadratic equations

10 a $2382.03 b $7658.36

1 E

d 206◦ T

c 17.242 km

√ c 2 5−5 5 3b2 2x2 10 2 2 2 5 a 24x y b 3a b c d a5 5y3 6 a i 37 200 ii 0.0000049 b i 7.30 × 10−5 ii 4.73 × 109 7 a i 10 2

Answers

Extended-response question

5 C

c x = 0, −5 f x=2

c x = −4, 5

ii (x + 2)2 + 3, does not factorise further  √  √  iii x + 3 − 5 x + 3 + 5 2 2 2 2 √ ii no solutions b i x = 3 ± 5√ iii x = − 3 ± 5 2 √ √ 57 − 3 ± 7 a x= b x = 2 ± 10 4

Extended-response question 1

a 4x2 + 40x d x = 2.2

b 44 m2

c x=3

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Answers

Chapter 6 Exercise 6A 1

a 40 mm

b 9.6 cm

c 1m

d 0.3 km g 1.27 cm

e 8m h 510.2 cm

f 0.297 km i 3.2 m

2 a 1 4

b 1 2

c 3 4

e 7 24 3 a 810 m

f 11 12

d 1 9

b 9.4 km

c 180 cm

4 a 36.6 cm 5 a 21.8 m

b 5.1 cm b 3.2 m

6 a 43.98 cm d 3.46 km

b 7.54 m

c x = P − 28.6 c x = P − 16.8

c 89.22 mm

7 a 75.40 m

b r= C 2p

c 5.57 m

8 a i 8 + 4p m b i 4+pm c i 2 + p km ii 3.0 km 3 d i 12 + 10p cm e i 10 + 70p mm 9

ii 20.6 m ii 7.1 m

f i 6 + 31p cm 12 9 a 3 b 8.8

ii 14.1 cm

e 3.87 10 57.6 m

f 2.4

11 a 12.25 e 19.77

b 53.03 f 61.70

13 a i 201 cm

ii 34.4 mm

c 0.009

d 2.65

c 1.37

d 62.83

c p + 1 km

14 r = 2n p √ 15 p 2x 16 a l = P − 2w or 1 P − w b l=5−w 2 2 c 0 −4 ii c = −4 iii c < −4 15 a 1 + 4k b i k > −1 ii k = − 1 iii k < − 1 4 4 4 16 a Discriminant from resulting equation is less than 0.

  − 1 , −2 2

b k≥2 17 a m = 2 or m = −6

b The tangents are on different sides of the parabola, where one has a positive gradient and the other has a negative

7 a x = 0, y = 0 and x = 3, y = 9 b x = 0, y = 0 and x = −2, y = 4

gradient.

c x = −3, y = 9 and x = 6, y = 36

d x = 0, y = 5 and x = 3, y = 8 e x = −6, y = 34 and x = −2, y = 22 f x = −2, y = −3 and x = 3, y = 17 g no solutions

c m > 2 or m < −6

Exercise 7H y

1

h no solutions i x = − 9 , y = 65 and x = −1, y = 8 2 2 j x = − 5 , y = − 25 and x = 3, y = 1 3 3 k x = −3, y = 6 l x = −1, y = 2

8 a x = −4, y = 16 and x = 2, y = 4 b x = −1, y = 1 and x = 2, y = 4 c x = −1, y = 1 and x = 1 , y = 1 3 9

2

−2

O

2

x

−2

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Answers

√ 2 a x=± 5 √ d y = ± 11 3 a (0, 0)

√ c x=± 3

b x = ±4 √ e y = ± 57 b r

y

12

f y = ±2

√10

√ c y=± 5

b r=3 √ √ 27 ± 3 3 d x=± = 2 2 e y

4 a (0, 0)

(1, 3)

O

−√10

x

√10

3

−3

O

(−1, −3) −√10

x

3

y

13

√6

−3 √ c y = ± 19 2

b r=5

5 a (0, 0) d x = ±3

−

2√6 √30 √5 5

O

−√6

y

e

5

O

5

y

x √5 (2, 1)

−5 √

6 a r=6 b r=9 c r = 12 √ √ √ e r = 14 f r = 20 = 2 5 7 a x2 + y2 = 4 c x2 + y2 = 10 000

d r= 5

b x2 + y2 = 49 d x2 + y2 = 2601

f x2 + y2 = 10 e x2 + y2 = 6 2 2 g x + y = 1.21 h x2 + y2 = 0.25 √ √ √ √ b (−1, 3), (−1, − 3) 8 a (1, 3), (1, − 3)  √   √  c 1 , 15 , 1 , − 15 2 2 2 2 √   √  15 1 d ,− , − 15 , − 1 e (0, −2) 2 2 2 2 f (2, 0), (−2, 0) 9 a x-intercepts: ±1, y-intercepts: ±1 b x-intercepts: ±4, y-intercepts: ±4 √ √ c x-intercepts: ± 3, y-intercepts: ± 3 √ √ d x-intercepts: ± 11, y-intercepts: ± 11 √ b r=2 c r=3 10 a r = 2 2 √ √ √ d r = 10 e r=2 3 f r=2 5

−√5 (−1, −2)

O

x

√5

−√5

√ Chord length = 3 2 units √ 15 a m = ± 3 √ √ b m > 3 or m < − 3 √ √ c − 3 − 3 or x < 1 3 2 4 3 √ √ 41 41 7 − 7 + c 0

b b2 − 4ac = 0

4 (x − 2)2 + (y + 3)2 = −15 + 9 + 4 = −2, which is impossible. 5 a k=1 b k1 3 3 3 √ √ 6 a k = ± 20 = ±2 5 √ √ b k > 2 5 or k < −2 5 √ √ c −2 5 < k < 2 5

7 a y = −(x + 1)(x − 3) b y = 3 (x + 2)2 − 3 4 c y = x2 − 2x − 3

  8 a = 2, b = −3, c = −8; TP 3 , − 73 4 8

9 20 √ 10 4 3 11

y

y value on the circle is 0, which is less than 1. c Exponential graph rises more quickly than the straight line

and this line sits below the curve. d Solving 2 − 1 = 1 gives a quadaratic with  < 0, x+3 3x thus no points of intersection. 12 a i (x + 2)2 + (y − 1)2 = 4, C(−2, 1), r = 2 ii (x + 4)2 + (y + 5)2 = 36, C(−4, −5), r = 6

iii (x − 3)2 + (y − 2)2 = 16, C(3, 2), r = 4 √ iv (x − 1)2 + (y + 3)2 = 15, C(1, −3), r = 15 √ v (x + 5)2 + (y + 4)2 = 24, C(−5, −4), r = 2 6 √ vi (x + 3)2 + (y + 3)2 = 18, C(−3, −3), r = 3 2 √  2   29 vii x + 3 + (y − 3)2 = 29 , C − 3 , 3 , r = 2 4 2 2 2    viii x + 5 + (y − 2)2 = 49 , C − 5 , 2 , r = 7 2 4 2 2   2  2   1 3 3 1 3 ix x − + y+ = ,C ,− ,r = 3 2 2 2 2 2 2  2  2   5 x x − 3 + y − 5 = 25 , C 3 , 5 , r = √ 2 2 2 2 2 2 b (x + 2)2 + (y − 3)2 = −2; radius can’t be negative.

(0, 8) y5 = 8x − 8

O

x

(1, 0)

y5 = 8 − 8x (0, −8)

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iv

1 B 6 A

2 D 7 C

3 E 8 D

4 D 9 A

5 A 10 E

11 C

12 D

13 B

14 A

15 C

y 10

Short-answer questions 1

a minimum at (1, −4) c −1 and 3

y

(−3, −8) 5 a

−2 O

y

1 O

x

2

x

2 − √3

−4

2 + √3

(2, −3) y

b

y

b

x

−1 O

−5

b x=1 d −3

2 a minimum at (2, 0) b maximum at (0, 5) c maximum at (−1, −2) d minimum at (3, 4) 3 a

Answers

Multiple-choice questions

16

O

−4

−3 − √17 2

x

x −3 + √17 2

− 3 , − 17 2 4

6 a 1 7 a i 5

y

c

O 2

b 0 ii (2, −3)

iv

c 2

d 0 iii 0.8 and 3.2

y

5 −2 O

4

x 0.8 O

−8

(1, −9)

4 a i maximum at (1, −3) iii no x-intercepts

iv

y

O

iii −1 and 4

3 , 25 2 4

4

y

iv

x

(2, −3)   ii 3 , 25 2 4

b i 4 ii −4

3.2

x

−1

(1, −3)

O

4

x

−4 8 a 100 − x b A = x(100 − x) b i minimum at (−3, −8)

c 0 < x < 100

ii 10

iii −1 and −5

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Answers

d

y

b

A

(50, 2500) x

O −1 O

100

e 2500 m2 f 50 m by 50 m 9 a x = 2, y = 10 and x = −6, y = 10

(1, −3)

x

y

c

b no solutions c x = 1 , y = 10 and x = −1, y = 2 3 9 10 Show b2 − 4ac = 0.

O

5

O

x

y

14 a

−5

(1, 15 )

1

y

11 a

x

5

(1, 2) −5

(−1, −2)

√7 −√7

x

O

y

b

O

x

√7

y

b

−√7

(−1, 3) y

12

(

3

O

−3 − 3 ,− 6 √5 √5

3 6 , √5 √5

)

(1, −3)

x

3

x

O

  √ √ √ √ 4,3 b ( 2, 2 2) and (− 2, −2 2) 3 16 a y 15 a

−3

2 + √3

y

13 a

(1, 4)

(−1, 2) −1 O 2 − √3

1 O

x

x

830 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

y

b

Chapter 8

3.5

y=3

Exercise 8A x

O c

y

−

5 3

x

O

−2

−2.5

y = −3

−3 x = −2

Extended-response questions 1

b 0 ≤ x ≤ 400

a (200, 30) c 30 ≤ h ≤ 80 d h

(200, 30) O

400

e 400 m

x

c yes

2 a 1 4

b 1 6

c 1 4

d 3 8

e 2 3

f 0

f 1

3 a 1 7

b 2 7

c 5 7

d 3 7

4 a 3 10

b 2 5

c 3 5

d 1 2

5 a 1 10

b 1 2

c 1 2

d 1 2

g 3 10 c 0.47 c 0.97

h 1 10 d 0.91 d 0.38

e 2 5 6 a 0.09 7 a 0.62 8 a 1 50

f 1 5 b 0.43 b 0.03 b 3 10

9 a 1 2

d 2 5 11 a i

f 30 m

7 10

ii 1 5

c 49 50

b 3 8

c 1 4

e 1

f 0

b 1 50

c 21 25

e 2 25

f

iii 1 20

iv 0

4 25

v 1 20

b 1 10 12 a 59 b 4, as 41 of 10 is closest to 4. 100

b i 150 g ii 5.6 g c after 2 years

c 8, as 41 of 20 is closest to 8. 100

A 450

O

b {H, T} e 1 2

10 a 6 25

g 80 m 2 a 450 g

d

a 2 d 1 2

d 5 24

(400, 80)

80

1

13 a 1 4

t

e 9.8 years

e 2 13

b 1 13

c 1 52

d 1 2

4 13

g 12 13

h 9 13

f

14 a 7 15 b 15; any multiple of 15 is a possibility as 3 and 5 must be factors. 15 a 625p b i 25p c i 1 25

ii 200p ii 8 25

v 24 vi 17 25 25 d No it doesn’t.

iii 400p iii 16 25 vii 1

iv 9 25 viii 17 25

831 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

Exercise 8B 1

a

A

b

B

A

B

b i 25 c i 2 5 7 a

ii 5 ii 1 5

F

iii 1 5

B

25 10 5 c

A

e

A

g

A

f

B

A

5

d

B

A

h

B

A

B

b i 25 ii 5 c i 7 ii 2 iii 8 iv 2 9 9 9 9 8 a A A

B

B

B

2

6

8

B

5 7

3 9

8 16

b i 2 v 7 c i 1 8 9 a

ii 6 vi 8 ii 9 16

A

A

4

2 a ∅ d ∅

3

b A∩B e E∩F

g A∪B∪C h A∩B∩C 3 a no b yes c no

A 4

B

c A∪B f W∪Z

B 3

B

3

6

1 5

3 5

4 10 iii 2 5 c a, c, e

b 10, 12

C

iv 7 10 d nothing

D

3 3

b i

9 13

5

25 10 10

B B 13 3 14 a 1 − a

iii 10 13

iv 4 13

A

A

3 4

3 1

6 5

7

4

11

A

A

2 2

7 1

9 3

4

8

12

b

b i A ∩ B = {2, 13} ii A ∪ B = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} c i 1 ii 7 iii 1 iv 1 2 10 5 6 a F N

1

ii 6 13

12 a B B

0

6

3

B

2

A 2

ii 3

10 a 4 11 a

b i A ∩ B = {2, 5, 8} ii A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} c i 7 ii 3 iii 1 10 10 d No, since A ∩ B π 0.

A

A 4

c i 2

0

5 a

iv 3 viii 16

2

B b

4 a

iii 5 vii 13 iii 5 16

B

1

i

v 1 9

b a+b

v 3 13

c 0

5

832 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

B

A

A ∩ B′ A ∩ B A′ ∩ B

A′ ∩ B′ 16 a

(A ∪ B)′

M

v

c n s

m b i 1 3 17 a

ii 2 3

iii 1 6

iv 2 3

v 1 3

L 5

S

1

2 2

2

b 5 24 b 0.2

8 a 0.3 9 a 3 8

b 0.1 b 5 32

10 a 4 13

b 4 13

14 Pr(A ∪ B ∪ C) = Pr(A) + Pr(B) + Pr(C)− Pr(A ∩ B) − Pr(A ∩ C) − Pr(B ∩ C) + Pr(A ∩ B ∩ C) 15 a 3 b 1 c 3 10 4 20 d 13 20

1

16 a 1 4

1

d

b 1 c i 3 5 18 a

ii 1 3

iii 13 15

iv 1 15

L L = Own State S = Interstate I = Overseas

18 - 2x S

2x

2 2 x y

7

I

3 b i 4 c i 5 19

d 7 13

13 a Pr(A) < Pr(A ∩ B) b Pr(A) + Pr(B) < Pr(A ∪ B)

I

1

c 7 13

e 49 f 10 g 10 h 25 52 13 13 26 11 a 0.4 b 0.45 12 Because Pr(A ∩ B) = 0 for mutually exclusive events.

E

w

6 a 1 8 7 a 0.1

ii 10

7 100

e 9 20

f 3 5

b 71 500

c 33 500

e 1 25

f

7 500

Exercise 8D 1

a i 2 b 2 9 2 a i 7 b 7 10

ii 9

ii 10 c 7 12

3 a 1 3

b 1 2 9 13

ii 3 13

iii 3 7

iv 1 3

Exercise 8C

b i 14 17

ii 4 17

iii 4 7

iv 2 7

1

c i 3 4

ii 5 8

iii 5 7

iv 5 6

ii 1 19

iii 7 38

iv 35 38

v 25 38

a i {4, 5, 6} ii {2, 4, 6} iii {2, 4, 5, 6} iv {4, 6} b No, A ∩ B π 0. c 2 3 2 a 0.8 b 0.8 c 0.7 d 1 3 0.05 4 a i 13 b i 1 4

ii 4 ii 3 4

d i

7 16

ii 1 8

iii 1 4

iv 2 7

5 a i

7 18

ii 1 9

iii 1 5

iv 2 7

ii 1 9

iii 1 5

iv 1 4

ii 7 17

iii 7 10

iv 7 8

ii 1 4

iii 2 3

iv 1 3

c i

c 4 13 d 3 52 5 a i {3, 6, 9, 12, 15, 18 } b i 1 ii 13 20 20

4 a i

b i 4 9

iii 1 iii 1 52

8 17

d i 3 4 6 a ii {2, 3, 5, 7, 11, 13, 17, 19 }

c 7 20

Answers

15

B B b 1 5

c 3 5

A 9

A 6

15

4 13

1 7

5 20

d 9 13

833 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

7 a

V 3

4 a

P 2

Like tennis Dislike tennis Total

6 4

c 2 5

b 4 8 a

3 29

17 1

20 30

Total

32

18

50

d 1 4

A 14

Like squash Dislike squash

tennis

squash

C 6

29

9

3

17

1

B B

b i

A

A

6 14

9 1

15 15

20

10

30

3 10

1 b 29 c 1 50

ii 7 15

c 2 5 d 3 10 9 a B B

A 2

A 2

4

3 5

1 3

4 8

ii 2 5

i 1 b B B

11 12 13 14 15

A 13

16

5 8

6 19

11 27

3 a 1 13

c 1 4

b 1 10 b 0.29

1

c 1 52

d 1 13

d 1 3

Like soft drink A Dislike soft drink A Total 15 5 20 20 35

0 5

20 40

a i

2nd

D

1st O

G

D O G

(D, D) (D, O) (D, G)

(O, D) (O, O) (O, G)

(G, D) (G, O) (G, G)

D O G

D X (D, O) (D, G)

1st O (O, D) X (O, G)

G (G, D) (G, O) X

ii

2nd

f 18 31

c 1 20 c 0.33

c 1 4

Exercise 8E

d 1 2

e 31 231

b 1 26

b 1 6

d 7 33

b 1 8

Progress quiz a 1 10 2 a 0.17

7 a 1 2 8 a

c 5

Dislike water B Total

a 1 b 1 3 2 Pr(A|B) = Pr(B|A) = 0 as Pr(A ∩ B) = 0 a 1 b 1 5 a Pr(A ∩ B) = Pr(A) × Pr(B|A) b 0.18 174 a 329 b c 81 329 329

1

f 28 33 b 0.17

iii 3 16

b 1 13

d 24 155

e 13 33 6 a 0.83

iii 1 2

ii 3 8

10 a 1 13

b 20

Like water B

A 3

i 6

5 a 7

d 2 5 d 0.67 e 1 13

e 3 5

b i 9 c i 1 3

ii 6 ii 5 9

d i 0

ii 2 3 b 6

2 a 9

iii 4 9

iv 8 9

v 2 9

iii 1 3

iv 1

v 1 3

f 12 13

834 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

3 a 1 (1, 1) (1, 2) (1, 3) (1, 4)

1 2 3 4

2nd roll

4 (4, 1) (4, 2) (4, 3) (4, 4)

Die 2

b 16 c 1 16 d i 1 4 4 a

ii 5 8

b 36 c i 2 d i 1 6

iii 13 16 1st toss H (H, H) (H, T)

H T

2nd toss

ii 3 4

2nd

5 a

S E T

2nd b i 1 6 v 1 6 a

2nd

S

1st E

T

X (S, E) (S, T)

(E, S) X (E, T)

(T, S) (T, E) X

ii 2 3

L E V E L

L X (L, E) (L, V) (L, E) (L, L)

iii 2 3

E (E, L) X (E, V) (E, E) (E, L)

b 20 c i 8 d i 2 5

ii 12 ii 3 5

iii 12 iii 3 5

e 1 5

b 21 9 a i 100 b i 1 10

iv 1 3

1st V (V, L) (V, E) X (V, E) (V, L)

1 2 3 4 5 6

2 3 4 5 6 7 8

3 4 5 6 7 8 9

ii 6 ii 1 6

Die 1 4 5 6 7 8 9 10

5 6 7 8 9 10 11

iii 15 iii 35 36

6 7 8 9 10 11 12

iv 1 12

e 1. Her guess is wrong. 6 8 a

T (T, H) (T, T)

b 4 c 1 4 d i 1 2 e 250

1 2 3 4 5 6 7

Answers

7 a 1st roll 2 3 (2, 1) (3, 1) (2, 2) (3, 2) (2, 3) (3, 3) (2, 4) (3, 4)

E (E, L) (E, E) (E, V) X (E, L)

1st L (L,C) (L, O) (L, L) (L, L) (L, E) (L, G) (L, E)

O (O, C) (O, O) (O, L) (O, L) (O, E) (O, G) (O, E)

C O L L E G E c 1 7 ii 90 ii 1 10

D (D, C) (D,O) (D, L) (D, L) (D, E) (D, G) (D, E)

iii 4 5

c 19 100

L (L, L) (L, E) (L, V) (L, E) X

10 a i 1 4 b i 2 5 11 a without 12 a 30 b i 1 15

ii 5 8 ii 1 10 b with

iii 2 3 c with

ii 1 15

d without

iii 2 15

iv 4 15

c 1 18 13 a 1st

2nd

b 16 c i 1 d i 1 16

2.5 5 10 20

ii 8 ii 1 8

2.5

5

10

20

5 7.5 12.5 22.5

7.5 10 15 25

12.5 15 20 30

22.5 25 30 40

iii 8 iii 1 4

iv 3 16

e 7 16

835 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

Exercise 8F

4 a

a i 2 5

ii 3 5

b i 2 5

ii 3 5

c i 1 4 2 a

ii 3 4

1

3 8

5 8

3 8

3 5 15 × = 8 8 64

3 8

M F, M

5 × 3 = 15 8 8 64

5 8

F

F

2 7

5 8

3 5 × 8 7

= 15 56

3 7

M

F, M

5 3 × 8 7

= 15 56

F, F

5 4 × 8 7

5 = 14

F F

1 4

1 4

1 4 1 4

1 4

4

1 4

1 4 1 4

1 4

1 4

Outcome 1

(1, 1)

2 3

(1, 2) (1, 3)

4 1

(1, 4) (2, 1)

2 3

(2, 2) (2, 3)

4 1

(2, 4) (3, 1)

2 3

(3, 2) (3, 3)

4 1

(3, 4) (4, 1)

2 3

(4, 2) (4, 3)

4

(4, 4)

b 16

Outcome

Probability

yellow

(A, yellow)

1 1 1 × = 2 4 8

c i

1 16

ii 1 4

d i

1 16

ii 1 4

3 4

orange

(A, orange)

1 3 3 × = 2 4 8

3 4

yellow

(B, yellow)

1 3 3 × = 2 4 8

(B, orange)

1 1 1 × = 2 4 8

iii 5 8

5 a

Outcome Probabilities

A

orange

3

1 4

Counter

B

2

1 4

1 4 1 4

1 4

3 = 28

M, F

1 4

d 3 8

3 2 × 8 7

F

1 4

1 2

M, M

5 7

b 3 4 Box

1 2

M

1 4

5 × 5 = 25 8 8 64

F, F

M

4 7

3 a 1 4 c

F

1 4 1 4

1 4

1 4

3 ×3= 9 8 8 64

M M, M M, F

5 8

2nd toss

1

M

b 3 8

1st toss

2 3

1 3

c i 2 9 6 a

4 7

i 1 7 b i

9 49

(R, R)

2 3 2 × = 3 5 5

2 5

W

(R, W)

2 2 4 × = 3 5 15

4 5

R

(W, R)

1 4 4 × = 3 5 15

1 5

W

(W, W)

1 1 1 × = 3 5 15

W

4 15

3 7

R

R

e 1 2 b i

3 5

ii 2 5

iii 8 15

ii 4 9

iii 4 9

Outcome Probabilities 1 3

M

(M, M)

3 1 1 × = 7 3 7

2 3

F

(M, F)

3 2 2 × = 7 3 7

1 2

M

(F, M)

4 1 2 × = 7 2 7

1 2

F

(F, F)

4 1 2 × = 7 2 7

ii 2 7

iii 4 7

iv 3 7

ii 16 49

iii 24 49

iv 25 49

M

F

836 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Outcome

Probability

Outcome

12 a

1 3 3 × = 2 4 8

3 4

white (Falcon, white)

1 4

silver

2 3

white (Commodore,white)

1 2 1 × = 2 3 3

red

1 1 1 × = 2 3 6

1 4

Falcon

1 2

1 2

1 1 1 × = 2 4 8

(Falcon, silver)

Commodore 1 3

b i 3 8 iv 7 24 8 a

1 2

1 2

W

(R, W)

1 2

R

(W, R)

1 2

W

(W, W)

1 45 c 62.2% 10 a i 0.17

1

B

(R, R, B)

R

(R, B, R)

B

(R, B, B)

R

(B, R, R)

2 3 2 3

B

(B, R, B)

R

(B, B, R)

1 3

B

(B, B, B)

1 3

B

1 2

R

B B

1 2

1 10

ii 3 10

1 3

2 3

iv 9 10

iii 0

1 3

R

2 3

W

iv 3 4 Probability

(R, R)

1 1 1 × = 2 3 6

R (R, W)

1 2 1 × = 2 3 3

R

(W, R)

1 2 1 × = 2 3 3

1 3

W

(W, W)

1 1 1 × = 2 3 6

W

iii 5 6

iv 5 6

Outcome

Probability

1 9

U

(U, U)

1 1 1 × = 5 9 45

8 9

N

(U, N)

1 8 8 × = 5 9 45

2 9

U

(N, U)

4 2 8 × = 5 9 45

7 9

N

(N, N)

4 7 28 × = 5 9 45

U

N

ii 16 45

iii 44 45

ii 0.11

iii 0.83

b i 0.1445 ii 0.0965 3 4 11 a b 7 7

1 3

P

(A, P)

1 1 1 × = 3 4 12

3 4

G

(A, G)

1 3 1 × = 3 4 4

1 2

P

(B, P)

1 1 1 × = 3 2 6

1 2

G

(B, G)

1 1 1 × = 3 2 6

3 4

P

(C, P)

1 3 1 × = 3 4 4

1 4

G

(C, G)

1 1 1 × = 3 4 12

B

1 3

C b 6 c i

2 3

1 10 1 5 1 5 1 10

Outcome Probability

1 3

iii 3 4 Outcome

0 1 10 1 10 1 5

v 9 10

1 4

A

ii 1 2

Probability

ii 2 5

13 a

W

ii 4 5

(R, R, R)

R

3 4

3 5

i

R

R

2 5

b i 1

1 2

9 a i 1 5 b

i

vi 1 3

(R, R)

ii 2 3

4 5

v 5 6 Outcome

R

i 1 6

1 5

iii 17 24

1 2

b

1 2

ii 1 6

R

i 1 4

1 2

(Commodore, red)

0

Answers

7 a

1 12

ii 1 6

iii 1 4

d 1 2 14 a i 7 ii 1 8 8 b $87.50 to player A, $12.50 to player B c i A $68.75, B $31.25 iii A $81.25, B $18.75

ii A $50, B $50 iv A $34.38, B $65.62

Exercise 8G 1

a i 1 2 b yes c 1 2

2 a i

ii 1 2

3 10

ii 1 3

b no c no 3 a with 4 a

b without

A

B

iii 0.8555

1

2

2 3

837 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

b i 3 ii 1 8 2 c not independent 5 a

A 2

5 63 64 6

B 2

7 3 5

1

1 12 9 true 10 8 9 8

1 b i 2 ii 2 3 3 c independent 6 a i 3,1 ii not independent 4 2 b i 1,1 4 4

Multiple-choice questions

ii independent

c i 1,1 3 3

ii independent

d i 2,0 7

ii not independent

b Pr(A) = 3 , Pr(A | B) = 1, not independent 10 4 c Pr(A) = 5 , Pr(A | B) = 3 , not independent 12 20 d Pr(A) = 1 , Pr(A | B) = 1, independent 9 9 8 a T G

8

i 15 17 b no 9 a 1 32 10 a

1 216

4 E

5 A

8 C

9 A

10 E

a 1 8

b 1 4

c 3 8

2 a 5 8

b 1 2

c 5 8

3 a i 2 5

ii 1 4

1

b i 3 5 4 a

iii 1 5

d 5 8

iv 1 10

e 1 2

v 1 20

ii 17 20

C

H 6

5 13

b

8 7

2 0

10 7

15

2

17

ii 7 17 b 31 32 1 216

H H

iii 4 5

c 1 72

d i 1 6 5 a 6 d 1 36

d 2

b 0.76

6 a i 13 b i 3 4

Problems and challenges b 0.192

c 0.144

b 1

c 4 7

4 a 1 12

b 1 2

c 3 4

C

6 12

5 13

11 25

18

18

36 iii 1 2

b 6 13 ii 4 ii 1 52

iii 1

b 0.5 b 1 5

4 ii 5 11 11 b No, Pr(A | B) π Pr(A). c i 1 ii 1 2 4 d Yes, Pr(A|B) = Pr(A).

9 a i

d 2 3

ii 5 36

c 4 13 d 10 13 7 a 0.1 8 a 2 5

1 a 0.16 2 0.593 75 3 a 7 8

C

c 13

c 31 32

11 False; Pr(A | B) = 0 but Pr(A) = 2. 9 12 a 6 b 22 c 49 13 a 0.24 14 5 6

3 C

7 D

12

T

b

2 B

6 B

2

T G G

1 E

Short-answer questions

7 a Pr(A) = 1 , Pr(A | B) = 1, independent 2 2

7

1 13 983 816

iii 1 5 iii 1 2

838 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

H (H, H) (H, E) (H, Y)

H E Y

2nd

1st P (P, H) (P, E) (P, Y)

A (A, H) (A, E) (A, Y)

P (P, H) (P, E) (P, Y)

Y (Y, H) (Y, E) (Y, Y)

d i 1 2

1 15

ii 2 15

11 a

2nd 1 4 1 4

1

1 4

1 4 1 4

1 4 1 4

2

1 4

1 4

3

1 4

1 4

ii 1 4

12

1 4

1 4 1 4

4

1 16

1 4

1 4 1 4

1 4

b i

1 4

1 4

Total 1

2

2 3

3 4

4 1

5 3

2 3

4 5

4 1

6 4

2 3

1 4

F

3 4

M

1 2

F

1 2

M

4 1

7 5

2 3

6 7

4

8

3 5

b 3 c 3 4 10 b 0.58

2 7 2 7

S S

(R, S)

W

(R, W)

1 3

R

(S, R)

S

(S, S)

W R

(S, W) (W, R)

S

(W, S)

1 6

1 2 1 3

1 3

W 1 3

i 1 21

(R, R)

S

S

3 7

R

W (W, W)

ii 10 21

iii 6 7

iv 16 21

Chapter 9 iv 1

Exercise 9A b D b E

e 7 10

c A c C

d B d D

e E e F

b categorical d numerical

5 a numerical and discrete b numerical and discrete c categorical and nominal d numerical and continuous e categorical and ordinal 6 a D b D is the most representative sample. A may pick out the keen students; B probably are good maths students who

ii 1 15

c

iv 4 9

1 2

c categorical 4 D d 3 5

1 3

R

f A 3 a numerical

a 3 7 15

iii 5 9

W (W, R) (W, S) (W, W)

Outcome

1 a C 2 a B

Extended-response questions

b i

ii 1 3

1 6

M

a 2 5 13 a 0.12

1st S (S, R) (S, S) (S, W)

R (R, R) (R, S) (R, W)

c 4 d 5 9 e

F

2 5

1

b i 1 9

5 6

iii 0

R S W

2nd

iii 13 15

1st

ii 3 4

2 a

b 15 c i

Answers

10 a

like maths; and C will have different-sized classes. 7 a For example, likely to be train passengers.

R

R

3 3

1 8

4 11

6

9

15

b For example, email will pick up computer users only. c For example, electoral roll will list only people aged 18 years and over.

839 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

8 Check with your teacher.

4 a Type of transport Car

9 a small survey, misinterpreted their data b Survey more companies and make it Australia-wide. c No, data suggest that profits had reduced, not necessarily that they were not making a profit. Also, sample size is too small. 10 a graph A

b graph B

Percentage frequency 40

Train Tram

6 8

15 20

Walking Bicycle

5 2

12.5 5

3 40

7.5 100

Bus Total

c The scale on graph A starts at 23, whereas on graph B it starts at 5. d Graph A because the scale expands the difference in column heights.

Frequency 16

b i 6

ii car

11 For example, showing only part of the scale, using different column widths, including erroneous data values.

5 a symmetrical c positively skewed

12–14 Research required.

6 a i 34.3 b i 19.4

Exercise 9B 1

a 10

c 1

d 1

Frequency

Percentage frequency

0– 10–

2 1

20 10

20– 30–40

5 2

50 20

10

100

Total b Class interval

Frequency

Percentage frequency

80– 85–

8 23

16 46

90– 95–100

13 6

26 12

Total

50

100

Frequency 5

Percentage frequency 25

9 6

45 30

20

100

3 a Class interval 0– 5– 10–15 Total

Frequency

50

8

40

6

30

4

20

2

10

0 1

5

10 Wins

Percentage frequency

Histogram of wins 10

Stem

iii 39 iii no mode

15

0 1 2 3 Nick’s goal scoring b

0

1 2 3 4 5 Jayden's goal scoring c Well spread performance. d Irregular performance, positively skewed. 8 a Sensor A Sensor B Sensor C frequency frequency frequency 0– 21 12 6 3– 0 1 11 6– 0 1 3 9– 0 1 1 12– 0 0 0 15– 0 2 0 18– 0 2 0 21– 0 1 0 24–26 0 1 0 Total 21 21 21 b

Frequency

Class interval

c

ii 38 ii 20

b negatively skewed d symmetrical

e 90%

2 a

0

iv 17.5%

7 a

b 1.4

b

iii 40%

v 42.5%

25 20 15 10 5 0

Sensor A

3 6 9 12 15 18 21 24 27

Leaf 01344556778999 012235

d 7.5

840 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Percentage cumulative frequency 5.4

40– 80–

1 12

3 15

8.1 40.5

120– 160–

18 3

33 36

89.2 97.3

1

37

Bill ($) 0–

0

Frequency

Frequency 2

Cumulative frequency 2

3 6 9 12 15 18 21 24 27

12 10 8 6 4 2

200–240

Sensor C

b 15 c

0

3 6 9 12 15 18 21 24 27

c i insensitive

ii very sensitive

100%

40

80%

30

60%

Mass

3 6

6 12

20– 25–

16 21

32 42

30–35 4 8 Total 50 100 b 50 c 32% d At least 25 g but less than 30 g.

ii $100

80

120 Bill ($)

160

200

240

iii $150

f approx. 20%

Exercise 9C 1

e 42%

Frequency

Percentage frequency

Strings Woodwind

21 8

52.5 20

Brass Percussion

7 4

17.5 10

b 40

40

a Min, lower quartile (Q1 ), median (Q2 ), upper quartile (Q3 ), max b Range is max – min; IQR is Q3 − Q1 . Range is the spread of all the data, IQR is the spread of the middle 50% of data. c An outlier is a data point (element) outside the vicinity of the rest of the data.

f 94% 10 a Section

Total

d i $130 e $180

Frequency Percentage frequency

10– 15–

10

20% 0

9 a

100

20

40%

iii moderately sensitive

Answers

14 a

Sensor B

Percentage cummulative frequency

Frequency

14 12 10 8 6 4 2

40 c 52.5%

100 d 47.5%

e 9.3%

f 65.6% 11 8 students scored between 20 and 30 and there are 32 students all together, so this class interval makes up 25% of the class. 12 No discrete information, only intervals are given and not individual values. 13 3 ≤ a ≤ 7, 0 ≤ b ≤ 4, c = 9

d If the data point is greater than Q3 + 1.5 × IQR or less than Q1 − 1.5 × IQR.

2 a 0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 8 b 2 c i 1

ii 3

d 2 e −2, 6

f yes; 8 3 a i 10.5

ii 7.5

iii 12

b 4.5 c 0.75, 18.75 d no 4 a min = 3, Q1 = 4, median = 8, Q3 = 10, max = 13; range = 10, IQR = 6 b min = 10, Q1 = 10.5, median = 14, Q3 = 15.5, max = 18; range = 8, IQR = 5 c min = 1.2, Q1 = 1.85, median = 2.4, Q3 = 3.05, max = 3.4; range = 2.2, IQR = 1.2 d min = 41, Q1 = 53, median = 60.5, Q3 = 65, max = 68; range = 27, IQR = 12

841 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

5 a median = 8, mean = 8

ii

b median = 4.5, mean = 4.2 c median = 4, mean = 4.3 6 a min = 0, max = 17 c Q1 = 10, Q3 = 15

10 12 14 16 18 20 22 24 26 28 30 32 d i Q1 = 0.03, Q3 = 0.05; no outliers

b median = 13 d IQR = 5 e 0

ii

f Road may have been closed that day. 7 a i min = 4, max = 14 ii 7.5 iii Q1 = 5, Q3 = 9 b i min = 16, max = 31

0.02

iv IQR = 4 ii 25

v no outliers

iii Q1 = 21, Q3 = 27 iv IQR = 6 8 a i min = 25, max = 128 ii 47

v no outliers

iii Q1 = 38, Q3 = 52.5 v yes; 128 vi 51.25

9 a no outliers c Outliers are 103, 182. 10 a IQR = 12 d 22

b Outlier is 2. d Outliers are 2, 8.

b No outliers.

0.07

36

40

44

48

52

60

0

1

2

3

4

5

70

80

90

100

110

120

64

c

d

130

6 a Same minimum of 1. b B b It is doubled.

c It is divided by 10. 13 a It stays the same.

b It doubles.

c i 5 ii 10 d Data points for B are more evenly spread than those for A. 7 a Q1 = 14.6, Q2 = 15.3, Q3 = 15.8 b 19.7 kg

c It is reduced by a scale factor of 10. 14 Answers may vary. Examples:

c

a 3, 4, 5, 6, 7 b 2, 4, 6, 6, 6

Box plot of lemur weights

c 7, 7, 7, 10, 10

13 14 15 16 17 18 19 20 Lemur weights

15 It is not greatly affected by outliers. 16 Answers will vary

8 a They have the same median and upper quartile. b B c i 4

Exercise 9D a 15

b 5

c 25

e 10

f 20

g 10

b 2

c 18

2 a 4 e It is.

0.06

0 2 4 6 8 10 12 14 16 18 20

c 24

11 1, 2, 3 12 a Increases by 5.

1

0.05

b

iv IQR = 14.5

b Median as it is not affected dramatically by the outlier. c A more advanced calculator was used.

0.04

0.03

5 a

ii 5

d Set B is more spread out.

d 20

9 a A d 20

10 a

b B

c B

Box plot of Set 1, Set 2 Set 2

3 a

Set 1 1 2 3 4 5 6 7 8

0 2 4 6 8 10 12 14 16 Spelling errors

b

5 6 7 8 9 10 11 12 13 14 15 16 17

b Yes, examiner 2 found more errors.

4 a i Q1 = 4, Q3 = 7; outlier is 13

11 Answers may vary. Examples:

ii

2 4 6 8 10 b i Q1 = 1.6, Q3 = 1.9; outlier is 1.1

12

14

ii

a i, ii Class results had a smaller spread in the top 25% and bottom 25% performed better. iii State results have a larger IQR. b The class did not have other results close to 0 but the school did.

1.0 1.2 1.4 1.6 1.8 2.0 c i Q1 = 19, Q3 = 23; outliers are 11 and 31

2.2

12 Answers will vary

842 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

b smaller b A

the data values in B. 4 a Gum Heights b Gum Heights 5 a mean = 6, s = 2.2

5

0

c

from the school. d Answers will vary. 8 a false b true

d Yes, one of the deviations would be calculated using the outlier. 11 a No, standard deviation reflects the spread of the data

ii 53.16 vi 21.16

b i 66% ii 96% c i Research required

iii 101.16

iv 37.16

50

Q3 = 6.1, IQR = 1.6 4 a i min = 30, max = 69 ii iii

median = 37.5 lower quartile = 34, upper quartile = 44

iv v

IQR = 10 Q1 − 1.5 × IQR = 19; Q3 + 1.5 × IQR = 59

The outlier is 69. b For example, school open day or grandparents day. 5 a Q1 = 16, Q3 = 22, IQR = 6

iii 100%

Q3 + 1.5 × IQR = 31 32 is an outlier.

ii One SD from the mean = 68% Two SDs from the mean = 95%

b

Three SDs from the mean = 99.7% Close to answers found.

12 14 16 18 20 22 24 26 28 30 32 6 2.7

Progress quiz 1

10 20 30 40 Time (in minutes) Leaf

d Median = 18 + 19 = 18.5 2 3 a Range = 32 − 4 = 28; Q2 = 17, Q1 = 12, Q3 = 23, IQR = 11 b Range = 6.6 − 4.2 = 2.4; Q2 = 5.2, Q1 = 4.5,

values from the mean not the size of the data values. b No. As for part a. 12 The IQRs would be the same, making the data more comparable.

0

Stem 0 689 1 24555689 2 03578 3 238 4 4 2 | 3 means 23

b There is less spread. Data values are closer to the mean. c Students at outer-suburb schools may live some distance

b mean = 5.25, s = 0.7 10 a no b no c yes

25

2 1

c mean = 8, s = 3.8 d mean = 32.5, s = 3.6 6 a mean = 2.7, s = 0.9 b mean = 14.5, s = 6.6

c true 9 a mean = 2, s = 1.0

50

4 3

b mean = 3.6, s = 2.6

7 a The outer-suburb school has more data values in the higher range.

75

6

Relative frequency

7

3 a A b The data values in A are spread farther from the mean than

13 a i 85.16 v 117.16

100

8

Frequency

1 a larger 2 a B

Travel time from home to school

b

Answers

Exercise 9E

Exercise 9F

a numerical and discrete

b categorical and nominal 2 a Class interval Frequency

1 a linear 2 a i 28◦ C Percentage frequency

0− 10−

3 8

15 40

20− 30−

5 3

25 15

40−50

1

5

b no trend ii 33◦ C

b 36◦ C c i 12 p.m. to 1 p.m.

c non-linear d linear iii 33◦ C iv 35◦ C ii 3 p.m to 4 p.m.

d Temperature is increasing from 8 a.m. to 3 p.m. in a generally linear way. At 3 p.m. the temperature starts to drop.

843 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Population

Answers

3 a

1000 900 800 700 600 500 400 0 2003 2005 2007 2009 2011 2013 2015 2017

b Generally linear in a positive direction. c i 500

ii 950

4 a

Price ($) 1.45 1.4 1.35 1.3 1.25 1.2 0

8 a i 5.8 km

c 8:30 p.m. 9 a The yearly temperature is cyclical and January is the next month after December and both are in the same season. b no c Northern hemisphere, as the seasons are opposite, June is summer. 10 a Increases continually, rising more rapidly as the years progress. b Compound interest—exponential growth. 11 a Graphs may vary, but it should decrease from room temperature to the temperature of the fridge. b No. Drink cannot cool to a temperature lower than that of the internal environment of the fridge. 12 a Innings

J

F M A M J J A S O N D

June and then continually decreased to a yearly low in November before trending upwards again in the final month.

Number of runs

b

c $0.21 Pass rate (%)

5 a

100 90 80 70 0 1994 1996 1998 2000 2002 2004

2

3

4

5

6

7

8

9 10

120 100 80 60 40 20 0

Score Moving average

0

2

4

6 8 Innings number

10

12

c Innings number. i The score fluctuates wildly. ii The graph is fairly constant with small increases and

b The pass rate for the examination has increased marginally over the 10 years, with a peak in 2001. c 2001 6 a linear

1

Score 26 38 5 10 52 103 75 21 33 0 Moving average 26 32 23 20 26 39 44 41 40 36

b The share price generally increased until it peaked in

decreases. d The moving average graph follows the trend of the score

d 11%

graph but the fluctuations are much less significant.

b i $650 000 ii $750 000 7 a i $6000 ii $4000

Exercise 9G

b 1 c

1 20

Southbank

18

Sales ($’000)

ii 1.7 km

b i Blue Crest slowly gets closer to the machine. ii Green Tail starts near the machine and gets further from it.

a unlikely e likely

2 a i

y

16

12

14

10

12

c unlikely

d likely

Scatter plot

8

10

6

City central

8

4

6

2

4

0

2 0

b likely f likely

Jul Aug Sept Oct Nov Dec

2

4

6

8

10

x

ii y generally increases as x increases.

d i The sales trend for City Central for the 6 months is fairly constant. ii Sales for Southbank peaked in August before taking a downturn. e about $5000

844 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Scatter plot

y

b positive

y

10

Scatter plot

150

8

140

6

130

4

120

2 0 0.0 0.5 1.0 1.5 2.0 ii y generally decreases as x increases.

3 a

y

x

2.5

110 100 0

2

1 2 3 4 5 c strong d (8, 1.0)

6

7

8

x

8

x

10

26 24 22 20 18 16 14 12 10 8

x

15.0 17.5 20.0 22.5 25.0 27.5 30.0 32.5 6 a none b weak negative c positive

Scatter plot

d strong positive

4 3

7 a yes b decrease

2

c i yes 8 a

ii car H

1

5.0 negative strong (14, 4) negative y

7.5

10.0

12.5

15.0

x

Scatter plot

Average diameter (cm)

Scatter plot

0

d 5 a

6

Scatter plot

y

0 b positive 4 a y

4

c none

Scatter plot

1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9

b c

Answers

b i

8.5

E

8.0 7.5 7.0

B A

6.5

D

6.0 20

24 22 20 18 16 14 12 10 8

C

25 30 35 Fertiliser (grams per week)

40

b D c Seems likely but small sample size does lead to doubt.

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9

x

845 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

9 a

13 Each axis needs a better scale. All data are between 6 and 8 hours sleep and show only a minimal change in exam marks. Also, very small sample size.

Words

Answers

Scatter plot 2250 2000 1750 1500 1250 1000 750 500 250

14 a i students I, T b i students H, C

ii students G, S ii students B, N

c students C, G, H, S, d students B, I, N, T e no

0

1

2

3 4 Photos

5

6

7

Exercise 9H 1

b Negative, weak correlation. 10 a

a

b

c

d

Scatter plot

Volume (dB)

4.5 4.0 3.5 3.0 2.5 2.0 1.5 0

200 400 600 800 1000120014001600 Distance (m)

b negative c As d increases, v decreases.

2 a y = 1x + 7 2 2

11 a i Weak, negative correlation. Scatter plot 35 Crime

30

3 a i 17

ii 23 4

b i 28 5

ii 14 5

4 a

Scatter plot

y

25

6

20

5 4

15 5

10

15

20 Police

25

3

30

2 1

ii no correlation

Scatter plot

Crime

b y = − 2 x + 17 3 3

26 25 24 23 22 21 20 19

0 b c d 5 a

7.5 10.0 12.5 15.0 17.5 20.0 22.5 25.0 27.5 Police b Survey 1, as this shows an increase in the number of police has seen a decrease in the incidence of crime. 12 The positive correlation shows that as height increases ability to play tennis increases.

0 1 2 3 4 5 6 7 positive correlation As above. All answers are approximate. i 3.2 ii 0.9 iii 1.8 iv 7.4 ≈ 4.5 b ≈6 c ≈ 0.5 d ≈ 50

x

6 a y = 3 x + 18 5 b i 42

ii 72

c i 30

ii 100

846 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

3 A a

Growth (cm)

Scatter plot 90 80 70 60 50 40 30 20

7 6 5 4 500

600 700 Rainfall (mm) ii ≈ 85 cm

c i ≈ 25 cm d i ≈ 520 mm

800

3 0 1 2 3 4 b y = 0.554762x + 3.45357

5

6

7

8

grey line. e i 7.3

b Too few data points. 10 a Too few data points to determine a correlation. b The data points suggest that the trend is not linear. 11 a i 50 ii 110 b It is possible to obtain scores of greater than 100%. 12 a Experiment 1

Scatter plot

ii 10.1

f i 7.1 B a y 4.5

ii 9.2

Scatter plot

4.0 3.5 3.0

200

2.5

180

2.0

160

0

5

10

15

20

25

x

b y = −0.077703x + 4.21014

140

c y = −0.086667x + 4.34333 d Least squares is the blue line, median–median is the

120 100 10

20

30 40 50 Age (years)

60

70

Experiment 2

Scatter plot

red line. e i 3.7 f i 3.7 C a y

ii 3.3 ii 3.3

Scatter plot

9 8 7 6 5 4 3 2

220 200 180 160 140 120 100

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 20

b i ≈ 140

30

40

50 60 Age (years)

70

80

b y = −3.45129x + 7.41988 c y = −2.85714x + 6.67381

e i −16.7 f i −13.3

e Research required.

ii −34.0 ii −27.6

4 a i y = −3.54774x + 43.0398 ii y = −3.28571x + 40.8333

Exercise 9I a i 12

90

x

d Least squares is the blue line, median–median is the red line.

ii ≈ 125

c i ≈ 25 ii ≈ 22 d experiment 2

1

x

c y = 0.42x + 4.17667 d Least squares is the black line, median–median is the

ii ≈ 720 mm

8 a y = 5x − 5 b 85 cm c 21 kg 9 a Data do not appear to have any correlation.

Max. heart rate (b.p.m.)

Scatter plot

8

400

Max heart rate (b.p.m.)

y

Answers

7 a, b

b $32397 c $1405

ii 3.26

b i 7 ii 2 2 a There is no linear correlation. b The correlation shown is not a linear shape.

d 8 years e 10 years

847 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

3 19

Scatter plot

160

4 1.1 5 a larger by 3

150 140 130

25.0 27.5 30.0 32.5 35.0 37.5 40.0 42.5 Temperature (°C)

8 Physics, Biology. The number of standard deviations from the mean shows the relative position of Emily’s mark within the spread of all results from each class. This number gives a meaningful comparison of results. Physics 1.67, Maths 1.11,

b y = −1.72461x + 190.569

d i 139 6 a, c

Number of break downs

d no change

6 y = x2 − 3x + 5 7 5.8 ≤ a < 6.2

120 110

ii 130

Biology 0.

iii 113

Scatter plot

Multiple-choice questions

35 30 25 20 15 10 5 0

1 D

2 B

3 D

4 C

5 A

6 C

7 D

8 E

9 C

10 B

Short-answer questions 1

20 40 60 80 100 120 140 160 Number of copies b y = 0.25x − 8 d 232 000 copies e The regression line suggests that the photocopier will be considered for scrap because of the number of copies made, as it’s likely to reach 200 000 copies before breaking down 50 times. 7 a y Scatter plot

a

Class interval

0

0–

2

12.5

5– 10–

7 5

43.75 31.25

15– 20–25

1 1

6.25 6.25

16

100

50

6 4

25

2 0

100 90 80 70

1970 1975 1980 1985 1990 1995 2000 Year b y = 2.01053x − 3881.6 c i 139 m ii 180 m d No, records are not likely to continue to increase at this rate. 8 a All deviations are used in the calculation of the least squares regression. b Outliers have little or no effect on the median values used to calculate the median–median regression. 9 A, as it has been affected by the outlier. 10 Research required.

1 66 kg 2 88%

Percentage frequency

8 Frequency

110

Problems and challenges

Frequency

Total b

120 Record (m)

b larger by 3

c no change e no change

% Frequency

Number of jackets

Answers

5 a, c

x

5 10 15 20 25 Number of hours of TV c It is positively skewed. Leaf e 8.5 hours d Stem 0 135667889 012346 1 4 2 iii 3 2 a i 10 ii Q1 = 2.5, Q3 = 5.5 iv 12 v 2

6 8 10 12 4 ii Q1 = 15, Q3 = 24 iii 9

10

20 25 30 15 ii Q1 = 2.1, Q3 = 2.6 iii 0.50

b i 17 iv none v

c i 2.4 iv 0.7 v

0.5

1.0

1.5

2.0

2.5

3.0

848 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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4 a

b true

y

c true

Chapter 10

d true

Scatter plot

25

Exercise 10A

20

1 x

0

1

2

3

4

5

15

2x 3x

1 1

2 3

4 9

8 27

16 81

32 243

10

4x 5x

1 1

4 5

16 25

64 125

256 625

1024 3125

10x

1

10

100

1000

5 1 b negative

2 c weak

3 d (3, 5)

4

x

5

2 a 4 3 a

5 a y = 3x + 7 5 5 b i 3.8

ii 7.4

c i 22 3

ii 17 2 3

e 1 27

7 a The Cats b The Cats c The Cats d The Cats 8 a non-linear b linear 9 a y = −3.75x + 25.65

b y = −3.8333x + 25.1667

Extended-response questions a i 14 ii 41 b i no outliers

ii no outliers

c

Tree 2 Tree 1 10

20

30 40 50 60 70 Number of flying foxes

80

90

d More flying foxes regularly take refuge in tree 1 than in tree 2, for which the spread is much greater. 2 a

y

Scatter plot

Number of shoppers

1200 1000

e

h 1 36 c 33 = 27 f 3−2 = 1 9

6 a 4 f 2

b 2 g 3

c 6 h 3

d 3 i 2

e 1 j 2

k 5 l 3 p undefined

m0

n 0

o 0

7 a −3 e −2

b −2 f −4

c −2 g −4

d −3 h −1

8 a 0.699 e −0.770

b 1.672 f −1.431

c 2.210

d −0.097

i −1 m −3

9 a 3

j −3 n −1

b 5

g 16

h 81

m3

n 2

s 2

t −1

k −5 o −2

l −1 p −2

c 6

d 4 i 1000 j 1 9 o 4 p 8

e 3 k 1 4 q 3

Time (min)

0

1

2

3

4

5

Population

1

2

4

8

16

32

f 2 1 343 r 10

l

b P = 2t c 256 d 13 min e log2 10 000 11 Yes. 0 < b < 1 ⇒ loga b < 0, when a > 1; e.g. log2 1 = −2. 4

1400

c d

g 1 64

b log3 81 = 4 d log4 16 = 2 f log5 1 = −3 125

1600

b

1 25

d 4 d 1 32

5 a log2 8 = 3 c log2 32 = 5 e log10 1 = −1 10

10 a

1800

800 600

f

10 000 100 000

c 3 c 1 4

4 a 24 = 16 b 102 = 100 d 2−2 = 1 e 10−1 = 0.1 4

6 a mean = 7, s = 2.5 b mean = 4, s = 3.0

1

b 4 b 1 2

1 10 000

Answers

3 a false

12 No. A positive number to any power is always > 0.

27.5 30.0 32.5 35.0 37.5 x Maximum daily temperatures positive correlation y = 92.8571x − 1491.67 i 737 ii 32.2◦ C y = 74.5585x − 1010.06 i 779 ii 33.7◦ C 25.0

13 a 16 b 26 c 6 14 loga 1 = 0 and dividing by 0 is not possible. 15 a 1 b 1 c 1 d 1 4 5 2 3 f 1 3

g 2 3

k 6 5

l 4 7

h 4 3

i 1 2

e 1 2 j 1 2

849 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

Exercise 10B 1

b Recall index law 2: am ÷ an = am−n

Now let x = am and y = an (1) so m = loga x and n = loga y (2) x From (1), x ÷ y = = am ÷ an = am−n y x So m − n = loga y From (2) m − n = loga x − loga y So loga x = loga x − loga y, as required. y c Recall index law 3: (am)n = amn Let x = am

a logb xy = logb x + logb y b logb x = logb x − logb y y c loga bm = m × loga b

d loga a = 1 e logc 1 = 0 f logb 1 = −loga b b 2 a 2

b 1

c 2

d 4

f 3

g 4

h −1

3 a 2 e 12

b 5 f 0

c 3

d −4

4 a loga 6 e logb 15

b loga 15 f logb 17

c loga 28

d logb 18

e 12 i −1

d logb 2

6 a loga 9 e loga 32

b loga 25 c loga 27 f loga 1000

d loga 16

7 a 0

b 0

c 0

f 1

g 0

h 3

d 1 i 1 3

k 1

l 1 2 b −3 f −5

10 a log3 20

9 a 1 f 2

e log3 8 11 a 3 2

So mn = loga xn From (1): nloga x = loga xn , as required.

Exercise 10C

b loga 3 c loga 10 5 a loga 2     3 7 e logb f logb 2 5

8 a −2 e −2

So m = loga x (1) xn = amn using index law 3

1

e 1 j 2 3

a log2 8 = 3 c log4 2 = 1 2 e log7 2 = x

2 a 3 f 0.1 3 a 0.845 e 0.780 4 a 1.465

c −3

d −1

b 1

c 3

d 2

b log10 48 f 0

c log10 2 d log7 2   3 g log2 h log5 6 4

b 5 2

c 4 3

d 3 2

e 1.177 e 2

e 1 3

b log5 25 = 2 d log3 10 = x f log1.1 7 = x b 4

c 2

d 8

b −0.222

c −0.125

d 1.277

c 1.594

d 6.871

f 0.897 b 3.459

e 2

f 2

5 a 1 e 6.579

b 1 f 1.528

c 3.969

d 1.727

6 a 2.585 e 1.559

b 1.893 f 6.579

c 1.209 g 3.322

d 1.129 h 1.262

i 0.356 7 a 2 days

j 3.969 k 3.106 b 2.548 days

8 a 14.21 years 9 a 10.48 years

b 23.84 years b 22.20 years

l 1.137 c 3.322 days c 47.19 years c 91.17 years

b 7 years 10 a A = 2000 × 1.1n 11 a F = 300 000 × 0.92n b 8.3 years

f 4 5 12 a loga 1 = loga 1 − loga x = 0 − loga x = −loga x as x required (using 2nd log law) b loga 1 = loga x−1 = −loga x as required (using x 3rd log law) 1 √ loga x as required 13 loga n x = loga x n = 1 loga x = n n (using 3rd log law) 14 a Recall index law 1: am × an = am + n Now let x = am and y = an (1) so m = loga x and n = loga y From (1), xy = am × an = am + n

(2)

So m + n = loga xy From (2), m + n = loga x + loga y So loga xy = loga x + loga y, as required.

12 a 69 years b 1386 years

13 a i

log10 7 log10 2

ii

log10 16 log10 3

iii

b i

1 log10 5

ii

3 log10 2

iii − 1 log10 3

c i 1.631

log10 1.3 log10 5

ii 1.167

iii −0.196

ii −2

iii 1

Exercise 10D 1

a 5 b i 3 c 2

iv −1

2 a linear b quadratic c quartic d quadratic e constant f linear g constant h quartic i cubic

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c −2

d −2

b 92 b 11

c 8 c 1

b 92 b 25

c −4 c −22

d 4 d −45

b 3

e −9 f 2 4 a, b, f are polynomials. 5 a 14 6 a −5 7 a 0 8 a −2

f −351

e 17

d 42 d −17

9 a −1 2

b −1

c 1 2

10 a 0 11 a i 30 m

b 4 ii 24 m

c −108 iii 0 m

13 a − 9 8

b − 20 27

c 5 8

d 27 64

f − 216 125

g −1 2

h −9 8

14 a 2k3 − k2 − 5k − 1 b 2b3 − b2 − 5b − 1 c 16a3 − 4a2 − 10a − 1

d −2a3 − a2 + 5a − 1 e −16a3 − 4a2 + 10a − 1 f −54k3 − 9k2 + 15k − 1

g 2a3 b3 − a2 b2 − 5ab − 1 h −2a3 b3 − a2 b2 + 5ab − 1 ii 2 vi −18

b i 3 ii −11 c a = 2 and b = −1

iv −13

iii 1

iii −22

a x2 + 2x 6x2

b x − 3x2

− 13x − 5 e 2 a x4 − 5x3 + 4x2 − 3 −3x8

x6

− − 6x + 3 c 3 a, b, c are true.

c x2 − 1

d x2 + 6x − 55

8x2

f − 26x + 15 b −x6 − 3x4 + x3 − x2 + 13

i −4x7 + 8x10 5 a x5 + x3 + 2x2 + 2 x5

x4

3x3

3x2

x5

− − + d − c e x5 + 2x4 + 2x3 − 2x2 − 3x

x3

−

f x5 − 2x4 + 5x3 − 4x2 g x6 − x5 + x4 − 4x3 + 2x2 − x + 2

c x4 − 4x3 + 6x2 − 4x + 1 7 a x5 + 3x4 − x3 − 9x2 − 2x + 8

c 7 c m+n

d 12 d 2m

b x5 + 2x4 − 3x3

c x3 + 4x2 + x − 6 d 6x3 + 23x2 − 5x − 4 e 15x3 − 11x2 − 48x + 20

f x5 + 3x4 − x3 − 3x2 − 2x − 6

Exercise 10F 1 a 1 b 3 c 0 2 a If 182 ÷ 3 = 60 remainder 2 then 182 = 3 × 60 + 2.

b If 2184 ÷ 5 = 436 remainder 4 then 2184 = 5 × 436 + 4.

c If 617 ÷ 7 = 88 remainder 1 then 617 = 7 × 88 + 1. 3 P(x) = (x − 1)(x2 + 2x) + 3

b 2x3 + 2x2 − x − 3 = (x + 2)(2x2 − 2x + 3) − 9 c 5x3 − 2x2 + 7x − 1 = (x + 3)(5x2 − 17x + 58) − 175

d − x3 + x2 − 10x + 4 =

(x − 4)(−x2 − 3x − 22) − 84 − 2x3 − 2x2 − 5x + 7 =

(x + 4)(−2x2 + 6x − 29) + 123 − 5x3 + 11x2 − 2x − 20 =

(x − 3)(−5x2 − 4x − 14) − 62

2x2

h x6 − 5x5 − x4 + 8x3 − 5x2 − 2x + 2 i x8 − x6 + x5 − 2x4 − x3 + 3x2 + x − 3

6 a x5 − 2x4 + 2x3 − 3x2 + 3x − 1 b x6 + 2x4 − 2x3 + x2 − 2x + 1

b 5 b m

e 2m f 3n 14 a x4 − x3 + x2 − x

f

h −x7 + x4 b x5 − x

12 a 3 13 a m

e

4 a x3 − 3x2 b x4 − x2 c 2x2 + 6x3 d x3 − x4 e x5 + 3x4 f −3x6 + 3x3 g −2x5 − 2x4

9 (x2 + x − 1)4 = x8 + 4x7 + 2x6 − 8x5 − 5x4 + 8x3 + 2x2 − 4x + 1

4 P(x) = (x + 1)(3x2 − 4x + 5) − 3 5 a 2x3 − x2 + 3x − 2 = (x − 2)(2x2 + 3x + 9) + 16

Exercise 10E 1

e −x6 − 2x4 − x2 + 4 f −x6 − x4 − 10x3 + 26x2 − 10x + 1

could use DOPS) 11 Yes. Multiplicative axiom ab = ba. d 1

15 a i 10 v −9

c 2x3 + 5x2 − 23x + 5 d −x5 + 5x4 − 2x3 + 5x2 − x + 1

10 (x2 − x − 1)2 − (x2 − x + 1)2 = x4 − 2x3 − x2 + 2x + 1 − (x4 − 2x3 + 3x2 − 2x + 1) = 4x − 4x2 as required (or

b Yes, when 5 < x < 7. 12 a 8 b n+1 c 1

e 16 27

b x4 + 2x3 − 3x2 − 4x + 4

c x6 + 4x5 + 2x4 − 12x3 − 15x2 + 8x + 16 8 a x3 + x2 − 4x + 1 b x3 − x2 + 6x − 1

Answers

3 a 4

− 2x + 4

6 a 6x4 − x3 + 2x2 − x + 2

= (x − 3)(6x3 + 17x2 + 53x + 158) + 476

b 8x5 − 2x4 + 3x3 − x2 − 4x − 6

= (x + 1)(8x4 − 10x3 + 13x2 − 14x + 10) − 16

7 a x2 − 2x + 3 −

5 b x2 + 2x + 2 − 1 x+2 x−1

c x3 − 3x2 + 9x − 27 + 79 x+3

d x3 + 4x2 + 15x + 60 + 240 x−4 8 −1, 1, 2

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Answers

9

2  6x − 7x − 3 x − 5 6x3 − 37x2 + 32x + 15

8 a x+1

6x3 − 30x2

−7x2 + 32x

b x − 3, x − 1 and x + 2 c x − 3, x − 2 and x + 1

2

−7x + 35x −3x + 15 −3x + 15

c − 41 b − 253 16 27 12 a x3 − x2 + 3x + 2 = (x2 − 1)(x − 1) + 4x + 1

b 2x3 + x2 − 5x − 1 = (x2 + 3)(2x + 1) − 11x − 4 c 5x4 − x2 + 2 = 5x(x3 − 2) − x2 + 10x + 2

Progress quiz b log10 1000 = 3 c loga a = 1

2 a 102 = 100 3 a x=4

b 23 = 8 b x=4

d x = 625 4 a 3

e x=5 f x = −6 b 2 c 2 e 1 f 1 2 b x = 29.060

d −2 5 a x = 1.771

c 70 = 1 c x=0

c x = 8.043 √ 6 The term involving x has a fractional index when written in index notation. 7 a 8 b 7 8 a 3 b 1 9 a x7 − 2x5 + x4

10 a x4 + x3 + 3x + 2

c 3k4 − 2k3 + k2 + 7k + 8 c 3 d 4x3 b x4 + 2x3 + 5x2 − 2x − 6 b −x4 + x3 − x + 2

d x7 + x5 + 4x4 + 2x2 + 4x c x8 + 4x5 + 4x2 11 P(x) = (x + 2)(2x2 − 5x + 13) − 27

Exercise 10G 1

a −1

b 41

2 a 3 3 0

b −2

4 a 3 e −127

b 11 f −33

d 96

b 1

d −3

13 a −2 b 23 14 a a = −1 and b = 2

11 a −8

a log2 32 = 5

d x − 5, x − 1 and x + 4 10 a −4 b −2 c −14

11 −38 12 a 5

0 Remainder of 0, as required. 10 a 4 b 13 8

1

b x − 1, x + 1 or x + 2 c x + 2

d x−2 9 a x − 2, x − 1 and x + 1

c −19

d −141

c 27 g −13

d 57 h −69

5 a 3 b 20 c 36 6 b, c and e are factors of P(x).

d 5

c 5

b a = 3 and b = −4

Exercise 10H 1

P(−1) = 0

2 a x = −3 or 1 c x = − 1 or 4 2 e x = −3 3 a −3, 1, 2 d −1,−1,3 2 3

b x = −2 or 2 d x = −3 or 4 f x = −4 or −3 b −7, −2, 1 c −4, 3, 4 2 1 e − ,− ,3 f −2,1,2 3 2 7 4 5

g − 12 , − 1 , − 11 h − 3 , − 2 , 1 11 2 3 5 19 2 4 a (x − 3)(x − 2)(x + 1); −1, 2, 3

b (x + 1)(x + 2)(x + 3); −3, −2, −1 c (x − 3)(x − 2)(x − 1); 1, 2, 3 d (x − 4)(x − 3)(x − 1); 1, 3, 4 e (x − 6)(x + 1)(x + 2); −2, −1, 6

f (x − 2)(x + 3)(x + 5); −5, −3, 2 √ √ b x = −2 5 a x = 1 or 1 + 5 or 1 − 5 6 a x = −1, 3 or 5 b x = −3, −2 or 1

7 a x = −4, 1 or 3 8 a 3 b 4

b −2, −1 or 3 c n

9 a x2 (x − 1); 0, 1 b x2 (x + 1); −1, 0 c x(x − 4)(x + 3); −3, 0, 4 d 2x3 (x + 1)2 ; −1, 0 10 0 = x4 + x2 = x2 (x2 + 1)

No solution to x2 + 1 = 0. Thus, x = 0 is the only solution. 11 The discriminant of the quadratic is negative, implying solutions from the quadratic factor are not real. x = 2 is the only solution. 12 a x = −4, −3, −2 or 1 b x = −2 or 3 c x = −3, −2, 1 or 3 d x = −2, 1, 1 or 2 2

7 b, d, f, g

852 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Exercise 10I

x-intercepts: −1, 3, 4 y

y

a

3 12

−1 O

−3

x

2

−1

y

b

3

x

4

c y-intercept: 10 x-intercepts: −2, 1, 5

3 O

−5

O

Answers

1

b y-intercept: 12

2

10

x

5

y

−2 O

1

x

5

2 a y-intercept is 12. x-intercepts are −1, 3, 4. b y-intercept is −21.

d y-intercept: 3

x-intercepts are −3, −1, 7. c y-intercept is 0.

x-intercepts: −3, 1, 2

y

x-intercepts are −2, 0, 4. d y-intercept is 0.

3

x-intercepts are −7, 0, 5. 3 a, b, c: x

–2

–1

y = x2

4

1

y = x3

–8

–1

y = x4

16

1

–1 2 1 4 –1 8 1 16

0 0 0 0

1 2 1 4 1 8 1 16

1

2

1

4

1

8

1

16

x-intercepts: −3, 0, 2

1

2

x

2

y

−1 O

5

x

y

6 −2

O

f y-intercept: 0 x-intercepts: −1, 0, 5

x

4 a y-intercept: 6 x-intercepts: −2, 1, 3

x

2

y

-3

10 5 O −1 −5 −10

1

e y-intercept: 0

y

−2

O

−3

O

1

3

x

g y-intercept: 0 x-intercepts: −3, 0, 1

y

−3

O

1

x

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Answers

h y-intercept: 0

b

y

y

x-intercepts: −1, 0, 3

80

−1 O

x

3

60 40

i y-intercept: 8 x-intercepts: −4, −2, 1

20

y 8 −4

O

−2

1

−2 8 a

j y-intercept: 3 2

3 2

y

5 a

−1

−1 1

O

1

x

x

−2

−2

−3

2

8 6 4 2 x −4

−2 −2 −4

d

O

−2

2

y

x 2

−2

−1

−1

−2 −1

−4

O1 2

x

−2 y

e

4

6 a y = (x − 2)(x + 1)(x + 4)

2

b y = (x + 3)(x − 1)(x − 3) c y = 1 x(x − 2)(x + 3) 2

−4 −3 −2 −1

d y = − 1 (x + 3)(x + 1)(x − 2) 2 7 a y

x

O

y

f

15 13

60

10

40 30 20 −2 −1O

y

x

y

c

y

b

1 O

−1−2O −4 −6 −8 −9 −10

x

O 0.5

−1

b

y 2

x-intercepts: −3, −1, 1 2 y

−3

x

O1 2 3 4

−4 −3 −2 −1

x

5

1 2

3 4 5

−20

x

O −5

5 (2, −3)

x

854 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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y

b

y

iv

Answers

y

9 a

15

O

3

x

−1 O

x

−1 O 3 5 y

c

y

d

4 12

x

−1 O

−4

c i y-intercept = (0, 8) ii y = (x − 2)(x − 1)(x + 1)(x + 4)

x

iii x-intercepts: (−4, 0), (−1, 0), (1, 0), (2, 0) iv y

8 −2O

x

3

y

e

−1 O

−4

1 2

x

y

f

d i y-intercept = (0, 225) ii y = (x − 5)(x − 3)(x + 3)(x + 5) iii x-intercepts: (−5, 0), (−3, 0), (3, 0), (5, 0)

2

−1

O

x

2

−2

O

2

y

iv

x

225

10 a i y-intercept = (0, −6)

ii y = (x − 1)(x + 2)(x + 3) iii x-intercepts: (−3, 0), (−2, 0), (1, 0) iv

−5

−3

O

3

5

x

y O − 3 −2

1

x

Problems and challenges 1

a 2

b 8 b −1.43

2 a 1.43 3 x=6 4 2p − q + 2 5 a x > 2.10

−6

6 a=2× b i y-intercept = (0, 15) ii y = (x − 5)(x − 3)(x + 1)

7

iii x-intercepts: (−1, 0), (3, 0), (5, 0)

1 34 ,

b = 1 log2 3 4

c 1 2 c −2.71

d 3

b x ≥ −2.81

log10 2 = log1.1 2 log10 1.1

8 −2 9 a = 5, b = −2

10 Proof using long division required. a (x3 − a3 ) ÷ (x − a) = x2 + ax + a2 b (x3 + a3 ) ÷ (x + a) = x2 − ax + a2

855 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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Answers

11 a 2 ≤ x ≤ 5 or x ≤ −1

4 A

5 D

6 D

7 A

8 B

9 E

10 E

Short-answer questions b log10 1000 = 3

b 4−2 = 1 16 b 4 e −3 h −4

b logb 21 f loga 1000

1

b

7 a −1 b 1 8 a x4 + 3x2 + 2

c i 7.27 2 a i 32

c 10−1 = 0.1

c logb 144 g 2

f −3 i −2

iii 16.89

d (x − 7)(x − 1)(x + 3) e x = 7, 1 or −3

d loga 10 h 1

f P(0) = 21 g y

21 −3 O 1

log10 2 log10 0.8

7

x

c −2 d −34 b x5 − x4 − 3x3

Semester review 2

9 a x3 + x2 + 2x + 3 = (x − 1)(x2 + 2x + 4) + 7 b x3 − 3x2 − x + 1 = (x + 1)(x2 − 4x + 3) − 2

Measurement

c 2x3 − x2 + 4x − 7 = (x + 2)(2x2 − 5x + 14) − 35 d −2x3 − x2 − 3x − 4 = −(x − 3)(2x2 + 7x + 24) − 76

10 a −3 b −39 11 b, c and d are factors.

c −91

d 41

b x = − 1 , 3 or 5 3 2 13 a (x − 1)(x + 2)(x + 3) = 0 x = −3, −2 or 1 b (x + 2)(x − 5)(x − 6) = 0 x = −2, 5 or 6

y

b

1 x

−1 −2

−5 −10

2 D

4 A

5 D

a 36 cm, 52 cm2

c 220 mm, 2100 mm2 2 a 188.5 m2 , 197.9 m3 m2 ,

4 5

3 B

b 1.3 m, 0.1 m2

y

−1O

1 B

1

5 4 1

Multiple-choice questions

Short-answer questions

12 a x = −2, 1 or 3

O

ii 6.17 ii 0

b There is no remainder; i.e. P(1) = 0. c x2 − 4x − 21

c 4

c x5 + x4 − 3x3 − x2 − x + 3 d x6 + 2x4 − 4x3 + x2 − 4x + 3

−1

x

O

a B b i $121 000 ii $15 369 iii $272 034

b x = log1.2 2

log10 13 log10 2

14 a

5

−10

j 4 loga 2 = loga 16

5 a x = log3 6 6 a

x

5

Extended-response questions

a log2 16 = 4 c log3 1 = −2 9

4 a loga 8 e loga 4 i 3 2

3

−5

3 E

d 0 g −1

10

−5 −2 O

2 B

3 a 3

y

5

1 C

2 a 34 = 81

d

10

Multiple-choice questions

1

y

c

b −4 < x < 1 or x > 4 12 y = 1 (x − 3)2 (x + 2) 9 13 16 14 y = − 1 x2 (x − 3)(x + 3) 10

x

b 50.3 cm2 , 23.7 cm3

m3

c 6.8 1.3 3  a 1.8 cm b 58.8 cm2 27 4 cm p

Extended-response question a 753.98 cm3 d 1.79 cm

b 206.02 cm3

c 17 cm

856 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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y (3, 8)

d

Answers

Parabolas and other graphs Multiple-choice questions 1 D

2 A

3 D

4 C

5 B

O 1

x

5

Short-answer questions 1

−10

a Dilated by a factor of 3.

y 4 a y = (x + 3)2 − 7

3

(1, 3) 2

x

O

y

b Reflected in x-axis and translated 2 units left.

O

−3 − √7

x

−3 + √7

y x

O

−2

b y=

−4

(−3, −7) 2 x−5 +7 2 4 y



8 c Translated 5 units up.

y

( 52 , 74 )

x

O (1, 6)

5

x

O 2 a −5 d

−5

b (1, −5)

c (−2, −9)

y

O 1

5 a Discriminant = 72, thus two x-intercepts. b y

−1.1 O x (1, −9) c (1, −9) and (−2, 9) 6 a y

−5

b

y √10

2

(−2, −9) −9 3 a Maximum at (3, 8).

x

3.1

b −10

c (1, 5)

−2

O −2

2

x −√10

O

x √10

−√10

857 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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y

Answers

c

d

y

c

y (−1, 5)

x = −2

(1, 3) 1 x

O

1

(1,

1) 5

O

x

5 −2 − 3 O − 52

x

y = −3

−3 e

y (1, 2)

Extended-response question x

O

a

h (200, 427.5)

(−1, −2) y

f

27.5

(−1, 6)

x

O 10

110 (60, −62.5)

x

O (1, −6) √ √ √ √ 7 a ( 3, 2 3), (− 3, −2 3)    c 1 , 4 , − 1 , −4 2 2 8 a y

b (4, 16)

b 27.5 m d 427.5 m

c 10 m and 110 m from start e 62.5 m

Probability Multiple-choice questions

−1 + 2√3

1 C

x 2 + √15

O

2 − √15

(2, −1)

b

a

x

A

(1, 17)

v

g

u

e

r

d

f

i

a

o

p

b h

j

k

x

2 a

A A b i 3 c i 1 4 3 a 0.18

z s c

n l

m

B 3

B 1

4

4 7

4 5

8 12

ii 4 ii 1 12 b 0.37

q

iii 9 26

b i

y=1

5 C

y

t

5 ii 3 26 26 c No, A ∩ B π ∅

9

4 D

B

w

y

1 O

3 B

Short-answer questions 1

−1 − 2√3

2 E

iii 5 iii 7 12

iv 19 26

iv 8 iv 3 4

858 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

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2

3

4

2 3 4 5

3 4 5 6

4 5 6 7

5 6 7 8

Die 2

b

5

b 16 c i 3 16

ii 5 8

5 a

A 4

4

10

2

iii 1 5

1 0

2

5

0

c i 14

10 15 20 Number of days

25

30

0

ii 50%

iii 20–24 days, those that maybe catch public transport to work or school each week day.

2 b i 2 3

20

3

B 4

30

6

Answers

1

1 2 3 4

Percentage frequency (%)

Die 1

Frequency

4 a

ii 2 3

2 a

c Yes they are since Pr(A | B) = Pr(A).

0

5

10

15

20

25

b

Extended-response question Outcome

5 8

b

5 7

250 g

3 7

200 g 450 g

4 7

250 g

Balance

200 g 400 g

200 g

3 8

a i

2 7

450 g

ii 15 28

9 14

500 g

6

7

8

9

b Balance fluctuated throughout the year but ended up with more money after 12 months. c May and June

iii 5 14

d increase of $500 y 4 a, c

c 3 5

6 5 4 3 2

Statistics Multiple-choice questions 1 E 2 B

0

3 C 4 B

x

1 2 3 4 5 6 7 8 9 10 11 12 13

b positive d i ≈ 3.2

5 A

ii ≈ 11.5

5 a i under 40 ii over 40 b Over 40: mean = 11, standard deviation = 7.3;

Short-answer questions 1

5

J F M A M J J A S O N D Months

250 g

3 28

4

2 3 2500 2000 1500 1000

3 a

Under 40: mean = 24.1, standard deviation = 12.6

a Class interval 0−

Frequency 2

Percentage frequency 10%

5− 10−

4 4

20% 20%

15− 20−

3 6

15% 30%

25 − 30 Total

1 20

5% 100%

Extended-response questions 1

a y = 1.50487x + 17.2287 b y = 1.55x + 16.0833

2 41 cm 3 22 mL

859 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

Logarithms and polynomials

Extended-response question

Multiple-choice questions

a P(−3) = 0 b P(x) = −(x + 3)(2x − 1)(x − 4)

1 E

2 A

3 D

4 C

5 D

P(x)

c

Short-answer questions 1

a 3 e 2

2 a x=3

(−5, 198)

b −2 f 2

c 3 g 1

b x=3

c x = 81

3 a i x = log3 30 b i x = 2.460

ii x = log2.4 4 ii x = 9.955

4 a i 6 ii 0 iii −49 iv −5 b i 2x6 + 6x5 − 11x4 − 25x3 + 34x + 24

ii 4x6 − 12x4 − 16x3 + 9x2 + 24x + 16 5 P(x) = (x − 3)(x2 − x − 1) + 4 6 a −24, not a factor c −40, not a factor

7 a x = −1, 3, −6 c x = −4, −2, 1

d 0 h 3

O −5

−3

−12

1 2

4

6

x

(6, −198)

b 0, a factor

b x = 0, 5 , − 2 2 3 d x = −1, 1, 2 2

860 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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861 Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Essential Mathematics for the Australian Curriculum Year 10 & 10A 2ed

Cambridge University Press ISBN 978-1-107-56890-7 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.