Student solutions manual algebra & trigonometry, third edition, James Stewart, Lothar Redlin, Saleem Watson [Compressed] 0840069235, 9780840069238

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Student Solutions Manual

Student Solutions Manual

Algebra & Trigonometry THIRD EDITION ■/

#

lames Stewart McMaster University and University of Toronto

lothar Redlin The Pennsylvania State University

Saleem Watson California State University, Long Beach

Prepared by *

Andrew Bulman-Fleming BROOKS/COLE t **

CENGAGE Learning'

Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States

•

BROOKS/COLE

t%

CENGAGE Learning*

©2012 Brooks/Cole, Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher.

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Printed in the United States of America

1 2 3 4 5 67 15 14 13 12 11

ISBN-13: 978-0-8400-6923-8 ISBN-10: 0-8400-6923-5 Brooks/Cole 20 Davis Drive Belmont, CA 94002-3098 USA Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan. Locate your local office at: www.cengage.com/global Cengage Learning products are represented in Canada by Nelson Education, Ltd. To learn more about Brooks/Cole, visit www.cengage.com/brookscole Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com

CONTENTS CHAPTER

P Prerequisites P.1

Real Numbers and Their Properties

P.2

The Real Number Line and Order

P.3

Integer Exponents

P.4

Rational Exponents and Radicals

P.5

Algebraic Expressions

P.6

Factoring

P.7

Rational Expressions

Chapter P Test ■

2

4 6

8

13 17

>

20

FOCUS ON MODELING: Modeling the Real World with Algebra

21

1 Equations and inequalities 1.1 Basic Equations 23 1.2 Modeling with Equations 25

23

1.3

Quadratic Equations

1.4

Complex Numbers

1.5

Other Types of Equations

1.6

Inequalities

1.7

Absolute Value Equations and Inequalities

Chapter 1 Test ■

29 34 36

40

Chapter 1 Review

CHAPTER

1

10

Chapter P Review

CHAPTER

1

51




Systems of equations and inequalities

11.1

vii

Sequences and series

13.1

Sequences and Summation Notation

13.2

Arithmetic Sequences

427

13.3

Geometric Sequences

429

13.4

Mathematics of Finance

433

13.5

Mathematical Induction

435

423

425 425

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

viii

Contents

13.6

The Binomial Theorem Chapter 13 Review Chapter 13 Test

■

chapter 14

440

442

446

FOCUS ON MODELING: Modeling with Recursive Sequences

Counting and probability

14.1

Counting Principles

14.2

Permutations and Combinations

14.3

Probability

14.4

Binomial Probability

14.5

Expected Value

449

449 451

454

Chapter 14 Test

458

460

Chapter 14 Review

■

446

461

464

FOCUS ON MODELING: The Monte Carlo Method

464

Cumulative Review Test: Chapters 13 and 14

465

©2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P

PREREQUISITES

P.1

REAL NUMBERS AND THEIR PROPERTIES (a) The natural numbers are {1, 2, 3,...}.

(b)

The numbers {..., —3, —2, —1,0} are integers but not natural numbers.

P

t

c

(c) Any irreducible fraction — with q ^ 1 is rational but is not an integer. Examples: |, — -fe,

(d)

i79Q

Any number which cannot be expressed as a ratio — of two integers is irrational. Examples are V2, \/3, 7r, and e. ’z2

[xy2z^

V

57. 6/l = « 3-IJ-2

2'

59.

z

■(?)’(•-¥)’-

2_9 = ;c4-2.3^4-2.3z34-3 = r£

X

(x2y2z)3 8a3h~4

2-5+3-3^-5+2-3c-3-3 _ aX9b

63.

= 4a3-(-5)b-4-5 =

4a*

2 a~5b5

b9

10 = M~1-2—3-3y2-2—(—2)3 _ £ (h3u

2)3

5-1(-3)x-(-2)(-3>>2-3

.

7)

=

63

67

IT 125 =

r —S * ^ r-*sq-* )

q2r*

q~^r~ ^ s~2

73. 69,300,000 = 6.93 x I07

75. 0.000028536 = 2.8536 x 10-5

77. 129,540,000 = 1.2954 x 108

79. 0.0000000014 = 1.4 x 10~9

81. 3.19 x 105 = 319,000

83. 2.670 x 10~8 = 0.00000002670

85. 7.1 x 1014 = 710,000,000,000,000

87. 8.55 x 10-3 = 0.00855

89. (a) 5,900,000,000,000 mi = 5.9 x 1012 mi

(b)

0.0000000000004 cm = 4 x 10“13 cm

(c) 33 billion billion molecules = 33 x 109 x 109 = 3.3 x 1019 molecules 91 . (7.2 x 10~9) (l.806 x 10~12) = 7.2 x 1.806 x 10~9 x 10~12 » 13.0 x 10-21 = 1.3 x 10~2°

1.295643 x 109

1.295643

(3.610 x 10-17) (2.511 x 106)

3.610 x 2.511

93.

(0.0000162)(0 .01582) 95.

(594621000) (0.0058)

(1.62

10-5) (1.582

X

x 109+l7-6 w 0.1429 x 1019 = 1.429 x 10 19

xio-2)

(5.94621 x 108) (5.8 x 10-3)

,.62 x 1. 582 5.94621 x 5.8

x 10—3-2—84-3 = 0.074 x 10-12

= 7.4 x 10-14 a"' 97. (a) — =

a ■ a ■ a.a (m factors) a ■a ■a

/a\«

/l6 = V42 = 4 (b)

(c)

^16

3

= 4P =

17. (a)

2

-vG) -

-m-i

(b) 4/256 = 4/^ = 4 2

(c)

2

4/T

1 2

-1/2

19. (a)

2-1

3

P1

2

(b) (-32)2/5 = [(—2)5]2/5 = (-2)2 = 4 (c) (-125)"1/3 = [(-5)3]"1/3

,,

/ 1 \2/5 _ / 1 \2/5 _ /l\5(2/5) _ /1\2 _ 1 (32)

(2V

(2)

(2)

l

4 Sfcv

©2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION P.4 Rational Exponents and Radicals

(b) (27)~4/3 = (33) 4/3 = (3)3'(“4/3) = 3-4 = 1 = 1

/3

1

75 2

1 7? 7? rut 1 /5)

3 (l + V5)

5 (jr - jfj

~~

5

3-7

-4

Jr ~s/2

2 (V? - y/i)

y V3 - yjy

1-5

r—2

X

-5

y

Jr — J2

\JX2 + 1 —

2(V2-7?)

2-7

V3 - Jy _y

\/2> + y/y

— 2 + -v/3

2(V2-V?)

■j2 + y/l~ ^2 + Jl' J2-Jl~

3

97.

(x + 2)2 (x — 13)

x+2

\/3 + *Jy

95.

(*-3)4

(x + 2)2 (3x — 9 — 2x — 4)

(1 + x)-1//2 [2 (1 + x) — x]

2 91.

-x2

2(1 +x)1//2 — x (1 +x)-1/2

(l+x)2/3

89.

1 - X2

(x + 2)2 (x - 3) [3 (x - 3) - (x + 2) (2)]

(x - 3)4

85.

V

l-X2

Vx2 + 1 + X x2 + 1 — x2 1 . = —=^=- = -■ Jx2 + l+X yx2 + l+x VX^ + l+X

99. V xz + 1 — x = - • —

16 + a |£ a a 101. ——— = + — = 1 + —, so the statement is true. 103. This statement is false. For example, take x = 2, then LHS =

2 2 1 -—- = - = -, while 4+2 6 3

4+x

„„„ 1 2 1 2 3 1 3 RHS = - + - = - + - = -, and - ± 2 x 2 2 2 3 2

105. This statement is false. For example, take x = 0 and y = l.Then substituting into the left side we obtain LHS =

x

0

1 1 1 1 = 0, while the right side yields RHS = --= --- = -, and 0^ 0+1 l+y 1 + 1 2 2

x+y

107. This statement is true: —J- = (—a)

109. (a) R

1

1

R\R2

J_

_L “ J_

J_ '

R\

R2

Ri

R\

= (—1) (a)

= (—1)

= —^.

R\R2

~rJri ~ r2 + R\

(b) Substituting R\ = 10 ohms and R2 = 20 ohms gives R =

(10) (20)

200

(20) + (10)

30

6.7 ohms.

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER P

Review

17

.

111

X

1 -9

2.80

2.90

2.95

2.99

2.999

3

3.001

3.01

3.05

3.10

3.20

5.80

5.90

5.95

5.99

5.999

?

6.001

6.01

6.05

6.10

6.20

X

x — 3

From the table, we see that the expression x2 -9 _ (jr - 3) (jc + 3)

jc — 3

x —3

jc2 -9 x —3

approaches 6 as x approaches 3. We simplify the expression:

= jc + 3, jc ^ 3. Clearly as jc approaches 3,x + 3 approaches 6. This explains the result in the

table.

113. Answers will vary. Counterexample

Algebraic Error 1

1 , 1 + , i1 . , a b a+b

1

(a + b)2 ^ a2 + b2

(1 +3)2 ^ l2 + 32

yja2 + b2 ^ a + b

V52 + 122 ^5 + 12

1 ^

1

a+b a (a3 + b3yP ± a + b

{7} + 23)1/3 ^ 2 + 2

am — # am/n an

35 4 # 35/2 32

a~l/n # 4 an

64-1/3 ^ ^ 643

_L

CHAPTER P REVIEW 1. (a) 16 is rational. It is an integer, and more precisely, a natural number. (b) — 16 is rational. It is an integer, but because it is negative, it is not a natural number. (c) Vl6 = 4 is rational. It is an integer, and more precisely, a natural number. (d) y/2 is irrational. (e) | is rational, but is neither a natural number nor an integer. (0 -§ = -4 is rational. It is an integer, but because it is negative, it is not a natural number.

5. Distributive Property.

3. Commutative Property for addition.

7. (a)

5

2 _ 5

4 _ 9

3

12

6

3 ~ 6 + 6~ 6

2

15 • 12

3-3

9

y: 12

8-5 15-5

2~\ 5-5

2

5

17. — 1 < x < 5

x e [5, oo)

x e (—1, 5]

-1 21. (a) A DC = {1,2}

19. (a) AU B = {—1,0, j, 1,2,3,4}

(b)5ni) = {i,lj

(b) ADB = {1}

23. |7 — 10| = |—3| = 3 27.

31.

25. |3-|-9|| = |3-9| = |-6|=6 w, 1 1 1 29. 216-1/3 =-ttt = -7= = 216*/3 4/2T6 6

9-3 _ o-2 _ I _ I _ 9_ _ _8_ _ J_ z j - g 9-72 72 “ 72

242

V242

33. 4/=T25 =

= VI2T = 11

V2

V

2

35. (a) |5 — 3| = |2| = 2

^1 (—5)3 = -5

37. -4 = x-2 xz

(b) |—5 - 3| = |—8| = 8

39. 43.

x2xrfl

(>T =

x2+m+3m _ x4w-|-2

41. xaxbxc = xa+^+c

,c+l (x2c“*)2 = jc(c+1)+2(2c-1) = xc+l+4c-2 _ x5c-l

45

^7 = ?l/3

(b) l/l* = 74/5

47. (a)

= x5/6

(b) (V*)9 = (*1,/2)

51.

49. (2x3y) (3x~ V2) = 4xV • 3x~V2 = 4 • 3x = 12 x5/

= x9/2

*4 (3x>2 = *4 • 9*2 = 9x4+2-3 = 9x3

53 .

]/(x3y)2 y4 =

= ^6 = x2^2

XJ

XJ X

X

2 - Vx

x (2 — v^x)

2 + V*

2 -f ^/x

2 — *Jx

4—x

55.

6- 1^2+2

8r!/25-3

57.

. Here simplify means to rationalize the denominator. 4r5/2

= 4r(l/2)-(-2)s-3-4 _ 4r5/25-7 _

2r 254

59. 78,250,000,000 = 7.825 x 1010 61.

ab

(0.00000293) (l.582 x 10~14)

(2.93 x 10~6) (l.582 x 10~14)

2.93 • 1.582

c

2.8064 x 1012

2.8064 x 1012

2.8064

x 10- -6-14-12

« 1.65 x 10~32

63. 2x2y — 6xy2 = 2xy (x — 3y) 65. x2 — 9x + 18 = (x — 6) (x — 3) 67. 3x2 — 2x — 1 = (3x 4- 1) (x — 1) 69. 4/2 - 13r - 12 = {At + 3) (r - 4) 71. 25 -16f2 = (5-40 (5 + 40 73. x6 — 1 = (x3 — 1^ (x3 + 1^ = (x — 1) (x2 + x + l) (x + 1) (x2 — X + 1^ 75. x-1/2 _ 2x1/2 + x3/2 = x-1/2

(1 - 2X + x2) = x"1/2 (1 - x)2

77. 4x3 — 8x2 + 3x — 6 = 4x2 (x — 2) + 3 (x — 2) = (4x2 + 3^ (x — 2)

©2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER P

Review

19

79. (x2 4-2) ^ 4-2x(x24-2) ^ 4-x2Vx2 + 2 = (x2 + 2) ^ ^(x24-2) + 2x ^x2 + 2^ + x2^ = Vx2 + 2 (x4 + 4x2 + 4 + 2x3 + 4x + x2) = Vx2 + 2 (x4 + 2x3 + 5x2 + 4x + 4^ = y/x2 + 2 (x2 + x + 2^

81. a2y - b2y = y (a2 — b2) = y (a - b) (a + b) 83. (x 4- l)2 — 2 (x 4- 1) 4- 1 = [(x + 1) — l]2 = x2. You can also obtain this result by expanding each term and then simplifying.

85. (2x + 1) (3x - 2) - 5 (4x - 1) = 6x2 - 4x + 3x - 2 - 20x + 5 = 6x2 - 21x 4- 3 87. (la2

-b)2

= (la2y - 2 (2a2) (b) + {b)2 = 4a4 - 4a2b + b2

89. (x — 1) (x — 2) (x — 3) = (x — 1) (x2 — 5x + 6) = x3 — 5x2 + 6x — x2 + 5x — 6 = x3 — 6x2 + 1 lx — 6 91. s/x (V* 4- l) {2y/x — l) = (x + *Jx) (2y/x — l) = 2Xy/x — x + 2x — s/x = 2x3^2 + x — x1/2 93. x2 (x — 2) + x (x — 2)2 = x3 — 2x2 + x (x2 — 4x + 4) = x3 — 2x2 4- x3 — 4x2 + 4x = 2x3 — 6x2 + 4x 95. 97. 99. 101. 103.

x2 — 2x — 3

(x — 3) (x + 1)

x —3

2x2 + 5x + 3 ~ (2x + 3) (x + 1) _ 2x + 3 x2 + 2x — 3

3x + 12

(x 4- 3) (x — 1)

3 (x + 4)

3 (x + 3)

x2 4-8x4-16

x —1

(x + 4)(x + 4)

(x - 1)

x+4

x2 — 2x — 15

x2 — x — 12

(x — 5) (x + 3)

(x — 1) (x + 1)

x + 1

x2 — 6x 4- 5

x2 — 1

(x — 5) (x — 1)

(x — 4) (x 4- 3)

x —4

1

x

x2 4- 1

x (x — 1)

x2 4- 1 - x2 4-x

x 4- 1

x — 1

x2 4- 1

(x — 1) (x2 4- l)

(x — 1) (x2 4- l)

(x — 1) (x2 4- l)

(x — 1) (x2 4- l)

_J__2_ _ _1_2 x — 1

x2 — 1

1

2 105. x x—2

1

x 4~ 1_2

x — 1 (x — 1) (x 4- 1) (x — 1) (x 4- 1) x 4- 1 - 2 x - 1 1 (x — 1) (x 4- 1)

1

_

(x — 1) (x 4- 1)

x + 1

x

2x 2x x —2

2 —x 2x

— 1 (x — 2)

1

-1

2x

x—2

2x

x —2

3 (x + h)2 - 5 (x + h) - (3x2 - 5x)

3x2 4- 6xh + 3h2 - 5x - 5h - 3x2 + 5x

6xh + 3h2-5h

h

h

107. h (6x 4- 3/2 - 5)

h 109. 111.

(x — 1) (x 4- 1)

x +5 . is defined whenever x 4- 10 ^ 0 x 4- 10

6x 4- 3/? — 5

x / —10, so its domain is {x | x ^ —10}.

■s/x

is defined whenever x > 0 (so that y/x is defined) and x2 — 3x — 4 = (x 4-1) (x — 4) ^ 0 x2 — 3x — 4 x^4. Thus, its domain is (x | x > 0andx / 4}.

x =2 —i and

113. This statement is false. For example, take x = 1 and y = 1. Then LHS = (x 4-y)3 = (1 4-1)3 = 23 = 8, while RHS = x3 + y3 = l3 + l3 = 1 4-1 = 2, and 8 / 2. 12 4-y

12

y

12

y

y

y

y

115. This statement is true: -=-I— =-Hi. 117. This statement is false. For example, take a = — 1. Then LFIS = Vo2 = -J(—l)2 = n/T = 1, which does not equal a = -1. The true statement is Va2 = \a\.

©2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

20

CHAPTER P Prerequisites

119. This statement is false. For example, take x = 1 and y = 1. Then LHS = x3 +y3 = 13 + 13 = 1 + 1 = 2, while RHS = (x+y)(x2+xy + y2) = (l + l)[l2 + (l)(l) + l2] = 2 (3) = 6.

121. Substituting for a and b, we obtain 2

a2 + b2 = (2 mn)2 + ^m2 — n2^j

= 4m2n2 + m4 — lm2n2 + n4 = m4 + 2m2n2 + «4 = (m2 + «2^

Since this last expression is c2, we have a2 + b2 = c2 for these values.

CHAPTER P TEST 1. (a) 5 is rational. It is an integer, and more precisely, a natural number. (b) V5 is irrational. (c) — | = —3 is rational, and it is an integer. (d) — 1,000,000 is rational, and it is an integer. 3. (a)

-•

-►

-4

0

-»

•

2

0

3

M,2)

[0,3]

(b)

■o2

0

-4

[-4, 2) n [0, 3] = [0, 2)

3

[-4, 2) U [0, 3] = [-4, 3]

(c) |-4-2| = |-6| = 6 5. (a) 186,000,000,000 = 1.86 x 1011

(b) 0.0000003965 = 3.965 x 10-7

7. (a) 3 (x + 6) + 4 (2x - 5) = 3x + 18 + 8x - 20 = 1 lx - 2 (b) (x + 3) (4x — 5) = 4x2 — 5x + 12x — 15 = 4x2 + 7x — 15 (c) (^y/a + Vb) (^y/a -

= (V«)2 -

(V5)

=a —b

(d) (2x + 3)2 = (2x)2 + 2 (2x) (3) + (3)2 = 4x2 + 12x + 9 (e) (x + 2)3 = (x)3 + 3 (x)2 (2) + 3 (x) (2)2 + (2)3 = x3 + 6x2 + 12x + 8 (0 x2 (x - 3) (x + 3) = x2 (x2 - 9) = x4 - 9x2 ^

x2 + 3x + 2 _ (x + 1) (x + 2) _ x + 2 x2 — x — 2

(x + l)(x-2)

2x2 — x — 1 x2 — 9 X2

x2 - 4

y

y

x + 3 _ (2x + 1) (x — 1)

x +3

x — 1

2x + 1

2x + 1

x —3

_ x+ 1

(x — 3) (x + 3) _

x +2

x2

2

x + 1 _

(x - 2) (x + 2)

x +2

(x - 2) (x + 2) + x +2

(x — 2) (x + 2)

(x — 2) (x + 2)

y

x

Z 1

x

y

xy _ y*

1

1

xy

y

x

r-

xz

x —y

(y-x)(y + x)

x-y

- (x + 1) (x - 2)

X

*2_(J2 ~2)

x

X

x-2

_

(x - 2) (x + 2)

1 x —2

- (x - y)

(y + x)

x —y

= -(y + x)

©2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Modeling the Real World with Algebra

21

FOCUS ON MODELING Modeling the Real World with Algebra 1. Ifx = $120 and T — 0.08x, then T = 0.08 (120) = 9.6. The sales tax is $9.60.

3. P = 0.8 (1000) - 500 = $300. 5. (a) M =

(b) 25 =

N _ 240

G

~

8

~

175 G

7. (a) V = 9.5S = 9.5 ^4 km3) = 38 km3

= 30 miles/gallon

~

175 ^

(b) 19 km3 = 9.55’

S = 2 km3

G = ~25 =1 gallons

9. (a)

(b) We know that P — 30 and we want to find d, so we solve the Depth (ft)

Pressure (lb/in2)

equation 30 = 14.7 + 0.45d 15.3 = 0.45d

0

0.45 (0) + 14.7 = 14.7

10

0.45(10)4-14.7 = 19.2

d

20

0.45 (20) 4- 14.7 = 23.7

is 34 ft.

30

0.45 (30) + 14.7 = 28.2

40

0.45 (40) + 14.7 = 32.7

50

0.45 (50) 4- 14.7 = 37.2

60

0.45 (60) + 14.7 = 41.7

15.3 0.45

= 34.0. Thus, if the pressure is 30 lb/in2, the depth

11. The number N of days in w weeks is N — lw.

13. The average A of two numbers, a and b, is A =

a+b

15. The cost C of purchasing x gallons of gas at $3.50 a gallon is C = 3.5x. 17. The distance d in miles that a car travels in t hours at 60 mi/h is d = 601. 19. (a) $12 + 3 ($1) = $12 + $3 = $15 (b) The cost C, in dollars, of a pizza with n toppings is C = 12 + n. (c) Using the model C = 12 4- n with C = 16, we get 16 = 12 + n

n = 4. So the pizza has four toppings.

21. (a) (i) For an all-electric car, the energy cost of driving x miles is Ce = 0.04x.

(ii) For an average gasoline powered car, the energy cost of driving x miles is Cg = 0.12x. (b)

(i) The cost of driving 10,000 miles with an all-electric car is Ce = 0.04 (10,000) = $400. (ii) The cost of driving 10,000 miles with a gasoline powered car is Cg = 0.12 (10,000) = $1200.

23. (a) A 10-minute call costs 1 + 0.1 (10) = 1 -I- 1 = $2. ' connect \ / cost \ , . ) + ( )x (minutes), so C = 1 + 0.1/. fee / \per minute/

(b) The cost is ^( _

(c) When C = 2.2, we have 2.2 = 1 + 0.1/

1.2 = 0.1/

t = 12. The call lasted 12 minutes.

_ / connect \ / cost \ (d) The cost is I _ ) + I ) x (minutes), so when the connection fee is F cents and the rate is r cents then \ fee / \ per minute / the cost (in cents) is given by C = F + rt.

© 2012 Cengage Learning. Ail Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1

EQUATIONS AND INEQUALITIES

1.1

BASIC EQUATIONS X

c

1. (a) - + 2x = 10 is equivalent to jx — 10 = 0, so it is a linear equation.

2

2

(b) - — 2x = 1 is not linear because it contains the term a multiple of the reciprocal of the variable, x x (c) x + 7 = 5 — 3x 4x — 2 = 0, so it is linear. 3. (a) This is true: If a = b, then a + x = b + x. (b) This is false, because the number could be zero. However, it is true that multiplying each side of an equation by a nonzero number always gives an equivalent equation. (c) This is false. For example, — 5 = 5 is false, but (—5)2 = 52 is true. 5. (a) When x = -2, LHS = 4 (-2) + 7 = -8 + 7 = -1 and RHS = 9 (-2) — 3 = —18 — 3 = —21. Since LHS # RHS, x = —2 is not a solution. (b) When x = 2, LHS = 4 (-2) + 7 = 8 + 7 = 15 and RHS = 9(2) — 3 = 18 — 3 = 15. Since LHS = RHS, x = *2 is a solution. 7. (a) When x = 2, LHS = 1 — [2 — (3 — (2))] = 1 - [2 - 1] = 1 - 1 = 0 and RHS = 4 (2) - (6 + (2)) = 8-8 = 0. Since LHS = RHS, x = 2 is a solution.

(b) Whenx = 4 LHS = 1 - [2 - (3 - (4))] = 1 - [2 - (-1)] = 1 -3 = -2 and RHS = 4 (4) - (6 + (4)) = 16- 10 = 6. Since LHS ^ RHS, x = 4 is not a solution. 9. (a) Whenx = -1, LHS = 2 (-1)1/3 -3 = 2 (-1) - 3 = -2 - 3 = -5. Since LHS # l,x = -1 is not a solution. (b) When x = 8 LHS = 2 (8)1/3 — 3 = 2 (2) — 3 = 4 — 3 = 1 = RHS. So x = 8 is a solution. 11. (a) When x = 0, LHS = -—7 = —7 = % = RHS. So x = 0 is a solution. 0—b —b b (b) When x = b, LHS = ^

13. 2x + 7 = 31

17. x — 3 = 2x T 6 21. 2 v — 2 =

-a = — -- — is not defined, so x = b is not a solution.

2x = 24

x = 12

15. jx — 8 = 1

—9 = x

= 15 — 2w

3y — 12 = 2y (multiply both sides by the LCD, 6)

o

2 —2x = 3 + 6x + 5

25. 4 (y — i) - y = 6 (5 - y)

4y - 2 - y = 30 - 6y

27. x — ^x —

6x — 2x — 3x — 30 = 0 (multiply both sides by 6)







—5w =

x = 18 15

10

= —3

y = 12

23. 2 (1 — x) = 3 (1 + 2x) + 5

—5 =0

jx = 9

2 —2x = 8 + 6x

3y - 2 = 30 - 6y

—6 = 8x 3 = 4 + 3x (multiply both sides by the LCD, 3x) — 1 = 3x « x = — i x 3x J 2 3 31. -=- => 2 (t — 1) = 3 (t + 6) [multiply both sides by the LCD, (t — 1) (t + 6)] 2t — 2 = 3t + 18 t+6 t — 1 —20 = t

33. r — 2 [1 -3(2r + 4)] = 61 13r = 39



r - 2 (1 - 6r - 12) = 61

r — 2 (—6r — 11) = 61

r + 12r + 22 = 61



r =3 23

©2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 1

24

Equations and Inequalities

x +5

35. Vlx + a/12 = 2 6 37.-5 = —1-4 39.

=>

—4 = 9x

2x = — 1

x = —A

4 —9 = x

3 (6) — (3x + 3) = 2 [multiply both sides by 6 (x + 1)]

— 3x = —13



18 — 3x — 3 = 2

x =

(2x — 7) 3 = 2 (2x + 4) (cross multiply)

2x+4

43.1-1-1 = 2z

5z

10

u — + j =4

3z + 3 = 100z

3 = 97 z

97

( u+ 1\ w = 4 I w--— I (cross multiply)

=>

6x — 21 = 4x + 8

29

2x = 29

x = -j-

10 (z + 1) - 5 (z + 1) - 2 (z + 1) = 10 (lOz) [multiply both sides by lOz (z + 1)]

z+ 1

3(z+l) = 100z

45.



3x + 3

2x -7

z

2 — 5x = 6 + 4x

1

1

x+ 1 —3x + 15 = 2

41.

3x + 6 = x + 5 (multiply both sides by a/3)



=z

u — 4u —2u —2

«

m = 2m—2

3 (x) = (x + 4) + 6x + 12 (multiply both sides by x (x + 4)] 3x = 7x + 16 x +4 x xl + 4x —4x = 16 x = —4. But substituting x = —4 into the original equation does not work, since we cannot divide by 0. Thus, there is no solution. 51. x2 = 49

=*>

53. x2 - 24 = 0

x = ±7

55. 8x2 - 64 = 0 57. x2 + 16 = 0

x2 = 24



59. (x + 2)2 =4

=>

x2 - 8 = 0

x = ±V24 = ±2^6 «

x2 = 8

=>

x = ±a/8 = ±2a/2

x2 = —16 which has no real solution.

(x + 2)2 =4

=>

x + 2 = ±2. If x + 2 = 2, then x = 0. If x + 2 = —2, then x = —4. The

solutions are —4 and 0. 61. x3=27

x = 271/3 = 3

63. 0 = x4 — 16 = ^x2 + 4^ ^x2 — 4^ = ^x2 + 4^ (x — 2) (x + 2) .x2 + 4 = 0 has no real solution. If x — 2 = 0, then x = 2. If x + 2 = 0, then x = —2. The solutions are ±2. 65. x4 + 64 = 0

x4 = —64 which has no real solution.

67. (x+2)4-81 =0

-

(x + 2)4 = 81

«

[(x + 2)4]‘/4 = +811/4

x + 2 = ±3. So x + 2 = 3, then x = 1.

If x + 2 = —3, then x = —5. The solutions are —5 and 1.

69. 3 (x - 3)3 = 375 71. %x = 5



(x - 3)3 = 125

(x - 3) = 1251/3 = 5

x =3 + 5 = 8

x = 53 = 125

73. 2x5/3 + 64 = 0

75. 3.02x + 1.48 = 10.92

2x5/3 = -64

77. 2.15x — 4.63 = x + 1.19

3.02x = 9.44

x5/3 = -32

1.15x = 5.82

79. 3.16 (x + 4.63) = 4.19 (x - 7.24)



x = (—32)3/5 = (~25)1/5 = (-2)3 = -8

x = —^3.13 3.02

x =

5 82

% 5.06

3.16x + 14.63 = 4.19x - 30.34

«=>

44.97 = 1.03x

44 97 x = y^- » 43.66

2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.2 Modeling with Equations

„

0.26*-1.94

81. ——= 1.76 3.03 - 2;44* 7.27

* =

85. P = 21 + 2w 87,

ax + b cx 4- d

=2

b2 — c2 — a2

3F r3 5 = — 4tt

a2x — (a 4- 1)* .= — (a — 1)

(a2 — (a 4- 1)^ * = —a + 1

-a 4- 1 * = —=-*az — a — 1

3V





= iVc2—a2

r — 0.032(250)-2.5

8 -2.5

10,000

10,000

97. (a) The shrinkage factor when w = 250 is S = - = - = 0.00055. So the beam shrinks 0.00055 x 12.025 « 0.007 rn, so when it dries it will be 12.025 — 0.007 = 12.018 m long. 0.032io - 2.5

(b) Substituting S = 0.00050 we get 0.00050 =-

5 = 0.032tu — 2.5

7.5 = 0.032io

7.5 , w =-% 234.375. So the water content should be 234.375 kg/m . 0.032 &7

99. (a) Solving for v when P = 10,000 we get 10,000 = 15.6o3

o

o3^ 641.02

(b) Solving for v when P = 50,000 we get 50,000 = 15.6o3

o3 « 3205.13

101. (a) 3 (0) 4- & — 5 = A: (0) — A: 4- 1

k — 5 = —k + l

2k = 6

(b) 3 (1) 4- & — 5 = £ (1) — & 4-1

o

3+k-5 = k-k+\

(c) 3 (2) 4- A; — 5 = k(2) — ^4-1

6 + k — 5 = 2k — k+\





8.6 km/h. v & 14.7 km/h.

k =3

k-

2 = 1

k= 3

A:-|-l=A:4-l.*=2isa solution for every value

of k. That is, * = 2 is a solution to every member of this family of equations.

1.2

MODELING WITH EQUATIONS

1. An equation modeling a real-world situation can be used to help us understand a real-world problem using mathematical methods. We translate real-world ideas into the language of algebra to construct our model, and translate our mathematical results back into real-world ideas in order to interpret our findings. 3. (a) A square of side * has area A = x2.

(b) A rectangle of length / and width w has area A = Iw. ,

(c) A circle of radius

\

r has area A = ■nr2.

1 wall 1 5. A painter paints a wall in * hours, so the fraction of the wall she paints in one hour is --= —. * hours * 7. If n is the first integer, then n 4- 1 is the middle integer, and n 4- 2 is the third integer. So the sum of the three consecutive integers is « 4- (« 4-1) 4- (« 4- 2) = 3» 4- 3.

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

26

CHAPTER 1

Equations and Inequalities

9. If s is the third test score, then since the other test scores are 78 and 82, the average of the three test scores is 78 + 82 + 5 3

160 + 5 =

'

11. If x dollars are invested at 2^% simple interest, then the first year you will receive 0.025.x dollars in interest. 13. Since w is the width of the rectangle, the length is three times the width, or 3 w. Then area = length x width = 3w x w = 3w2 ft2. 1 h

3

60 mm

*

15. Since distance = ratex time we have distance = 5 x (45 mm) ——— = js mi17. If x is the quantity of pure water added, the mixture will contain 25 oz of salt and 3 + x gallons of water. Thus the 25 concentration is-. 3 +x

19. If d is the number of days and m the number of miles, then the cost of a rental is C = 65d + 0.20m. In this case,

60 x = — = 200. So 200 mL of 60% acid solution must be mixed with 100 mL of 30% solution to get 300 mL of 50% acid 0.3 6 solution.

47. Let x be the number of grams of silver added. The weight of the rings is 5 x 18 g = 90 g. 5 rings

Pure silver

Mixture

90

X

90+ x

0.90

0

0.75

0.90 (90)

Ox

0.75 (90 + x)

Grams Rate (% gold) Value So 0.90 (90) + Ox = 0.75 (90+ x)

81 = 67.5 + 0.75x

0.75x = 13.5

x =

= 18. Thus

18 grams of silver must be added to get the required mixture.

49. Let x be the number of liters of coolant removed and replaced by water. 60% antifreeze

60% antifreeze (removed)

Water

Mixture

3.6

X

X

3.6

0.60

0.60

0

0.50

0.60 (3.6)

—0.60x

Ox

0.50(3.6)

Liters Rate (% antifreeze) Value

_a T/r

so 0.60 (3.6) — 0.60x + Ox = 0.50 (3.6)

2.16 — 0.6x = 1.8

— 0.6x = — 0.36

x =- =0.6.

0.6

—

Thus 0.6 liters must be removed and replaced by water.

©2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

28

CHAPTER 1 Equations and Inequalities

51. Let c be the concentration of fruit juice in the cheaper brand. The new mixture that Jill makes will consist of 650 mL of the original fruit punch and 100 mL of the cheaper fruit punch. Original Fruit Punch

Cheaper Fruit Punch

Mixture

mL

650

100

750

Concentration

0.50

c

0.48

0.50 • 650

100 c

0.48 ■ 750

Juice So 0.50-650+ 100c = 0.48-750

325 + 100c = 360

o

100c = 35

c = 0.35. Thus the cheaper

brand is only 35% fruit juice.

53. Let / be the time in minutes it would take Candy and Tim if they work together. Candy delivers the papers at a rate of jq of the job per minute, while Tim delivers the paper at a rate of ^ of the job per minute. The sum of the fractions of the job that each can do individually in one minute equals the fraction of the job they can do working together. So we have y = ^ + ^

560 = 81 + 7/

560 = 15/

/ = 37 j minutes. Since j of a minute is 20 seconds, it

would take them 37 minutes 20 seconds if they worked together.

55. Let / be the time, in hours, it takes Karen to paint a house alone. Then working together, Karen and Betty can paint a house 11 in |/ hours. The sum of their individual rates equals their rate working together, so —F g = — 1

6+/ =9

«

3l

13 —h g = — 1

t

/ = 3. Thus it would take Karen 3 hours to paint a house alone.

57. Let / be the time in hours that Wendy spent on the train. Then y — / is the time in hours that Wendy spent on the bus. We construct a table: Rate

Time

Distance

By train

40

/

40/

By bus

60

li 2

‘

The total distance traveled is the sum of the distances traveled by bus and by train, so 300 = 40/ + 60 ^y — /j

300 = 40/ + 330 — 60/

—30 = —20/

/ =

= 1-5 hours. So the time spent on the train is

5.5 — 1.5 = 4 hours. i

'

59. Let r be the speed of the plane from Montreal to Los Angeles. Then r + 0.20r = 1,20r is the speed of the plane from Los Angeles to Montreal. Rate Montreal to L.A.

r

L.A. to Montreal

1.2 r

Time

Distance

2500

2500

r 2500

2500

1.2 r

T, ... . Qi 2500 2500 The total time is the sum of the times each way, so 94 =-- 6 r 1.2 r 55- 1.2r = 2500-6- 1.2 + 2500-6

66r = 18,000 + 15,000

55 2500 2500 — =-- 6 r 1.2 r

66r = 33,000

r =

DO

= 500.

Thus the plane flew at a speed of 500 mi/h on the trip from Montreal to Los Angeles.

61. Let x be the distance from the fulcrum to where the mother sits. Then substituting the known values into the formula given, we have 100 (8) = 125x

800 = 125x

x = 6.4. So the mother should sit 6.4 feet from the fulcrum.

©2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.3 Quadratic Equations

29

63. Let / be the length of the lot in feet. Then the length of the diagonal is / + 10. We apply the Pythagorean Theorem with the hypotenuse as the diagonal. So l* 2 * * * * 7 * 9 * 11 + 502 = (/ + 10)2 «

«

/2 + 2500 =/2 + 20/+ 100

20/= 2400

/ = 120. Thus the length of the lot is 120 feet.

65. Let h be the height in feet of the structure. The structure is composed of a right cylinder with radius 10 and height

and a

cone with base radius 10 and height Ih. Using the formulas for the volume of a cylinder and that of a cone, we obtain the equation 14007T = 7r (10)2 (jh^ + j7r (10)2 (5^) 9 both sides by-) J

126 = Ih

14007T = ^y^/t + Aypl/z

126 = 6h + h (multiply

h = 18. Thus the height of the structure is 18 feet.

100tt

6

67. Pythagoras was bom about 569 Euclid was bom about 325

BC

Archimedes was bom in 287

1.3



BC

in Samos, Ionia and died about 475

and died about 265

BC

BC

BC.

in Alexandria, Egypt,

in Syracuse, Sicily and died in 212

BC

in Syracuse.

QUADRATIC EQUATIONS —b ± \/b2 — 4a c

1. (a) The quadratic formula states that x =

(b) In the equation ^x2 — x — 4 = 0, a =

l)2 — 4 |

b = —1, and c = —4. So, the solution of the equation is

l(-4)

(«

-(-^V 2|

*

2a

1 ±3

-2 or 4.

1

(1)1

3. For the quadratic equation ax2 + bx + c = 0 the discriminant is D = b2 — 4ac. If D > 0, the equation has two real solutions; if D = 0, the equation has one real solution; and if D < 0, the equation has no real solution. 5. x2+x = 12

x2 4-x — 12 — 0

(x + 4) (x — 3) = 0

4=>

x + 4 = 0 orx — 3 = 0. Thus, x — —4 or x = 3.

7. x2 — lx + 12 = 0

(x — 4) (x — 3) = 0

, 9. 3x2 — 5x — 2 = 0

«

(3x + 1) (x — 2) = 0

1 3x + 1 = 0 or x — 2 = 0. Thus, x = — - or x = 2.

o 11. 2 y2 + ly + 3 = 0

(2 y + 1) (y + 3) = 0

1 2y + 1 = 0 or y + 3 = 0. Thus, y = — - or y = —3.

13. 6x2 + 5x = 4

6x2 + 5x — 4 = 0

x — 4 = 0 or x — 3 = 0. Thus, x = 4 or x =3.

(2x — 1) (3x + 4) = 0

x2 — 5x — 500 = 0

2x — 1 = 0 or 3x + 4 = 0. If 2x — 1 = 0,

then x = j; if 3x + 4 = 0, then x = — |.

15. x2 = 5 (x + 100)

x2 = 5x + 500

(x — 25) (x + 20) = 0

x — 25 = 0 or

x + 20 = 0. Thus, x = 25 or x = —20.

17. x2+2x—5 = 0

19. x2 — 6x — 11 = 0

x2+2x = 5

x2-6x

= 11

x2+2x+l = 5+1



(x + l)2 = 6

x2 — 6x+9=ll+9

o

=>

x+l=±v/6

(x — 3)2 = 20

=>

x = —1±\/6.

x — 3 = ±2^5

«•

x = 3 ± 2V5. 21. x2+x —1=0

x2+x = ^

x2+x + | = | + ^

(x + 5)

= 1

=>

x + 2=±l«

x = — 2±1.

Sox = -2 — l= —|orx = -2 + l = 2-

23.

x2

+ 22x + 21 = 0

«

x2 + 22x = -21

x2 + 22x + 112 =-21 + 112 =-21 + 121

=>

x + ll=±10

x = —11 ± 10. Thus,x = —1 orx = —21.

o

(x + ll)2 = 100

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 1

30

Equations and Inequalities

25. 2x2 + 8x + 1 = 0 =>

x2+4x + A=0

x + 2 = ±J 4

27. 4xz—x = 0

x = -2 ±

x2 — 4x = 0

sox = i ~ i =Oorx = ^

«

x2+4x = -A

x2 + 4x + 4 = -A+4

vT4

1 1 2 1 . xz-4x +A = 64 64 “

(* “ 7 “ 64

8

x

~

^8

(x + 3) (x — 5) = 0

x + 3 = 0 or x - 5 = O.Thus, x = -3 or x = 5.

31. x2 - 7x + 10 = 0

(x - 5) (x - 2) = 0

x - 5 = 0 orx - 2 = O.Thus, x = 5 orx = 2.

o

(x

^

x

~

8

^

8’

|

29. x2 - 2x - 15 = 0 33. 2x2+x-3 = 0

(x+2)2 = J

- l)(2x + 3) = 0

x - 1 = 0 or 2x + 3 = 0. Ifx - 1 = 0, thenx = 1; if2x + 3 = 0,

*3

then x = — 4.

35. x2 + 3x + 1 = 0 X

x2 + 3x = — 1

-j , IT . 75 + J=±yjl=±—

37. x2 + 12x - 27 = 0 x + 6 = ±377

x2 + 3x + \ = — 1 + | = f

(* + 3)

= f

=>

o

(x + 6)2 = 63

=>

y \/5 x = -|±—.

x2 + 12x = 27

x2 + 12x.+ 36 = 27 + 36

x = —6 ± 3-s/7.

39. 3x2 + 6x — 5 = 0

x + l=±V§ 41. z2 — |z +



x2 + 2x - f = 0

x2 + 2x = 4

x2 + 2x + 1 = 4 + 1

(x + l)2 = f

=>

x =-1 + ^71.

=0

0

=>

43. 4x2 + 16x - 9 = 0



z =

(2x — 1) (2x + 9) = 0

o

2x - 1 = 0 or 2x + 9 = 0. If 2x - 1 = 0, then x = ±; if

2x + 9 = 0, then x = — |.

45. to2 = 3

(10 — 1)

tn2 — 3to + 3 = 0

=>

-(-3)±7(-3)2-4(l)(3)

3 ±79^12

2(1)

2

w =

3 + 7=3 0. . Since

2

the discriminant is less than 0, the equation has no real solution.

47. x2 — 75x + 1=0 49.107-16^ x =

+

5

=

=>

0

x =

-b ± Vb2 - 4ac

_4(1H1)

75 ±75^4

75 ±1

2(1)

2a

=>

-b ± Vb2 - 4ac

— (— 16) ± ^(— 16)2 -4(10) (5)

16 + 7256-200

16 + 756

8 ± 714

2a

2(10)

20

20

10

51. 3x2 +2x + 2 = 0=>x=

-b ± Vb2 - 4ac _ 2a

-(2) ±7(2)2 -4 (3) (2) _ 2 (3)

-2 + 74=^24 6

-2 + 7=20

. . Since

the discriminant is less than 0, the equation has no real solution.

53. x2-0.01 lx- 0.064 = 0

=>

- (-0.011) ± yj(-0.011)2 - 4 (1) (-0.064) x =

0.011 ± 70.000121 +0.256

0.011 +0.506

2(1)

, _ 0.011 +0.506 0.011 -0.506 Thus, x &-1- = 0.259 orx %-—-= -0.248.

2

2

55. x2 -2.450x + 1.501 = 0

=>

- (-2.450) ± J(-2.450)2 -4(1) (1.501) x =

2.450 ± 76.0025 - 6.004

2.450 ± 7-0.0015

2(1)

Thus, there is no real solution.

©2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.3 Quadratic Equations

57. 2.232x2 — 4.112x = 6.219 - (-4.112)

2.232x2 - 4.112x - 6.219 = 0

±J(~4.112)2 - 4 (2.232) (-6.219)

* “

2 (2.232)

31

=>

4.112 ± V16.9085 + 55.5232

4.112 d= 8.511

4.464

4.464

“

. Thus.

4.112-8.511 4:112 + 8.511 x - -——-= —0.985 orx =-——-= 2.828. 4.464 4.464

59. h

=

\et2 jgt2

+ vot

jgt2

+ t>ot —

-(wo)±yW-4(V) (-A)

h

0.

Using the quadratic formula,

-V0±y/v2 + 2gh

g

:Gg) 61. 4 = 2jc2 + 4x/z

=

2x2 + 4x/t -4=0. Using the quadratic formula.

- (4A) ± 7(4h)2 - 4 (2) (-4) 2(2)

_4/t + Vl6/t2 + 84 ~

4

~~

2(-2/t±V4/t2 + 24)

_2 h±^h^U

4

2

111

63. -- H-- = 5+as+oc

c (s + 6) + c

~4/* ± ^4 (4/r2 +- 24)

+ a) = (s + a) (s + b)

1

_4/z ± 2^2 + 24

~~

C5 + Zrc +

4

9

+ ac = 5 +

+ aZ? >

«

s2 + (a + b — 2c) s + (ah — ac — be) = 0. Using the quadratic formula, — (a + b — 2c) ± yj(a + b — 2c)2 — 4 (1) (ab — ac — be)

S =

2(1) — (a + b — 2c) ± Ja2 + b2 + 4c2 + 2ab — 4ac — 4hc — 4ah + 4ac + 4hc

=

2 — (a + b — 2c) ± \la2 +b2 + 4c2 — 2ab 2

65. D = b2 — 4ac = (—6)2 — 4 (1) (1) = 32. Since D is positive, this equation has two real solutions. 67. D — b2 — 4ac = (2.20)2 — 4 (1) (1.21) = 4.84 — 4.84 = 0. Since D = 0, this equation has one real solution. 69. D = b2 — 4ac = (5)2 — 4 (4) ^^ = 25 — 26 = —1. Since D is negative, this equation has no real solution. 71. D = b2 — 4ac = (r)2 —4(1) 73. a2x2 + 2ax + 1 = 0

(—5)

= r2 + 4s. Since D is positive, this equation has two real solutions.

(ax + l)2 = 0

75. ax2 — (2a + l)x + (a + 1) = 0



ax + 1 = 0. So ax + 1 = 0 then ax = — 1

[ax — (a + 1)] (x — 1) = 0



x = — —. a

ax — (a + 1) = 0 or x — 1 = 0. If

a+ 1 ax — (a + 1) = 0, then x = ——; if x — 1 = 0, then x = 1. a

77. We want to find the values of k that make the discriminant 0. Thus k2 — 4 (4) (25) = 0

k2 = 400

k = ±20.

79. Let n be one number. Then the other number must be 55 — n,since n + (55 — n) = 55. Because the product is 684, we have («) (55 — n) = 684 -(-55)±y( 55)2

4(1 )(684)

55±V3025-2736

n = -"2(1)- - n = 55217 =

«

55n — n2 — 684 55±-s/289

2- - -2- “

55±17

2

n2 — 55n + 684 = 0 on _

55+17

72

' b0 " “ ~T~ ~ T = 36 or

= 19. In either case, the two numbers are 19 and 36.

81. Let w be the width of the garden in feet. Then the length is w + 10. Thus 875 = w (w + 10) «

=>

w2 + 10u> — 875 = 0

(w + 35) (u> — 25) = 0. So id + 35 = 0 in which case w = —35, which is not possible, or to — 25 = 0 and so

to = 25. Thus the width is 25 feet and the length is 35 feet.

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32

CHAPTER 1

Equations and Inequalities

83. Let w be the width of the garden in feet. We use the perimeter to express the length / of the garden in terms of width. Since the perimeter is twice the width plus twice the length, we have 200 = 2w + 21

21 = 200 — 2w

— 100m + 2400 - 0

/ = 100 — w. Using the formula for area, we have 2400 = w (100 — to) = 100tn — w2



to = 60. If to = 40, then

(to — 40) (to — 60) = 0. So to — 40 = 0

to = 40, or to — 60 = 0

w2

/ = 100 - 40 = 60. And if w — 60, then / = 100 - 60 = 40. So the length is 60 feet and the width is 40 feet.

1

85. The shaded area is the sum of the area of a rectangle and the area of a triangle. So A = y (1) + 5 (y) (y) = 5T + T- We are given that the area is 1200 cm2, so 1200 = ^y2 +y

y2 + 2y — 2400 = 0

{y + 50) (y — 48) = 0. y is

positive, so y = 48 cm.

87. Let x be the length of one side of the cardboard, so we start with a piece of cardboard x by x. When 4 inches are removed from each side, the base of the box is x — 8 by x — 8. Since the volume is 100 in3, we get 4 (x — 8)2 = 100 x2 — 16x + 64 = 25

x2 — 16x + 39 = 0



(x — 3) (x — 13) = O.So x = 3 or x = 13. But x = 3 is not

possible, since then the length of the base would be 3 — 8 = —5, and all lengths must be positive. Thus x = 13, and the piece of cardboard is 13 inches by 13 inches.

89. Let

10

be the width of the lot in feet. Then the length is w + 6. Using the Pythagorean Theorem, we have

w2 + (m + 6)2 = (174)2 w2 + 6w — 15120 = 0



w2 + w2 + 12w + 36 = 30,276