Student Solutions Manual for Larson's Intermediate Algebra: Algebra within Reach, 6th 9781285419855, 1285419855

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Student Solutions Manual

Intermediate Algebra Algebra Within Reach SIXTH EDITION

Ron Larson The Pennsylvania State University, The Behrend College

Prepared by Ron Larson The Pennsylvania State University, The Behrend College

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CONTENTS Chapter 1

Fundamentals of Algebra .......................................................................1

Chapter 2

Linear Equations and Inequalities........................................................17

Chapter 3

Graphs and Functions...........................................................................54

Chapter 4

Systems of Equations and Inequalities ................................................90

Chapter 5

Polynomials and Factoring.................................................................145

Chapter 6

Rational Expressions, Equations, and Functions ..............................171

Chapter 7

Radicals and Complex Numbers .......................................................208

Chapter 8

Quadratic Equations, Functions, and Inequalities.............................243

Chapter 9

Exponential and Logarithmic Functions ...........................................286

Chapter 10

Conics..................................................................................................318

Chapter 11

Sequences, Series, and the Binomial Theorem .................................350

iii

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C H A P T E R 1 Fundamentals of Algebra Section 1.1

The Real Number System ......................................................................2

Section 1.2

Operations with Real Numbers..............................................................3

Section 1.3

Properties of Real Numbers ...................................................................6

Mid-Chapter Quiz..........................................................................................................8 Section 1.4

Algebraic Expressions............................................................................9

Section 1.5

Constructing Algebraic Expressions ...................................................12

Review Exercises ..........................................................................................................14 Chapter Test ...............................................................................................................15

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C H A P T E R 1 Fundamentals of Algebra Section 1.1 The Real Number System

{

}

6, − 43 , 0, 85 , 1,

1. −6, −

13.

2, 2, π , 6

{

{

4, − 19 , 0,

(a) Natural numbers: (b) Integers:

{

2, π

1 2

on the real number

21. Distance = 18 − ( −32) = 18 + 32 = 50

{ 4}

23. Distance = 0 − ( −8) = 0 + 8 = 8

}

4, 0

4, { (d) Irrational numbers: { 11}

is to the right of

19. Distance = 7 − ( −12) = 7 + 12 = 19

11, 5.5, 5.543

(c) Rational numbers: 4.2,

5 8

17. Distance = 10 − 4 = 6

} }

3, 11

because

number line.

}

(c) Rational numbers: −6, − 43 , 0, 58 , 1, 2, 6

3. −4.2,

1 2

15. − 23 > − 10 because − 23 is to the right of − 10 on the real 3 3

(b) Integers: {−6, 0, 1, 2, 6}

{

>

line.

(a) Natural numbers: {1, 2, 6}

(d) Irrational numbers: − 6,

5 8

− 19 ,

0,

3, 11

}

5.5, 5.543

25. Distance = 35 − 0 = 35 27. Distance = ( −6) − ( −9) = ( −6) + 9 = 3 29. 10 = 10

5. (a) The point representing the real number 3 lies between 2 and 4.

31. −225 = 225

3 0

1

2

3

33. − − 34 = − 34

4

(b) The point representing the real number between 2 and 3.

5 2

lies

greater than 2.

5 2 0

1

2

3

37. 47 > −27 because 47 = 47 and −27 = 27, and 47

4

(c) The point representing the real number − 72 lies between − 4 and −3.

−3

−2

−1

− 5.2

7.

4 5

−5

−4

< 1 because

line.

−3

4 5

−2

−1

39. Label: The weight on the elevator = x

41. Label: Contestant’s weight = x

0

(d) The point representing the real number −5.2 lies between −6 and −5, but closer to −5. −6

is greater than 27. Inequality: x ≤ 2500

−7 2 −4

35. −6 > 2 because −6 = 6 and 2 = 2, and 6 is

0

is to the left of 1 on the real number

9. −5 < 2 because −5 is to the left of 2 on the real number line. 11. −5 < −2 because −5 is to the left of −2 on the real number line.

Inequality: x > 200 43. Label: Person’s height = x

Inequality: x ≥ 52 45. Label: Balance of checking account = x

Inequality: 200 ≤ x ≤ 700 47. The number line shows − 2.5 < 2 because − 2.5 is to the

left of − 2. 49. The fractions are converted to decimals and plotted on a number line to determine the order.

2

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Section 1.2 51. {−5, − 4, − 3, − 2, −1, 0, 1, 2, 3}

− 53

55. a = −1, b =

−2

1 2

−7

and − 53 from 0 is 53 .

−4

4.25 −2

0

2

4

6

79. You have more than 30 coins and fewer than 50 coins in a jar. 81. Because −4 = 4 and 4 = 4, the two possible values

7 0

4

of a are −4 and 4.

8

The distance of both −7 and 7 from 0 is 7.

83. Because −2 − 3 = −5 = 5 and 8 − 3 = 5 = 5, the

two possible values of a are −2 and 8.

67. The opposite of 5 is −5 . −5

85. Sample answers: −3, −100, − 14

5 0

2

4

6

The distance of both −5 and 5 from 0 is 5.

−3 0

2, π , − 3 3

91. Sample answers: − 12 , π , −

3 5

5

87. Sample answers:

89. Sample answers: 34 , 112 , 0.16

69. The opposite of − 53 is 53 .

−1

5 3

77. u ≥ 16

65. The opposite of − 7 is 7.

−2

2

75. x < 0

63. −π = π

−4

1

0

The distance of both − 4.25 and 4.25 from 0 is 4.25.

61. − 3.5 = −3.5

−6

−1

− 4.25 −6

59. − −85 = −85

−4

5 3

73. The opposite of − 4.25 is 4.25.

< −2

−8

3

is − 53 .

The distance of both

1 2

57. a = − 92 , b = −2, − 92

5 3

71. The opposite of

53. {5, 7, 9}

−1
−6

9.

−6 −7

2.

3 4


− 23

3. −14 + 9 − 15 = (−14 + 9) − 15 = −5 − 15 = −20 2 3

( )

+ − 76 =

4 6

( )

+ − 76 = − 63 = − 12

5. −2( 225 − 150) = −2(75) = −150 6. ( −3)( 4)( −5) = ( −12)(−5) = 60

( )( )

7 − 8 = 7. − 16 21

÷

15 8

=

5 18

⋅

8 15

=

4 27

3

2. d = −4.4 − 6.9 = −11.3 = 11.3

4.

5 18

1 6

27 ⎛ 3⎞ 9. ⎜ − ⎟ = − 125 ⎝ 5⎠

10.

42 − 6 16 − 6 10 + 13 = + 13 = + 13 = 2 + 13 = 15 5 5 5

11. (a) Associative Property of Multiplication

(b) Multiplicative Inverse Property 12. −6( 2 x − 1) = −6( 2 x) + −6 ⋅ ( −1) = −12 x + 6 13. 3x 2 − 2 x − 5 x 2 + 7 x − 1 = −2 x 2 + 5 x − 1

14. x( x + 2) − 2( x 2 + x − 13) = x 2 + 2 x − 2 x 2 − 2 x + 26 = ( x 2 − 2 x 2 ) + ( 2 x − 2 x) + 26 = − x 2 + 26 15. a(5a − 4) − 2( 2a 2 − 2a) = 5a 2 − 4a − 4a 2 + 4a = a 2

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16

Chapter 1

F undamentals of Algebra

16. 4t − ⎡⎣3t − (10t + 7)⎤⎦ = 4t − [3t − 10t − 7] = 4t − [−7t − 7] = 4t + 7t + 7 = 11t + 7 17. Evaluating an expression is solving the expression when values are provided for its variables.

(a) When x = −1:

(b) When x = 3:

7 + ( x − 3) = 7 + ( −1 − 3) 2

= 7 + ( − 4)

2

7 + ( x − 3) = 7 + (3 − 3)

2

= 7 + 16 = 23

2

2

= 7 + 02 = 7+0 = 7

18. Verbal Model: 17 ⋅ Length of each piece = Total length

Equation:

17 ⋅ n = 102 n = 6

Each piece should be 6 inches. 19. Verbal Model:

Volume of 1 cord = Length ⋅ Width ⋅ Height

Equation:

V = 4⋅4⋅8 V = 128 cubic feet

Verbal Model:

Volume of 5 cords = 5 ⋅ Volume of 1 cord

Equation:

V = 5 ⋅ 128 = 640

There are 640 cubic feet in 5 cords of wood. 20. The product of a number n and 5 is decreased by 8 is translated into the algebraic expression 5n − 8. 21. Verbal Description:

Labels:

The sum of two consecutive even integers, the first of which is 2n 2n = first even integer 2n + 2 = second even integer

Algebraic Description:

2n + ( 2n + 2) = 4n + 2

22. Perimeter = 2l + 2(0.6l ) = 2l + 1.2l = 3.2l

Area = l (0.6l ) = 0.6l 2 When l = 45: Perimeter = 3.2( 45) = 144 Area = 0.6( 45) = 1215 2

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C H A P T E R 2 Linear Equations and Inequalities Section 2.1

Linear Equations...................................................................................18

Section 2.2

Linear Equations and Problem Solving...............................................22

Section 2.3

Business and Scientific Problems ........................................................27

Mid-Chapter Quiz........................................................................................................33 Section 2.4

Linear Inequalities ................................................................................36

Section 2.5

Absolute Value Equations and Inequalities.........................................41

Review Exercises ..........................................................................................................43 Chapter Test ...............................................................................................................51

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C H A P T E R 2 Linear Equations and Inequalities Section 2.1 Linear Equations 1. (a)

11.

x = 0 3(0) − 7 = 2

2 x + 10 = 24

−7 ≠ 2

Not equivalent

Not a solution

13. 3( 4 − 2t ) = 5

x = 3

(b)

12 − 6t = 5

?

3(3) − 7 = 2

Equivalent

9−7 = 2

15.

2 = 2

2 x = 10

x = 4

2x 10 = 2 2 x = 5

?

4 + 8 = 3( 4) 12 = 12

Not equivalent

Solution x = −4

(b)

17.

?

−4 + 8 = 3(−4)

19. 3x − 12 = 0

x = −4

(−4)

3 x = 12

?

−1 ≠ 3 x = 12

1 4

(12)

21.

?

3 = 3

3x + 15 = 0

Original equation

3x + 15 − 15 = 0 − 15 Subtract 15 from each side. 3 x = −15 3x −15 = 3 3 x = −5 9.

6x + 4 = 0 6x + 4 − 4 = 0 − 4

= 3

Solution 7.

0 = 0

?

Check: 3( 4) = 12

12 = 12

3x 12 = 3 3 x = 4

= 3

Not a solution (b)

?

Check: 3 − 3 = 0

x = 3

Not a solution

1 4

x −3 = 0 x −3+3 = 0+3

4 ≠ −12

5. (a)

2x − 7 = 3 2x − 7 + 7 = 3 + 7

Solution 3. (a)

x + 5 = 12 2( x + 5) = 2(12)

?

Combine like terms. Divide each side by 3.

? ⎛ 2⎞ Check: 6⎜ − ⎟ + 4 = 0 ⎝ 3⎠ ?

6 x = −4

−4 + 4 = 0

−4 6x = 6 6 4 x = − 6 2 x = − 3

0 = 0

Simplify.

4 x = x + 10 4 x − x = x − x + 10 3 x = 10 Equivalent

18

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Section 2.1

23.

3t + 8 = −2 3t + 8 − 8 = −2 − 8

25.

? ⎛ 10 ⎞ Check: 3⎜ − ⎟ + 8 = −2 ⎝ 3⎠

−10 + 8 = −2

3t −10 = 3 3 10 t = − 3

−2 = −2

3( x − 4) = 7 x + 6

31.

3x − 12 = 7 x + 6

?

3t = −10

3 x − 7 x − 12 = 7 x − 7 x + 6 −4 x − 12 = 6 −4 x − 12 + 12 = 6 + 12 −4 x = 18 18 −4 x = −4 −4

7 − 8 x = 13 x

x = −

7 − 8 x + 8 x = 13 x + 8 x 7 = 21x

27.

? ⎛1⎞ = 13⎜ ⎟ ⎝ 3⎠ ? 13 = 3 ? 13 = 3 13 = 3

8 ⎞ ? 63 12 ⎛ 9 3⎜ − − ⎟ = − + 2⎠ 2 2 ⎝ 2 ⎛ 17 ⎞ ? 51 3⎜ − ⎟ = − 2 ⎝ 2⎠ 51 51 − = − 2 2 33.

t −

t −

2 5

=

3 2

2 5

2 5

=

3 2

+

t =

3 x − 1 = 2 x + 14 3x − 2 x − 1 = 2 x + 14 − 2 x x − 1 = 14

35.

x − 1 + 1 = 14 + 1 x = 15 ?

Check: 3(15) − 1 = 2(15) + 14

8 x − 64 = 24

?

Check: 8(11 − 8) = 24 ?

8(3) = 24 24 = 24

19 10

19 10

−

?

2 5

=

4 10

=

15 10 3 2

=

? ?

=

3 2 3 2 3 2 3 2

10 −3t = −3 −3 10 t = − 3

8( x − 8) = 24

8x 88 = 8 8 x = 11

2 5

−

−3t = 10

44 = 44

8 x = 88

+

19 10

2t − 5t = 10

?

8 x − 64 + 64 = 24 + 64

Check:

t t − =1 5 2 t⎞ ⎛t 10⎜ − ⎟ = (1)10 ⎝5 2⎠

45 − 1 = 30 + 14

29.

9 2

⎛ 9 ⎞ ? ⎛ 9⎞ Check: 3⎜ − − 4 ⎟ = 7⎜ − ⎟ + 6 ⎝ 2 ⎠ ⎝ 2⎠

7 21x = 21 21 1 = x 3

⎛1⎞ Check: 7 − 8⎜ ⎟ ⎝ 3⎠ 8 7 − 3 21 8 − 3 3 13 3

19

Linear Equations

Check:

−

10 10 − 3 − 3 5 2 10 10 + −15 6 2 5 − + 3 3 3 3 1

?

= 1 ?

= 1 ?

= 1 ?

= 1 = 1

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20

37.

Chapter 2

Linear Eq uations and Inequalities

8x x − = −3 5 4 x⎞ ⎛ 8x − ⎟ = 20( −3) 20⎜ 4⎠ ⎝5

43.

4( 2 x − 4) = 3( x + 3) − 24

32 x − 5 x = −60 27 x = −60

8 x − 16 = 3 x + 9 − 24

27 x −60 = 27 27 20 x = − 9

8 x − 16 = 3 x − 15 8 x − 3 x − 16 = 3 x − 3 x − 15 5 x − 16 = −15

⎛ 20 ⎞ ⎛ 20 ⎞ 8⎜ − ⎟ ⎜ − ⎟ 9⎠ ⎝ 9⎠ − Check: ⎝ 5 4 ⎛ 160 ⎞ ⎛ 20 ⎞ ⎜− ⎟ ⎜− ⎟ ⎝ 9 ⎠ − ⎝ 9⎠ 5 4 32 5 − + 9 9 27 − 9 −3 39.

2 1 (2 x − 4) = ( x + 3) − 4 3 2 ⎡2 ⎤ ⎡1 ⎤ 6 ⎢ ( 2 x − 4)⎥ = ⎢ ( x + 3) − 4⎥ 6 ⎣3 ⎦ ⎣2 ⎦

5 x − 16 + 16 = −15 + 16 5x = 1

?

= −3

5x 1 = 5 5 1 x = 5

?

= −3 ?

= −3

Check:

?

= −3

⎤ ? 1⎛ 1 2⎡ ⎛1⎞ ⎞ 2⎜ ⎟ − 4⎥ = ⎜ + 3⎟ − 4 3 ⎢⎣ ⎝ 5 ⎠ 2 5 ⎝ ⎠ ⎦ 2⎛ 2 20 ⎞ ? 1 ⎛ 1 15 ⎞ ⎜ − ⎟ = ⎜ + ⎟ − 4 3⎝ 5 5⎠ 2⎝ 5 5⎠

= −3

2 ⎛ 18 ⎞ ? 1 ⎛ 16 ⎞ ⎜− ⎟ = ⎜ ⎟ − 4 3⎝ 5 ⎠ 2⎝ 5 ⎠ ? 12 8 20 − = − 5 5 5 12 12 − = − 5 5

0.3 x + 1.5 = 8.4 10(0.3 x + 1.5) = (8.4)10 3x + 15 = 84 3 x + 15 − 15 = 84 − 15 3 x = 69

45.

3x 69 = 3 3 x = 23

4y = 4y + 3 ?

4y − 4y = 4y + 3 − 4y

Check: 0.3( 23) + 1.5 = 8.4

0 = 3

?

6.9 + 1.5 = 8.4

0 ≠ 3

8.4 = 8.4 41.

4y − 3 = 4y 4y − 3 + 3 = 4y + 3

1.2( x − 3) = 10.8 1.2 x − 3.6 = 10.8

No solution 47.

4( 2 x − 3) = 8 x − 12 8 x − 12 = 8 x − 12

10(1.2 x − 3.6) = (10.8)10

8 x − 12 − 8 x = 8 x − 12 − 8 x

12 x − 36 = 108

−12 = −12

12 x − 36 + 36 = 108 + 36

Infinitely many solutions

12 x = 144 12 x 144 = 12 12 x = 12 ?

Check: 1.2(12 − 3) = 10.8 ?

1.2(9) = 10.8 10.8 = 10.8

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Section 2.1 49. The fountain reaches its maximum height when the velocity of the stream of water is zero.

12 x + 36 = 7 x + 21 12 x + 36 − 7 x = 7 x + 21 − 7 x

0 + 32t = 48 − 32t + 32t

5 x + 36 = 21

32t = 48

5 x + 36 − 36 = 21 − 36

32t 48 = 32 32 3 t = seconds = 1.5 seconds 2

5 x = −15 −15 5x = 5 5 x = −3

51. Equivalent equations are equations that have the same set of solutions.

?

Check: 12 ⎣⎡( −3) + 3⎤⎦ = 7 ⎡⎣( −3) + 3⎤⎦ ?

12[0] = 7[0]

53. Dividing by zero cannot be done because it is undefined.

0 = 0

55. 6( x + 3) = 6 x + 3

−9 y + 9 y − 4 = −9 y + 9 y

Contradiction

2x 3

−4 = 0

+ 4 =

1x 3

+ 12

− 13 x + 4 =

1x 3

− 13 x + 12

1x 3 1x 3

−9 y − 4 = −9 y

61.

6 x + 18 ≠ 6 x + 3

2x 3

−4 ≠ 0 No solution

+ 4 = 12

7( x + 6) = 3( 2 x + 14) + x

63.

+ 4 − 4 = 12 − 4 1x 3

21

12( x + 3) = 7( x + 3)

59.

0 = 48 − 32t

57.

Linear Equations

7 x + 42 = 6 x + 42 + x 7 x + 42 = 7 x + 42

= 8

7 x + 42 − 7 x = 7 x + 42 − 7 x

( )

3 13 x = 3(8)

42 = 42

x = 24

Infinitely many solutions

Conditional equation

65.

38h + 162 = 257 38h + 162 − 162 = 257 − 162 38h = 95 38h 95 = 38 38 5 = 2.5 hours h = 2 So, the repair work took 2.5 hours.

67. (a)

t

1

1.5

2

Width

250

200

Length

250

Area

62,500

3

4

5

166.7 125

100

83.3

300

333.4 375

400

416.5

60,000

55,577.8

46,875 40,000

34,694.5

(b) Because the perimeter is fixed, as t increases the length increases and the width and area decrease. The maximum area occurs when the length and width are equal. 69. False. Multiplying both sides of an equation by zero does not yield an equivalent equation because this does not follow the Multiplication Property of Equality.

71. No. An identity in standard form would be written as ax + b = 0, with a = b = 0. But, the equation would not be in standard form.

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Chapter 2

22 73.

Linear Equations and Inequalities

3x = 3x + 1 3x − 3x = 3x − 3 x + 1 0 ≠1 The equation is a contradiction. Let x = 1 pound, then 3 pounds ≠ 4 pounds.

75.

5w + 3 = 28 5w + 3 − 3 = 28 − 3 5w = 25 5w 25 = 5 5 w = 5

The equation is a conditional equation. Let w be the number of full (5-day) work weeks. So, 28 days are worked.

77.

5 6

−

2 3

=

5 6

−

4 6

=

1 6

79. −12 − (6 − 5) = −12 − 1 = −13 81.

2(1) 2 = =1 1+1 2 2(5) 10 5 = = 5+1 6 3

83. 3( −1) − 7 = −3 − 7 = −10 = 10 3(1) − 7 = 3 − 7 = −4 = 4

85.

n 4

87.

1n 2

−5

Section 2.2 Linear Equations and Problem Solving 1. Verbal Model:

26 ⋅ Amount of each paycheck + Bonus = Income for year

Labels:

Amount of each paycheck = x Bonus = 2800 Income for year = 37,120

Equation:

26 x + 2800 = 37,120 26 x = 34,320 x = 1320

Each paycheck will be $1320. 3. Percent: 30%

Parts out of 100: 30 Decimal: 0.30 Fraction:

30 100

=

3 10

11. Verbal Model:

Labels:

Parts out of 100: 7.5

Percent = p Equation:

=

3 40

7. Percent: 66 23 %

Parts out of 100: 66 23 Decimal: 0.66. . . Fraction:

2 3

9. Percent: 100%

Parts out of 100: 100 Decimal: 1.00 Fraction: 1

a = p⋅b a = (0.35)( 250)

Decimal: 0.075 75 1000

Compared number = a Base number = b

5. Percent: 7.5%

Fraction:

Compared = Percent ⋅ Base number number

a = 87.5

So, 35% of 250 is 87.5. 13. Verbal Model:

Compared Percent Base = number (decimal form) ⋅ number Labels:

Compared number = a Percent = 0.425 Base number = 816

Equation:

a = 0.425(816) a = 346.8

So, 346.8 is 42.5% of 816.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.2

15. Verbal Model:

Labels:

Compared Base = Percent ⋅ number number Compared number = a

Linear Equations and Problem Solving

23. Verbal Model:

Labels:

Equation:

a = (0.125)(1024) a = 128

Labels:

Compared Base = Percent ⋅ number number Compared number = a Percent = p Base number = b

Equation:

a = p⋅b

42 = b 1.2 35 = b So, 42 is 120% of 35. 25. Verbal Model:

Percent Compared Base = number (decimal form) ⋅ number Labels:

So, 600 is 0.4% of 150,000. 19. Verbal Model:

Labels:

Percent = p Base number = b Equation:

Base number = b Equation:

a = p⋅b a = ( 2.50)(32)

So, 22 is 0.8% of 2750. 27. Verbal Model:

Labels:

21. Verbal Model:

Labels:

Compared Base = Percent ⋅ number number Compared number = a Percent = p Base number = b

Equation:

Compared Base = Percent ⋅ number number Compared number = a Percent = p

a = 80

So, 80 is 250% of 32.

22 = 0.008b 22 = b 0.008 2750 = b

Compared Base = Percent ⋅ number number Compared number = a

Compared number = 22 Percent = 0.008

a = (0.004)(150,000) a = 600

a = p⋅b 42 = (1.2)(b)

So, 128 is 12.5% of 1024. 17. Verbal Model:

Compared number = a Base number = b

Base number = b a = p⋅b

Compared Base = Percent ⋅ number number Percent = p

Percent = p Equation:

23

Base number = b Equation:

a = p⋅b 496 = ( p)(800) 496 800

= p

0.62 = p p = 62% So, 496 is 62% of 800.

a = p⋅b 84 = (0.24)(b) 0 = b 0.24 350 = b

So, 84 is 24% of 350.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

24

Chapter 2

Linear Eq uations and Inequalities

29. Verbal Model:

Labels:

Compared Base = Percent ⋅ number number

41. (a) Unit price =

1.69 = $0.4225 per ounce 4

(b) Unit price =

2.39 = $0.3983 per ounce 6

Compared number = a Percent = p

The 6-ounce tube is a better buy.

Base number = b Equation:

a = p⋅b

43.

2.4 = ( p )( 480) 2.4 = p 480 0.005 = p

3 x = 12 x = 4

p = 0.5%

45.

So, 2.4 is 0.5% of 480. 31. Verbal Model:

Labels:

Compared Base = Percent ⋅ number number

y =

Compared number = a Base number = b

47.

a = p⋅b 2100 = ( p)(1200) 2100 1200

Labels:

Commission = a Percent = p Price of home = b

Equation:

a = p⋅b 12,250 = p ⋅ 175,000 12,250 p ⋅ 175,000 = 175,000 175,000

x 4 = 7 5.5 5.5 ⋅ x = 7 ⋅ 4 x = 5.09

So, 2100 is 175% of 1200. Commission = Percent ⋅

216 7

5.5 x = 28

= p

175% = p

33. Verbal Model:

y 6 = 36 7 7 ⋅ y = 36 ⋅ 6 7 y = 216

Percent = p Equation:

2 x = 6 3 3⋅ x = 6⋅2

49.

Price of home

2 x = 6 4 4⋅ x = 6⋅2 4 x = 12 x = 3

51. Verbal Model: Tax 1 Tax 2 = Assessed value 1 Assessed value 2

Labels:

Assessed value 1 = 160,000 Tax 2 = 1650

0.07 = p

Assessed value 2 = 110,000

So, it is a 7% commission. 120 meters 12 2 35. = = 180 meters 18 3 37.

40 milliliters 0.04 liter 4 1 = = = 1 liter 1 100 25

2.32 39. (a) Unit price = = $0.16 per ounce 14.5

(b) Unit price =

Tax 1 = x

1650 x = 160,000 110,000

Proportion:

110,000 ⋅ x = 1650 ⋅ 160,000 110,000 x = 264,000,000 x = 2400 So, the tax is $2400.

0.99 = $0.18 per ounce 5.5

The 14 12 -ounce bag is a better buy.

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Section 2.2

53.

Linear Equations and Problem Solving

63. Verbal Model:

5 x = 105 360 360 ⋅ 5 = 105 ⋅ x

Compared = Percent ⋅ Base number number

1800 = 105 x

Labels:

1800 = x 105 17.1 ≈ x

Compared number = 66 Percent = p Base number = 220

Equation:

So, about 17.1 gallons of fuel are used.

a = p⋅b 66 = p ⋅ 220

55. The ratio of a to b is a b if a and b have the same units.

66 220

Examples: Price earnings ratio, gear ratio

p = 30% So, 66 is 30% of 220.

59. Verbal Model:

Compared = Percent ⋅ Base number number

Base number = 225 Equation:

Total price Total units $1.10 = ≈ $0.06 per ounce 20 ounces

65. Unit price =

Compared number = a Percent = 0.20

Total price Total units $2.29 = ≈ $0.11 per ounce 20 ounces

67. Unit price =

a = p⋅b = (0.20)(225) = 45

So, 45 is 20% of 225.

69.

61. Verbal Model:

12 = 2 y 6 = y

Compared number = 13 Percent = 0.26 Base number = b

Equation:

a = p⋅b

13 = 0.26 ⋅ b 13 0.26

= b

50 = b

So, 13 is 26% of 50.

y y − 2 = 6 4 4 y = 6( y − 2)

4 y = 6 y − 12

Compared = Percent ⋅ Base number number Labels:

= p

0.30 = p

57. Not always. You need to cross-multiply to solve for an unknown quantity.

Labels:

25

71.

z −3 z +8 = 3 12 12( z − 3) = 3( z + 8)

12( z − 3)

=

3( z + 8)

3 3 4( z − 3) = z + 8 4 z − 12 = z + 8 3z − 12 = 8 3 z = 20 z =

20 3

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26

Chapter 2

73. Verbal Model:

Labels:

Linear Eq uations and Inequalities Number Number of = Percent ⋅ laid off employees

81. Verbal Model:

Total defective units Total units

Number laid off = a Percent = p

Labels:

Total units = 200,000 Defective units = 1

a = p⋅b

25 = ( p)(160)

Sample = 75

( p)(160) 25 = 160 160 25 = p 160 15.625 = p

Labels:

Tip =

Percent

(decimal form)

75 ⋅ x = 200,000 ⋅ 1 75 x = 200,000 x = 2667

of ⋅ Cost meal

Tip = a Percent = 0.15 Cost of meal = 32.60

Equation:

a = 0.15(32.60) a = 4.89

So, you should leave a $4.89 tip. 77. Verbal Model:

Labels:

Defective = Percent ⋅ Total parts parts

Total parts = b Equation:

So, the expected number of defective units is 2667. 83.

h 6 = 86 11 11 ⋅ h = 86 ⋅ 6 11h = 516 516 11 h ≈ 46.9 h =

So, the height of the tree is about 46.9 feet. 85. To change percents to decimals divide by 100. To change decimals to percents multiply by 100.

a = p⋅b 3 = (0.015)(b)

(0.015)(b) 3 = 0.015 0.015 3 = b 0.015 200 = b So, the sample contained 200 parts. Tax $12.50 125 1 = = = 79. Pay $625 6250 50

42 100

Examples: 42% =

Defective parts = a Percent = p

x 1 = 200,000 75

Proportion:

So, 15.625% of the workforce was laid off. 75. Verbal Model:

Defective units Sample

Total defective units = x

Number of employees = b Equation:

=

= 0.42

0.38 = (0.38)(100)% = 38% 87. Mathematical modeling is the use of mathematics to solve problems that occur in real-life situations. For examples review the real-life problems in the exercise set. 4 ⋅ 89. − 15

15 16

60 = − 240 = − 14

91. (12 − 15) = ( −3) = −27 3

3

93. Commutative Property of Addition 95. Distributive Property 97. 2 x − 5 = x + 9

2 x = x + 14 x = 14 99. 2 x +

3 2

=

3 2

2x = 0 x = 0

101. −0.35 x = 70 x = −200

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Section 2.3

Business and Scientific Problems

27

Section 2.3 Business and Scientific Problems 1. Verbal Model:

Labels:

Equation:

Selling = Cost + Markup price Selling price = 64.33

5. Verbal Model:

Labels:

Selling = Cost + Markup price Selling price = 26,922.50

Cost = 45.97

Cost = x

Markup = x

Markup = 4672.50

64.33 = 45.97 + x

Equation:

x = 64.33 − 45.97

26,922.50 = x + 4672.50 26,922.50 − 4672.50 = x

x = 18.36 The markup is $18.36.

22,250.00 = x The cost is $22,250.00.

Verbal Model:

Markup Markup = ⋅ Cost rate

Verbal Model:

Markup =

Labels:

Markup = 18.36

Labels:

Markup = 4672.50

Equation:

Markup rate = x

Markup rate = x

Cost = 45.97

Cost = 22,250.00

18.36 = x ⋅ 45.97 18.36 = x 45.97 0.40 ≈ x

The markup rate is 40%. 3. Verbal Model:

Labels:

Equation:

Selling = Cost + Markup price Selling price = 250.80

4672.50 = x 22,250.00 0.21 = x

7. Verbal Model:

Labels:

250.80 = x + 98.80 152.00 = x

Verbal Model:

Markup Markup = ⋅ Cost rate

Labels:

Markup = 98.80

Equation:

98.80 = x ⋅ 152.00 98.80 = x 152.00 0.65 = x

Markup = x

x = 0.852 ⋅ 225.00 x = 191.70

The markup is $191.70.

Verbal Model:

Selling = Cost + Markup price

Labels:

Selling price = x Cost = 225.00

Markup rate = x Cost = 152.00

Markup ⋅ Cost rate

Cost = 225.00

250.80 − 98.80 = x The cost is $152.00.

Markup =

Markup rate = 85.2%

Markup = 98.80 Equation:

4672.50 = x ⋅ 22,250.00

The markup rate is 21%.

Cost = x

Equation:

Markup ⋅ Cost rate

Markup = 191.70

Equation:

x = 225.00 + 191.70 x = 416.70

The selling price is $416.70.

The markup rate is 65%.

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28

Chapter 2

9. Verbal Model:

Labels:

Equation:

Linear Eq uations and Inequalities Sale = List price − Discount price Sale price = 25.74

13. Verbal Model:

Labels:

Percent = 0.60

Discount = x

List price = x

25.74 = 49.95 − x

Equation:

The discount is $24.21. Discount =

Discount List ⋅ rate price

Discount = 24.21 Discount rate = x

Equation:

The list price is $45.00.

Verbal Model:

Discount =

Labels:

Discount = x Discount rate = 0.40

24.21 = x ⋅ 49.95

List price = 45.00

Equation:

The discount is $18.00.

Sale = List price − Discount price Sale price = x

15. Verbal Model:

Labels:

Discount = 189.00

Sale = List price − Discount price Sale price = 831.96 List price = x

List price = 300.00

Equation:

x = 0.40 ⋅ 45.00 x = 18.00

The discount rate is 48.5%.

Labels:

Discount = 323.54

Equation:

x = 300.00 − 189.00

831.96 = x − 323.54 831.96 + 323.54 = x

x = 111.00

1155.50 = x

The sale price is $111.00.

The list price is $1155.50. Discount List ⋅ rate price

Verbal Model:

Discount =

Labels:

Discount = 189.00

Equation:

Discount List ⋅ rate price

List price = 49.95 24.21 = x 49.95 0.485 ≈ x

11. Verbal Model:

27.00 = 0.60 x 27.00 = x 0.60 45.00 = x

x = 24.21

Labels:

Sale price = 27.00

List price = 49.95

x = 49.95 − 25.74

Verbal Model:

Sale = Percent ⋅ List price price

Discount List ⋅ rate price

Verbal Model:

Discount =

Labels:

Discount = 323.54

Discount rate = x

Discount rate = p

List price = 300.00

List price = 1155.50

189.00 = x ⋅ 300.00 189.00 = x 300.00 0.63 = x

The discount rate is 63%.

Equation:

323.54 = p ⋅ 1155.50 323.54 = p 1155.50 0.28 = p

The discount rate is 28%.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.3

17. Verbal Model:

Labels:

Sale = List price − Discount price Sale price = 45

Business and Scientific Problems

23. Verbal Model:

Amount of Amount of = Amount of + solution 2 solution 1 final solution

Labels:

List price = 75 45 = 75 − x

Quarts of solution 2 = 24 − x Percent of final solution = 45%

45 − 75 = − x 30 = x

Quarts of final solution = 24

The discount is $30. 19. Verbal Model:

Labels:

Equation:

Sale = List price − Discount price

24 − x = 16

8 quarts of solution 1 and 16 quarts of solution 2 are needed.

Discount = x 16 = 20 − x

x = 20 − 16

25. Verbal Model:

x = 4 Verbal Model:

Discount =

Labels:

Discount = 4

Discount List ⋅ rate price

Labels: Number of pounds of seed 1 = x Number of pounds of seed 2 = 100 − x Cost per pound of seed 2 = 20

List price = 20

Number of pounds of final seed mix = 100

4 = x ⋅ 20 4 20

Cost of Cost of = Cost of final + seed 2 seed 1 seed mix

Cost per pound of seed 1 = 12

Discount rate = x

Equation:

0.15 x + 0.60( 24 − x) = 0.45( 24) 0.15 x + 14.4 − 0.60 x = 10.8 −0.45 x = −3.6 x = 8

Sale price = 16 List price = 20

Equation:

Percent of solution 1 = 15% Quarts of solution 1 = x Percent of solution 2 = 60%

Discount = x

Equation:

Cost per pound of final seed mix = 14

= x

Equation:

0.20 = x

12 x + 20(100 − x) = 14(100) 12 x + 2000 − 20 x = 1400

The discount rate is 20%.

−8 x = −600

21. Verbal Model:

x = 75

Amount of Amount of = Amount of + solution 2 solution 1 final solution

Labels:

Equation:

29

Percent of solution 1 = 20% Gallons of solution 1 = x Percent of solution 2 = 60% Gallons of solution 2 = 100 − x Percent of final solution = 40% Gallons of final solution = 100 0.20 x + 0.60(100 − x) = 0.40(100) 0.20 x + 60 − 0.60 x = 40 −0.40 x = −20

75 pounds of seed 1 and 25 pounds of seed 2 are needed. 27. Verbal Model:

Labels:

Distance = Rate ⋅ Time Distance = x Rates = 480 and 600 4 3

Time = Equation:

( 43 ) + 600( 43 )

x = 480

x = 1440

The planes are 1440 miles apart after 113 hours.

x = 50 100 − x = 50

50 gallons of solution 1 and 50 gallons of solution 2 are needed.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

30

Chapter 2

Linear Eq uations and Inequalities

29. Verbal Model:

Distance = Rate ⋅ Time

Labels:

33. Verbal Model:

Labels:

Distance = 317

Interest = I Principal = 5000

Rate for first part of trip = 58 Time for first part of trip = x

Rate = 6.5%

Rate for second part of trip = 52

Time = 6

Time for second part of trip = 5 34 − x Equation:

(1st part of trip) ( 2nd part of trip)

58 x = 58 ⋅ x

(

Interest = Principal ⋅ Rate ⋅ Time

)

(

52 5 34 − x = 52 ⋅ 5 34 − x

(

317 = 58 x + 52

)

5 34

− x

)

I = (5000)(0.065)(6)

Equation:

I = 1950 The interest is $1950. 35. Verbal Model:

Labels:

Interest = Principal ⋅ Rate ⋅ Time Interest = 500

317 = 58 x + 299 − 52 x

Principal = P

18 = 6 x

Rate = 7%

3 = x

Time = 2

The first part of the trip took 3 hours, and the second part took 5 34 − 3 = 2 34 hours. 31. (a) Your rate =

1 5

Friend's rate =

500 = P(0.14) 500 = P 0.14 3571.43 ≈ P

job per hour 1 8

job per hour

The principal required is $3571.43.

(b) Verbal Model: Work done Work done Work = by first + by second done person person

Labels:

Work done = 1 Rate for you =

37. E = IR

E = R I 39.

1 5

Rate for friend =

S = L 1− r

1 8

Time for friend = t

( 15 )(t ) + ( 18 )(t ) 1 = ( 15 + 18 )t t 1 = ( 13 40 ) 1=

1 13 40

1 hours. It will take 313

41.

40 13

h = 36t + 1 2 at 2 2( h − 36t − 50) = at 2

1 2 at + 50 2

h − 36t − 50 =

2h − 72t − 100 = at 2

= t

1 = 313

S = L − rL S = L(1 − r )

Time for you = t

Equation:

500 = ( P)(0.07)( 2)

Equation:

2h − 72t − 100 = a t2

= t

S = 2π r 2 + 2π rh

43.

S − 2π r 2 = 2π rh

(S

− 2π r 2 )

= h 2π r S − r = h 2π r

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.3 45. Verbal Model:

Labels:

Equation:

Perimeter = 2 Width + 2 Height Perimeter = 40

Business and Scientific Problems

51. Verbal Model:

Labels:

Selling price = Cost + Markup Selling price = 157.14

Height = x

Cost = 130.95

Width = x − 4

Markup = x

40 = 2 x + 2( x − 4)

Equation:

157.14 = 130.95 + x

40 = 2 x + 2 x − 8

x = 157.14 − 130.95

40 = 4 x − 8

x = 26.19

48 = 4 x 12 = x The height is 12 inches. 47. The formula for the volume V of a cube with side length

31

The markup is $26.19. 53. Verbal Model:

Labels:

Markup = Markup rate ⋅ Cost

Markup = 37.33 Markup rate = p

s is V = s 3.

Cost = 46.67

49. Perimeter is measured in linear units, such as inches, yards, and meters. Area is measured in square units, such as square feet, square centimeters, and square miles. Volume is measured in cubic units, such as cubic inches, cubic feet, and cubic meters.

Equation:

37.33 = p ⋅ 46.67 37.33 = p 46.67 0.80 ≈ p

The markup rate is 80%. 55. Verbal Model:

Labels:

Total cost = Cost per minute ⋅ Number of minutes

Total cost = 0.70 Cost per minute = 0.02 Number of minutes = x

Equation:

0.70 = 0.02 x 35 = x

The length of the call is 35 minutes. Verbal Model:

Discount = Discount rate ⋅ List price

Labels:

Discount = x Discount rate = 20% List price = 0.70

Equation:

x = 0.20 ⋅ 0.70 x = 0.14

Verbal Model:

Selling price = List price − Discount

Labels:

Selling price = x List price = 0.70 Discount = 0.14

Equation:

x = 0.70 − 0.14 x = 0.56

The call would have cost $0.56.

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32

Chapter 2

Linear Eq uations and Inequalities

57. Verbal Model:

Labels:

Selling price

Cost + Markup = Cost = x

63. Verbal Model:

Labels:

Distance = 5000 Rate = 17,500

Markup = 0.10 x Selling price = 59.565 Four tires cost: 3($79.42) = $238.26 Each tire costs:

Time = t 5000 = 17,500 ⋅ t

Equation:

5000 = t 17,500

$238.26 = $59.565 4

2 hour = t 7 17.14 ≈ t

x + 0.10 x = 59.565

Equation:

1.10 x = 59.565

About 17.14 minutes are required.

x = 54.15 The cost to the store for each tire is $54.15. 59. Verbal Model:

Labels:

Total Adult Children = + sales sales sales

Total sales = 2200

65. Verbal Model:

Labels:

Equation:

2200 = 6(3 x) + 4 x 2200 = 18 x + 4 x

( 301 )(t ) + (151 )(t ) 1 + 1 t 1 = ( 30 15 ) 1 3 30

100 = x

3 t 30

= t

10 = t It will take 10 minutes.

61. Verbal Model: Original Some Final Pure = antifreeze antifreeze − antifreeze + antifreeze solution solution solution

67. Common formula: V = π r 2 h

( )

2

Equation: V = π 3 12 12 V = 147π

Labels:

V ≈ 461.8 cubic centimeters

Number of gallons of original antifreeze = 5 Percent of antifreeze in original mix = 40% Number of gallons of antifreeze withdrawn = x Number of gallons of pure antifreeze = x

69. The sale price of an item is the list price minus the discount rate times the list price. 71. No, it quadruples. The area of a square of side s is s 2 . If

Percent of pure antifreeze = 100%

the length of the sides is 2s, the area is ( 2 s ) = 4 s 2 . 2

Number of gallons of final solution = 5 Percent of antifreeze in final solution = 50% 0.40(5) − 0.40 x + 1.00 x = 0.50(5) 2 − 0.40 x + 1.00 x = 2.5 0.60 x = 0.5 x = 5 6

1 15

1 =

1 =

2200 = 22 x 100 children’s tickets are sold.

1 30

Time for each pump = t

Number of children’s tickets = x

Equation:

Work done = 1 Rate of larger pump =

Price of adult tickets = 6 Price of children’s tickets = 4

Work done Work done Work = by smaller + by larger done pump pump Rate of smaller pump =

Number of adult tickets = 3x

Equation:

Distance = Rate ⋅ Time

gallon must be withdrawn and replaced.

5 6

73. (a) 21 + ( −21) = 0, so it is −21.

( 211 ) = 1, so it is

(b) 21

1. 21

75. (a) −5 x + 5 x = 0, so it is 5x.

1 ⎛ 1⎞ (b) −5 x⎜ − ⎟ = 0, so it is − . 5 x 5x ⎝ ⎠ 77. 2 x( x − 4) + 3 = 2 x 2 − 8 x + 3 79. x 2 ( x − 4) − 2 x 2 = x 3 − 4 x 2 − 2 x 2 = x 3 − 6 x 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Mid-Chapter Quiz for Chapter 2 81.

52 = 0.40 x

33

83. 117 = p(900)

130 = x

13% = p

Mid-Chapter Quiz for Chapter 2 1.

4x − 8 = 0 4x − 8 + 8 = 0 + 8

5.

4x = 8 4x 8 = 4 4 x = 2

1 3 x + 6 = x −1 4 2 ⎛1 ⎞ ⎛3 ⎞ 4⎜ x + 6 ⎟ = 4⎜ x − 1⎟ 4 2 ⎝ ⎠ ⎝ ⎠ x + 24 = 6 x − 4 x − x + 24 = 6 x − 4 − x 24 = 5 x − 4

Check:

24 + 4 = 5 x − 4 + 4

?

4( 2) − 8 = 0

28 = 5 x

8−8 = 0

28 5x = 5 5 28 = x 5

0 = 0 2. −3( z − 2) = 0 −3( z − 2)

0 = −3 −3 z − 2 = 0

Check: ? 3 28 1 ⎛ 28 ⎞ ⎛ ⎞ ⎜ ⎟ + 6 = ⎜ ⎟ −1 4⎝ 5 ⎠ 2⎝ 5 ⎠

z − 2+ 2 = 0+ 2

7 30 ? 42 5 + = − 5 5 5 5 37 37 = 5 5

z = 2

Check: ?

−3( 2 − 2) = 0 ?

−3(0) = 0 0 = 0 3.

2( y + 3) = 18 − 4 y 2 y + 6 = 18 − 4 y 2 y + 4 y + 6 = 18 − 4 y + 4 y 6 y + 6 − 6 = 18 − 6

6.

2b b + 5 2 4b 5b + 10 10 9b 10 9b

Check: ?

2( 2 + 3) = 18 − 4( 2) ?

2(5) = 18 − 8 10 = 10 4. 5t + 7 = 7(t + 1) − 2t

5t + 7 = 7t + 7 − 2t 5t + 7 = 5t + 7 7 = 7 Infinitely Many

= 3 = 3 = 30

b =

6 y = 12 6y 12 = 6 6 y = 2

= 3

30 10 = 9 3

Check:

⎛ 10 ⎞ ⎛ 10 ⎞ 2⎜ ⎟ ⎜ ⎟ ⎝3⎠ + ⎝3⎠ 5 2 ⎛ 20 ⎞ ⎛ 10 ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 3⎠ + ⎝3⎠ 5 2 20 10 + 15 6 4 5 + 3 3 9 3 3

?

= 3 ?

= 3 ?

= 3 ?

= 3 ?

= 3 = 3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

34

Chapter 2

7.

Linear Eq uations and Inequalities

4− x 5 +5 = 5 2 ⎛4 − x ⎞ ⎛5⎞ + 5⎟ = 10⎜ ⎟ 10⎜ ⎝ 5 ⎠ ⎝ 2⎠

9.

0.25 x + 6.2 = 4.45 x + 3.9 0.25 x − 0.25 x + 6.2 = 4.45 x + 3.9 − 0.25 x 6.2 = 4.2 x + 3.9 6.2 − 3.9 = 4.2 x − 3.9

2( 4 − x ) + 50 = 25

2.3 = 4.2 x

8 − 2 x + 50 = 25

2.3 4.2 x = 4.2 4.2 0.55 ≈ x

−2 x + 58 = 25 −2 x + 58 − 58 = 25 − 58 −2 x = −33 −2 x −33 = −2 −2 33 x = 2

10.

0.42 x + 6 = 5.25 x − 0.80 0.42 x + 6 − 5.25 x = 5.25 x − 0.80 − 5.25 x −4.83x + 6 = −0.80 −4.83 x + 6 − 6 = −0.80 − 6 −4.83x = −6.80

Check: 33 ? 2 +5 = 5 5 2 8 33 − ? 2 2 +5 = 5 5 2 ? 5 25 1 − ⋅ +5 = 2 5 2 5 10 ? 5 − + = 2 2 2 5 5 = 2 2

−4.83x −6.80 = −4.83 −4.83 x ≈ 1.41

4−

8.

11 5 = 12 16 11 11 5 11 3x + − = − 12 12 16 12 15 44 3x = − 48 48 29 3x = − 48 3x 29 = − ÷3 3 48 29 1 ⋅ x = − 48 3 29 x = − 144 3x +

11. 0.45 is 45 hundredths, so 0.45 = 9 20

which reduces to

and because percent means hundredths, 0.45 = 45%.

12. Verbal Model:

Labels:

Compared Base = Percent ⋅ number number Compared number = a Percent = p Base number = b a = p⋅b

Equation:

500 = ( 2.50)(b) 500 = b 2.50 200 = b 500 is 250% of 200. Total price Total units $4.85 = = $0.40 per ounce 12 ounces

13. Unit price =

14. Verbal Model:

Labels:

Number defective Total defective = Sample Shipment Number defective = 1 Sample = 150

Check:

⎛ 29 ⎞ 3⎜ − ⎟+ ⎝ 144 ⎠ 29 − + 48

45 100

Total defective = x

11 5 = 12 16 ?

44 ? 5 = 48 16 15 ? 5 = 48 16 5 5 = 16 16

Shipment = 750,000 Equation:

1 x = 150 750,000 750,000 = 150 x 5000 = x

The expected number of defective units is 5000.

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35

Mid-Chapter Quiz for Chapter 2 17. Verbal Model:

15. Store computer:

Verbal Model: Labels:

Equation:

Discount =

Discount List ⋅ rate price

Discount = x

Amount of Amount of = Amount of + solution 2 solution 1 final solution Labels:

Percent of solution 1 = 25%

Discount rate = 0.25

Gallons of solution 1 = x

List price = 1080

Percent of solution 2 = 50%

x = (0.25)(1080)

Gallons of solution 2 = 50 − x Percent of final solution = 30%

x = 270.00 Verbal Model:

Selling = List price − Discount price

Labels:

Selling price = x

Gallons of final solution = 50

0.25 x + 0.50(50 − x) = 0.30(50)

Equation:

0.25 x + 25 − 0.50 x = 15 25 − 0.25 x = 15

List price = 1080

−0.25 x = −10

Discount = 270 Equation:

x = 810

Mail-order catalog computer:

Verbal Model:

Selling = List price + Shipping price

Labels:

Selling price = x

Equation:

x = 40

x = 1080 − 270

50 − x = 10 40 gallons of solution 1 and 10 gallons of solution 2 are required. 18. Verbal Model:

Labels:

Distance = 300

List price = 799

Rate of first part = 62

Shipping = 14.95

Time for first part = x

x = 799 + 14.95

Rate of second part = 46 Time for seond part = 6 − x

x = 813.95

The store computer is the better buy. 16. Verbal Model:

Labels:

Total Overtime = Regular wages + wages wages Total wages = 616 Regular wages = 40(12.25) Overtime wages = x(18) Number of hours = x

Equation:

Distance = Rate ⋅ Time

616 = 40(12.25) + x(18) 616 = 490 + 18 x

Equation: 300 = 62 x + 46(6 − x) 300 = 62 x + 276 − 46 x 24 = 16 x 1.5 hours = x (first part of trip at 62 miles/hour ) 4.5 hours = 6 − x (second part of trip at 46 miles/hour ) 19. Verbal Model:

Labels:

Work time + Part done = Part by you done by friend Work done = 1

126 = 18 x

Time for each portion = t

7 = x

Per hour work rate for you =

You worked 7 hours of overtime.

1 3

Per hour work rate for friend = Equation:

1 =

( 13 )t + ( 15 )t

1 =

8 t 15

15 8

It will take

15 8

1 5

= t

hours, or 1.875 hours

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

36

Chapter 2

Linear Eq uations and Inequalities

20. Perimeter of square I = 20

4s = 20 s = 5 Perimeter of square II = 32 4 s = 32 s = 8 Length of side of square III = 5 + 8 = 13 Area = s 2 = 132 = 169

The area of square III is 169 square inches.

Section 2.4 Linear Inequalities 1. The length of the interval [− 3, 5] is 5 − (− 3) = 8.

21.

3. The length of the interval ( − 9, 2] is 2 − (− 9) = 11.

3 x − 2 < 12

3 x < 10

3x − 2 + 2 < 12 + 2

3x 10 < 3 3 10 x < 3

3x < 14 3x 14 < 3 3 14 x < 3

5. The length of the interval (− 3, 0) is 0 − ( − 3) = 3. 7.

x 0

1

2

3

4

5

The inequalities are not equivalent.

9. x > 3.5

23.

3.5 x 0

1

2

3

4

5

7 x − 6 ≤ 3 x + 12

4 x ≤ 18

7 x − 6 + 6 ≤ 3 x + 12 + 6

4x 18 ≤ 4 4 9 x ≤ 2

7 x ≤ 3 x + 18

1 2

11. x ≥

7 x − 3 x ≤ 3 x + 18 − 3 x 4 x ≤ 18

1 2

4x 18 ≤ 4 4 9 x ≤ 2

x −2

−1

0

1

2

3

13. −5 < x ≤ 3 −5

3

The inequalities are equivalent.

x −6

−4

−2

0

2

4

25. 15. 4 > x ≥ 1

x − 4 ≥ 0 x − 4+ 4 ≥ 0+ 4 x ≥ 4

x 0

1

2

3

4

5 x

17.

3 2

0

≥ x > 0 27.

3 2

x −1

0

1

1

2

2

3

4

5

6

x + 7 ≤ 9 x + 7 −7 ≤ 9−7

3

x ≤ 2

19. 3.5 < x ≤ 4.5 3.5

x 0

4.5

1

2

3

x 2

3

4

5

29. 2 x < 8

2x 8 < 2 2 x < 4 x −2

0

2

4

6

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.4 41.

31. −9 x ≥ 36

3x − 11 > − x + 7 4 x − 11 > 7 4 x − 11 + 11 > 7 + 11 4 x > 18

x −4

−3

−2

−1

0

4x 18 > 4 4 9 x > 2

− 34 x < −6

33.

37

3x − 11 + x > − x + 7 + x

−9 x 36 ≤ −9 −9 x ≤ −4 −5

Linear Inequalities

− 43 ⋅ − 43 x > −6 ⋅ − 43 x > 8

9 2

x

x 0

35.

2

4

6

8

0

2

4

10

43.

5 − x ≤ −2

−3 x + 7 < 8 x − 13 −3 x − 8 x + 7 < 8 x − 8 x − 13

5 − x − 5 ≤ −2 − 5

−11x + 7 < −13

− x ≤ −7

−11x + 7 − 7 < −13 − 7

−1 ⋅ x ≥ −7 ⋅ −1

−11x < −20

x ≥ 7

−11x −20 > −11 −11 20 x > 11

x 5

37.

6

7

8

9

2 x − 5.3 > 9.8 2 x − 5.3 + 5.3 > 9.8 + 5.3

20 11

2 x > 15.1

x 0

2x 15.1 > 2 2 x > 7.55

39.

7

8

−30 ≥ 7 y + 40

9

−40 − 30 ≥ 7 y + 40 − 40 −70 ≥ 7 y

5 − 3x − 5 < 7 − 5

−10 ≥ y

−3 x < 2

y

2 −3x > −3 −3 2 x > − 3

− 15

47.

− 10

5

0

0 < 2x − 5 < 9 0 + 5 < 2x − 5 + 5 < 9 + 5

−2 3

5 < 2 x < 14

x

−1

3

−3 y − 30 ≥ 4 y + 40

5 − 3x < 7

−2

2

3 y − 3 y − 30 ≥ 3 y + 4 y + 40

x 6

1

−3( y + 10) ≥ 4( y + 10)

45.

7.55 5

6

0

1

5 2x 14 < < 2 2 2 5 < x < 7 2 5 2

7 x

0

2

4

6

8

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

38

Chapter 2

49.

Linear Eq uations and Inequalities 53.

8 < 6 − 2 x ≤ 12 8 − 6 < 6 − 6 − 2 x ≤ 12 − 6

2x − 4 ≤ 4

and

2x + 8 > 6

2x − 4 + 4 ≤ 4 + 4

and

2x + 8 − 8 > 6 − 8

2 < −2 x ≤ 6

2x ≤ 8

and

2 x > −2

2 6 −2 x > ≥ −2 −2 −2 −1 > x ≥ −3

2x 8 ≤ 2 2 x ≤ 4

and

2x −2 > 2 2 x > −1

−3 ≤ x < −1

and

−1 < x ≤ 4 x

−4

51.

−3

−2

−1

x

0

−2 −1

−1 < −0.2 x < 1

55.

−1 −0.2 x 1 > > −0.2 −0.2 x −0.2 5 > x > −5 −5

57.

4

or

x − 5 ≥ 10

and

x − 5 + 5 ≥ 10 + 5 x ≥ 15

5 3 x −1≥ 9 − x 2 2 5 3 3 3 x + x −1≥ 9 − x + x 2 2 2 2 4x − 1 ≥ 9

3 x + 11 < 3 3x + 11 − 11 < 3 − 11

4x − 1 + 1 ≥ 9 + 1

3 x < −8

4 x ≥ 10

3x 8 < − 3 3 8 x < − 3

4x 10 ≥ 4 4 5 x ≥ 2

7.2 − 1.1x > 1

− 83

5 2 x

−4 −3 −2 −1

0

1

2

3

4

1.2 x − 4 > 2.7

or

7.2 − 1.1x − 7.2 > 1 − 7.2

1.2 x − 4 + 4 > 2.7 + 4

−1.1x > −6.2

1.2 x > 6.7

−1.1x −6.2 < −1.1 −1.1 62 x < 11

1.2 x 6.7 > 1.2 1.2 67 x > 12 −∞ < x < ∞

61. Verbal Model:

x −2

−1

63. Verbal Model:

Transportation Total money + Other costs ≤ for trip costs Transportation costs = 1900 Other costs = C Total money = 4500 Inequality:

5

There is no solution.

8

7 x + 11 < 3 + 4 x

Labels:

4

−3x −3 < −3 −3 x −3

x 0

2

8 − 3x > 5

5

−4

1

8 − 3x − 8 > 5 − 8

−5 < x < 5 −8

0

1900 + C ≤ 4500 1900 + C − 1900 ≤ 4500 − 1900 C ≤ 2600

Label:

Inequality:

0

1

2

90 ≤ Perimeter ≤ 120 Perimeter = 2( x + 22) 90 ≤ 2( x + 22) ≤ 120 2( x + 22) 90 120 ≤ ≤ 2 2 2 45 ≤ x + 22 ≤ 60 45 − 22 ≤ x + 22 − 22 ≤ 60 − 22 23 ≤ x ≤ 38

C must be no more than $2600.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.4 65. The graph of x < − 2 contains a parenthesis. A

parenthesis is used when the endpoint is excluded.

77.

67. Yes, because dividing by a number is the same as multiplying by its reciprocal.

x − 4 x +3 ≤ 3 8 ⎛x − 4 ⎞ ⎛ x⎞ + 3⎟ ≤ ⎜ ⎟ 24 24⎜ ⎝ 3 ⎠ ⎝8⎠ 8 x − 32 + 72 ≤ 3x 8 x + 40 ≤ 3x

x −1

0

1

2

3

8 x − 3 x + 40 ≤ 3x − 3 x

70. Matches graph (e).

5 x + 40 ≤ 0 5 x + 40 − 40 ≤ 0 − 40

x −2

−1

0

1

2

5 x ≤ −40

71. Matches graph (d).

5x 40 ≤ − 5 5 x ≤ −8

x −2

−1

0

1

2

39

8( x − 4) + 72 ≤ 3x

69. Matches graph (a). −2

Linear Inequalities

3

72. Matches graph (b).

x − 12 − 10 − 8 − 6 − 4 − 2

0

x −2

−1

0

1

2

3x 2x − 4 < −3 5 3 ⎛ 3x ⎞ ⎛ 2x ⎞ 15⎜ − 4⎟ < ⎜ − 3⎟15 5 3 ⎝ ⎠ ⎝ ⎠ 9 x − 60 < 10 x − 45

79.

73. Matches graph (f ) . x −3

−2

−1

0

1

2

74. Matches graph (c).

9 x − 10 x − 60 < 10 x − 45 − 10 x

x −2

75.

−1

0

1

2

− x − 60 < −45

x x > 2− 4 2 x⎞ ⎛ x⎞ ⎛ 4⎜ ⎟ > ⎜ 2 − ⎟ 4 2⎠ ⎝ 4⎠ ⎝ x > 8 − 2x

− x − 60 + 60 < −45 + 60 − x < 15 x > −15 − 15 x

x + 2x > 8 − 2x + 2x

− 20 − 16 − 12 − 8

3x > 8

−4 ≤ 2 − 3x − 6 < 11 −4 ≤ −4 − 3x < 11 −4 + 4 ≤ −4 − 3 x + 4 < 11 + 4 0 ≤ −3 x < 15

8 3

0 15 −3 x ≥ > −3 −3 −3 0 ≥ x > −5

x 0

1

0

−4 ≤ 2 − 3( x + 2) < 11

81.

3x 8 > 3 3 8 x > 3

−4

2

3

−5 x −6

−4

−2

0

2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

40

Chapter 2

Linear Eq uations and Inequalities

2x − 3 < 3 2 −6 < 2 x − 3 < 6

83.

−3
−2 −3 −3 < x − 4 < 6

85.

99. x > a or x < b; The solution set includes all values on the real number line. ?

1>

101. 4 = −5

4 < 5

−3 + 4 < x − 4 + 4 < 6 + 4 1 < x < 10 1

93. The multiplication and division properties differ. The inequality symbol is reversed if both sides of the inequality are multiplied or divided by a negative real number.

?

103. −7 = 7

7 = 7

10 x

0

3

6

9

12

?

87. Verbal Model:

3(9) = 27

18 ≠ 27

27 = 27

Temp in Temp in Temp in > > Miami Washington New York The average temperature in Miami, therefore, is greater than (>) the average temperature in New York. 89. Verbal Model:

?

105. 3(6) = 27

Not a solution

Solution

?

107. 7( 2) − 5 = 7 + 2 ?

Operating < $12,000 cost

Inequality:

Operating cost = 0.35m + 2900 0.35m + 2900 < 12,000

109.

42 − 5 = 14

9 = 9

37 ≠ 14

2 x − 17 = 0 2 x = 17 2x 17 = 2 2 17 x = 2

0.35m < 9100 0.35m 9100 < 0.35 0.35 m < 26,000

91. Verbal Model:

Labels:

Second plan > First plan First plan: $12.50 per hour Second plan: $8 + $0.75n per hour where n represents the number of units produced.

Inequality:

Not a solution

2 x − 17 + 17 = 0 + 17

0.35m + 2900 − 2900 < 12,000 − 2900

The maximum number of miles is 26,000.

?

14 − 5 = 9

Solution Label:

?

7(6) − 5 = 7 + 6

111. 32 x = −8

32 x −8 = 32 32 1 x = − 4

8 + 0.75n > 12.5 0.75n > 4.5 n > 6

If more than 6 units are produced per hour, the second payment plan yields the greater hourly wage.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.5

Absolute Value Equations and Inequalities

41

Section 2.5 Absolute Value Equations and Inequalities 19. 5 − 2 x + 10 = 6

1. 4 x + 5 = 10, x = −3

5 − 2 x = −4

?

4( −3) + 5 = 10

No solution

?

−12 + 5 = 10 ?

21.

−7 = 10 7 ≠ 10

x − 2 = 0 3

Not a solution

x − 2 = 0 3 x − 2 = 0

3. 6 − 2w = 2, w = 4 ?

6 − 2( 4)

x − 2 + 6 = 6 3

= 2

x = 2

?

6−8 = 2

23. 3 2 x − 5 + 4 = 7

?

−2 = 2

3 2x − 5 = 3

2 = 2

2x − 5 = 1

Solution

2x − 5 = 1

5. x = 4

x = 4 or x = −4 7. t = −45

or

2 x − 5 = −1

2x = 6

2x = 4

x = 3

x = 2

25. 2 x + 1 = x − 4

No solution

2 x + 1 = x − 4 or x + 1 = −4

9. h = 0

x = −5

h = 0

2 x + 1 = − ( x − 4) 2x + 1 = − x + 4 3x + 1 = 4 3x = 3

11. 5 x = 15

5 x = 15

x =1

or

x = 3

5 x = −15

27. x + 8 = 2 x + 1

x = −3

x + 8 = 2x + 1

13. x + 1 = 5

or

8 = x +1

x +1 = 5

or

x = 4

x + 1 = −5

x + 8 = −2 x − 1

7 = x

3 x + 8 = −1

x = −6

3 x = −9

x = −3

15. 4 − 3 x = 0 4 − 3x = 0

29. 3 x + 1 = 3x − 3

−3 x = −4 x =

17.

3x + 1 = 3x − 3

4 3

or

1 ≠ −3

3 x + 1 = −(3 x − 3) 3 x + 1 = −3 x + 3 6x + 1 = 3

2s + 3 = 5 5 2s + 3 = 5 5 2 s + 3 = 25

x + 8 = − ( 2 x + 1)

6x = 2

or

2s + 3 = −5 5 2s + 3 = −25

2s = 22

2 s = −28

s = 11

s = −14

x =

1 3

The only solution is x = 13.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

42

Chapter 2

Linear Eq uations and Inequalities

31. 4 x − 10 = 2 2 x + 3 4 x − 10 = 2( 2 x + 3)

4 x − 10 = −2( 2 x + 3)

or

4 x − 10 = 4 x + 6

45.

1 1 y ≤ ≤ 3 3 3 −1 ≤ y ≤ 1

−

4 x − 10 = −4 x − 6 8 x − 10 = −6

−10 ≠ 6

8x = 4 x =

4 8

=

1 2

y 1 ≤ 3 3

47. 2 x + 3 > 9

2x + 3 < −9

The only solution is x = 12 . 33. x = 2

2 < 3

2 x < −12

2x > 6

x < −6

x > 3

−7 ≤ 2 x − 1 ≤ 7

Solution

−6 ≤ 2 x ≤ 8

35. x = 9

−3 ≤ x ≤ 4

9 −7 ≥ 3

51. 3 x + 10 < −1

2 ≥ 3

No solution

2 ≥ 3

Absolute value is never negative.

Not a solution 53.

y < 4 −4 < y < 4

a + 6

≥ 16 2 a + 6 ≥ 32 a + 6 ≥ 32 or a + 6 ≤ −32

39. x ≥ 6

a ≥ 26 or

x ≥ 6 or x ≤ −6

a ≤ −38

55. 0.2 x − 3 < 4

41. x + 6 > 10

−4 < 0.2 x − 3 < 4

x + 6 > 10

x + 6 < −10

or

x > 4 43.

2x + 3 > 9

49. 2 x − 1 ≤ 7

2 < 3

37.

or

−1 < 0.2 x < 7

x < −16

−1 7 < x < 0.2 0.2 −5 < x < 35

2 x < 14 −14 < 2 x < 14 −7 < x < 7

57.

t − 42.238 ≤ 0.412 −0.412 ≤ t − 42.238 ≤ 0.412 −0.412 + 42.238 ≤ t − 42.238 + 42.238 ≤ 0.412 + 42.238 41.826 ≤ t ≤ 42.65 The fastest time is 41.826 seconds and the slowest time is 42.65 seconds. 41.826

42.65 t

40

41

42

43

44

45

59. The solutions of x = a are x = a and x = − a. For

example, to solve x − 3 = 5 : x − 3 = 5 or x = 8

61. The statement, “The distance between x and zero is a, ”

can be represented using absolute values as x = a.

x − 3 = −5 x = −2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 2 1 2

63. 4 x + 1 =

4x + 1 =

81.

1 2

or

4 x + 1 = − 12

h − 68.5 ≤1 2.7 h − 68.5 ≤1 2.7 − 2.7 ≤ h − 68.5 ≤ 2.7 −1 ≤

4 x = − 32

4 x = − 12

x = − 83

x = − 18

65.8 ≤ h ≤ 71.2

65. x − 4 = 9

The heights h lie on the interval 65.8 inches ≤ h ≤ 71.2 inches.

67. x ≤ 2

83. The graph of x − 4 < 1 can be described as all real

numbers that are within one unit of four.

69. x − 19 > 2

85. 2 x − 6 ≤ 6 because:

71. x − 4 ≥ 2

−6 ≤ 2 x − 6 ≤ 6

3x − 2 +5 ≥ 5 73. 4

0 ≤ 2 x ≤ 12 0 ≤ x ≤ 6

3x − 2 ≥ 0 4

87. 4( n + 3)

−∞ < x < ∞ Absolute value is always positive. 75. x < 3

89. x − 7 > 13 x > 20

91. 4 x + 11 ≥ 27

4 x ≥ 16

77. 2 x − 3 > 5 79. (a)

43

x ≥ 4

s − x ≤

3 16

(b) 5 18 − x ≤

3 16

3 − 16 ≤ 5 81 − x ≤

3 16

3 − 16 ≤

3 16

41 8

− x ≤

85 79 − 16 ≤ − x ≤ − 16 85 16

≥ x ≥

79 16

15 5 ≤ x ≤ 516 416

Review Exercises for Chapter 2 1. (a) 45 − 7(3) = 3

45 − 21 = 3

3. (a)

28 7 140 35

24 = 3

+ +

Not a solution (b) 45 − 7(6) = 3 45 − 42 = 3

= 12

196 35

= 12

336 35 48 5

= 12

? ?

≠ 12

Not a solution

3 = 3 Solution

?

28 5

(b)

35 7

+

35 5

?

= 12

5 + 7 = 12 Solution

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

44

Chapter 2 5.

Linear Eq uations and Inequalities

3 x + 21 = 0

15.

3x + 21 − 21 = 0 − 21

5 x + 4 − 4 = 19 − 4

3 x = −21

5 x = 15

3x −21 = 3 3 x = −7

5x 15 = 5 5 x = 3

Check:

Check: ?

?

5(3) + 4 = 19

3( −7) + 21 = 0 −21 + 21 = 0 7.

5 x + 4 = 19

?

15 + 4 = 19 19 = 19

5 x − 120 = 0 5 x − 120 + 120 = 0 + 120 5 x = 120

17.

17 − 7 x = 3 17 − 7 x − 17 = 3 − 17

5x 120 = 5 5 x = 24

−7 x = −14 −7 x −14 = −7 −7 x = 2

Check: ?

5( 24) − 120 = 0

Check:

120 − 120 = 0 9.

?

17 − 7( 2) = 3 ?

x + 4 = 9

17 − 14 = 3

x − 4+ 4 = 9− 4 x = 5 Check:

3 = 3 19.

7 x − 5 = 3 x + 11 7 x − 3 x − 5 = 3 x − 3 x + 11

?

5+ 4 = 9

4 x − 5 = 11

9 = 9

4 x − 5 + 5 = 11 + 5 4 x = 16

11. −3 x = 36

4x 16 = 4 4 x = 4

−3 x 36 = −3 −3 x = −12 Check:

Check: ?

−3( −12) = 36 36 = 36 13.

− 18 x = 3

(−8)(− 18 x)

?

7( 4) − 5 = 3( 4) + 11 ?

28 − 5 = 12 + 11 23 = 23

= (3)( −8)

x = −24

Check: ?

− 18 ( −24) = 3 3 = 3

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Review Exercises for Chapter 2 21.

3( 2 y − 1) = 9 + 3 y

29.

1.4t + 2.1 = 0.9t 1.4t + 2.1 − 0.9t = 0.9t − 0.9t

6 y − 3 = 9 + 3y

0.5t + 2.1 = 0

6 y − 3y − 3 = 9 + 3y − 3y

0.5t + 2.1 − 2.1 = 0 − 2.1

3y − 3 = 9

0.5t = −2.1

3y − 3 + 3 = 9 + 3

0.5t −2.1 = 0.5 0.5 t = −4.2

3 y = 12 3y 12 = 3 3 y = 4

Check: ?

1.4( −4.2) + 2.1 = 0.9( −4.2)

Check: ?

3( 2( 4) − 1) = 9 + 3( 4)

?

−5.88 + 2.1 = −3.78

?

−3.78 = −3.78

3(7) = 9 + 12 21 = 21 23. 4 y − 4( y − 2) = 8

31. Verbal Model:

4y − 4y + 8 = 8

Labels:

8 = 8

Pay per week = 320 Pay for training = 75 2635 = 320 x + 75

Equation:

4(3x − 5) = 6( 2 x + 3)

2560 = 320 x

12 x − 20 = 12 x + 18

8 = x

12 x − 12 x − 20 = 12 x − 12 x + 18

The internship is 8 weeks long.

−20 = 18 No solution 27.

4 1 3 x − = 5 10 2 1⎤ ⎡4 ⎡3⎤ 10 ⎢ x − ⎥ = ⎢ ⎥10 5 10 ⎣ ⎦ ⎣2⎦ 8 x − 1 = 15 8 x − 1 + 1 = 15 + 1

33.

35.

Percent

Parts out of 100

Decimal

68%

68

0.68

Percent

Parts out of 100

Decimal

60%

60

0.6

8 x = 16 8x 16 = 8 8 x = 2

37. Verbal Model:

Labels:

Check:

4 1 ? 3 = ( 2) − 5 10 2 8 1 ? 3 − = 5 10 2 16 1 ? 3 − = 10 10 2 15 ? 3 = 10 2 3 3 = 2 2

Total = Pay per ⋅ x + Pay for pay training week

Total pay = 2635

Infinitely many solutions 25.

45

Fraction 17 25

Fraction 3 5

Compared Base = Percent ⋅ number number

Compared number = a Percent = 1.30 Base number = 50

Equation:

a = p⋅b a = 1.30 ⋅ 50 a = 65

So, 65 is 130% of 50.

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46

Chapter 2

39. Verbal Model:

Labels:

Linear Eq uations and Inequalities Compared Base = Percent ⋅ number number

Total price Total units $9.79 = = $0.25 per ounce 39 ounces

47. (a) Unit price =

Compared number = 645 Percent = 0.215

Total price Total units $1.79 = = $0.22 per ounce 8 ounces

(b) Unit price =

Base number = b Equation:

645 = 0.215 ⋅ b 645 = b 0.215 3000 = b

1 So, 645 is 21 % of 3000. 2

41. Verbal Model:

Labels:

Compared Base = Percent ⋅ number number

The 8-ounce can is the better buy. 49.

51.

Compared number = 250

28 8 7 y = 2

Base number = 200 250 = p ⋅ 200 250 200

= p

1.25 = p So, 250 is 125% of 200. 43. Verbal Model:

Labels:

⋅ Sales Commission = Percent rate

Commission = 9000 Percent rate = x Sales = 150,000

Equation:

9000 = x ⋅ 150,000 9000 = x 150,000 0.06 = x

y 7 = 8 4 8 y = 28 y =

Percent = p Equation:

Tax 9.90 1.10 0.1 1 = = = = Pay 396 44 4 40

53.

b 5+b = 6 15 15b = 6(5 + b) 15b = 30 + 6b 9b = 30 30 9 10 b = 3 b =

Proportion:

It is a 6% commission. 45. Verbal Model:

Labels:

Percent rate = 1.6% Sample = x Equation:

57. Verbal Model:

Labels:

The sample contained 375 parts.

Tax 1 Tax 2 = Assessed value 1 Assessed value 2 Tax 1 = 1680 Assessed value 1 = 105,000

6 = 0.016 ⋅ x 6 = x 0.016 375 = x

2 x = 6 9 6 x = 18 x = 3

Defective = Percent ⋅ Sample parts rate

Defective parts = 6

Leg 1 Leg 1 = Leg 2 Leg 2

55. Verbal Model:

Tax 2 = x Assessed value 2 = 125,000

1680 x = 105,000 125,000

Proportion:

125,000 ⋅ 1680 = 105,000 ⋅ x 210,000,000 = 105,000 x 2000 = x The tax is $2000.

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Review Exercises for Chapter 2

59. Verbal Model:

Labels:

47

Flagpole’s height Lamp post’s height = Length of flagpole’s shadow Length of lamp post’s shadow Flagpole’s height = h Length of flagpole’s shadow = 30 Lamp post’s height = 5 Length of lamp post’s shadow = 3 h 5 = 30 3 3 ⋅ h = 30 ⋅ 5

Proportion:

3h = 150 h = 50 The flagpole’s height is 50 feet. 61. Verbal Model:

Labels:

Selling = Cost + Markup price Selling price = 149.93

63. Verbal Model:

Labels:

Sale price = 53.96

Cost = 99.95

List price = 71.95

Markup = x 149.93 = 99.95 + x

Equation:

Discount = x Equation:

53.96 = 71.95 − x x = 71.95 − 53.96

149.93 − 99.95 = x

x = 17.99

49.98 = x The markup is $49.98.

Sale = List − Discount price price

The discount is $17.99.

Verbal Model:

Markup = Markup ⋅ Cost rate

Verbal Model:

List Discount = Discount ⋅ price rate

Labels:

Markup = 49.98

Labels:

Discount = 17.99

Equation:

Markup rate = x

Discount rate = x

Cost = 99.95

List price = 71.95

49.98 = x ⋅ 99.95 49.98 = x 99.95 0.50 ≈ x

The markup rate is about 50%.

Equation:

17.99 = x ⋅ 71.95 17.99 = x 71.95 0.25 ≈ x

The discount rate is about 25%.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

48

Chapter 2

Linear Eq uations and Inequalities

65. Verbal Model:

Labels:

Sales tax = Tax rate ⋅ Cost

Labels:

Sales tax = x

Equation:

69. Verbal Model:

Rate = 1500 mph

Cost = 2795

Time = 2 13 hours Equation:

x = 167.7

Verbal Model:

Total bill = Sales tax + Cost

Labels:

Total bill = x

The distance is 3500 miles. 71. Verbal Model:

Sales tax = 167.70

Labels:

Cost = 2795

Equation:

The total bill is $2962.70.

d r 100 100 + t = 48 40 55 t = 4.583 or 12

Amount Total Down = − financed Bill payment Amount financed = x Downpayment = 800 x = 2962.70 − 800 x = 2162.70

The amount financed is $2162.70.

Verbal Model:

Average Total = Total distance ÷ time speed

Labels:

Average speed = r Total distance = 200 miles

67. Verbal Model:

Total time = 4.583 hours

Amount of Amount of = Amount of + solution 2 solution 1 final solution Percent of solution 1 = 30% Liters of solution 1 = x Percent of solution 2 = 60% Liters of solution 2 = 10 − x Percent of final solution = 50%

Equation:

The average speed is 43.6 miles per hour. 73. Verbal Model:

Labels:

−0.30 x = −1 10 − x =

3 13 liters of solution 1 and 6 23 liters of solution 2 are required.

Work done = 1 1 4.5 1 Rate of person 2 = 6 Time = t

0.30 x + 6 − 0.60 x = 5

6 23

Work Work done + Work done = by person 1 done by person 2

Rate of person 1 =

0.30 x + 0.60(10 − x) = 0.50(10)

x = 3 13

r = 200 ÷ 4.583 r ≈ 43.6

Liters of final solution = 10

Equation:

d = rt t =

Total bill = 2962.70

Labels:

Distance = 100 miles Time = t

x = 2962.70

Equation:

Distance = Rate ⋅ Time Rates = 48 mph and 40 mph

x = 167.70 + 2795

Labels:

d = 1500 ⋅ 2 13 d = 3500

The sales tax is $167.70.

Verbal Model:

Distance = d

Tax rate = 0.06 x = 0.06 ⋅ 2795

Equation:

Distance = Rate ⋅ Time

Equation:

t t + 4.5 6 27 = 6t + 4.5t 1=

27 = 10.5t 27 = t 10.5 2.57 ≈ t The time required is about 2.57 hours.

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Review Exercises for Chapter 2 75. Verbal Model:

Labels:

Interest = Principal ⋅ Rate ⋅ Time

81. Verbal Model:

Interest = i

Perimeter = 2 ⋅ Width + 2 ⋅ Length

Principal = $1000

Labels:

Rate = 0.085

Perimeter = 64 Width = 53 l

Time = 4 Equation:

Length = l

i = 1000 ⋅ 0.085 ⋅ 4

i = 340 The interest is $340.

77. Verbal Model:

Labels:

64 =

Interest = Principal ⋅ Rate ⋅ Time

64 =

16 l 5

Interest = $20,000

20 = l

Rate = 0.095

83.

x −4

−3

−2

−1

0

−6

−5

1

2

20,000 = p ⋅ 0.095 ⋅ 4 20,000 = p 0.38 52,631.58 ≈ p

The principal required is about $52,631.58. 79. Verbal Model:

+ 2l

The dimensions are 20 feet × 12 feet.

Principal = p

Equation:

( )

64 = 2 53 l + 2l

Equation:

6 l 5

Time = 4

49

85.

x −8

−7

−4

x − 5 ≤ −1

87.

x − 5 + 5 ≤ −1 + 5 x ≤ 4

Interest = Principal ⋅ Rate ⋅ Time

x

Labels:

Interest = 4700 Principal 1 = p Rate 1 = 0.085

0

2

3

4

5

5

6

7

89. −6 x < −24

−6 x −24 > −6 −6 x > 4

Principal 2 = 50,000 − p Rate 2 = 0.10 Time = 1

Equation:

1

4700 = 0.085 p + 0.10(50,000 − p )

x 1

2

3

4

8

4700 = 0.085 p + 5000 − 0.10 p −300 = −0.015 p

91. 5 x + 3 > 18

−300 = p −0.015 20,000 = p

5 x > 15 x > 3

30,000 = 50,000 − p

x 0

1

2

3

4

The smallest amount you can invest is $30,000. 93.

8 x + 1 ≥ 10 x − 11 8 x − 10 x + 1 ≥ 10 x − 10 x − 11 −2 x + 1 ≥ −11 −2 x + 1 − 1 ≥ −11 − 1 −2 x ≥ −12 −2 x −12 ≤ −2 −2 x ≤ 6 x −2

0

2

4

6

8

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50

95.

Chapter 2

Linear Eq uations and Inequalities

1 1 − y < 12 3 2 2 − 3 y < 72

Cost per Money 105. Verbal Model: minute ⋅ Number of ≥ Remaining minutes

Labels:

−3 y < 70

Number of minutes = x

70 3

y > −

Cost per minute = 0.10 Money remaining = 12.50

Equation:

− 70

0.10 x ≤ 12.50

x ≤ 125

3

y − 25

− 24

− 23

You can talk for no more than 125 minutes.

− 22

107. x = 6

−4(3 − 2 x) ≤ 3( 2 x − 6)

97.

x = 6 or x = −6

−12 + 8 x ≤ 6 x − 18 −12 + 8 x − 6 x ≤ 6 x − 6 x − 18

109. 4 − 3x = 8

−12 + 2 x ≤ −18

4 − 3x = 8

−12 + 12 + 2 x ≤ −18 + 12

4 − 4 − 3x = 8 − 4

2 x ≤ −6 2x −6 ≤ 2 2 x ≤ −3 x −5 −4 −3 −2 −1

99.

0

1

2

−3 x 4 = −3 −3 4 x = − 3

−3x −12 = −3 −3

−14 ≤ 2 x < −4

5x + 4 = 4

−14 2x −4 ≤ < 2 2 2 −7 ≤ x < −2

5x + 4 − 4 = 4 − 4

−7 x

101.

−3x = −12

−4

−2

5 x = −8 −8 5x = 5 5 8 x = − 5

3x − 4 = x + 2

or

2x = 6

−1

3 x − 4 = −( x + 2) 3x − 4 = − x − 2

x = 3

x

103.

5x = 0

113. 3 x − 4 = x + 2

−16 < x < −1 −4

5 x + 4 − 4 = −4 − 4

x = 0

x +1 > 0 −3 −15 < x + 1 < 0

−8

5 x + 4 = −4

or

5x 0 = 5 5

0

5 >

− 20 − 16 − 12

x = 4

5x + 4 = 4

−6 − 8 ≤ 2 x + 8 − 8 < 4 − 8

−6

4 − 4 − 3 x = −8 − 4

−3 x = 4

111. 5 x + 4 − 10 = −6

−6 ≤ 2 x + 8 < 4

−8

4 − 3 x = −8

or

4x = 2 x =

0

5x − 4 < 6

and

5x − 4 + 4 < 6 + 4

3x + 1 > −8 3 x > −9

5x 10 < 5 5 x < 2

3x −9 > 3 3 x > −3

=

1 2

115. x − 4 > 3

3 x + 1 − 1 > −8 − 1

5 x < 10

2 4

x − 4 < −3 x 3 x > 7

3 x < 12 −12 < 3x < 12 −4 < x < 4

−3 < x < 2 −3 x −4

−2

0

2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test for Chapter 2

51

123. (1, 5)

119. 2 x − 7 < 15

−15 < 2 x − 7 < 15

1< x < 5

−8 < 2 x < 22

1−3 < x −3 < 5−3

−4 < x < 11

−2 < x − 3 < 2 x −3 < 2

121. b + 2 − 6 > 1 b + 2 > 7

125. t − 78.3 ≤ 38.3

b + 2 < −7

or

b < −9

b + 2 > 7

−38.3 ≤ t − 78.3 ≤ 38.3

b > 5

40 ≤ t ≤ 116.6 The minimum temperature is 40 degrees Fahrenheit and the maximum temperature is 116.6 degrees Fahrenheit. 116.6 t 0

20

40

60

80

100 120

Chapter Test for Chapter 2 1.

6 x − 5 = 19 6 x − 5 + 5 = 19 + 5 6 x = 24 6x 24 = 6 6 x = 4

2.

4.

x 2x = + 4 3 2 ⎛ 2x ⎞ ⎛ x ⎞ 6⎜ ⎟ = ⎜ + 4 ⎟6 3 2 ⎝ ⎠ ⎝ ⎠ 4 x = 3 x + 24 4 x − 3 x = 3 x + 24 − 3 x x = 24

5 x − 6 = 7 x − 12 5 x − 7 x − 6 = 7 x − 7 x − 12 −2 x − 6 = −12 −2 x − 6 + 6 = −12 + 6

5. Verbal Model:

Labels:

Compared Base = Percent ⋅ number number

Compared number = a

−2 x = −6

Percent = 1.25

−2 x −6 = −2 −2 x = 3

Base number = 3200

3. 15 − 7(1 − x) = 3( x + 8) 15 − 7 + 7 x = 3x + 24 8 + 7 x = 3x + 24 8 + 7 x − 3 x = 3x + 24 − 3x 8 − 8 + 4 x = 24 − 8

Equation:

a = 4000

So, 125% of 3200 is 4000. 6. Verbal Model:

Labels:

Compared Base = Percent ⋅ number number

Compared number = a Percent = p

4 x = 16 4x 16 = 4 4 x = 4

a = 1.25 ⋅ 3200

Base number = b Equation:

32 = p ⋅ 8000 32 8000

= p

0.004 = p So, 32 is 0.4% of 8000.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

52

Chapter 2

Linear Eq uations and Inequalities List price − Discount = Sale price

7. Verbal Model:

Labels:

List price = x Discount = 0.20 x

Equation:

11. Verbal Model:

Amount of food 1 + Amount of food 2 = Mixture Labels:

Cost for food 1 = 2.60

Sale price = 8900

Pounds of food 1 = x

x − 0.20 x = 8900

Cost for food 2 = 3.80 Pounds of food 2 = 40 − x

0.80 x = 8900

Cost for mixture = 3.35

x = 11,125

Pounds of mixture = 40

The list price is $11,125. Total price $2.49 249 8. = = = $0.2075 per ounce Total units 12 ounces 1200 Total price $2.99 299 = = = $0.1993 per ounce Total units 15 ounces 1500 The 15-ounce can is the better buy. It has a lower unit price. Tax = Tax rate ⋅ Assessed value

9. Verbal Model:

Labels:

Tax = 1650

2.60 x + 3.80( 40 − x) = 3.35( 40)

Equation:

−1.20 x + 152 = 134 −1.20 x = −18 x = 15 There are 15 pounds at $2.60 and 25 pounds at $3.80. 12. Verbal Model:

Distance of car 1 + 10 miles = Distance of car 2 Labels:

Time = x Distance of car 1 = 40 x

Tax rate = p Assessed value = 110,000 1650 = p ⋅ 110,000

Equation:

Distance of car 2 = 55 x Equation:

10 = 15 x

0.015 = p Verbal Model:

Tax = Tax rate ⋅ Assessed value

Labels:

Tax = x Tax rate = 0.015 Assessed value = 145,000

Equation:

x = 0.015 ⋅ 145,000 x = 2175

2 3

Labels:

10 15

= x

2 3

= x

hour, or 40 minutes, must elapse.

13. Verbal Model:

Labels:

Interest = Principal ⋅ Rate ⋅ Time Interest = 300 Principal = p

The tax is $2175. 10. Verbal Model:

40 x + 10 = 55 x

Total Cost of + Cost of bill = parts labor

Total bill = 165 Cost of parts = 85 Number of half hours of labor = x

Rate = 0.075 Time = 2 Equation:

300 = p ⋅ 0.075 ⋅ 2 2000 = p

The principal required is $2000.

Cost of labor = 16 x Equation:

165 = 85 + 16 x 80 = 16 x 5 half hours = x

The repairs took 2 12 hours.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test for Chapter 2 14. (a)

16. t ≥ 8

3x − 6 = 9 3x − 6 = 9

and −3x + 6 = 9

3 x = 15

−3x = 3

3x 15 = 3 3 x = 5

−3x 3 = −3 −3 x = −1

17. (a)

x −3 ≤ 2 −2 ≤ x − 3 ≤ 2 1≤ x ≤ 5

(b) 5 x − 3 > 12

(b) 3 x − 5 = 6 x − 1

5 x − 3 > 12 or 5 x − 3 < −12

3x − 5 = 6 x − 1 and −3 x + 5 = 6 x − 1

(c)

−4 = 3 x

6 = 9x

−4 3x = 3 3 4 − = x 3

6 = x 9 2 = x 3

(c)

x > 3

x < −

9 5

x + 2 < 0.2 4

−8.8 < x < −7.2 −

15. (a) 3x + 12 ≥ −6

x −8

3x ≥ −18

−6

−4

−2

44 36 < x < − 5 5

0

x ≥ −6

18. Verbal Model:

(b) 9 − 5 x < 5 − 3x

x

−2 x < −4

0

1

2

3

4

x > 2 1− x < 2 4 0 ≤1− x < 8

5 x < −9

x + 2 < 0.2 4 −0.8 < x + 8 < 0.8

There is no solution.

0 ≤

5 x > 15

−0.2
−7 −7 < x ≤ 1

(d)

−7 < 4( 2 − 3x) ≤ 20 −7 < 8 − 12 x ≤ 20

5 4

−15 < −12 x ≤ 12 1≤ x
0, y < 0 is in Quadrant IV.

41.

( x, y ), xy

> 0 is in Quadrant I or III.

(−3, − 4) shifted 2 units right and

5

(1, − 3) shifted 2 units right and

+ ( 40 − 10)

= 7

2

= 7 y 6

(3, 5)

4 2

= 5 61 ≈ 39.05

x −4

The pass is about 39.05 yards long.

−2

47. d = 10 − 3 = 7 = 7 y

⎛ x + x2 y1 + y2 ⎞ Midpoint = ⎜ 1 , ⎟ 2 ⎠ ⎝ 2

3

(3, 2) 2

⎛1 + 6 6 + 3 ⎞ = ⎜ , ⎟ 2 ⎠ ⎝ 2

(10, 2) 1

⎛7 9⎞ = ⎜ , ⎟ ⎝ 2 2⎠

x 4

31. The graph of an ordered pair of real numbers is a point on a rectangular coordinate system.

3 4

8

12

Horizontal line 49. d1 =

35. ( −3, − 5) is in Quadrant III.

(

(3, −2)

Vertical line

29. Let ( x1 , y1 ) = (1, 6) and ( x2 , y2 ) = (6, 3).

33. Yes, the order of the points does not affect the computation of the midpoint.

6

4

−4

⎛ x + x2 y1 + y2 ⎞ Midpoint = ⎜ 1 , ⎟ 2 ⎠ ⎝ 2 ⎛ −2 + 4 0 + 8⎞ , = ⎜ ⎟ 2 ⎠ ⎝ 2 = (1, 4)

2 −2

27. Let ( x1 , y1 ) = ( − 2, 0) and ( x2 , y2 ) = ( 4, 8).

37. − 89 ,

5 units up = (3, 2)

45. d = 5 − ( −2)

1525

=

5 units up = ( −1, 1)

2

Let ( x, y ) = (10, 10) and ( x2 , y2 ) = (35, 40). 2

39.

43. ( −2, −1) shifted 2 units right and 5 units up = (0, 4)

2

2

57

The Rectangular Coordinate System

) is in Quadrant II.

(−2

− 0) + (0 − 5)

d2 =

(0

d3 =

(−2

P =

29 +

2

− 1) + (5 − 0) 2

2

2

= =

− 1) + (0 − 0) = 2

2

4 + 25 = 1 + 25 =

29 26

9 = 3

26 + 3 ≈ 13.48

51. Let the point (2009, 36.5) represent Apple’s sales for the year 2009 and (2011, 108.2) represent the sales for the year 2011. The midpoint of these points is

⎛ x + x2 y1 + y2 ⎞ ⎛ 2009 + 2011 36.5 + 108.2 ⎞ Midpoint = ⎜ 1 , , ⎟ = ⎜ ⎟ = (2010, 72.35). 2 ⎠ ⎝ 2 2 ⎝ 2 ⎠ Apple’s net sales for the year 2010 is estimated to be $72.35 billion. 53. No. The distance between two points is always positive.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

58

Chapter 3

Grap hs and Functions

55. ( −3, 4) is not a solution point of y = 4 x + 15 because

65. Verbal model: Discount = Discount rate ⋅ List price

4 ≠ 4( −3) + 15

Labels: Discount = 80 − 52 = 28 List price = 80

4 ≠ 3. 57.

Discount rate = p

y

Equation:

8

(7, 3)

4 2

(2, 1)

−8 −6 −4 −2

(− 3, − 5) − 4

(2, − 1)

−6

6

8

The discount is $28 and the discount rate is 35%.

(7, − 3)

67. Verbal model: Sale price = List price − Discount

−8

Labels: Sale price = x

When the sign of the y-coordinate is changed, the point is on the opposite side of the x-axis from the original point. 59.

61.

List price = 112.50 Discount = 31.50 Equation: x = 112.50 − 31.50

3 x = 4 28 3 ⋅ 28 = 4 x

x = 81

The sale price is $81.

84 = 4 x

Verbal model: Discount = Discount rate ⋅ List price

21 = x

Labels: Discount = 31.50 Discount price = p

a 4 = 27 9 9a = 4 ⋅ 27

List price = 112.50 Equation:

9a = 108

31.50 = p(112.50) 31.50 = p 112.50 0.28 = p

a = 12 63.

= p

0.35 = p

x 4

28 = p(80) 28 80

6

(− 3, 5)

z +1 z = 10 9 9( z + 1) = 10 z

The discount rate is 28%.

9 z + 9 = 10 z 9 = z

Section 3.2 Graphs of Equations 1.

y

x

−2

−1

0

1

2

y = 3x

−6

−3

0

3

6

(−2, − 6)

(−1, − 3)

(0, 0)

(1, 3)

(2, 6)

Solution point

9 6 3 −9

−6

x

−3

3

6

9

3

4

5

−9

3.

x

−2

−1

0

1

2

y = 4− x

6

5

4

3

2

(−2, 6)

(−1, 5)

(0, 4)

(1, 3)

(2, 2)

Solution point

y 5 4 3 2 1 −1

x 1

2

−1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.2

59

Graphs of Equations

5. 2 x − y = 3 y

− y = −2 x + 3 y = 2x − 3

3 2

x

−2

−1

0

1

2

y = 2x − 3

−7

−5

−3

−1

1

(−2, − 7)

(−1, − 5)

(0, − 3)

(1, −1)

(2, 1)

Solution point

1 −3

−2

−1

x 1

2

3

−1 −2 −3

7. 3x + 2 y = 2

y

2 y = −3x + 2

3

y = − 32 x + 1

1

x y =

−2 − 32 x

+1

5 2

4

0

1

1

− 12

(−1, 52 )

(−2, 4)

Solution point

9.

−1

2

−3

−2

−2

(0, 1)

(1, − 12 )

(2, − 2)

−1

x 2

3

−1 −2 −3

y

x

−2

−1

0

1

2

y = − x2

−4

−1

0

−1

−4

Solution point

(−2, − 4)

(−1, −1)

(0, 0)

(1, −1)

(2, − 4)

2 1 x –3

2

–2

3

–1 –2 –3 –4

11.

x

−2

−1

0

1

2

y = x2 − 3

1

–2

–3

–2

1

Solution point

(−2, 1)

(−1, – 2)

(0, − 3)

(1, − 2)

(2, 1)

y 2 1 −3

−2

−1

x 1

2

3

−1 −2

−4

2

13. − x − 3 x + y = 0

y

2

y = x + 3x

3 2

x

−2

−1

0

1

2

y = x 2 + 3x

−2

−2

0

4

10

Solution point

(−2, − 2)

(−1, − 2)

(0, 0)

(1, 4)

(2, 10)

1 −4

−2

x

−1

1

2

−2 −3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

60

Chapter 3

Grap hs and Functions

15. x 2 − 2 x − y = 1 y

− y = − x2 + 2x + 1 y = x2 − 2 x − 1

4 3

x

−2

−1

0

1

2

y = x2 − 2 x − 1

7

2

−1

−2

−1

(−2, 7)

Solution point

(−1, 2)

(0, −1)

(1, − 2)

(2, −1)

2 1 −2

−1

x 1

2

3

4

1

2

3

1

2

3

−1 −2

17.

y

x

−2

−1

0

1

2

y = x

2

1

0

1

2

Solution point

(−2, 2)

(−1, 1)

(0, 0)

(1, 1)

(2, 2)

5 4 3 2 1 −3

19.

−2

−1

x −1

y

x

−2

−1

01

y = x +3

5

4

3

4

5

Solution point

(−2, 5)

( −1, 4)

(0, 3)

(1, 4)

(2, 5)

2

5 4 3 2 1 −3

−2

x

−1 −1

21.

y

x

−2

−1

0

1

2

y = x +3

1

2

3

4

5

5

Solution point

(−2, 1)

(−1, 2)

(0, 3)

(1, 4)

(2, 5)

3

6 4

1 x –7 –6 –5 –4 –3 –2 –1

1

–2

23. y = 3 − x

y

y = 3−0 y = 3

4

(0, 3)

3

0 = 3− x x = 3

(1, 2)

2

(3, 0)

1

(3, 0)

y = 3−1 y = 2

(0, 3)

(1, 2)

−1

1

2

3

x

4

−1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.2 25. y = 2 x − 3

31.

y = 2(0) − 3

1

( 32 , 0)

= x

−2

−1

y = 2(3) − 3

4 y = 12 3 x + 4(0) = 12

x 2

−1

(3, 3)

−3

3

4

3x + 0 = 12 3x = 12

(0, −3)

3(1) + 4 y = 12

4(0) + y = 3

3 + 4 y = 12

(0, 3)

y = 3

4y = 9

4x + 0 = 3 y

( )

3 4

3 , 4

0

5 4

4(1) + y = 3

(1, −1)

y = −1

(1, 94 (

1

(0, 3)

(4, 0) x

−3 −2 −1

1

2

3

4

−2 −3

2 1

) 34 , 0)

−1

33.

x 2

x + 5 y = 10 0 + 5 y = 10

(1, − 1)

−1

29.

(0, 3)

2

y

−2

(1, 94 )

9 4

y =

4x = 3

3

(4, 0)

x = 4

4x + y = 3

x =

(0, 3)

y = 3

( 32 , 0 (

−2

y = 3 27.

0 + 4 y = 12

2

3 = 2x 3 2

3 x + 4 y = 12

(3, 3)

3

0 = 2x − 3

61

3(0) + 4 y = 12

y

(0, − 3)

y = −3

Graphs of Equations

(0, 2)

y = 2

2x − 3 y = 6

x + 5(0) = 10

2(0) − 3 y = 6

(10, 0)

x = 10

−3 y = 6

(0, − 2)

y = −2

5 + 5 y = 10 5y = 5

2 x − 3(0) = 6

(5, 1)

y =1

2x = 6

y

(3, 0)

x = 3

6

2(1) − 3 y = 6

4

−3 y = 4

(0, 2)

(

y = − 43

1, − 43

)

(5, 1)

(10, 0) x

2

4

6

8

10

−2 y

−4 1

(3, 0) x 1

2

3

4

−1

(1, − 43 (

(0, − 2) −3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

62 35.

Chapter 3

Grap hs and Functions 39. (a)

5 x − y = 10 5(0) − y = 10 0 − y = 10 − y = 10

369

F

0

4 8 12 16

5 x = 10

(2, 0)

5(1) − y = 10

12

F

Force (in pounds)

5 x − (0) = 10 x = 2

0

(b)

(0, −10)

y = −10

x

20 18 16 14 12 10 8 6 4 2 x

5 − y = 10

2

4

6

8

10

12

Length (in inches)

−y = 5

(1, − 5)

y = −5

41. To complete the graph of the equation, connect the points with a smooth curve.

y 2

43. False. To find the x-intercept, substitute 0 for y in the equation and solve for x.

(2, 0) x

−6 −4 −2 −2

4

−4

6

8

45. y = x 2 − 9

(1, −5)

−6

0 = x2 − 9

−8 −10

0 = ( x − 3)( x + 3)

(0, −10)

−12

(3, 0), ( −3, 0)

x = ±3 y = 02 − 9

(0, − 9)

y = −9

37. (a) The annual depreciation is 40,000 − 5000 = 5000. 7

y

40,000 − 5000(1) = 35,000

3

(−3, 0)

40,000 − 5000( 2) = 30,000

(3, 0)

−6

x

6

40,000 − 5000(3) = 25,000

−3

So, y = 40,000 − 5000t , 0 ≤ t ≤ 7. (b)

9 (0, −9)

Value (in dollars)

y 40,000 30,000

47.

20,000

y = 9 − x2 0 = 9 − x2

10,000 t 1 2 3 4 5 6 7

Time (in years)

x2 = 9

(3, 0), (−3, 0)

x = ±3 y = 9−0

(c) y = 40,000 − 5000(0)

2

(0, 9)

y = 9

y = 40,000, (0, 40,000)

y 10

The y-intercept represents the value of the delivery van when purchased.

(0, 9)

6 4 2

(− 3, 0)

(3, 0) x

–6

–4

–2

2

4

6

–2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.2 49. y = x( x − 2)

53.

y = x + 2

y = 02 − 2(0)

(0, 0)

4

0 = x + 2

0 = x − 2x 0 = x( x − 2)

(0, 0), ( 2, 0)

y

3

55. y = x − 3

2

0 = x −3

3

y = −(0)(0 + 4)

(0, 0)

0 = − x ( x + 4)

(0, 0), (−4, 0)

x = 0, − 4 y = −( −2)( −2 + 4) y = 2( 2)

(−2, 4)

y = 4

1

−1

y

6 5

(0, 3)

y = 3

51. y = − x( x + 4)

(− 2, 0)

7

y = −3

(1, − 1)

−4

(3, 0)

y = 0−3

x

y = 0

(0, 2) x

−5

(−4, 2)

x = 3 (2, 0)

1

(− 4, 2)

y = −4 + 2 = 2

(0, 0)

3

(−2, 0)

−2 = x

x = 0, 2

−1

5

y = x + 2

2

−1

y

(0, 2)

= 2

y = 0

63

Graphs of Equations

y = 6−3

3

y = 3

1

(0, 3)

(6, 3)

2 x

(6, 3)

y = 3

–1 –1

1

(3, 0)

5

6

7

57. The scales on the y-axes are different. From graph (a) it appears that sales have not increased. From graph (b) it appears that sales have increased dramatically. 59.

y

y

Profit

(−2, 4) 4 3

(−4, 0) −5

t

(0, 0) −3

−2

−1

x −1 −2

Time

1

61. A horizontal line has no x-intercepts unless y = 0. The y-intercept is (0, b), where b is any real number.

So, a horizontal line has one y-intercept. 63.

1 7

65.

5 4

67. x − 8 = 0 x = 8

Check: ?

8 − 8=0 0 = 0

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

64

Chapter 3

Grap hs and Functions

69. 4 x + 15 = 23

71.

y

4x = 8

8

x = 2

6

(3, 7) (0, 5)

(3, 5)

4

Check:

2

?

4( 2) + 15 = 23

−4

x

−2

2

8 + 15 = 23

4

6

−2

Section 3.3 Slope and Graphs of Linear Equations 8−0 8 = = 2 4−0 4

1. m =

7. m =

The slope is 2.

5−

3 4

The slope is −

y

− 92

=

17 4

= −

9 4 −9 ⋅ 2 18 ⋅ = = − 2 17 17 17

18 . 17

y

(4, 8)

8

− 52 − 2

6 2

4

( 34 , 2 (

1

2

(0, 0)

x 2

4

6

8

−1

x 1

2

−2

3. m =

5−3 2 = =1 −2 − ( − 4) 2

The slope is 1.

5

3 2

11. The slope of the line through (0, 0) and (5, 2) is

1 −3

−2

−1

x 1

m =

−1

10 − 7 3 3 = = − 5. m = −2 − 3 −5 5 3 The slope is − . 5

2−0 5−0 =

2 . 5

13. The slope of the line through (2, 4) and (3, 2) is

m =

y

2− 4 3− 2 2 1 = − 2. = −

12

(− 2, 10) 8

(3, 7)

15. (a) m =

6 4

3 4

⇒ L3

(b) m = 0 ⇒ L2

2 −6 −4 −2

9. The slope of the line through (3, 2) and (3, 4) is undefined.

Because division by zero is not defined, the slope of a vertical line is not defined.

4

(− 4, 3)

−5 − 4

5

5, − 2

−3

4− 2 2 = 3−3 0

y

(−2, 5)

5

4

−1

x 2

4

6

8

(c) m = −3 ⇒ L1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.3

Slope and Graphs of Linear Equations

27. x + y = 0

17. y = 2 x − 1

y = −x

slope = 2 y -intercept = (0, −1)

slope = −1 y -intercept = 0

y 2

y

1

−2

65

(1, 1)

2

x

−1

1

1

2

(0, 0)

(0, −1)

−1

−2

−1

x

1

2

−1 −2

19. y = − x − 2

slope = −1

29. 3x − y − 2 = 0

y -intercept = (0, − 2)

− y = − 3x + 2 y = 3x − 2

y

−2

slope = 3

x

−1

1

2

3

y -intercept = −2

−1 (0, −2)

−2

y

−3

3

(2, −4)

−4

2

−5

1 x

21. y = − 12 x + 4

−3

−2

−1

slope = − 12

1

−1

2

3

(0, − 2)

−2

y -intercept = (0, 4)

31. 3x + 2 y − 2 = 0

y

2 y = −3 x + 2

6

y = − 32 x + 1

(0, 4) 2

slope = − 32

(8, 0) x

−2

2

4

y -intercept = 1

6

−2

y

−4

23. y = 3 x − 2

2

m = 3; y -intercept = (0, − 2) 25. 4 x − 6 y = 24

−2

x

−1

2

− 6 y = − 4 x + 24

−1

2x 3

−2

y = m =

2; 3

− 4

y -intercept = (0, − 4)

(0, 1)

1

33.

y

L2

4 3

L1

1 −3

−2

x 1

2

3

−1 −2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

66

Chapter 3

Grap hs and Functions

35. L1 : y =

1x 2

− 2

L2 : y =

1x 2

+ 3

1 2

m1 =

and m2 =

6 − ( −1) 7 70 5 = = − = − 84 6 − 4.2 − 4.2 −8.4

49. m = 1 2

Line falls.

y

(−4.2, 6) 6

m1 = m2 so the lines are parallel.

4

37. L1 : y =

3 x 4

−3

2

L2 : y = − 43 x + 1 3 4

m1 =

−6

−4

x

−2

2 −2

and m2 = − 34

6

(4.2, −1)

−4

m1 ⋅ m2 = −1 so the lines are perpendicular.

51.

39. L1 : y = − 4 x + 6

L2 : y = − 14 x − 1

−2 7 −5 = x − 4 3 −2( x − 4) = 6 −2 x + 8 = 6

m1 = − 4 and m2 = − 14

−2 x = −2

m1 ≠ m2 and m1 ⋅ m2 ≠ −1, so the lines are neither

x =1

parallel nor perpendicular. 41. Let w represent hourly wages and let t represent the year. The two given data points are represented by (t1 , w1 ) and (t2 , w2 ).

(t1 , w1 ) = (2005, 17.05)

3 3− y = 2 9 − ( −3) 3(12) = 2(3 − y ) 36 = 6 − 2 y

(t2 , w2 ) = (2010, 20.43) Use the formula for slope to find the average rate of change. Rate of change =

53.

30 = −2 y −15 = y 8−4 4 = = 2 2−0 2 5 − ( −1) 6 L2 : m2 = = = 2 3−0 3 m1 = m2 so the lines are parallel.

w2 − w1 t2 − t1

55. L1 : m1 =

20.43 − 17.05 2010 − 2005 = 0.676 =

From 2005 to 2010, the average rate of change in the hourly wage of health service employees was about $0.68. 43. The rise is the numerator of the fraction − 23 , so the rise is − 2. The run is the denominator of the fraction − 23 , so

the run is 3.

−2 − 2 −4 2 = = − 6−0 6 3 4−0 4 2 = = L2 : m2 = 8−2 6 3

57. L1 : m1 =

m1 ≠ m2 and m1 ⋅ m2 ≠ −1, so the lines are neither parallel nor perpendicular.

45. A line with a negative slope falls from the left to right.

1 1 − 8 2 −1 1 = 47. m = 4 8 ⋅ = 3 −3 8 6 12 18 + − 4 2

Line rises.

y 2

(− 32 , 18 ( −2

1

( 34 , 14 ( x

−1

1

2

−1 −2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.4 3 x − 5 y − 15 = 0

59.

65.

3(0) − 5 y − 15 = 0 −5 y = 15

(0, − 3)

y = −3 3x − 5(0) − 15 = 0 3 x = 15

(5, 0)

x = 5 y 2

(5, 0) x 2

–2

Equations of Lines

67

3 h = 4 15 45 = 4h 45 = h 4 The maximum height in the attic is 45 feet = 11.25 feet. 4

67. Yes, any pair of points on a line can be used to calculate the slope of the line. When different pairs of points are selected, the change in y and the change in x are the lengths of the sides of similar triangles. Corresponding sides of similar triangles are proportional.

–2 –4

69. Two lines with undefined slope are parallel. Because the slopes are undefined, the lines are vertical lines, so they are parallel.

(0, −3)

–6

61.

71. No, it is not possible for two lines with positive slopes to be perpendicular to each other. Their slopes must be negative reciprocals of each other.

−4 x − 2 y + 16 = 0 −4 x − 2(0) + 16 = 0 −4 x = −16

(4, 0)

x = 4 −4(0) − 2 y + 16 = 0

(0, 8)

y = 8 y

4h = −24

h = 6

81. x + 4 = 5

2

(4, 0) −2

4h = 24

or

h = −6

4

63. −

x = −8 and x = 8 79. 4h = 24

(0, 8)

6

−4

75. 85 ≤ z ≤ 100 77. x = 8

−2 y = −16

8

73. x < 0

x 2

4

8 −2000 = 100 x −8 x = −200,000 x = 25,000

The change in your horizontal position is 25,000 feet.

x + 4 = −5

x + 4 = 5

or

x = −9

x =1

83. 6b + 8 = −2b

6b + 8 = −2b

or

8 = −8b −1 = b

6b + 8 = 2b 8 = −4b −2 = b

Section 3.4 Equations of Lines 1. y − 0 = − 12 ( x − 0)

y = − 12 x 3. y + 4 = 3( x − 0)

5. y − 6 = − 34 ( x − 0)

y − 6 = − 34 x y = − 34 x + 6

y + 4 = 3x y = 3x − 4

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

68

Chapter 3

Grap hs and Functions

7. y − 8 = −2⎡⎣ x − ( −2)⎤⎦

y − 8 = −2( x + 2) y − 8 = −2 x − 4

⎛3 ⎞ ⎛9 ⎞ 15. Let ( x1 , y1 ) = ⎜ , 3⎟ and let ( x2 , y2 ) = ⎜ , 4 ⎟. ⎝2 ⎠ ⎝2 ⎠ m =

y = −2 x + 4 9. y − ( −7) =

− ( −4)⎤⎦

5 ⎡x 4⎣

(x

+ 4)

y + 7 =

5 4

y + 7 =

5 x 4

+5

y =

5 x 4

− 2

=

1 6 2

1 3

y − y1 = m( x − x1 ) y −3 =

11. Let ( x1 , y1 ) = (0, 0) and let ( x2 , y2 ) = (5, 2). 2−0 2 m = = 5−0 5

y − y1 = m( x − x1 ) 2 y − 0 = ( x − 0) 5 2 y = x 5 5 y = 2x − 2x + 5 y = 0 The general form of the equation of the line is − 2 x + 5 y = 0. 13. Let ( x1 , y1 ) = (1, 4) and let ( x2 , y2 ) = (5, 6). 6−4 5 −1 1 = 2

m =

1⎛ 3⎞ ⎜x − ⎟ 3⎝ 2⎠

1 1 x − 3 2 6 y − 18 = 2 x − 3 y −3 =

− 2 x + 6 y − 15 = 0 2 x − 6 y + 15 = 0 The general form of the equation of the line is 2 x − 6 y + 15 = 0. 17. x = −1 because every x-coordinate is −1. 19. y = −5 because every y-coordinate is −5. 21. x = −7 because both points have an x-coordinate of −7. 23. Because the line is horizontal and passes through the point (0, − 2), every point on the line has a y-coordinate

of − 2. So, an equation of the line is y = − 2. 25. 6 x − 2 y = 3

y − y1 = m( x − x1 ) y − 4 =

=

4−3 − 32

9 2

1 (x 2

− 1)

2y − 8 = x − 1 − x + 2y − 7 = 0 The general form of the equation of the line is − x + 2 y − 7 = 0.

slope = 3

−2 y = −6 x + 3 y = 3x −

3 2

(a) y − 1 = 3( x − 2) y − 1 = 3x − 6 y = 3x − 5 (b) y − 1 = − 13 ( x − 2) y − 1 = − 13 x +

2 3

y = − 13 x +

2 3

y = − 13 x +

5 3

+

3 3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.4 27. 5 x + 4 y = 24

33. m =

4 y = −5 x + 24 y = − 54 x + 6

C = 20( 400) + 5000

y − 4 = − 54 ( x + 5) y = − 54 x −

25 4

y = − 54 x −

9 4

(b) y − 4 =

4 ⎡x 5⎣

(x

= 13,000 The cost is $13,000. +

16 4

35. y − y1 = m( x − x1 ) y − 1 = 4( x − 1)

− (−5)⎤⎦ + 5)

y − 4 =

4 5

y − 4 =

4x 5

+ 4

y =

4x 5

+8

6000 − 5000 1000 = = 20 50 − 0 50

C = 20 x + 5000

(a) y − 4 = − 54 ⎡⎣ x − (−5)⎤⎦ 25 4

69

C − 5000 = 20( x − 0)

slope = − 54

y − 4 = − 54 x −

Equations of Lines

The slope is 4. The line passes through (1, 1). 37. Yes. The horizontal line passing through ( − 4, 5) is

y = 5. 39.

29. 4 x − y − 3 = 0

− y = −4 x + 3

41.

y = 4x + 3 slope = 4 (a) y − ( −3) = 4( x − 5) y + 3 = 4 x − 20 y = 4 x − 23 (b) y − ( −3) = − 14 ( x − 5) y + 3 = − 14 x +

5 4

y = − 14 x −

7 4

31. (5, 1500), (6, 1560)

1560 − 1500 60 m = = = 60 6−5 1 N − 1500 = 60(t − 5) N − 1500 = 60t − 300 N = 60t + 1200

x y + =1 3 2

x + −5 6 6x − − 5

43. m =

y =1 −7 3 3y =1 7

200,000 − 500,000 −300,000 = = 100,000 2−5 −3

S − 500,000 = 100,000(t − 5) S − 500,000 = 100,000t − 500,000 S = 100,000t S = 100,000(6) = 600,000 The total sales for the sixth year are $600,000. 45. (a)

(0, 7400), (4, 1500) m =

7400 − 1500 5900 = = −1475 0−4 −4

V − 7400 = −1475(t − 0) V − 7400 = −1475t V = −1475t + 7400 (b) V = −1475( 2) + 7400 V = −2950 + 7400 V = 4450 The photocopier has a value of $4450 after 2 years.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

70

Chapter 3

Grap hs and Functions

47. (0, 0), ( 40, 5) m =

5−0 5 1 = = 40 − 0 40 8

Distance from deep end

0 8 16 24 32

Depth of water

9

8

7

40 6

5

4

1 ( x − 0) 8 1 y = x 8 8y = x

y −0 =

x − 8y = 0 Depth of water = 9 − y

(a) 9 − y

(b)

(c)

9 − y

9 − y

8 = 9− y

9 = 9− y

7 = 9− y

−1 = − y

0 = y x − 8(0) = 0 x = 0

−2 = − y

1= y

2 = y

x − 8(1) = 0

x − 8( 2) = 0

x = 8 (d) 9 − y

(e)

6 = 9− y

x = 16 (f )

9 − y

5 = 9− y

9 − y

4 = 9− y

−3 = − y

−4 = −y

−5 = − y

3 = y

4 = y

5 = y

x − 8(3) = 0

x − 8( 4) = 0

x − 8(5) = 0

x = 24

x = 32

49. Point-slope form: y − y1 = m( x − x1 )

x = 40 57.

Slope-intercept form: y = mx + b General form: ax + by + c = 0 51. The variable y is missing in the equation of a vertical line because any point on a vertical line is independent of y. 53.

2 = 2a − 2 4 = 2a

y

2 = a

4 3

59.

2 1 x

−4 −3 −2 −1 −1

4− 2 = 2 a −1 2 = 2 a −1 2 = 2( a − 1)

1

2

3

3− a 1 = 2 −2 − ( − 4 ) 2(3 − a ) = 1( 2)

4

6 − 2a = 2 −2a = −4 a = 2

55.

y

61.

5 4 3 2

a = 3

1 −4 −3 −2 −1 −2 −3

a −0 3 = − 0−5 5 5a = 15

x 1

2

3

4

63.

a − ( −2 ) = −1 − 1 − ( −7 ) a + 2 = −6 a = −8

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

71

Mid-Chapter Quiz for Chapter 3

Mid-Chapter Quiz for Chapter 3 1. Quadrant I or II. Because x can be any real number and y is 4, the point ( x, 4) can only be located in quadrants in

4.

y 8 6

which the y-coordinate is positive.

(− 4, 3)

4 2

2. 4 x − 3 y = 10

x

−8 −6 −4 −2 −2

?

(a) 4( 2) − 3(1) = 10

2

−6

?

Not a solution 2

⎡⎣6 − (−4)⎤⎦ + ( −7 − 3)

?

d =

?

=

102 + ( −10)

=

100 + 100

=

200 =

(b) 4(1) − 3( −2) = 10 4 + 6 = 10 10 = 10

Solution

?

(c) 4( 2.5) − 3(0) = 10

10 = 10

Solution

?

( )

(d) 4( 2) − 3 − 23 = 10

5.

y = 6 Solution

y

3x = 6 x = 2

(1, 72 (

y = 3

2

(3, 2)

1

(2, 0)

4

(1, 3)

2

(2, 0)

(1, 3)

x −4

−2

4

6

8

−2

x 1

2

3

6. y = 6 x − x 2 y = 6(0) − 02

d =

(0, 6)

6

3(1) + y − 6 = 0

3

−3 −2 −1 −1

(0, 6)

3x + 0 − 6 = 0

y

4

2 ⋅ 100 = 10 2

3(0) + y − 6 = 0

8 + 2 = 10

(− 1, 5)

2

3x + y − 6 = 0

?

10 = 10

2

⎛ − 4 + 6 3 + ( −7) ⎞ ⎛ 2 − 4 ⎞ M = ⎜ , ⎟ = ⎜ , ⎟ = (1, − 2) 2 ⎝ 2 ⎠ ⎝2 2 ⎠

?

10 − 0 = 10

3.

8

(6, − 7)

−8

5 ≠ 10

6

(1, − 2)

−4

8 − 3 = 10

4

(−1 − 3)

=

16 + 9

=

25

2

+ (5 − 2)

2

= 5 ⎛ −1 + 3 5 + 2 ⎞ ⎛ 7 ⎞ M = ⎜ , ⎟ = ⎜1, ⎟ 2 ⎠ ⎝ 2⎠ ⎝ 2

y

(0, 0)

= 0

(3, 9) 9

y = 6(6) − 62

6

(6, 0)

= 0 y = 6(3) − 3

2

3

(0, 0) −6

−3

(6, 0) 3

9

x 12

−3

= 18 − 9 = 9

(3, 9)

7. y = x − 2 − 3

y

y = 0− 2 −3 = −1

(0, −1)

y = 5− 2 −3 = 0

(−1, 0) 1

(5, 0) x

(5, 0)

y = 2− 2 −3 = −3

4 3

(0, −1) −2 −3 −4

1 2 3

5 6

(2, −3)

(2, − 3)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

72

Chapter 3

8. m =

Grap hs and Functions

5−0 5 9. m = = 6−3 3

10. m =

12. 6 x − 4 y = 12

8−8 0 = = 0 Line is horizontal. 7 − ( −3) 10

−4 y = −6 x + 12 3 x −3 2

y =

Line rises.

3 2 y -intercept = (0, − 3)

m =

−1 − 7 −8 4 = = − 4 − ( −2) 6 3

y

Line falls.

2 1

11. 3x + 6 y = 6 1 y = − x +1 2

−3

2

3

4

(0, −3)

−5

1 13. y = 3 x + 2; y = − x − 4 3 m1 = 3

y 3

1 3 m1 ⋅ m2 = −1 m2 = −

2

(0, 1) x 1

1

−2

1 m = − ; y -intercept = (0, 1) 2

−1

x

−3 −2 −1 −1

6 y = −3 x + 6

2

4

The lines are perpendicular.

−1 −2

(−9) − 3 = −12 = 2 ⇒ y − 3 = 2 x ( ( −2) − 4 −6 5 − ( −5) 10 = = = 2 ⇒ y = 2x − 5

14. L1 : m1 =

L2 : m2

− 4) ⇒ y = 2 x − 5

5−0 5 The lines are neither parallel nor perpendicular because they are the same line. 15.

1 ( x − 6) 2 1 y + 1 = ( x − 6) 2 2( y + 1) = x − 6

y − ( −1) =

2y + 2 = x − 6 x − 2y − 8 = 0 16. The total depreciation over the 10-year period is $124,000 − $4000 = $120,000. $120,000 = $12,000. 10

$124,000 − ( 2)$12,000 = $100,000 $124,000 − (3)$12,000 = $88,000 $124,000 − ( 4)$12,000 = $76,000 So, y = 124,000 − 12,000t , 0 ≤ t ≤ 10.

V 150,000

Value (in dollars)

The annual depreciation is

125,000 100,000 75,000 50,000 25,000 t 2

4

6

8

10

12

Time (in years)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.5

Graphs of Linear Inequalities

73

Section 3.5 Graphs of Linear Inequalities 1. x − 2 y < 4

5. y > 0.2 x − 1 ?

(a) 0 − 2(0) < 4

?

(a) 2 > 0.2(0) − 1 2 > −1

0 < 4

(0, 0) is a solution. ?

(b) 2 − 2( −1) < 4

(0, 2) is a solution. ?

(b) 0 > 0.2(6) − 1 0 > 0.2

2+ 2 < 4 4 < 4

(2, −1) is not a solution. ?

(c) 3 − 2( 4) < 4

(6, 0) is not a solution. ?

(c) −1 > 0.2( 4) − 1 −1 > −0.2

3−8 < 4 −5 < 4

(3, 4) is a solution.

(4, −1) is not a solution. ?

(d) 7 > 0.2( −2) − 1 7 > −1.4

?

(d) 5 − 2(1) < 4

(−2, 7) is a solution.

5− 2 < 4 3 < 4

7. x ≥ 6

(5, 1) is a solution.

y 6

3. 3x + y ≥ 10

4 2

?

(a) 3(1) + 3 ≥ 10 9 ≥ 10

−2

x 2

4

8

10

4

6

−2 −4

(1, 3) is not a solution.

−6

?

(b) 3( −3) + 1 ≥ 10

−8 ≥ 10

9. y < 5

(−3, 1) is not a solution.

y 8

?

(c) 3(3) + 1 ≥ 10

6

10 ≥ 10

4

(3, 1) is a solution. ?

(d) 3( 2) + 15 ≥ 10 21 ≥ 10

(2, 15) is a solution.

2 −6

−4

11. y >

−2

x 2 −2

1x 2 y

2 1 x −2

1

2

−1 −2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

74

Chapter 3

Grap hs and Functions

13. y ≥ −2; (b)

27. 2 x + y ≥ 4

y ≥ − 2x + 4

14. x < −2; (a)

Because the origin (0, 0) does not satisfy the inequality,

15. x − y < 0; (d)

the graph consists of the half-plane lying on or above the line.

16. x − y > 0; (e)

y 7

17. x + y < 4; (f )

6

18. x + y ≤ 4; (c)

4 3 2

19. y ≥ 3 − x

1

y

x

−4 −3 −2 −1 −1

1

3

4

5 4

29. − x + 2 y < 4

3

2y < x + 4

2 1 −1

y
6

y > − 12 x + 6

y ≥ −x + 4

Because the origin (0, 0) does not satisfy the inequality,

y

the graph consists of the half-plane lying above the line.

3 2

y

1

12 −3

−2

−1

x 1

2

10

3

−1

8

−2 −3

4

25. Because the origin (0, 0) does not satisfy the inequality,

the graph consists of the half-plane lying below the line.

2 −6

−4

−2

x 2

4

6

y 8 6 4 2 −8 −6 −4 −2 −2

x 2

4

6

8

−8

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.5

−4y ≤ − x + 8 1x 4

− y ≤ 2x + 2 y ≥ −2 x − 2

− 2

y

Because the origin (0, 0) satisfies the inequality, the graph consists of the half-plane lying on or above the line.

3 2 1

y

−3

1 −3

−2

−1

75

41. 3x − 2 ≤ 5 x + y

33. x − 4 y ≤ 8

y ≥

Graphs of Linear Inequalities

−2

x

−1

1

2

3

x 1

2

3

−2

−1

−3 −3

43. 0.2 x + 0.3 y < 2

−4

2 x + 3 y < 20

−5

3 y < −2 x + 20

35. 3x + 2 y ≥ 2

20 3

y < − 23 x +

2 y ≥ −3 x + 2

y

y ≥ − 32 x + 1

10

y

8

3

6 4

1 −2

−1

2 x 2

−1

3

−2

4

x

2

4

6

8

10

−2

−2

45. y − 1 > − 12 ( x − 2)

−3

37. 5 x + 4 y < 20

y − 1 = − 12 x + 1

4 y < −5 x + 20 − 54 x

y


1x 3

x y + ≤1 3 4 4 x + 3 y ≤ 12

3 y ≤ −4 x + 12

−3

4 y ≤ − x + 4 3

y 8

y

6 4

5

2 −8 −6 −4 −2 −2

4 x 2

8

3

−4

2

−6

1

−8

−1

x 1

2

3

4

5

−1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

76

Chapter 3

Grap hs and Functions 61. (a) Verbal model:

49. (a) 11x + 9 y ≥ 240

Cost Cost of Cost of cheese + extra + for ≤ 32 pizzas toppings drinks

9 y ≥ −11x + 240 y ≥

(b)

− 11 x 9

+

80 ; 3

x ≥ 0, y ≥ 0

Cost for drinks = y (dollars)

20 15

Inequality: 20 + 0.60 x + y ≤ 32

10

0.60 x + y ≤ 12,

5 x

(b)

Time at grocery store (in hours)

x ≥ 0, y ≥ 0

y 20

Number of drinks

Time mowing lawns (in hours)

Cost for extra toppings = 0.60 x (dollars)

25

5 10 15 20 25 30

( x, y ): ( 2, 25), ( 4, 22), (10, 15) 51. False. Another way to determine if ( 2, 4) is a solution of

15 10 5

the inequality is to substitute x = 2 and y = 4 into the inequality 2 x + 3 y > 12. 53. The solution of x − y > 1 does not include the points on the line x − y = 1. The solution of x − y ≥ 1 does include the points on the line x − y = 1. 55. m =

Cost of cheese pizzas = 2(10) = $20

Labels:

y 30

x 5

10

15

20

Number of toppings

(c)

(6, 6) ?

0.60(6) + 6 ≤ 12 ?

3.6 + 6 ≤ 12

2−5 3 = − 3+1 4

9.6 ≤ 12

(6, 6) is a solution of the inequality.

3 ( x − 3) 4 4 y − 8 > −3 x + 9

y −2 > −

63. (a) 2 w + t ≥ 70

t ≥ −2 w + 70,

3 x + 4 y > 17

(b)

57. y < 2

x ≥ 0, y ≥ 0

t

Number of ties

100

59. (a) 10 x + 15 y ≤ 1000

15 y ≤ −10 x + 1000 y ≤

(b)

− 23 x

+

200 , 3

x ≥ 0, y ≥ 0

80 60 40 20 w 20

y

40

60

80

100

Number of wins

Number of tables

100

( w, t ): (10, 50), ( 20, 30), (30, 10)

80 60

65. On the real number line, the solution of x ≤ 3 is an unbounded interval.

40 20 x 20

40

60

80 100

Number of chairs

On a rectangular coordinate system, the solution of x ≤ 3 is a half-plane. 67. To write a double inequality such that the solution is the graph of a line, use the same real number as the bounds of the inequality, and use ≤ for one inequality symbol and ≤ for the other symbol. For example, the solution of 2 ≤ x + y ≤ 2 is x + y = 2. Graphically, x + y = 2 is a line in the plane.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.6 69. The boundary lines of the inequalities are parallel lines. Therefore, a = c. Because every point in the solution set of y ≤ ax + b is a solution of y < cx + d , the inequality y ≤ ax + b lies in the half-plane of y < cx + d . Therefore, y = ax + b lies under y = cx + d and b < d (assuming b > 0 and d > 0 ).

y

77.

5

x + 2 = −3 or

x + 2 = 3

3

) 12 , 32 )

2

) 2, 12 )

1

x 1

2

y =

2x 3

y −7 =

10 8

y =

6 4

(3, 3)

+ 2

x 2

4

6

2(x 3 2x 3

− 5) +

11 3

(b) y − y1 = m( x − x1 )

2

−4

5

(a) y − y1 = m( x − x1 )

y

−8 −6 −4 −2 −2

4

−3 y = −6 − 2 x

8 − 3 x = −10 or 8 − 3x = 10

(− 7, 4)

3

−1

79. 2 x − 3 y = −6

73. 8 − 3 x = 10

75.

) 12 , 72 )

4

−1

71. x + 2 = 3

77

Relations and Functions

y − 7 = − 32 ( x − 5)

8

(5, − 1)

y = − 32 x +

−6

29 2

Section 3.6 Relations and Functions 1. Domain = {−2, 0, 1}

3. Domain = {0, 2, 4, 5, 6}

y

(1, 4)

Range = {−1, 0, 1, 4}

y

Range = {−3, 0, 5, 8}

4

6

3

2

(0, 1)

(0, 0)

(− 2, 0) −3

−2

−4

x

(0, − 1)

(6, 5)

(5, 5)

4

2 1

(2, 8)

8

1

−2

2

x −2

2

4

6

8

(4, −3)

−4

5. (1, 1), ( 2, 8), (3, 27), ( 4, 64), (5, 125), (6, 216), (7, 343) 7.

{(2008, Philadelphia Phillies), (2009, New York Yankees), (2010, San Francisco Giants), ( 2011, St. Louis Cardinals)}

9.

{(3, 9), (1, 3), (2, 6), (8, 24), (7, 21)}

13. x 2 + y 2 = 25 ?

11. (a) Yes, this relation is a function because each number in the domain is paired with exactly one number in the range.

(b) No, this relation is not a function because the 1 in the domain is paired with 2 different numbers in the range. (c) Yes, this relation is a function because each number in the domain is paired with exactly one number in the range. (d) No, this relation is not a function because each number in the domain is not paired with a number.

?

02 + 52 = 25

02 + ( −5) = 25

25 = 25

25 = 25

2

There are two values of y associated with one value of x, which implies y is not a function of x. 15. y = x + 2 ?

3 = 1+ 2 3 = 3

?

−3 = 1 + 2 3 = 3

There are two values of y associated with one value of x, which implies y is not a function of x.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

78

Chapter 3

Grap hs and Functions 33. Domain: All real numbers r such that r > 0

17. y 2 = x y =

x or −

Range: All real numbers A such that A > 0

x

Because ( 4, 2) and ( 4, − 2) are both solutions, y is not a function of x. 19. y = x y = −3 = 3

Because one value of x corresponds to one value of y, y is a function of x. 21. f ( x) = 12 x − 7

(a) f (3) = 12(3) − 7 = 29 (b) f

37. Negative numbers are excluded from the domain because you cannot produce a negative number of video games. 39. No, this relation is not a function because −1 in the domain is matched to 2 numbers (6 and 7) in the range. 41. Yes, this relation is a function as each number in the domain is matched to exactly one number in the range.

( 32 ) = 12( 32 ) − 7 = 11

(c) f ( a) + f (1) = ⎡⎣12( a) − 7⎤⎦ + ⎡⎣12(1) − 7⎤⎦ = 12a − 7 + 12 − 7

43. g ( x) = 2 − 4 x + x 2

(a) g ( 4) = 2 − 4( 4) + 42 = 2 − 16 + 16 = 2 (b) g (0) = 2 − 4(0) + 02 = 2

= 12a − 2

(d) f ( a + 1) = 12( a + 1) − 7

(c) g ( 2 y ) = 2 − 4( 2 y ) + ( 2 y ) = 2 − 8 y + 4 y 2 2

= 12a + 12 − 7

(d) g ( 4) + g (6) = ⎡⎣2 − 4( 4) + 42 ⎤⎦ + ⎡⎣2 − 4(6) + 62 ⎤⎦

= 12a + 5 23. f ( x ) =

35. The domain of a function is the set of inputs of the function, or the set of all the first components of the ordered pairs. The range of a function is the set of outputs of the function, or the set of all the second components of the ordered pairs.

= ( 2 − 16 + 16) + ( 2 − 24 + 36)

3x x −5

(a) f (0) =

= 2 + 14 = 16

3(0) 0−5

45. The domain of f ( x ) =

= 0

⎛5⎞ 3⎜ ⎟ 3 15 15 3 ⎛5⎞ 3 = = − (b) f ⎜ ⎟ = ⎝ ⎠ ⋅ = 5 3 5 − 15 − 10 2 ⎝ 3⎠ −5 3 ⎡ 3( 2) ⎤ ⎡ 3(−1) ⎤ (c) f ( 2) − f ( −1) = ⎢ ⎥ −⎢ ⎥ ⎣ 2 − 5 ⎦ ⎣ −1 − 5 ⎦ −3 6 1 5 = − = −2 − = − −3 −6 2 2 (d) f ( x + 4) =

3( x + 4) x + 4−5

=

t +3 27. The domain of f (t ) = is all real numbers t (t + 2) t such that t (t + 2) ≠ 0. So, t ≠ 0 and t ≠ −2. 29. The domain of g ( x) =

x such that 2 x − 1 ≥ 0. So, x ≥ 12 . 47. The domain of f (t ) = t − 4 is all real numbers t. 49. Verbal Model:

Total cost = Variable costs + Fixed costs Labels:

Fixed costs = 495 Number of units = x Function:

Volume = Length ⋅ Width ⋅ Height Labels:

Volume = V Length = ( 24 − 2 x) Width = ( 24 − 2 x)

x such that x + 4 ≥ 0. So, x ≥ −4.

Range: All real numbers C such that C > 0

C ( x) = 3.25 x + 495, x > 0

51. Verbal Model:

x + 4 is all real numbers

31. Domain: All real numbers r such that r > 0

Total cost = C ( x) Variable costs = 3.25 x

3x + 12 x −1

25. The domain of f ( x ) = x 2 + x − 2 is all real numbers x.

2 x − 1 is all real numbers

Height = x Function:

V = x( 24 − 2 x) , x > 0 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.7 y

67.

128,160 L

53. S ( L) =

79

Graphs of Functions

6 5

128,160 (a) S (12) = = 10,680 pounds 12

4 3 2

128,160 (b) S (16) = = 8010 pounds 16

1 x

−4 −3 −2 −1 −1

55. (a) w( 20) = 12( 20) = $240

1

2

3

4

−2

w(35) = 12(35) = $420 w(50) = 18(50 − 40) + 480 = $660

69. 5 x − 3 y = 0

−3 y = −5 x

w(55) = 18(55 − 40) + 480 = $750

5x 3

y =

(b) When the worker works between 0 and 40 hours, the wage is $12 per hour. When more than 40 hours are worked, the worker earns time-and-a-half for the additional hours. 57. Because the prices of shoes vary, the store may sell more or fewer shoes depending on the prices. So, the use of the word function is not mathematically correct.

y 5 4 3 2 1 x

−5 −4 −3 −2 −1

1 2 3 4 5

−3 −4 −5

71.

y 7 6

59. If set B is a function, no element of the domain is matched with two different elements in the range. A subset of B would have fewer elements in the domain, and therefore, a subset of B would be a function. 61. Statement: The distance driven on a trip is a function of the number of hours spent in the car.

Domain: The number of hours in the car Range: The distance driven

5 4 2 1 −4 −3 −2 −1 −1

x 1

2

3

4

73. 2 x + 3 y ≤ 6

y

3y ≤ 6 − 2x

The function has an infinite number of ordered pairs.

y ≤ 2−

63. Multiplication Property of Zero

7 6

2x 3

5 4 3 2

65. Distributive Property

1 −4 −3 −2 −1 −1

x 1

2

3

Section 3.7 Graphs of Functions 1. f ( x) = 2 x − 7

x

0

f ( x)

3. h( x) = − ( x − 1) 2

1234

5

–7 –5 –3 –1 1 3

x

–1

h( x)

–4 –1 0 –1 –4

y

0

1

x

−1

x

−2

3

y

2 −4

2

2

4

6

1

2

3

−1

−2 −4 −6

−3 −4

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

80

Chapter 3

Grap hs and Functions 15. Basic function: f ( x) = c The constant c is 7.

5. f ( x) = x + 3

x f ( x)

–6

–3 0 3 6

3

0

3

69

y

19. Yes, y =

8

4

−6

−4

1 x 3 passes 3

the Vertical Line Test and is a

function of x.

6

−8

17. Basic function: f ( x) = x 2 The graph is upside down and shifted 1 unit to the right and 1 unit downward.

21. No, y is not a function of x by the Vertical Line Test. 23. (a) Vertical shift 2 units upward

x

−2

2 −2

y

7 6

7.

y

5 4

5

h(x) = 3 − x x≥0

4

3 1

h(x) = 2x + 3 3 x2

5 4

−4

−3

−2

−1

x 1 −1

(d) Horizontal shift 4 units to the right

g(x) = 1 3 x≤1 2

y 8

−2

−1

x 1

2

3

4

6

−1

4

Domain: −∞ < x ≤ 1 and 2 < x < ∞

2

Range: y = 1 and 2 < y < ∞

x 2

4

6

8

13. Basic function: f ( x) = x

The graph is shifted 1 unit downward.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.7 25. (a) Vertical shift 3 units upward

Graphs of Functions

81

30. y = x3 ; (b)

y

31. y = − x3 ; (c)

5 4

32. y = ( − x) ; (c) 3

2

33. The basic function related to h( x) = ( x − 1) 2 + 2 is

1 −3

f ( x) = x 2 .

x

−2

1

−1

3

2

(b) Vertical shift 5 units downward

35. The part of h( x) = ( x − 1) 2 + 2 that indicates a vertical

shift is ( x − 1) 2 + 2. The expression is of the form f ( x) + c.

y

−4

x

−2

2

4

37.

−2

y 10

−4

8 6 2

−8

x −10

−6 −4 −2

(c) Horizontal shift 3 units to the right

4

6

−4 −6

y

y is a function of x.

3 2

39.

1 −1

2

y 3

x 1

3

4

5

2

−1

1

−2

x

−3

−2 −1

(d) Horizontal shift 2 units to the left

1

2

3

4

−1 −2 −3

y

y is not a function of x.

3 2

41. Basic function: y = x3

1 x −4

−2

−1

1 −1 −2 −3

2

Transformation: Horizontal shift 2 units right 3 Equation: y = ( x − 2)

43. Basic function: y = x 2

27. y = x 2 ; (d)

Transformation: Reflection in the x-axis, horizontal shift 1 unit left, vertical shift 1 unit upward

28. y = − x 2 ; (a)

2 Equation: y = −( x + 1) + 1

29. y = (− x) 2 ; (d)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

82

Chapter 3

Grap hs and Functions

45. (a) y = f ( x ) + 2

(b) y = − f ( x )

(c) y = f ( x − 2) y

y

y

4

2

5

(4, 4)

(0, 1)

3

1

4

x

3

1

(3, 3)

3

−1

2 1

(6, 2)

2

(1, 0)

(1, 2)

2

3

4

(5, 1)

(3, 0)

x −1

(4, −2)

2

3

4

5

6

(2, − 1)

−2

−3

x 1

1

5

1

−2

(0, 1)

4

(3, −1)

5

(d) y = f ( x + 2)

(e) y = f ( x ) − 1

(f ) y = f ( − x )

y

y 3

y

2

3

(4, 1)

(2, 2) 2

(−4, 2)

1

2

(3, 0) 1

(− 1, 0) −3

x

(1, 1)

1 x

−2

1

(− 2, −1)

2

−1

−1

−2

−2

−3

3

4

5

(−3, 1)

(1, −1)

−5

−3

(0, −2)

47. (a)

(−1, 0) −2

x

−1 −1

(0, −1)

−2

49. Use the Vertical Line Test to determine if an equation represents y as a function of x. If the graph of an equation has the property that no vertical line intersects the graph at two (or more) points, the equation represents y as a function of x.

100 −

A = l ⋅w

51. The graph of g ( x) = f ( − x) is a reflection in the y-axis

A = l (100 − l )

of the graph of f ( x).

Let l = length. 100 − l = width

53. The sum of four times a number and 1

P = 2l + 2 w

55. The ratio of two times a number and 3

200 = 2l + 2 w

57. 2 x + y = 4

100 = l + w

y = 4 − 2x

100 − l = w (b)

−4

A

59. − 4 x + 3 y + 3 = 0

3000

3y = 4x − 3

2500

y =

2000 1500

4x 3

−1

61. y < 2 x + 1

1000

?

1 < 2(0) + 1

500 l 20

40

60

1 < 1; (0, 1) is not a solution.

80

(c) When l = 50, the largest value of A is 2500.

(50, 2500) is the highest point on the graph of A giving the largest value of the function. The figure is a square.

63.

2x − 3y > 2 y ?

2(6) − 3( 2) > 2( 2) ?

12 − 6 > 4 6 > 4; (6, 2) is a solution.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 3

83

Review Exercises for Chapter 3 1.

⎛1 + 7 4 + 2 ⎞ ⎛8 6⎞ , 13. M = ⎜ ⎟ = ⎜ , ⎟ = ( 4, 3) ⎝ 2 ⎝ 2 2⎠ 2 ⎠

y 5 4 3

(

1

−4, 2

2

(

⎛ 5 + ( −3) −2 + (5) ⎞ ⎛ 2 3 ⎞ ⎛ 3 ⎞ , 15. M = ⎜ ⎟ = ⎜ , ⎟ = ⎜1, ⎟ ⎝ 2 2 ⎠ ⎝ 2 2⎠ ⎝ 2⎠

(0, 2)

1 x

−5 −4 −3 −2 −1 −1

1

2

−2

3

4

17.

(2, −3)

−3

3y − 2x − 3 = 0 3 y − 2(0) − 3 = 0

−4

3y = 3

3. Quadrant I

3(0) − 2 x − 3 = 0

?

(2, 3):

7. (a)

3 = 4−

1 2

−2 x = 3

( 2)

y

3 = 3

(2, 3) is a solution. (−1, 5):

2

?

5 = 4 − 12 ( −1) ?

5 ≠ 4 12

(0, 1)

(− 32 , 0( −3

1 2

5 = 4+

(−6, 1):

?

1 = 4−

1 2

−2

(−6)

19. y = x 2 − 1

y = 02 − 1

?

2

0 = x −1

(−6, 1) is not a solution. (d) (8, 0):

0 = 4

(0, −1)

= −1

1 ≠ 7

− 12

1 −1

1 = 4+3

?

x

−1

(−1, 5) is not a solution. (c)

(− 32 , 0)

x = − 32

?

3 = 4 −1

(b)

(0, 1)

y =1

5. Quadrant I or IV

0 = ( x − 1)( x + 1)

(8)

(1, 0), (−1, 0)

x = 1, x = −1

?

0 = 4−4

y

0 = 0

4

(8, 0) is a solution. 9. d =

(4

− 4) + (3 − 8)

=

0 + 25

=

25

2

3 2 1

2

(− 1, 0) −3

−2

(1, 0) 1

−1 −2

2

3

x

(0, − 1)

= 5

11. d =

(−5 − 1)2

=

36 + 9

=

45

+ ( − 1 − 2)

2

= 3 5

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

84 21.

Chapter 3

Grap hs and Functions 27. y = 2 x + 1 − 5

y = x −2 y = 0 −2

y -intercept: y = 2(0) + 1 − 5

(0, − 2)

= −2

= 1−5

0 = x −2 2 = x

x-intercepts: 0 = 2 x + 1 − 5

( 2, 0), (−2, 0)

±2 = x

5 = 2x + 1 5 = 2 x + 1 or −5 = 2 x + 1

y 3

4 = 2x

−6 = 2 x

2

2 = x

−3 = x

(2, 0), (−3, 0)

1

(− 2, 0) −3

−2

(0, − 4)

= −4

(2, 0)

−1

1

x

2

3

y −2

(0, −2)

3

−3

2 1

(−3, 0)

23. 8 x − 2 y = −4

(2, 0)

−4 −3 −2 −1

1

x =

− 12

(0, −4)

(0, 2) 29. (a) The annual depreciation is

(

− 12 ,

0

35,000 − 15,000 = 4000. 5

)

35,000 − 4000(1) = 31,000

y

35,000 − 4000( 2) = 27,000

4

35,000 − 4000(3) = 23,000

3

(0, 2)

2

(

1 − 2, 0

So, y = 35,000 − 4000t.

( x

−4 −3 −2 −1

1

2

3

(b)

4

y

(− 1, − 2)

Value (in dollars)

−3 −4

25. y = 5 − x

y -intercept: y = 5 − x

t

x-intercepts: 0 = 5 − x x = −5, x = 5 y

−8

−4 −2 −2

(0, 5) (5, 0) x 2

4

3

4

5

Time (in years)

(−5, 0), (5, 0)

y = 35,000

(0, 35,000) The y-intercept represents the value of the SUV when purchased.

8

2

2

(c) y = 35,000 − 4000(0)

x = 5

(− 5, 0)

40,000 35,000 30,000 25,000 20,000 15,000 10,000 5,000 1

(0, 5)

y = 5

6

x

4

−3

−2 y = −4

8 x = −4

3

−2

y -intercept: 8(0) − 2 y = −4 y = 2 x-intercept: 8 x − 2(0) = −4

2

8

31. m =

−1 − 2 −3 3 = = 7 −3 − 4 −7

−4 −6 −8

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 3 47. m = − 4; ( x1 , y1 ) = (1, − 4)

3 3 − 0 33. m = 4 4 = = 0 4 − ( −6) 10

35. m =

y − y1 = m( x − x1 ) y − (− 4) = − 4( x − 1) y + 4 = − 4x + 4

0−6 −6 −3 3 = = = − 8−0 8 4 4

y = − 4x

37. 5 x − 2 y − 4 = 0

y

−2 y = −5 x + 4 5 x 2

y =

49. m =

3

1 −2

1; 4

( x1 , y1 ) = (− 6, 5)

y − y1 = m( x − x1 )

2

− 2

x

−1

1

2

3

4

−1 −2

39. 6 x + 2 y + 5 = 0 4

5 2

2

− (− 6))

1 4

y −5 =

1 (x 4

y −5 =

1 x 4

+

3 2

y =

1x 4

+

13 2

m =

3

−4 −3 −2 −1

(x

y −5 =

+ 6)

51. ( x1 , y1 ) = (− 6, 0); ( x2 , y2 ) = (0, − 3)

y

2 y = −6 x − 5 y = −3 x −

85

x 1

2

3

−3 − 0 0 − (− 6)

= −

4

1 2

y − y1 = m( x − x1 ) −3

1 ( x − (− 6)) 2 1 y = − ( x + 6) 2 2 y = − ( x + 6)

y −0 = −

−4

41. L1 : y =

3 x 2

+1

L2 : y =

2 x 3

−1

m1 =

3 , 2

m2 =

2 3

m1 ≠ m2 , m1 ⋅ m2 ≠ −1

The lines are neither parallel nor perpendicular. 43. L1 : y =

3 x 2

− 2

L2 : y = − 23 x + 1 m1 =

3 , 2

m2 = − 32

m1 ⋅ m2 = −1

The lines are perpendicular. 45. Let p represent the price of milk and let t represent the year. The two data points are (t1 , p1 ) and (t2 , p2 ).

(t1 , p1 ) = (2000, 2.79) (t2 , p2 ) = (2010, 3.32)

2y = − x − 6 x + 2y + 6 = 0

53. ( x1 , y1 ) = (− 2, − 3); ( x2 , y2 ) = (4, 6)

m = =

6 − (− 3) 4 − (− 2) 3 2

y − y1 = m( x − x1 ) 3 ( x − (− 2)) 2 3 y + 3 = ( x + 2) 2 2 y + 6 = 3( x + 2)

y − (− 3) =

2 y + 6 = 3x + 6 − 3x + 2 y = 0 3x − 2 y = 0

The slope formula can be used to find the average rate of change.

55. y = −9

3.32 − 2.79 m = = 0.053 2010 − 2000

57. x = −5

From 2000 to 2010, the average rate of change of the price of a gallon of whole milk is $0.053.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

86

Chapter 3

Grap hs and Functions

59. 3x + y = 2

65. y > −2

y

y = −3x + 2

2 1

(

(a) y +

4 5

= −3 x −

y +

4 5

= −3 x +

3 5

)

9 5

(

3 5

4 5

=

1 3

y +

4 5

=

1x 3

−

y =

1 x 3

−1

x −

2

3

4

−4 −5

)

−6

3 15

67. x − 2 ≥ 0

y

x ≥ 2

3 2

61. (a) Let (t1 , S1 ) = (7, 32,000). Because the annual

1

salary increased by about $1050 per year, the slope is m = 1050. S − S1 = m(t − t1 ) S − 32,000 = 1050t − 7350 S = 1050t + 24,650 The equation that represents the salary is S = 1050t + 24,650.

63. 5 x − 8 y ≥ 12

(a) 5( −1) − 8( 2) ≥ 12

4

−2

69. 2 x + y < 1

y

y < −2 x + 1

3

1

−2

x

−1

1

2

−1

71. −( x − 1) ≤ 4 y − 2

y

−x + 1 ≤ 4 y − 2

4

−x + 3 ≤ 4 y

3

− 14 x +

?

3

1 −1

(b) For t = 15, the equation S = 1050t + 24,650 has the value S = 1050(15) + 24,650 = 40,400. Your annual salary in the year 2015 is predicted to be $40,400. (c) For t = 10, the equation S = 1050t + 24,650 has the value S = 1050(10) + 24,650 = 35,150. Your annual salary in the year 2010 is estimated to be $35,150.

x

−1

S − 32,000 = 1050(t − 7)

3 4

2

≤ y

−5 − 16 ≥ 12 −21 ≥/ 12

1

−3

y = −3 x + 1 (b) y +

x

−4 −3 −2 −1 −1

−2 −1

Not a solution

x −1

1

2

3

−2

?

(b) 5(3) − 8( −1) ≥ 12 15 + 8 ≥ 12 23 ≥ 12

Solution

73. (a) P = 2 x + 2 y

2 x + 2 y ≤ 800

?

(c) 5( 4) − 8(0) ≥ 12

x + y ≤ 400,

20 − 0 ≥ 12 20 ≥ 12

Solution

?

y

400

(d) 5(0) − 8(3) ≥ 12

300

0 − 24 ≥ 12 −24 ≥/ 12

(b)

x ≥ 0, y ≥ 0

200

Not a solution

100 x 100 200 300 −100

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 3 75. Domain: {−3, −1, 0, 1}

89. Verbal model:

Perimeter = 2 Length + 2 Width

Range: {0, 1, 4, 5}

150 = 2Length + 2 x 150 − 2 x = Length 2 75 − x = Length

y

(−3, 5)

5

(−3, 4)

4

(1, 4) 3 2 1

(0, 1)

Verbal model:

Area = Length ⋅ Width

Labels:

Area = A Length = 75 − x

(−1, 0) −3

−2

−1

x 1

2

3

Width = x

−1

79. Yes, this relation is a function because each number in the domain is matched with only one number in the range.

Domain: 0 < x
0 (a) f (3) = 1 − 32 = −8

( )

−6

93. g ( x) = 6 − 3x, − 2 ≤ x ≤ 4

( )

(b) f − 23 = −3 − 23 = 2 (c) f (0) = −3(0) = 0

x

–2

–1 0 1

g ( x)

12

9

(d) f ( 4) − f (3) = (1 − 42 ) − (1 − 32 )

–6

4

Domain: −∞ < x < ∞

1

−2 −1

2

3

x

4

−4

3 x + 10

−8

95.

y

3 x ≥ −10

4

x ≥ − 10 3

3

− 10 3

2

Range: f ( x) ≥ 0

1

Domain: x ≥

0

12

85. h( x) = 4 x 2 − 7

3x + 10 ≥ 0

3

4

y

= −15 − ( −8) = −15 + 8 = −7

87. f ( x) =

6

2

−1

y = 2 + (x − 1) 2 x≥1 y = 2 − (x − 1) 2 x 2x

= k 2 − 3k ( 2 + k )

2 1

2

= k − 6k − 3k

2

1

2

3

4

1

2

3

4

2

= −2 k − 6 k −3

2 3k

−4

5 k D x = x = −2k 2 − 6k D −13k 2k − 15k = = −2k 2 − 6k −2k ( k + 3) =

x

−4 − 3 − 2 −1

83. − x + 3 y > 12

y

3 y > x + 12 1x 3

y >

13 13 = 2( k + 3) 2k + 6

6 5

+ 4

3 2 1

k y =

2

Dy 2+ k 5 = −2k 2 − 6k D 5k − 2( 2 + k ) 5k − 4 − 2k = = −2k 2 − 6k −2k 2 − 6k −1( 4 − 3k ) 3k − 4 = = 2 −2k − 6k −1( 2k 2 + 6k ) =

−4 − 3 − 2 −1 −1 −2

85. The graph of h( x) has a vertical shift c units upward. 87. f ( x) − 2 y

4 − 3k 2k 2 + 6k

4 3 2

(b) −2k 2 − 6k = 0 −2k = 0 k = 0

(2, 1)

(3, 1)

1

−2k ( k + 3) = 0

x

−4 − 3 − 2 −1

k + 3 = 0 k = −3

77. A determinant is a real number associated with a square matrix. 79. The determinant is zero. Because two rows are identical, each term is zero when expanding by minors along the other row. Therefore, the sum is zero.

x

1

(−1, −2) (−2, −3)

2

3

4

(1, −2) −3 −4

89. f ( − x) y

(−2, 3)

4 3

(−3, 3)

2 1

(−1, 0) −4 −3 −2 −1

−2

(1, 0) x 1

3

4

(2, −1)

−3 −4

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 4.6

Systems of Linear Inequalities

125

Section 4.6 Systems of Linear Inequalities 1. ⎧⎪x + y ≤ 3 ⇒ ⎨ ⎪⎩ x − y ≤ 1

7. ⎧ x − 2 y > 4 ⎨ ⎩2 x + y > 6

y ≤ −x + 3 y ≥ x −1

y

1x 2

− 2

y > −2 x + 6

y

x−y=1

4

x+y=3

y
−1 ⎨ ⎩x + y < 3

y ≥ −x + 2

1x 2

Dotted lines at y =

y > −x − 1

⇒

y < −x + 3

y y

4

x+y=3

4

3 2

2

1

1

x

−1

1

4

5

6

7

−3 −4

x

−4 −3 −2 −1

x+y=2

1

2

3

4

−2

2x − 4y = 6

−3 −4

Solid lines at y =

1x 2

−

3 2

Dotted lines at y = − x − 1 and y = − x + 3. Shade above y = − x − 1 and below y = − x + 3.

above each line. 5. ⎧x + 2 y ≤ 6 ⎨ ⎩x − 2 y ≤ 0

y ≤ − 12 x + 3

⇒

y ≥

11. ⎧⎪ y ≥ 43 x + 1 ⎨ ⎪⎩ y ≤ 5 x − 2

1x 2

y

x + 2y = 6

x + y = −1

and y = − x + 2. Shade

y

6 5

5

4

4 3

2

−4 −3 −2

1

x −1

1

2

3

4

−3 −2

−2

−3

Solid lines at y = − 12 x + 3 and y = − 12 x

x −1 −2

x − 2y = 0

y =

4 y = 3x + 1

2

1

+ 3 and above y =

1 x. 2

1 x. Shade 2

below

1

2

3

5

y = 5x − 2

Solid lines at y = above y =

4

4x 3

4x 3

+ 1 and y = 5 x − 2. Shade

+ 1 and below 5 x − 2.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

126

Chapter 4

Systems of Equations and Inequalities

13. ⎧x > − 4 ⎨ ⎩x ≤ 2

19. ⎧x < 3 ⎨ ⎩x > − 3 y

x = −4

y

x = −3

x=2

4

4

3

3

2

2

1 −5

x=3

1 x

− 3 − 2 −1 −1

1

−4

3

x

− 2 −1 −1

−2

−2

−3

−3

−4

−4

1

2

4

Dotted line at x = − 4. Solid line at x = 2. Shade to

Dotted line at x = 3. Dotted line at x = − 3. Shade to

the right of x = − 4 and to the left of x = 2.

the left of x = 3 and to the right of x = − 3. 21. ⎧x + y ≤ 4 ⎪ ≥ 0 ⎨x ⎪ y ≥ 0 ⎩

15. ⎧x < 3 ⎨ ⎩x > −2 y 4

x = −2

6 5

1 −1

1

2

(0, 4)

4

x

x+y=4

3

4

2

−2

y=0

1

−3 −4

17. ⎧x ≤ 5 ⎨ ⎩x > − 6 y

1

x 2

3

4

5

6 4

6

−2

Solid line at y = − x + 4. x = 0 is the y-axis and y = 0 is the x-axis. Shade below y = − x + 4 in the first quadrant. 23. ⎧4 x − 2 y > 8 ⎪ ≥ 0 ⎨ x ⎪ y ≤ 0 ⎩

x=5

8

2 −4 −2 −2

(4, 0)

(0, 0)

−2 −1

Dotted lines at x = 3 and x = −2. Shade to the right of x = −2 and to the left of x = 3.

−8

y ≥ 0

x=0

x=3

2

x = −6

x ≥ 0

⇒

y

3

−4 −3

y ≤ −x + 4

y < 2x − 4 ⇒

x ≥ 0 y ≤ 0

y x 2

4

6

8

4 2

−4 −6 −8

Solid line at x = 5. Dotted line at x = − 6. Shade to the left of x = 5 and to the right of x = − 6.

−6 −4 −2

x=0 (0, −4)

y=0 (2, 0) 4

6

x 8 10

4x − 2y = 8

−8 −10 −12

Dotted line at y = 2 x − 4. x = 0 is the y-axis and y = 0 is the x-axis. Shade below y = 2 x − 4 in the fourth quadrant.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 4.6

Line 2: points (1, − 2) and (3, 0) m =

y

x=2

4

y=x+2

y = x −3

(2, 4)

2

Line 3: points (1, 4) and (3, 0)

x

−4

4 −2

(−9, −5)

m =

(2, −5)

−4 −6

27. ⎧ x ⎪ ⎪ x − 2y ⎨ ⎪3 x + 2 y ⎪ x + y ⎩

x ≥ 1

≥ 1 ≤ 3

y ≥

⇒

≥ 9

0−4 −4 = = −2 3−1 2

y − 0 = −2( x − 3)

y = −5

Dotted line at y = −5. Solid lines at x = 2 and y = x + 2. Shade above y = −5, to the left of x = 2 and below y = x + 2.

y

0+ 2 2 = =1 3−1 2

y − 0 = 1( x − 3)

6

−6

127

31. Line 1: vertical x = 1

25. ⎧ y > −5 ⎪ ⎨x ≤ 2 ⎪y ≤ x + 2 ⎩

−8

Systems of Linear Inequalities

y ≥

1x − 3 2 2 − 32 x + 92

y = −2 x + 6 System of linear inequalities:

x ≥1

Region on and to the right of x = 1

y ≥ x −3

Region on and above line y = x − 3.

y ≤ −2 x + 6

Region on and below line y = −2 x + 6.

y ≤ −x + 6

≤ 6

33. Line 1: x = −2

x=1

Line 2: x = 2

7

Line 3: points ( −2, 3) and ( 2, 1)

(1, 5) x + y = 6 4 3

m =

x − 2y = 3

(1, 3)

2

(3, 0)

1

(5, 1)

−1

4

5

6

7

3x + 2y = 9

1x 2

Solid lines at x = 1, y =

− 32 , y = − 32 x + 92 , and

y = − x + 6. Shade to the right of x = 1, above y =

1x 2

−

3 2

and y = − 32 x +

9 2

and below

y = − x + 6. 29. ⎧ x − y ⎪ ⎪2 x + 5 y ⎨ ⎪ x ⎪ y ⎩

8

≤ ≥

0

≥

0

y − ( −3) = −

x=0 2x + 5y = 25 4 2

y=0 x 2 4

−5 − ( −3) −2 1 = = − 2 − ( −2) 4 2

1 ( x − (−2)) 2 1 y + 3 = − x −1 2 1 y = − x −4 2 System of linear inequalities:

≤ 25

y

−4 −6 −8

Line 4: points ( −2, − 3) and ( 2, − 5) m =

10 8

−4 − 2

1 ( x − 2) 2 1 y −1 = − x +1 2 1 y = − + 2 2 y −1 = −

x 3

1−3 1 −2 = = − 2 − ( −2) 4 2

8 10 12

x−y=8

Solid lines at all lines. Shade below 2 x + 5 y = 25, above y = 0 and x − y = 8, and to the right of x = 0.

x ≥ −2:

Region on and to the right of x = −2

x ≤ 2:

Region on and to the left of x = 2

1 1 y ≤ − x + 2: Region on and below y = − x + 2 2 2 1 1 y ≥ − x − 4: Region on and above y = − x − 4 2 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

128

Chapter 4

Systems of Equations and Inequalities Number of hours Number Number of hours Number ⋅ + ⋅ ≤ 12 in assembly of tables in assembly of chairs

35. Verbal Model:

Number of hours Number Number of hours Number ⋅ + ⋅ ≤ 16 in finishing of tables in finishing of chairs

Labels:

Number of tables = x

y

Number of chairs = y 3 y 2 3 y 4

1x + 4 x 3

+

x

Number of chairs

System:

12

≤ 12 ≤ 16 ≥ 0

10 8 6 4 2

y ≥ 0

x 2

4

6

8

10 12

Number of tables

37. A system of linear inequalities in two variables consists of two or more linear inequalities in two variables. 39. Not necessarily. Two boundary lines can sometimes intersect at a point that is not a vertex of the region. 41. ⎧ x + y ≤ 1 ⎪ ⎨− x + y ≤ 1 ⎪ y ≥ 0 ⎩

y ≤ −x + 1 ⇒

45. ⎧−3 x + 2 y < 6 ⎪ ⎨ x − 4 y > −2 ⎪ ⎩ 2x + y < 3

⇒

y < − 32 x + 3 y
2

⇒

x ≥ 0

y

− 4) − (1)( 4 x − 4 y ) = 0

67. ⎧x + y < 5 ⎪ > 2 ⎨x ⎪ y ≥ 0 ⎩

y ≤ −3 x + 180

⇒

y ≥ 0

100

150

Solid lines at y = − 12 x + 80 and y = −3x + 180. x = 0 is the y-axis and y = 0 is the x-axis. Shade

y

below y = − 12 x + 80 and y = −3x + 180. Shade to

x=2

6 5 4

the right of x = 0 and above y = 0.

x+y=5

3 2

y=0

1 −1

x 1

3

4

5

6

7

−2

Dotted lines at y = − x + 5 and x = 2. y = 0 is the x-axis. Shade below y = − x + 5 and above y = 0. Shade to the right of x = 2. Cartons of soup for + soup kitchen

Cartons of soup for homeless ≤ 500 shelter Cartons of soup ≥ 150 for soup kitchen

y = 220

300 200 100 x 50

40

30

0

0

0

0

0

Cartons of soup for soup kitchen = x

x + y = 500

400

20

Labels:

x = 150

500

10

Cartons of soup for homeless ≥ 220 shelter

y

Number of cartons for homeless shelter

71. Verbal Model:

Number of cartons for soup kitchen

Cartons of soup for homeless shelter = y System:

x + y ≤ 500 x

≥ 150 y ≥ 220

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test for Chapter 4

139

Chapter Test for Chapter 4 1. (a)

(2, 1)

4. 2 x − 2 y = −2

2x − 2 y = 2

3x +

−x + 2 y = 0

?

−2 + 2(1) = 0

?

y = −3 x + 9

?

4−2=2

−2 + 2 = 0

2 = 2

0 = 0

Substitute into first equation. 2 x − 2( −3x + 9) = −2

(2, 1) is a solution. (b)

2 x + 6 x − 18 = −2 8 x = 16

(4, 3) 2x − 2 y = 2

x = 2

−x + 2 y = 0

?

y = −3( 2) + 9 = 3

?

2( 4) − 2(3) = 2

− 4 + 2(3) = 0

?

The solution is ( 2, 3).

?

8−6=2

−4 + 6 = 0

2 = 2

2 ≠ 0

(4, 3) is not a solution. 2.

5.

3x − −3 x +

y

x − 2y = −1

8 6

2x + 3y = 12

2

(3, 2) x 2

4

6

8

−2

The solution is (3, 2). 3. 4 x −

9

Solve for y.

?

2( 2) − 2(1) = 2

y =

y =

1

4 x − 3 y = −5 4x − y = 1 − y = −4 x + 1 y = 4x − 1 4 x − 3( 4 x − 1) = −5 4 x − 12 x + 3 = −5 −8 x + 3 = −5 −8 x = −8 x =1 4(1) − y = 1 − y = −3 y = 3

The solution is (1, 3).

4 y = −14 y =

8

−3 y =

−6

y =

2

3 x − 4( 2) = −14 3x

=

−6

x

=

−2

The solution is ( −2, 2). 6. ⎧ x + 2 y − 4 z = 0 ⎪ ⎨3 x + y − 2 z = 5 ⎪ ⎩3 x − y + 2 z = 7 ⎧x + 2 y − 4 z = 0 ⎪ −5 y + 10 z = 5 ⎨ ⎪ −7 y + 14 z = 7 ⎩ ⎧x + 2 y − 4 z = 0 ⎪ y − 2 z = −1 ⎨ ⎪ 7 y + 14 z = 7 − ⎩ ⎧x + 2 y − 4 z = 0 ⎪ y − 2 z = −1 ⎨ ⎪ 0 = 0 ⎩

Let a = z. (a is any real number.) y = 2z − 1 x = −2 y + 4 z = −2( 2 z − 1) + 4 z = − 4z + 2 + 4z x = 2

A solution is ( 2, 2a − 1, a).

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

140

Chapter 4

Systems of Equations and Inequalities

⎡ 1 0 −3 # −10⎤ ⎢ ⎥ 0⎥ ⎢0 −2 2 # ⎢ 1 −2 0 # −7⎥ ⎣ ⎦

7.

− R1 + R3

⎡ 1 0 −3 # −10⎤ ⎢ ⎥ 1 −1 # 0⎥ ⎢0 ⎢ 3⎦⎥ ⎣0 −2 3 #

2 R2 + R3

⎡ 1 0 −3 # −10⎤ ⎢ ⎥ 0⎥ ⎢0 1 −1 # ⎢0 0 1 # 3⎥⎦ ⎣

z = 3

y −3 = 0

− 12 R2

10.

R1 ⎡1 0 3 # 2⎤ ⎢ ⎥ ⎢1 −3 0 # 2⎥ − 13 R3 ⎢⎣3 −2 1 # 12⎥⎦ −1R1 + R2 −3R1 + R3

− 12 R3

1 # −3⎤ ⎡ 1 −3 ⎢ ⎥ − 3 2 5 # 18⎥ ⎢ ⎢0 1 1 # −1⎥⎦ ⎣

− R2 + R3

⎡ 1 0 3 # 2⎤ ⎢ ⎥ ⎢0 1 1 # 0⎥ ⎢0 0 3 # −3⎥ ⎣ ⎦

1 # −3⎤ ⎡ 1 −3 ⎢ ⎥ − 0 11 8 # 27⎥ ⎢ ⎢0 1 1 # −1⎥⎦ ⎣

1R 3 3

⎡ 1 0 3 # 2⎤ ⎢ ⎥ ⎢0 1 1 # 0⎥ ⎢ ⎥ ⎣0 0 1 # −1⎦

−3R3 + R1

1 # −3⎤ ⎡ 1 −3 ⎢ ⎥ 0 1 1 # −1⎥ ⎢ ⎢0 11 −8 # 27⎥ ⎣ ⎦

The solution is (5, 1, −1).

− 13 R2

x = −1

The solution is ( −1, 3, 3).

−3R1 + R2

R2 R3 3R2 + R1 −11R2 + R3

1 R − 19 3

− R3 + R2

4 # −6⎤ ⎡1 0 ⎢ ⎥ 1 # −1⎥ ⎢0 1 ⎢0 0 −19 # 38⎥ ⎣ ⎦

y = 1

1R 4 1

x + 4(−2) = −6 x = 2

The solution is ( 2, 1, − 2). ⎡2 −7 # 7⎤ 9. ⎢ ⎥ ⎣3 7 # 13⎦ D =

2 −7 3

7

= 14 + 21 = 35

1R 3 2

3R1 + R3

13 7 Dx 49 + 91 140 = = = = 4 D 35 35 35 2 7

D 3 13 26 − 21 5 1 y = y = = = = D 35 35 35 7

( 17 ).

The solution is 4,

1 # −1⎤ ⎡ 1 14 2 ⎢ ⎥ ⎢ 0 3 1 # 8⎥ ⎢−3 1 −3 # 5⎥ ⎣ ⎦ 1 # −1⎤ ⎡ 1 14 2 ⎢ 8⎥ 1 # ⎢0 1 3 3⎥ ⎢0 7 − 3 # 2⎥ 4 2 ⎣ ⎦

5 # − 5⎤ − 14 R2 + R1 ⎡ 1 0 12 3 ⎢ ⎥ 8 1 # ⎢0 1 3 3⎥ ⎢ 25 # − 8 ⎥ − 74 R2 + R3 ⎣0 0 − 12 3⎦

7 −7 x =

⎡ 1 0 0 # 5⎤ ⎢ ⎥ ⎢0 1 0 # 1⎥ ⎢0 0 1 # −1⎥ ⎣ ⎦

⎡ 4 1 2 # − 4⎤ ⎢ ⎥ 8⎥ ⎢ 0 3 1 # ⎢−3 1 −3 # 5⎥⎦ ⎣

11.

⎡ 1 0 4 # −6⎤ ⎢ ⎥ ⎢0 1 1 # −1⎥ ⎢0 0 1 # −2⎥ ⎣ ⎦ y + ( −2) = −1

z = −2

⎡ 1 0 3 # 2⎤ ⎢ ⎥ ⎢0 −3 −3 # 0⎥ ⎢0 −2 −8 # 6⎥ ⎣ ⎦ ⎡ 1 0 3 # 2⎤ ⎢ ⎥ ⎢0 1 1 # 0⎥ ⎢0 1 4 # −3⎥ ⎣ ⎦

x − 3(3) = −10

y = 3

8.

1 # 12⎤ ⎡ 3 −2 ⎢ ⎥ ⎢ 1 −3 0 # 2⎥ ⎢−3 0 −9 # −6⎥ ⎣ ⎦

R − 12 25 3 5R + R − 12 3 1

− 13 R3 + R2

⎡1 0 ⎢ ⎢0 1 ⎢0 0 ⎣

5 12 1 3

# − 53 ⎤ ⎥ 8 # 3⎥ ⎥ 1 # 32 25 ⎦

⎡ 1 0 0 # − 11⎤ 5 ⎢ ⎥ 56 ⎢0 1 0 # 25 ⎥ ⎢0 0 1 # 32 ⎥ 25 ⎦ ⎣

(

The solution is − 11 , 5

56 32 , 25 25

).

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test for Chapter 4 2 −2 0

12. −1 2

13.

−1

2 8

8 1

( x1, y1 ) x1

−1 3

3 1 = 0

2 −2 2

8

+1

2 −2 −1

141

= 0( −14) − 1( 20) + 1( 4) = 0 − 20 + 4 = −16

3

= (0, 0), ( x2 , y2 ) = (5, 4), ( x3 , y3 ) = (6, 0)

y1 1

0 0 1

x2

y2 1 = 5 4 1 = (1)

x3

y3 1

6 0 1

5 4 6 0

= (1)( −24) = −24

Area = − 12 ( −24) = 12 y < 12 x + 32 14. ⎧ x − 2 y > −3 ⎪ 2 22 ⎨2 x + 3 y ≤ 22 ⇒ y ≤ − 3 x + 3 ⎪ y ≥ 0 y ≥ 0 ⎩ 1x 2

Dotted line at y = Shade below y =

1x 2

3 2

+ +

3 2

and solid line at y = − 23 x + and y = − 23 x +

22 3

22 . 3

y = 0 is the x-axis.

and above y = 0.

y 10

2x + 3y = 22

6

x − 2y = −3

4 2

x

−2 −2

2

4

6

8 10

y=0

−4 −6

15. Verbal Model:

2 ⋅ Length + 2 ⋅ Width = 68 Width =

Labels:

8 9

⋅ Length

Length = x Width = y

System:

2 x + 2 y = 68 y =

8 x 9

Solve by substitution.

( )

2 x + 2 89 x = 68 18 x + 16 x = 612 34 x = 612 x = 18 y =

8 9

(18)

y = 16 16 feet × 18 feet

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

142

Chapter 4

Systems of Equations and Inequalities

16. Verbal Model:

Investment 1 + Investment 2 + Investment 3 = $25,000 4.5% ⋅ Investment 1 + 5% ⋅ Investment 2 + 8% ⋅ Investment 3 = $1275 Investment 1 + 4000 = Investment 3 + 10,000

Labels:

Investment 1 = x Investment 2 = y Investment 3 = z

System:

x +

y +

z = 25,000

0.045 x + 0.05 y + 0.08 z = 1275 y + 4000 = z + 10,000 ⇒ y − z = 6000 Using determinants and Cramer’s Rule: 1 1 1 25,000⎤ ⎡ ⎢ ⎥ 0.045 0.05 0.08 1275⎥ ⎢ ⎢ 0 1 −1 6000⎥⎦ ⎣

25,000

1

1

1

1

D = 0.045 0.05 0.08 = −0.04 0

1

−1

1

1275 0.05 0.08 6000

1 −0.04 1 25,000

Dx = D

x =

0.045 Dy = D

y =

=

−520 = $13,000 −0.04

1

1275 0.08

0

6000 −1 −360 = = $9000 −0.04 −0.04 1 1 25,000

0.045 0.05 Dz = D

z =

−1

0

1 −0.04

1275 6000

=

−120 = $3000 −0.04

Use your graphing calculator to find each determinant. $13,000 at 4.5%, $9,000 at 5%, $3,000 at 8% 30 ⋅ Reserved seat tickets + 40 ⋅ Floor seat tickets ≥ 300,000 Reserved seat tickets ≤ 9000 Floor seat tickets ≤ 4000 Labels:

Reserved seat tickets = x Floor seat tickets = y

30x + 40y = 300,000 y = 4000

5,000 x

0

00 ,0

0 ,0

15

10

≤ 9000

10,000

0 00

x

x = 9000

15,000

5,

30 x + 40 y ≥ 300,000

System:

y

Number of floor tickets

17. Verbal Model:

Number of reserved tickets

y ≤ 4000

Cumulative Test for Chapters 1–4 1. (a) −2 > −4

(b)

2 3

>

1 2

(c) −4.5 = − −4.5 2. “The number n is tripled and the product is decreased by 8,” is expressed by 3n − 8.

3. t (3t − 1) − 2(t + 4) = 3t 2 − t − 2t − 8

= 3t 2 − 3t − 8 4. 4 x( x + x 2 ) − 6( x 2 + 4) = 4 x 2 + 4 x3 − 6 x 2 − 24 = 4 x3 + 4 x 2 − 6 x 2 − 24 = 4 x3 − 2 x 2 − 24

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Cumulative Test for Chapters 1–4 5. 12 − 5(3 − x) = x + 3

7. x − 2 ≥ 3

12 − 15 + 5 x = x + 3

x − 2 ≤ −3

−3 + 5 x = x + 3 −3 + 5 x − x = x + 3 − x 3 − 3 + 4x = 3 + 3

x ≥ 5

8. −12 ≤ 4 x − 6 < 10

−6 ≤ 4 x < 16 − 64 ≤ x < 4

4x 6 = 4 4

6.

x−2 ≥ 3

x ≤ −1

4x = 6

− 32 ≤ x < 4

3 2

x =

or

143

9. 1150 + 0.20(1150) = 1150 + 230 = 1380

Your new premium is $1380.

x + 2 7 = 4 8 x + 2⎤ ⎡ ⎡7 ⎤ 8⎢1 − = ⎢ ⎥8 4 ⎥⎦ ⎣ ⎣8 ⎦ 1−

10.

8 − 2( x + 2) = 7

9 13 = 4.5 x 9 x = 13( 4.5) x =

8 − 2x − 4 = 7

13( 4.5)

9 x = 6.5

4 − 2x = 7 4 − 4 − 2x = 7 − 4 −2 x = 3 3 −2 x = − 2 −2 3 x = − 2 Weekly cost for Weekly cost for > Company B Company A

11. Verbal Model:

Labels:

Numbers of miles driven in one week = m

(miles)

Weekly cost for Company A = 240

(dollars)

Weekly cost for Company B = 100 + 0.25m

(dollars)

100 + 0.25m > 240

Inequality:

0.25m > 140 m > 560 So, the car from Company B is more expensive if you drive more than 560 miles in a week. 12. No, x − y 2 = 0 does not represent y as a function of x. 13. f ( x ) =

x − 2

6−0 6 3 = = 4+ 4 8 4

(−4 − 4)2

d =

D = x ≥ 2 x − 2 ≥ 0

2 ≤ x < ∞

=

64 + 36

=

100

+ (0 − 6)

2

= 10

14. (a) f ( x) = x 2 − 2 x

f (3) = 32 − 2(3) = 9 − 6 = 3 (b)

15. m =

⎛ −4 + 4 0 + 6⎞ Midpoint = ⎜ , ⎟ = (0, 3) 2 ⎠ ⎝ 2

f ( x) = x 2 − 2 x f ( −3c) = ( −3c) − 2( −3c) = 9c 2 + 6c 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

144

Chapter 4

Systems of Equations and Inequalities

16. (a) 2 x − y = 1

19. ⎧ x + y = 6 ⎨ ⎩2 x − y = 3

− y = −2 x + 1 y = 2x − 1

Solve for y. y = 6− x

m = 2 y − 1 = 2( x + 2)

Substitute into second equation. 2 x − (6 − x ) = 3

y − 1 = 2x + 4

2x − 6 + x = 3

2x − y + 5 = 0

3x = 9

(b) 3x + 2 y = 5

x = 3

2 y = −3 x + 5

y = 6−3

y = − 32 x + 5 m =

= 3

2 3

The solution is (3, 3).

(x

+ 2)

y −1 =

2 3

y −1 =

2 x 3

+

4 3

y =

2x 3

+

7 3

20. ⎧2 x + y = 6 ⎨ ⎩3x − 2 y = 16

4 x + 2 y = 12 3 x − 2 y = 16

3y = 2x + 7

7x

2x − 3y + 7 = 0 17.

4 x + 3 y − 12 = 0

8+ y = 6

3 y = 12

y = −2

y

y = 4

The solution is ( 4, − 2).

8

(0, 4)

4

4 x + 3(0) − 12 = 0

2

4 x = 12

x –6

–4

–2

2 –2

x = 3

6

21. ⎧2 x + y − 2 z = 1 ⎪ − z = 1 ⎨ x ⎪3 x + 3 y + z = 12 ⎩

–4

(3, 0) 2

R1 R2

y 2

x 1 −1 −2

2

3

4

5

⎡2 1 −2 # 1⎤ ⎢ 1 0 −1 # 1⎥ ⎢ ⎥ ⎢⎣3 3 1 # 12⎥⎦ ⎡ 1 0 −1 # 1⎤ ⎢2 1 −2 # 1⎥ ⎢ ⎥ ⎢⎣3 3 1 # 12⎥⎦

−3R2 + R3

⎡ 1 0 −1 # 1⎤ ⎢0 1 0 # −1⎥ ⎢ ⎥ ⎢⎣0 3 4 # 9⎥⎦ ⎡ 1 0 −1 # 1⎤ ⎢0 1 0 # −1⎥ ⎢ ⎥ ⎢⎣0 0 4 # 12⎥⎦

1R 4 3

⎡ 1 0 −1 # 1⎤ ⎢0 1 0 # −1⎥ ⎢ ⎥ ⎢⎣0 0 1 # 3⎥⎦

1 −1

4

=

2( 4) + y = 6

4(0) + 3 y − 12 = 0

18. y = 2 − ( x − 3)

= 28

x

−2 R1 + R2 −3R1 + R3

−3 −4

R3 + R1

⎡ 1 0 0 # 4⎤ ⎢0 1 0 # −1⎥ ⎢ ⎥ ⎢⎣0 0 1 # 3⎥⎦

The solution is ( 4, −1, 3).

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

C H A P T E R 5 Polynomials and Factoring Section 5.1

Integer Exponents and Scientific Notation........................................146

Section 5.2

Adding and Subtracting Polynomials................................................148

Section 5.3

Multiplying Polynomials....................................................................151

Mid-Chapter Quiz......................................................................................................155 Section 5.4

Factoring by Grouping and Special Forms .......................................156

Section 5.5

Factoring Trinomials ..........................................................................158

Section 5.6

Solving Polynomial Equations by Factoring.....................................161

Review Exercises ........................................................................................................165 Chapter Test .............................................................................................................169

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

C H A P T E R 5 Polynomials and Factoring Section 5.1 Integer Exponents and Scientific Notation 1. (u 3v)( 2v 2 ) = 2 ⋅ u 3 ⋅ v1 + 2 = 2u 3v3

33.

3. − 3( x3 ) = − 3( x3 i 2 ) = − 3x 6

1 = x6 x −6

2

35.

5. ( −5 z 2 ) = ( −5) ⋅ ( z 2 ) = −125 z 2 ⋅ 3 = −125 z 6 3

3

3

7. ( 2u ) ( 4u ) = 2 u ⋅ 4u = 16 ⋅ 4 ⋅ u 4

9.

4

4

4 +1

= 64u

5

37.

39.

= 3 ⋅ m5 −1 ⋅ n6 − 3 = 3m 4 n3

( 2) a 2 2

2

⎛ 2a ⎞ 4a = − 2 11. −⎜ ⎟ = − 2 9y ⎝ 3y ⎠ (3) y 2

xn y 2n 13. 3 2 = ( x n − 3 )( y 2 n − 2 ) = x n − 3 y 2 n − 2 x y

15.

(−2 x 2 y)

3

3

=

9 x2 y2

9 x2 y 2

( −8) x 6 y 3

=

2 2

9x y

= −

17. ( −3) = 1 0

⎛2⎞ 21. ⎜ ⎟ ⎝3⎠

−1

=

25. 7x

27. (8 x)

2

(5u ) 0 (5u )

⎛x⎞ 41. ⎜ ⎟ ⎝ 10 ⎠

−1

43. ( 2 x 2 )

47.

= (5u )

−2

−3

=

1

=

(2 x ) 2

=

51.

1

( 4 x)

3

=

1 64 x3

31. y 4 ⋅ y −2 = y 4 + (−2) = y 2

−4

=

1

(5u )

4

=

1 625u 4

2

=

1 4x4

3 ( y3 ) = − y3 i 3 = − y9 ⎛ y3 ⎞ = −⎜ ⎟ = − 53 ⋅ x3 125 x3 (5 x )3 ⎝ 5x ⎠ 3

−3

6 x3 y −3 6 x3 − (−2) y −3 −1 x5 y −4 x5 = = = −2 12 x y 6⋅2 2 2 y4 −2

1 1 = − 103 1000

1 8x

= (5u )

−4 − 0

10 x

=

7 = 4 x

−1

29. ( 4 x)

= (5 y ) = 52 y 2 = 25 y 2

3 2

23. −10−3 = − −4

1

(5 y ) − 2

⎛ 3u 2v −1 ⎞ 49. ⎜ 3 −1 3 ⎟ ⎝3 u v ⎠

1 1 = 52 25

19. 5−2 =

8x6 − 2 y3 − 2 8x4 y = − 9 9

1 = t2 t −2

=

⎛ 5x ⎞ 45. − ⎜ 3 ⎟ ⎝y ⎠

(−2) ( x 2 ) y 3

3

t

0

−2

−4

27 m5n 6 27 m5 n6 = ⋅ ⋅ 3 9mn 9 m n3

2

(4t )

5x0 y

(6 x)0 y 3

=

⎛ 3u 2 − (−1)v −1 − 3 ⎞ = ⎜ ⎟⎟ ⎜ 33 ⎝ ⎠ ⎛ u 3v −4 ⎞ = ⎜ 2 ⎟ ⎝ 3 ⎠

−2

⎛ 32 ⎞ = ⎜ 3 −4 ⎟ ⎝u v ⎠

2

=

34 u v

=

81v8 u6

−2

6 −8

5y 5 = 5 y 1− 3 = 5 y − 2 = 2 y3 y

53. 0.00031 = 3.1 × 10−4 55. 3,600,000 = 3.6 × 106 57. 9,460,800,000,000 = 9.4608 × 1012

146

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Section 5.1 59. 7.2 × 108 = 7.2 × 100,000,000 = 720,000,000

Integer Exponents and Scientific Notation

67.

61. 1.359 × 10−7 = 0.0000001359

147

64,000,000 6.4 × 107 = 0.00004 4.0 × 10−5 = 1.6 × 107 − (−5) = 1.6 × 1012

63. 1.5 × 107 = 15,000,000 65. ( 4,500,000)( 2,000,000,000) = (4.5 × 106 )(2 × 109 )

69. (7900)(5,700,000,000) = (7.9 × 103 )(5.7 × 109 )

= 45.03 × 103 + 9

= ( 4.5)( 2) × 1015

= 45.03 × 1012

= 9 × 1015

= 4.503 × 1013 ≈ 4.50 × 1013

71.

73.

(0.0000565)( 2,850,000,000,000) 0.00465

=

(5.65 × 10−5 )(2.85 × 1012 ) = (5.65)(2.85) × 1010 ≈ 3.4629032 × 1010 ≈ 3.46 × 1010 4.65 × 10−3

4.65

83. ( 2 x3 y −1 )

1.99 × 1030 1.99 = × 106 5.98 × 1024 5.98

−3

(4 xy −6 ) = (2−3 x−9 y 3 )(4 xy −6 ) 4 x −9 + 1 y 3 + (−6) 23 −8 −3 4x y = 8 1 = 2 x8 y 3

≈ 0.3327759 × 106

=

≈ 3.33 × 105 75. The value of c is 2.4 since the number has 24 to the left of the zeros. 77. To write a small number in scientific notation, write the number in the form c × 10− n , where 1 ≤ c < 10, by moving the decimal point n places to the right. 2

⎡ −5u 3v 2 ⎤ ⎡ ( −5)2 ⋅ u 3 2 ⋅ (v)2 ⎤ ( ) ( ) ⎢ ⎥ ⎥ = ⎢ 79. ⎢ 10u 2v ⎥ ⎢ ⎥ 10u 2v ⎣ ⎦ ⎣ ⎦

2

6 2 2

⎡ 25 u 6 v 2 ⎤ = ⎢ ⋅ 2 ⋅ ⎥ v⎦ ⎣10 u

3

2

2 n −1 4 n − 2 n −1

y

(2a −2b4 ) b 2 (10a3b)

2

=

23 a −6b12 ⋅ b 102 a 6b 2

=

8a −6 − 6b12 + 1 − 2 100

=

2a −12b11 25

=

2b11 25a12

91. ( 2 × 109 )(3.4 × 10−4 ) = ( 2)(3.4)(105 ) = 6.8 × 105 93. (5 × 104 ) = 52 × 108 = 25 × 108 = 2.5 × 109 2

x2n + 4 y 4n = x 2 n + 4 − 5 y 4 n − (2 n +1) x 5 y 2 n +1 = x

2

−1 2 87. ⎡( x −4 y −6 ) ⎤ = ( x 4 y 6 ) = x8 y12 ⎢⎣ ⎥⎦

2

⎡5 ⎤ = ⎢ ⋅ u 6 − 2 ⋅ v 2 −1 ⎥ 2 ⎣ ⎦

81.

0

89.

⎡ 25u v ⎤ = ⎢ 2 ⎥ ⎣ 10u v ⎦

⎡5 ⎤ = ⎢ u 4 v⎥ ⎣2 ⎦ 25 8 2 uv = 4

85. u 4 (6u −3v 0 )(7v) = u 4 (6u −3 )(1) = 6u 4 + (−3) = 6u

= x

2 n −1 2 n −1

y

95.

3.6 × 1012 3.6 = × 1012 − 5 = 0.6 × 107 = 6.0 × 106 6 × 105 6

97. 70,000,000,000,000,000,000,000 = 7 × 1022 stars

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

148

Chapter 5

P olynomials and Factoring

99. ( 43,783)(309,000,000) = ( 4.3783 × 104 )(3.09 × 108 )

= 13.528947 × 1012 ≈ 1.35 × 1013 = $13,500,000,000,000 101. Scientific notation makes it easier to multiply or divide very large or very small numbers because the properties of exponents make it more efficient. 103. False.

105. a m ⋅ b n = ab m + n is false because the product rule can be applied only to exponential expressions with the same base. 107. ( a m ) = a m + n is false because the power-to-power rule n

applied to this expression raises the base to the product of the exponents. 109. 3x + 4 x − x = (3 + 4 − 1) x = 6 x

1 1 1 = = = 27, which is greater than 1. 3 1 3−3 ⎛1⎞ ⎜ ⎟ 27 ⎝ 3⎠

111. a 2 + 2ab − b 2 + ab + 4b 2 = a 2 + ( 2 + 1)ab + ( −1 + 4)b 2 = a 2 + 3ab + 3b 2 y

113.

y≥x

4 3 2

y2

−3 −4

x > 2 ⎧x > 2 ⎪ ⎨x − y ≤ 0 ⇒ y ≥ x ⎪y < 0 y < 0 ⎩

Section 5.2 Adding and Subtracting Polynomials 1. Standard form: 4 y + 16

7. Standard form: t 5 − 14t 4 − 20t + 4

Degree: 1

Degree: 5

Leading coefficient: 4

Leading coefficient: 1

3. Standard form: x 2 + 2 x − 6

9. Standard form: −4

Degree: 2

Degree: 0

Leading coefficient: 1

Leading coefficient: −4

5. Standard form: −42 x 3 − 10 x 2 + 3 x + 5

Degree: 3 Leading coefficient: −42

11. 5 + ( 2 + 3 x) = (5 + 2) + 3 x = 3x + 7 13. ( 2 x 2 − 3) + (5 x 2 + 6) = ( 2 x 2 + 5 x 2 ) + (−3 + 6)

= 7 x2 + 3

15. (5 y + 6) + ( 4 y 2 − 6 y − 3) = 4 y 2 + (5 y − 6 y ) + (6 − 3) = 4 y 2 − y + 3 17. ( 2 − 8 y ) + ( −2 y 4 + 3 y + 2) = (−2 y 4 ) + (−8 y + 3 y ) + ( 2 + 2) = −2 y 4 − 5 y + 4 19. ( x3 + 9) + ( 2 x 2 + 5) + ( x3 − 14) = ( x3 + x3 ) + 2 x 2 + (9 + 5 − 14) = 2 x3 + 2 x 2 21.

5 x 2 − 3x + 4 − 4 −3 x 2 2 − 3x 2x

23. 4 x3 − 2 x 2 + 8 x 4 x2 + x − 6 3 + 4x 2 x2 + 9x − 6

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 5.2 25.

−3 p

+ 2p − 7

2p

2

− 2p − 5

149

29. ( 4 − y 3 ) − ( 4 + y 3 ) = ( 4 − y 3 ) + (−4 − y 3 )

5 p2 − 4 p + 2 2

Adding and Subtracting Polynomials

= ( 4 − 4) + ( − y 3 − y 3 ) = −2 y 3

27. −3.6b 2 + 2.5b −2.4b 2 − 3.1b + 7.1 6.6b 2 0.6b 2 − 0.6b + 7.1 31. (3 x 2 − 2 x + 1) − ( 2 x 2 + x − 1) = (3x 2 − 2 x + 1) + (−2 x 2 − x + 1) = (3x 2 − 2 x 2 ) + ( −2 x − x) + (1 + 1) = x 2 − 3x + 2

33. (6t 3 − 12) − ( −t 3 + t − 2) = (6t 3 − 12) + (t 3 − t + 2) = (6t 3 + t 3 ) − t + (−12 + 2) = 7t 3 − t − 10 x2 − x + 3 ⇒ x2 −

35. −

(x

− 2) ⇒

−

37.

x + 3

+ 6x4 +

4 ⇒ −3 x 7

2x2 − 4 x + 5

2 − ( − 4 x 2 + 5 x − 6) ⇒ − 4 x − 5 x + 6 −2 x 2 − 9 x + 11

x + 2

x2 − 2x + 5

39. −3x 7

2 x2 − 4 x + 5 ⇒

+ 6 x4 + 4

7 5 4 −(8 x 7 + 10 x5 − 2 x 4 − 12) ⇒ −8 x − 10 x + 2 x + 12 −11x 7 − 10 x5 + 8 x 4 + 16

41. −( 2 x3 − 3) + ( 4 x3 − 2 x) = −2 x3 + 3 + 4 x3 − 2 x = ( −2 x3 + 4 x3 ) + ( −2 x) + (3) = 2 x3 − 2 x + 3

43. ( 4 x5 − 10 x3 + 6 x) − (8 x5 − 3 x3 + 11) + ( 4 x5 + 5 x3 − x 2 ) = ( 4 x5 − 10 x3 + 6 x) + (−8 x5 + 3x3 − 11) + ( 4 x5 + 5 x3 − x 2 ) = ( 4 x 5 − 8 x5 + 4 x 5 ) + (−10 x 3 + 3 x3 + 5 x3 ) − x 2 + 6 x − 11 = −2 x3 − x 2 + 6 x − 11

45. (5n 2 + 6) + ⎡⎣( 2n − 3n 2 ) − ( 2n 2 + 2n + 6)⎤⎦ = (5n 2 + 6) + ⎡⎣2n − 3n 2 − 2n 2 − 2n − 6⎤⎦ = (5n 2 + 6) + ⎡⎣( −3n 2 − 2n 2 ) + ( 2n − 2n) − 6⎤⎦ = (5n 2 + 6) + (−5n 2 + 0 − 6) = (5n 2 − 5n 2 ) + (6 − 6) = 0

47. (8 x3 − 4 x 2 + 3 x) − ⎡⎣( x3 − 4 x 2 + 5) + ( x − 5)⎤⎦ = (8 x3 − 4 x 2 + 3 x) − ⎡⎣ x3 − 4 x 2 + x⎤⎦

= (8 x3 − 4 x 2 + 3 x) + (− x3 + 4 x 2 − x) = (8 x3 − x3 ) + ( −4 x 2 + 4 x 2 ) + (3 x − x) = 7 x3 + 2 x

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

150

Chapter 5

P olynomials and Factoring

49. 3( 4 x 2 − 1) + (3 x3 − 7 x 2 + 5) = 12 x 2 − 3 + 3 x3 − 7 x 2 + 5

= 3x3 + 5 x 2 + 2 51. 2(t 2 + 12) − 5(t 2 + 5) + 6(t 2 + 5) = 2t 2 + 24 − 5t 2 − 25 + 6t 2 + 30 = ( 2t 2 − 5t 2 + 6t 2 ) + ( 24 − 25 + 30) = 3t 2 + 29

53. 15v − 3(3v − v 2 ) + 9(8v + 3) = 15v − 9v + 3v 2 + 72v + 27 = (3v 2 ) + (15v − 9v + 72v) + 27 = 3v 2 + 78v + 27

59. P = ( 4 x + 1) + ( 2 x + 8) + (6 x − 8)

55. 5s − ⎡⎣6 s − (30s + 8)⎤⎦ = 5s − [6s − 30 s − 8]

= ( 4 x + 2 x + 6 x) + (1 + 8 − 8)

= (5s − 6 s + 30s ) + (8)

= 12 x + 1

= 29 s + 8

When x = 6:

57. h(t ) = −16t 2 + 40t + 500

P = 12(6) + 1

h(1) = −16(1) + 40(1) + 500 = 524 feet 2

= 73 units

h(5) = −16(5) + 40(5) + 500 = 300 feet 2

h(6) = −16(6) + 40(6) + 500 = 164 feet 2

61. P = (3 x + 1) + ( x) + ( x) + ( 2 x + 4) + ( 2 x + 4) + ( x) + ( x) + (3x + 1) = (3 x + x + x + 2 x + 2 x + x + x + 3 x) + (1 + 4 + 4 + 1) = 14 x + 10

When x = 4: P = 14( 4) + 10 = 66 units 63. The like terms are 3x 2 and x 2 , and 2 x and − 4 x, since they have the same variables with the same exponents. 65. The step in adding polynomials vertically that corresponds to grouping like terms when adding polynomials horizontally is aligning the terms of the polynomials by degrees. 67.

( 23 x

3

) (

− 4 x + 1 + − 53 + 7 x −

1 x3 2

) = ( 23 x = ( 46 x =

69.

( 14 y

2

) (

− 5 y − 12 + 4 y −

3 2 y 2

) = ( 14 y = ( 14 y =

7 2 y 4

1 x3 6 2 2

3 3

) + (−4 x + 7 x) + (1 − 53 ) − 63 x ) + 3 x + ( 55 − 53 )

−

1 x3 2 3

+ 3x +

2 5

3 y ) ) ( 2 + 32 y ) + ( −5 y − 4 y ) − 12

− 5 y + −12 − 4 y +

2

2

− 9 y − 12

71. ( 2 x 2 r − 6 x r − 3) + (3 x 2 r − 2 x r + 6) = ( 2 x 2 r + 3 x 2 r ) + (−6 x r − 2 x r ) + ( −3 + 6) = 5 x 2 r − 8 x r + 3 73. Area = 12( x + 6) − 7 x = 12 x + 72 − 7 x = 5 x + 72

(

) (

)

75. Area of region = 6 ⋅ 32 x + 6 ⋅ 92 x or 6 ⋅ ⎡⎣ 32 x +

9 x⎤ 2 ⎦

x⎤ = 36 x = 9 x + 27 x or 6⎡⎣12 2 ⎦

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 5.3 77. The free-falling object was dropped.

−16(0) + 100 = 100 feet 2

79. The free-falling object was thrown upward.

h(0) = −16(0) + 40(0) + 12 = 12 feet 2

81. Verbal Model:

Equation:

Profit = Revenue − Cost

89. B has 2 rows and 3 columns so the order is 2 × 3. 91. A is a square matrix because it has an equal number of rows and columns. 93.

⎡6 −3 ⎢ 1 ⎣2

0⎤ ⎥ 4⎦

1R 6 1

⎡1 − 1 2 ⎢ 1 ⎣⎢2

0⎤ ⎥ 4⎦⎥

−2 R1 + R2

⎡1 − 1 2 ⎢ 2 ⎢⎣0

0⎤ ⎥ 4⎥⎦

1R 2 2

⎡1 − 1 2 ⎢ 1 ⎢⎣0

0⎤ ⎥ 2⎥⎦

P = 14 x − (8 x + 15,000) P = 6 x − 15,000 P = 6(5000) − 15,000 P = $15,000

83. The degree of the term ax k is k. The degree of a polynomial is the degree of its highest-degree term.

(b) A trinomial is a polynomial is always true. Every trinomial is a polynomial.

151

87. No, not every trinomial is a second-degree polynomial. For example, x 3 + 2 x + 3 is a trinomial of third-degree.

P = R −C

85. (a) A polynomial is a trinomial is sometimes true. A polynomial is a trinomial when it has three terms.

Multiplying Polynomials

⎡6 −3 95. B = ⎢ ⎣2 1 97.

0⎤ 6x − 3y = 0 ⎥ ⇒ 4⎦ 2x + y = 4

A = (3)( 4) − (1)( 2) = 12 − 2 = 10

Section 5.3 Multiplying Polynomials 1. (3x − 4)(6 x) = (3 x)(6 x) − ( 4)(6 x) = 18 x 2 − 24 x 3. 2 y (5 − y ) = ( 2 y )(5) − ( 2 y )( y ) = −2 y 2 + 10 y

5. 4 x( 2 x 2 − 3x + 5) = ( 4 x)( 2 x 2 ) − ( 4 x)(3 x) + ( 4 x)(5)

= 8 x3 − 12 x 2 + 20 x 7. ( − 3x )( 2 + 5 x) = ( − 3x)( 2) + ( − 3 x)(5 x) = − 6 x − 15 x 2 = −15 x 2 − 6 x

9. −2m 2 (7 − 4m + 2m 2 ) = −2m 2 (7) − 2m 2 ( −4m) − 2m 2 ( 2m 2 ) = −14m 2 + 8m3 − 4m 4 = −4m 4 + 8m3 − 14m 2 11. − x3 ( x 4 − 2 x3 + 5 x − 6) = − x3 ( x 4 ) − x3 ( −2 x3 ) − x3 (5 x) − x3 (−6) = − x 7 + 2 x 6 − 5 x 4 + 6 x3 13. 15.

(x

+ 2)( x + 4) = x 2 + 4 x + 2 x + 8 = x 2 + 6 x + 8

23.

2x + 1 2

× 7 x − 14 x + 9

( x − 3)( x − 3) = x 2 − 3x − 3x + 9 = x 2 − 6 x + 9

18 x + 9 − 28 x 2 − 14 x

17. ( 2 x − 3)( x + 5) = 2 x 2 + 10 x − 3 x − 15

14 x3 + 7 x 2

2

= 2 x + 7 x − 15 19. (5 x − 2)( 2 x − 6) = 10 x 2 − 30 x − 4 x + 12

= 10 x 2 − 34 x + 12 21. ( 2 x − 1)( x + 2) = 2 x + 4 x − x − 2 2

3

14 x3 − 21x 2 + 4 x + 9 − x2 + 2x − 1

25.

2x + 1

× 2

− x + 2x − 1

2

3

− 2 x + 4x2 − 2x − 2 x3 + 3x 2

− 1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

152

Chapter 5

P olynomials and Factoring 37. (6 x − 9 y )(6 x + 9 y ) = (6 x) − (9 y ) = 36 x 2 − 81y 2 2

t2 + t − 2

27.

2

2

× t − t + 2

( )

39. ( x + 5) = x 2 + 2( x)(5) + (5) = x 2 + 10 x + 25 2

2t 2 + 2t − 4 3

2

2

− t − t + 2t 4

t + t 3 − 2t 2 t4

41.

(x

− 10) = ( x) − 2( x)(10) + 102 = x 2 − 20 x + 100 2

2

− t 2 + 4t − 4 43. ( 2 x + 5) = ( 2 x) + 2( 2 x)(5) + (5) 2

29.

x

2

1 1

2

= 4 x 2 + 20 x + 25

1 1

45. ( k + 5) = ( k + 5) ( k + 5) 3

1

2

= ( k 2 + 10k + 25)( k + 5)

x+3 x

= k 2 ( k + 5) + 10k ( k + 5) + 25( k + 5) = k 3 + 5k 2 + 10k 2 + 50k + 25k + 125

x+2

31.

x

= k 3 + 15k 2 + 75k + 125

1 1 1

1

47. (u + v) = (u + v)(u + v)(u + v) 3

= (u 2 + uv + uv + v 2 )(u + v)

x

= (u 2 + 2uv + v 2 )(u + v)

2x + 1

u2 +

x

2uv + v 2 u + v

2

u v + 2uv 2 + v3

x+3

3

u + 2u 2v +

33. ( x − 8)( x + 8) = ( x) − (8) = x 2 − 64 2

2

uv 2

u 3 + 3u 2v + 3uv 2 + v3

35. ( 2 + 7 y )( 2 − 7 y ) = ( 2) − (7 y ) = 4 − 49 y 2 2

49. (a) Verbal Model:

Function:

2

Volume = Length ⋅ Width ⋅ Height

V ( n) = n ⋅ ( n + 2) ⋅ ( n + 4) = n( n 2 + 6n + 8) = n3 + 6n 2 + 8n

(b) V (3) = 33 + 6(3) + 8(3) = 27 + 54 + 24 = 105 cubic inches 2

(c) Verbal Model:

Area = Length ⋅ Width

Function:

A( n) = n ⋅ ( n + 2) = n 2 + 2n

(d) Function:

A( n + 5) = ( n + 5)( n + 5 + 2) = ( n + 5)( n + 7) = n 2 + 7 n + 5n + 35 = n 2 + 12n + 35

A( n + 5) = ( n + 5) + 2( n + 5) 2

= n 2 + 10n + 25 + 2n + 10 = n 2 + 12n + 35

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 5.3

51. (a) Verbal Model:

Equation:

Multiplying Polynomials

153

Number of Price ⋅ units sold per unit

Revenue =

R = x(175 − 0.02 x) = 175 x − 0.02 x 2 = −0.02 x 2 + 175 x

(b) R = −0.02(3000) + 175(3000) = −180,000 + 525,000 = 345,000 cents = $3450 2

53. The binomials are x − y and x + 2. 55. To multiply ( x + y ) and ( x − y ) without using the FOIL Method, use the special product formula for the sum and difference

of two terms. 57.

(x

− 1)( x 2 − 4 x + 6) = ( x − 1)( x 2 ) + ( x − 1)(−4 x) + ( x − 1)(6) = x3 − x 2 − 4 x 2 + 4 x + 6 x − 6 = x3 − 5 x 2 + 10 x − 6

59. (3a + 2)( a 2 + 3a + 1) = (3a + 2)( a 2 ) + (3a + 2)(3a) + (3a + 2)(1) = 3a 3 + 2a 2 + 9a 2 + 6a + 3a + 2 = 3a 3 + 11a 2 + 9a + 2

61. ( 2u 2 + 3u − 4)( 4u + 5) = ( 4u + 5)( 2u 2 ) + ( 4u + 5)(3u ) + ( 4u + 5)( −4) = 8u 3 + 10u 2 + 12u 2 + 15u − 16u − 20 = 8u 3 + 22u 2 − u − 20

63. u 2v(3u 4 − 5u 2 + 6uv3 ) = u 2v(3u 4 ) + u 2v( −5u 2 ) + u 2v(6uv3 ) = 3u 6v − 5u 4v + 6u 3v 4 65. ( 2t − 1)(t + 1) + 1( 2t − 5)(t − 1) = 2t 2 + 2t − t − 1 + 2t 2 − 2t − 5t + 5 = 4t 2 − 6t + 4 67. ⎡⎣u − (v − 3)⎤⎡ ⎦⎣u + (v − 3)⎤⎦ = (u ) − (v − 3) 2

2

2 = u 2 − ⎡v 2 − 2(v)(3) + (3) ⎤ ⎣ ⎦

= u 2 − (v 2 − 6v + 9) = u 2 − v 2 + 6v − 9

69. (6 x m − 5)( 2 x 2 m − 3) = 6 x m ( 2 x 2 m − 3) + ( −5)( 2 x 2 m − 3) = 12 x m + 2 m − 18 x m − 10 x 2 m + 15 = 12 x3m − 10 x 2 m − 18 x m + 15 71. Verbal Model:

Function: 73. Verbal Model:

Function:

Area of Area of Area of = − shaded region outside rectangle inside rectangle A( x) = 3 x(3 x + 10) − x( x + 4) = 9 x 2 + 30 x − x 2 − 4 x = 8 x 2 + 26 x Area of Area of Area of = − shaded region larger triangle smaller triangle A( x) =

1 2

( 2 x)(1.6 x)

=

1x 2

(0.8 x)

75. A = length ⋅ width = ( x + a )( x + a ) = ( x + a )

= 1.6 x 2 − 0.4 x 2 = 1.2 x 2 2

77. (a) Verbal Model:

A = sum of area of pieces = x 2 + ax + ax + a 2

Perimeter = 2 Length + 2 Width

Sum of the parts is equal to the whole, so

P = 2 ⎡⎣ 52 ( 2 x)⎤⎦ + 2( 2 x) = 10 x + 4 x = 14 x

(x

+ a ) = x 2 + 2ax + a 2 . 2

This special product is the square of a binomial.

(b) Verbal Model: Area = Length ⋅ Width A =

5 2

( 2 x) ⋅ 2 x

= 10 x 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

154

Chapter 5

P olynomials and Factoring

79. Interest = 5000(1 + r )

2

= 5000(1 + r )(1 + r )

(

= 5000 1 + 2r + r 2

)

= 5000 + 10,000r + 5,000r 2 = 5000r 2 + 10,000r + 5000

81. (a)

(x

− 1)( x + 1) = x 2 − 1

(b)

(x

− 1)( x 2 + x + 1) = x3 + x 2 + x − x 2 − x − 1 = x3 − 1

(c)

(x

− 1)( x3 + x 2 + x + 1) = x 4 + x3 + x 2 + x − x3 − x 2 − x − 1 = x 4 − 1

(x

− 1)( x 4 + x3 + x 2 + x + 1) = x5 − 1

83. When two polynomials are multiplied together, an understanding of the Distributive Property is essential because the Distributive Property is used to multiply each term of the first polynomial by each term of the second polynomial.

91. To graph the equation y + 2 x = 0, plot a few points

and draw the graph. Use the Vertical Line Test to then determine that this equation is not a function.

85. The degree of the product of two polynomials of degrees m and n is m + n. 87. y = 5 −

x

y + 2x

point

0

0

(0, 0)

−1

2

(−1, 2)

−1

−2

(−1, − 2)

−2

4

(−2, 4)

−2

−4

(−2, − 4)

1x 2 y 8 7 6 4 3 2 1

y 3 x

−5 − 4 − 3 −2 − 1

2

1 2 3 4 5

−2 −4

Function

−3

−2

x

−1

1

2

−2

89. y − 4 x + 1 = 0

−3

y = 4x − 1 y

93. 2−5 =

4

1 1 = 25 32

3 2

95.

1 x −4 −3 −2

−1 −2

1

2

3

42 = 42 − (−1) = 43 = 64 4−1

4

97. (63 + 3−6 ) = 1 0

Function

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Mid-Chapter Quiz for Chapter 5

155

Mid-Chapter Quiz for Chapter 5 1. Standard form: − 2 x 4 + 4 x3 − 2 x + 3

⎛ 3a −2b5 ⎞ 9. ⎜ −4 0 ⎟ ⎝ 9a b ⎠

Degree: 4

−2

Leading coefficient: − 2

=

8

(b) 664,000,000 = 6.64 × 10

3. (5 y 2 )( − y 4 )( 2 y 3 ) = (5)(−1)( 2) y 2 + 4 + 3 = −10 y 9 4. ( −6 x)( −3x 2 ) = ( −6 x)( −3) ( x 2 ) 2

=

=

= −54 x5

⎛ 5 x 0 y −7 ⎞ 10. ⎜ −2 4 ⎟ ⎝ 2x y ⎠

5. ( −5n 2 )( −2n3 ) = 10n5 6. (3m

7.

) (−2m ) = (9m )(−2m ) 4

6 x −7

(−2 x 2 )

−3

=

6

6( −2 x 2 )

=

−3

= −18m

3

6( −2) ( x 2 )

−2

x7

6( −8) x 6

2

⎛ 5x ⎞ ⎛ 5 x ⎞⎛ 5 x ⎞ 25 x1 + 1 25 x 2 = ⎜ 2 ⎟ = ⎜ 2 ⎟⎜ 2 ⎟ = = 2+2 16 y 4 ⎝ 4y ⎠ ⎝ 4 y ⎠⎝ 4 y ⎠ 16 y

9 a 4b10

⎛ 2 x −2 y 4 ⎞ = ⎜ 0 −7 ⎟ ⎝ 5x y ⎠

3

=

8 x −2 + (−2) + (−2) y 4 + 4 + 4 125 x 0 + 0 + 0 y −7 + (−7) + (−7)

=

8 x −6 y12 125 x 0 y −21

=

8 x −6 − 0 y12 − (−21) 125

=

8 x −6 y 33 125

=

8 y 33 125 x 6

3

x7 48 = − x ⎛ 4 y2 ⎞ 8. ⎜ ⎟ ⎝ 5x ⎠

9a −8b0 a −4b10

⎛ 2 x −2 y 4 ⎞⎛ 2 x −2 y 4 ⎞⎛ 2 x −2 y 4 ⎞ = ⎜ 0 −7 ⎟⎜ 0 −7 ⎟⎜ 0 −7 ⎟ ⎝ 5 x y ⎠⎝ 5 x y ⎠⎝ 5 x y ⎠

10

x7 3

=

4

9a −2 + (−2)b5 + 5

= 9a −4b −10

= ( −6)(9) x1 + 4

2

81a −4 + (−4)b 0 + 0

= 9a −8 − (−4)b 0 −10

2

= ( −6 x)(9) x 4

3

2

⎛ 9a −4b 0 ⎞⎛ 9a −4b 0 ⎞ = ⎜ −2 5 ⎟⎜ −2 5 ⎟ ⎝ 3a b ⎠⎝ 3a b ⎠

2. (a) 0.00000054 = 5.4 × 10 − 7

2

⎛ 9a −4b 0 ⎞ = ⎜ −2 5 ⎟ ⎝ 3a b ⎠

11. ( 2t 3 + 3t 2 − 2) + (t 3 + 9) = 3t 3 + 3t 2 + 7

12. (3 − 7 y ) + (7 y 2 + 2 y − 3) = 7 y 2 − 5 y 13. (7 x3 − 3x 2 + 1) − ( x 2 − 2 x3 ) = 7 x3 − 3 x 2 + 1 − x 2 + 2 x3 = 9 x3 − 4 x 2 + 1 14. (5 − u ) − 2 ⎡⎣3 − (u 2 + 1)⎤⎦ = (5 − u ) − 2 ⎡⎣3 − u 2 − 1⎤⎦ = (5 − u ) − 2 ⎡⎣2 − u 2 ⎤⎦ = 5 − u − 4 + 2u 2 = 2u 2 − u + 1 15. 7 y ( 4 − 3 y ) = 28 y − 21 y 2 16. ( k + 8)( k + 5) = k 2 + 5k + 8k + 40 = k 2 + 13k + 40 17. ( 4 x − y )(6 x − 5 y ) = 24 x 2 − 20 xy − 6 xy + 5 y 2 = 24 x 2 − 26 xy + 5 y 2 18. 2 z ( z + 5) − 7( z + 5) = 2 z 2 + 10 z − 7 z − 35 = 2 z 2 + 3 z − 35

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

156

Chapter 5

P olynomials and Factoring

19. (6r + 5)(6r − 5) = 36r 2 − 25 20. ( 2 x − 3) = ( 2 x − 3)( 2 x − 3) = 4 x 2 − 12 x + 9 2

21.

(x

22.

( x2

+ 1)( x 2 − x + 1) = x3 − x 2 + x + x 2 − x + 1 = x3 + 1 − 3 x + 2)( x 2 + 5 x − 10) = x 2 ( x 2 + 5 x − 10) − 3 x( x 2 + 5 x − 10) + 2( x 2 + 5 x − 10) = x 4 + 5 x3 − 10 x 2 − 3x3 − 15 x 2 + 30 x + 2 x 2 + 10 x − 20 = x 4 + 2 x3 − 23 x 2 + 40 x − 20 Area of Area of Area of = − shaded region large triangle small triangle

23. Verbal Model:

Equation:

(x

A =

1 2

=

1 2

=

1 x2 2

( x2

+ 2) − 2

1 x2 2

+ 4 x + 4) − + 2x + 2 −

1 2 x 2

1 x2 2

= 2x + 2

24. h

( 32 ) = −16( 32 )

( 32 ) + 100 = −16( 94 ) + 16(3) + 100 = −4(9) + 48 + 100 = −36 + 148 = 112 feet

2

+ 32

h(3) = −16(3) + 32(3) + 100 = −16(9) + 96 + 100 = −144 + 196 = 52 feet 2

25. P = R − C = 24 x − (5 x + 2250) = 24 x − 5 x − 2250 = 19 x − 2250

P(1500) = 19(1500) − 2250 = 28500 − 2250 = $26,250

Section 5.4 Factoring by Grouping and Special Forms 17. x 2 + 25 x + x + 25 = ( x 2 + 25 x) + ( x + 25)

1. 8t 2 + 8t = 8t (t + 1)

= x( x + 25) + 1( x + 25)

3. 28 x 2 + 16 x − 8 = 4(7 x 2 + 4 x − 2)

= ( x + 25)( x + 1)

5. 7 − 14 x = −7( −1 + 2 x) = −7( 2 x − 1)

19. y 2 − 6 y + 2 y − 12 = ( y 2 − 6 y ) + ( 2 y − 12)

7. 2 y − 2 − 6 y 2 = −2(− y + 1 + 3 y 2 ) = −2(3 y 2 − y + 1) 9. (a) 45l − l 2 = l ( 45 − l )

= x 2 ( x + 2) + 1( x + 2)

(b) When l = 26:

= ( x + 2)( x 2 + 1)

Width = 45 − (26) = 19 feet 11. 2 y ( y − 4) + 5( y − 4) = ( y − 4)( 2 y + 5)

23. 3a 3 − 12a 2 − 2a + 8 = (3a 3 − 12a 2 ) + ( −2a + 8) = 3a 2 ( a − 4) − 2( a − 4)

13. 5 x(3x + 2) − 3(3x + 2) = (3 x + 2)(5 x − 3) 15. 2(7 a + 6) − 3a (7 a + 6) = (7 a + 6)( 2 − 3a

= ( y − 6)( y + 2)

21. x3 + 2 x 2 + x + 2 = ( x3 + 2 x 2 ) + ( x + 2)

So, the width is 45 − l.

2

= y ( y − 6) + 2( y − 6)

2

)

= ( a − 4)(3a 2 − 2)

25. x 2 − 9 = x 2 − 32 = ( x + 3)( x − 3) 27. 1 − a 2 = 12 − a 2 = (1 − a )(1 + a)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 5.4 29. 4 z 2 − y 2 = ( 2 z − y )( 2 z + y ) 33.

(x

Factoring by Grouping and Special Forms

157

31. 36 x 2 − 25 y 2 = (6 x) − (5 y ) = (6 x − 5 y )(6 x + 5 y ) 2

2

− 1) − 16 = ⎡⎣( x − 1) − 4⎤⎡ ⎦⎣( x − 1) + 4⎤⎦ = ( x − 5)( x + 3) 2

35. 81 − ( z + 5) = 92 − ( z + 5) = ⎡⎣9 − ( z + 5)⎤⎡ ⎦⎣9 + ( z + 5)⎤⎦ = [9 − z − 5][9 + z + 5] = ( 4 − z )(14 + z ) 2

2

37. ( 2 x + 5) − ( x − 4) = ⎡⎣( 2 x + 5) − ( x − 4)⎤⎡ ⎦⎣( 2 x + 5) + ( x − 4)⎤⎦ 2

2

= [2 x + 5 − x + 4][2 x + 5 + x − 4] = ( x + 9)(3 x + 1) 55. 3x 4 − 300 x 2 = 3x 2 ( x 2 − 100)

39. x3 − 8 = x3 − 23 = ( x − 2)( x 2 + 2 x + 4)

= 3x 2 ( x − 10)( x + 10)

41. y 3 + 64 = y 3 + 43 = ( y + 4)( y 2 − 4 y + 16)

(

43. 8t − 27 = ( 2t ) − 3 = ( 2t − 3) 4t + 6t + 9 3

3

3

2

57. 6 x 6 − 48 y 6 = 6( x 6 − 8 y 6 )

)

3 3 = 6 ⎡( x 2 ) − ( 2 y 2 ) ⎤ ⎥⎦ ⎣⎢

45. 27u 3 + 1 = (3u ) + 13 3

= 6( x 2 − 2 y 2 )( x 4 + 2 x 2 y 2 + 4 y 4 )

= (3u + 1)(9u 2 − 3u + 1) 47. x3 + 27 y 3 = x3 + (3 y )

59. The greatest common factor of 2 x3 − 18 x is 2x.

3

61. The completely factored form of 2 x3 − 18 x is 2 x( x + 3)( x − 3).

= ( x + 3 y )( x 2 − 3 xy + 9 y 2 ) 49. 8 − 50 x 2 = 2( 4 − 25 x 2 )

63. 4 x 2 n − 25 = ( 2 x n ) − 52 2

2 = 2 ⎡22 − (5 x) ⎤ ⎣ ⎦

= ( 2 x n − 5)( 2 x n + 5)

= 2[2 − 5 x][2 + 5 x]

65. 81 − 16 y 4 n = 92 − ( 4 y 2 n )

51. 8 x3 + 64 = 8( x3 + 8)

2

= (9 − 4 y 2 n )(9 + 4 y 2 n )

= 8( x3 + 23 )

2 = ⎡32 − ( 2 y n ) ⎤ (9 + 4 y 2 n ) ⎣⎢ ⎦⎥

= 8( x + 2)( x − 2 x + 4) 2

= (3 − 2 y n )(3 + 2 y n )(9 + 4 y 2 n )

53. y 4 − 81 = ( y 2 ) − 92 2

67. 2 x3r + 8 x r + 4 x 2 r = 2 x r ( x 2 r + 4 + 2 x r )

= ( y 2 − 9)( y 2 + 9)

= 2 x r ( x 2 r + 2 x r + 4)

= ( y − 3)( y + 3)( y + 9) 2

69. 3x3 + 4 x 2 − 3x − 4 = (3 x3 + 4 x 2 ) + ( −3 x − 4)

= x 2 (3 x + 4) − 1(3 x + 4)

= ( 3 x 3 − 3 x ) + ( 4 x 2 − 4) or

= 3 x( x 2 − 1) + 4( x 2 − 1)

= ( x 2 − 1)(3 x + 4)

= ( x 2 − 1)(3x + 4)

= ( x − 1)( x + 1)(3 x + 4)

= ( x − 1)( x + 1)(3 x + 4)

71. S = 2π r 2 + 2π rh = 2π r ( r + 2h)

75. P + Prt = P(1 + rt )

73. R = 800 x − 0.25 x 2 = x(800 − 0.25 x)

R = xp p = 800 − 0.25 x

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

158

Chapter 5

P olynomials and Factoring

77. (a) Total volume = a 3

Volume of I = a 2 ( a − b) Volume of II = ab( a − b) Volume of III = b 2 ( a − b) Volume of IV = b 3 (b) Volume of I + Volume of II + Volume of III = a 2 ( a − b) + ab( a − b) + b 2 ( a − b) = ( a − b)( a 2 + ab + b 2 )

(c) Subtract Volume of IV from the total volume. This is equal to the sum of the remaining volumes: a 3 − b3 = ( a − b)( a 2 + ab + b 2 ) The equation provides the formula for a difference of two cubes. 79. A polynomial is in factored form when the polynomial is written as a product of polynomials.

−1 3 0 87. −2 0 6 = −1

0 4 2

81. Check a result after factoring by multiplying the factors to see if the product is the original polynomial.

3 4 2 1

4 2

−3

−2 6 0 2

+0

= ( −1)[0 − 24] − (3)[−4 − 0] = 24 + 12 = 36

83. The polynomial (a) x 2 + y 2 cannot be factored at all because it is prime. 85.

0 6

89.

= (3)(1) − ( 2)( 4) = 3 − 8 = −5

(x

+ 7)( x − 7) = x 2 − 7 2 = x 2 − 49

91. ( 2 x − 3) = ( 2 x) − 2( 2 x)(3) + 32 = 4 x 2 − 12 x + 9 2

2

Section 5.5 Factoring Trinomials 1. x 2 + 4 x + 4 = x 2 + 2( 2 x) + 22 = ( x + 2)

19. x 2 − 5 x + 6 = ( x − 3)( x − 2)

2

3. 25 y 2 − 10 y + 1 = (5 y ) − 2(5 y ) + 1 = (5 y − 1) 2

5. 9b 2 + 12b + 4 = (3b) + 2(3b)( 2) + 22 = (3b + 2) 2

7. 36 x 2 − 60 xy + 25 y 2 = (6 x) − 2(6 x)(5 y ) + (5 y ) 2

= (6 x − 5 y )

2

9. 3m3 − 18m 2 + 27 m = 3m( m 2 − 6m + 9)

= 3m( m 2 − 2(3m) + 32 ) = 3m( m − 3)

2

11. 20v 4 − 60v3 + 45v 2 = 5v 2 ( 4v 2 − 12v + 9)

= 5v 2 ⎡( 2v) − 2( 2v)(3) + 32 ⎤ ⎣ ⎦

21. y 2 + 7 y − 30 = ( y + 10)( y − 3)

2

2

2

23. t 2 − 6t − 16 = (t − 8)(t + 2) 25. x 2 + 10 x + 24 = ( x + 4)( x + 6) 27. z 2 − 6 z + 8 = ( z − 4)( z − 2) 29. Area = x 2 + x − 12 = ( x − 3)( x + 4)

The length is x + 4. 31. x 2 − 15 x − 16 = ( x + 1)( x − 16) 33. x 2 + x − 72 = ( x − 8)( x + 9)

2

= 5v 2 ( 2v − 3) 13. a + 6a + 8 = ( a + 4)( a + 2) 2

15. y − y − 20 = ( y + 4)( y − 5) 2

17. x + 6 x + 5 = ( x + 1)( x + 5) 2

2

35. x 2 − 20 x + 96 = ( x − 12)( x − 8) 37. 5 x 2 + 18 x + 9 = ( x + 3)(5 x + 3) 39. 2 y 2 − 3 y − 27 = ( y + 3)( 2 y − 9) 41. 5 z 2 + 2 z − 3 = (5 z − 3)( z + 1)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 5.5

= − (12 x 2 + 3 x − 8 x − 2)

45. 6 x 2 − 5 x − 25 = (3 x + 5)( 2 x − 5)

= − ⎡⎣3x( 4 x + 1) − 2( 4 x + 1)⎤⎦ = − (3 x − 2)( 4 x + 1)

2

47. 3x + 10 x + 8 = 3 x + 6 x + 4 x + 8

= (3x + 6 x) + ( 4 x + 8) 2

57. 3x3 − 3x = 3x( x 2 − 1) = 3 x( x 2 − 12 ) = 3 x( x + 1)( x − 1)

= 3 x( x + 2) + 4( x + 2) = (3 x + 4)( x + 2)

59. 3x3 − 18 x 2 + 27 x = 3 x( x 2 − 6 x + 9)

= 3 x( x − 3)( x − 3)

49. 5 x 2 − 12 x − 9 = 5 x 2 − 15 x + 3x − 9

= 3 x( x − 3)

= 5 x( x − 3) + 3( x − 3) = (5 x + 3)( x − 3)

2

61. 10t 3 + 2t 2 − 36t = 2t (5t 2 + t − 18)

= 2t (5t − 9)(t + 2)

51. 15 x 2 − 11x + 2 = 15 x 2 − 6 x − 5 x + 2

= (15 x − 6 x) + (−5 x + 2) 2

63. x3 + 2 x 2 − 16 x − 32 = ( x3 + 2 x 2 ) + ( −16 x − 32)

= 3 x(5 x − 2) − 1(5 x − 2)

= x 2 ( x + 2) − 16( x + 2)

= (3 x − 1)(5 x − 2) 2

159

55. 2 + 5 x − 12 x 2 = − (12 x 2 − 5 x − 2)

43. 2t 2 − 7t − 4 = ( 2t + 1)(t − 4)

2

Factoring Trinomials

= ( x + 2)( x 2 − 16)

2

53. 10 y − 7 y − 12 = 10 y − 15 y + 8 y − 12

= ( x + 2)( x − 4)( x + 4)

= 5 y( 2 y − 3) + 4( 2 y − 3) = (5 y + 4)( 2 y − 3) 65. Verbal Model:

Equation:

Area of shaded region = Area of rectangle − Area of squares

Area = (8 ⋅ 18) − 4 ⋅ x 2 = 144 − 4 x 2 = 4(36 − x 2 ) = 4(6 + x)(6 − x)

67. When factoring x 2 + 5 x + 6, you know that the signs of the factors of 6 are positive because 6 is positive, so the factors must have like signs and match the sign of 5. 69. To factor x 2 + 6 x + 9, use the method for factoring a perfect square trinomial. 71. a 2 + 2ab + b 2 = ( a + b)

75. 4 x 2 + bx + 9 = ( 2 x) + bx + 32 2

(a) b = 12

( 2 x)2

2

= ( 2 x + 3)

2

or

2

(b) b = −12

( 2 x)2

73. x 2 + bx + 81 = x 2 + bx + 92

− 12 x + 32 = ( 2 x) − 2( 2 x)(3) + 32 2

= ( 2 x − 3)

(a) b = 18 x 2 + 18 x + 92 = x 2 + 2(9 x) + 92 = ( x + 9)

2

2

or

77. c = 16

x 2 + 8 x + c = x 2 + 2( 4 x) + c

(b) b = −18 x 2 − 18 x + 92 = x 2 − 2(9 x) + 92 = ( x − 9)

+ 12 x + 32 = ( 2 x) + 2( 2 x)(3) + 32

2

= x 2 + 2( 4 x) + 42 = ( x + 4)

2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

160

Chapter 5

P olynomials and Factoring

79. c = 9 y − 6 y + c = y − 2(3 y ) + c 2

2

= y 2 − 2(3 y ) + 32 = ( y − 3)

2

99. An example of a prime trinomial is x 2 + x + 1.

81. x 2 + bx + 8

b = 6

x 2 + 6 x + 8 = ( x + 4)( x + 2)

b = −6

x 2 − 6 x + 8 = ( x − 4)( x − 2)

b = 9

x 2 + 9 x + 8 = ( x + 8)( x + 1)

b = −9

x 2 − 9 x + 8 = ( x − 8)( x − 1)

83. b = 20:

97. To factor x 2 − 5 x + 6 begin by finding the factors of 6 whose sum is −5. They are −2 and −3. The factorization is ( x − 2)( x − 3).

x 2 + 20 x − 21 = ( x + 21)( x − 1)

b = −20:

x 2 − 20 x − 21 = ( x − 21)( x + 1)

b = 4:

x 2 + 4 x − 21 = ( x + 7)( x − 3)

b = − 4:

x 2 − 4 x − 21 = ( x − 7)( x + 3)

101. No, x( x + 2) − 2( x + 2) is not in factored form. It is

not yet a product. x( x + 2) − 2( x + 2) = ( x + 2)( x − 2) 103. For each pair of factors of 12, the signs must be the same to yield a positive product. So, it would be better to try 3, 4 and −3, − 4 instead of −3, 4 and 3, − 4. 105. Verbal Model:

Labels:

Compared number = a Percent = p Base number = b

Equation:

a = p⋅b

85. 4 w2 − 3w + 8 is prime. 87. 60 y 3 + 35 y 2 − 50 y = 5 y(12 y 2 + 7 y − 10)

a = 1.25 ⋅ 340

= 5 y(3 y − 2)( 4 y + 5) 89. 54 x3 − 2 = 2( 27 x3 − 1) = 2(3 x − 1)(9 x 2 + 3 x + 1) 2 2 91. 49 − ( r − 2) = −1⎡−49 + ( r − 2) ⎤ ⎣ ⎦

a = 425 107. Verbal Model:

Labels:

= −1⎡( r − 2) − 49⎤ ⎣ ⎦ 2

2 = −1⎡( r − 2) − 7 2 ⎤ ⎣ ⎦

= −⎡⎣( r − 2) + 7⎤⎡ ⎦⎣( r − 2) − 7⎤⎦ = −( r + 5)( r − 9)

Compared Base = Percent ⋅ number number

Equation:

Compared Base = Percent ⋅ number number Compared number = a Percent = p Base number = b

a = p⋅b 725 = p ⋅ 2000 725 2000

= p

36.25% = p

93. x8 − 1 = ( x 4 ) − 12 = ( x 4 − 1)( x 4 + 1) 2

2 = ⎡( x 2 ) − 12 ⎤ ( x 4 + 1) ⎢⎣ ⎥⎦

= ( x 2 − 1)( x 2 + 1)( x 4 + 1) = ( x − 1)( x + 1)( x 2 + 1)( x 4 + 1) 95. (a) 8n3 + 24n 2 + 16n = 8n( n 2 + 3n + 2)

= 8n( n + 1)( n + 2) (b) Factor 8 and then distribute a factor of 2 into each binomial factor. 8n( n + 1)( n + 2) = ( 2 ⋅ 2 ⋅ 2)n( n + 1)( n + 2) = 2n( 2n + 2)( 2n + 4)

(c) If n = 10,

2n = 2(10) = 20 2n + 2 = 2(10) + 2 = 22 2n + 4 = 2(10) + 4 = 24

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 5.6

109. Verbal Model:

Solving Polynomial Equations by Factoring

161

Amount of Amount of Amount of + = solution 1 solution 2 final solution

Labels:

Percent of solution 1 = 20% Liters of solution 1 = x Percent of solution 2 = 60% Liters of solution 2 = 10 − x Percent of final solution = 30% Liters of final solution = 10 L

Equation:

0.20 x + 0.60(10 − x) = 0.30(10) 0.20 x + 6 − 0.60 x = 3 −0.40 x = −3 x = 7.5 L 10 − x = 2.5 L

111. Verbal Model:

Labels:

Amount of Amount of Amount of + = solution 1 solution 2 final solution

Percent of solution 1 = 60% Gallons of solution 1 = x Percent of solution 2 = 90% Gallons of solution 2 = 120 − x Percent of final solution = 85% Gallons of final solution = 120 gal

Equation:

0.60 x + 0.90(120 − x) = 0.85(120) 0.60 x + 108 − 0.90 x = 102 −0.30 x = −6 x = 20 gal 120 − x = 100 gal

Section 5.6 Solving Polynomial Equations by Factoring 1. No, x( x − 1) = 2 is not in the correct form to apply the

11. 25( a + 4)( a − 2) = 0

Zero-Factor Property.

a + 4 = 0 a = −4

3. No, ( x + 2) + ( x − 1) = 0 is not in the correct form to

apply the Zero-Factor Property.

2t + 5 = 0

the Zero-Factor Property.

t = − 52

15.

x = 0

(y

3t + 1 = 0 t = − 13

− 3)( 2 x + 1)( x + 4) = 0 x = 3

x = 4 9.

(x

x −3 = 0

x − 4 = 0

a = 2

13. ( 2t + 5)(3t + 1) = 0

5. Yes, ( x − 1)( x + 2) = 0 is in the correct form to apply 7. x( x − 4) = 0

a − 2 = 0

2x + 1 = 0 x = − 12

x + 4 = 0 x = −4

− 3)( y + 10) = 0

y −3 = 0 y = 3

y + 10 = 0 y = −10

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

162

Chapter 5

P olynomials and Factoring

17. x 2 + 2 x = 0

−13 x + 36 = − x 2

33.

x ( x + 2) = 0

x 2 − 13 x + 36 = 0

x = 0

(x

x + 2 = 0 x = −2

− 9)( x − 4) = 0

x −9 = 0

x − 4 = 0

x = 9

x = 4

19. x 2 − 8 x = 0 x( x − 8) = 0

x = 0

35. m 2 − 8m + 18 = 2

m 2 − 8m + 16 = 0

x −8 = 0

(m

x = 8

(x

m = 4

+ 5)( x − 5) = 0

x +5 = 0

x −5 = 0

x = −5 23.

37. x 2 + 16 x + 57 = −7

x 2 + 16 x + 64 = 0

x = 5

(x

x 2 − 3 x − 10 = 0

(x

2

x +8 = 0 x = −8

x + 2 = 0

x = 5

x = −2

39. 4 z 2 − 12 z + 15 = 6

4 z 2 − 12 z + 9 = 0

25. x 2 − 10 x + 24 = 0

(2 z

− 6)( x − 4) = 0

x − 4 = 0

x = 6

x = 4

27. 4 x + 15 x − 25 = 0

z =

x( x − 12)( x − 7) = 0

x +5 = 0

4x = 5 x =

x = 0

x = −5

5 4

t (6t 2 − t − 1) = 0

7 − x = 0

1 + 2x = 0

7 = x

− 12 = x

t (3t + 1)( 2t − 1) = 0 t = 0 t = 0

3 y2 − 2 = − y 3 y2 + y − 2 = 0

2 3

3t + 1 = 0 t = − 13

2t − 1 = 0 t =

1 2

45. z 2 ( z + 2) − 4( z + 2) = 0

− 2)( y + 1) = 0 y =

x = 7

6t 3 − t 2 − t = 0

− x)(1 + 2 x) = 0

3y − 2 = 0

x = 12

x −7 = 0

6t 3 = t 2 + t

43.

29. 7 + 13x − 2 x = 0

(3 y

x − 12 = 0

x = 0

2

31.

3 2

x( x 2 − 19 x + 84) = 0

− 5)( x + 5) = 0

4x − 5 = 0

2

41. x3 − 19 x 2 + 84 x = 0

2

(4 x

− 3) = 0

2z − 3 = 0

x −6 = 0

(7

+ 8) = 0

− 5)( x + 2) = 0

x −5 = 0

(x

2

m − 4 = 0

x 2 − 25 = 0

21.

− 4) = 0

y +1 = 0 y = −1

(z

+

( z + 2)( z 2 2)( z − 2)( z

z + 2 = 0 z = −2

− 4) = 0 + 2) = 0 z − 2 = 0 z = 2

z + 2 = 0 z = −2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 5.6 49.

c3 − 3c 2 − 9c + 27 = 0

51.

− 9) = 0

c −3 = 0

c −3 = 0

c = 3

c = 3

Labels:

1 2

Equation:

c + 3 = 0

3 x − 108 = 0 3( x 2 − 36) = 0 3( x + 6)( x − 6) = 0

− 3)( x − x) = 0 3

x −3 = 0

x +6 = 0

− 1) = 0

x = −1

59. h = −16t 2 + 400

0 = −16t 2 + 400

8 x 4 + 12 x3 − 32 x 2 − 48 x = 0

0 = −16(t 2 − 25)

4 x3 ( 2 x + 3) − 16 x( 2 x + 3) = 0

(2 x (2 x

+

0 = −16(t + 5)(t − 5)

+ 3)( 4 x3 − 16 x) = 0

(2 x + 3) 4 x( x 2 3)( 4 x)( x − 2)( x

2x + 3 = 0 x = − 32 55. Verbal Model:

Labels:

t +5 = 0

− 4) = 0

t −5 = 0

t = −5

+ 2) = 0

4x = 0

x−2 = 0

x = 0

x = 2

t = 5 seconds

Reject

x+ 2 = 0

The tool reaches the ground after 5 seconds.

x = −2

61. h = −16t 2 + 30t

(

Length ⋅ Width = Area

0 = −16t t −

Length = w + 7

( w + 7) ⋅ w

Equation:

= 540

w = −27 Reject

t −

30 16

= 0

t =

30 16

t ≈ 1.9 About 1.9 seconds

27)( w − 20) = 0

w + 27 = 0

)

t = 0

w2 + 7 w − 540 = 0

(w +

30 16

−16t = 0

Width = w

20 feet × 27 feet

= 9 inches

The base is 6 inches. The height is 9 inches.

x+1 = 0

x =1

x = 6 inches 3x 2

Reject

x −1 = 0

x = 3

x −6 = 0

x = −6

+ 1) = 0

x = 0

− 27 = 0

2

x ( x − 3) − x( x − 3) = 0

( x − 3) x( x 2 ( x − 3) x( x − 1)( x

3 x 2

⋅ x ⋅ 23 x = 27

3 2 x 4

c = −3

x 4 − 3x3 − x 2 + 3x = 0

(x

⋅ Base ⋅ Height = Area

Height =

+ 3) = 0

163

Base = x

3

53.

1 2

57. Verbal Model:

c 2 (c − 3) − 9(c − 3) = 0

(c − 3)(c 2 (c − 3)(c − 3)(c

Solving Polynomial Equations by Factoring

w − 20 = 0 w = 20 feet w + 7 = 27 feet

63. Yes. It has the form ax 2 + bx + c = 0, with a ≠ 0. 65. The factors 2 x and x + 9 are set equal to zero to form two linear equations. 67.

8x2 = 5x 8x2 − 5x = 0 x(8 x − 5) = 0

x = 0

8x − 5 = 0 8x = 5 x =

5 8

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

164

Chapter 5

P olynomials and Factoring 81. From the graph, the x-intercepts are (0, 0) and (3, 0).

x( x − 5) = 36

69.

x 2 − 5 x = 36

The solutions of the equation 0 = x3 − 6 x 2 + 9 x are 0 and 3.

2

x − 5 x − 36 = 0

(x

x3 − 6 x 2 + 9 x = 0

− 9)( x + 4) = 0

x −9 = 0

x( x 2 − 6 x + 9) = 0

x + 4 = 0

x = 9

x( x − 3)( x − 3) = 0

x = −4

x = 0

71. x( x + 2) − 10( x + 2) = 0

(x

+ 2)( x − 10) = 0

x + 2 = 0

(x

73.

⎡⎣ x − ( −2)⎤⎦ [ x − 6] = 0 ( x + 2)( x − 6) = 0

x = 10

− 4)( x + 5) = 10

x 2 − 4 x − 12 = 0

2

x + x − 20 − 10 = 0

85. Verbal Model:

x 2 + x − 30 = 0

(x

Number + Its square = 240

Labels:

+ 6)( x − 5) = 0

x + 6 = 0

Number = x Its square = x 2

x −5 = 0

x = −6

x + x 2 = 240

Equation:

x = 5

x 2 + x − 240 = 0

81 − ( x + 4) = 0 2

75.

(x

⎣⎡9 − ( x + 4)⎤⎦ ⎡⎣9 + ( x + 4)⎤⎦ = 0

(5 − x)(13 + x) 5− x = 0

( t − 2) (t − 2)2 − 16 + 4)(t − 2 − 4) (t + 2)(t − 6)

(t

− 2

t + 2 = 0 t = −2

87. h = −16t 2 + 80

16 = −16t 2 + 80 = 16

0 = −16t 2 + 64

= 0

0 = −16(t 2 − 4)

= 0

0 = −16(t + 2)(t − 2)

= 0

t −6 = 0 t = 6

(3, 0). The solutions of the equation

t + 2 = 0

t −2 = 0

t = −2

t = 2

Reject

79. From the graph, the x-intercepts are ( −3, 0) and 0 = x 2 − 9 are

The object reaches the balcony after 2 seconds. 89. Verbal Model:

Equation:

3 and −3. 0 = ( x − 3)( x + 3) 0 = x −3 3 = x

x = 15

Reject

x = 5 77.

x − 15 = 0

x = −16

x = −13

2

+ 16)( x − 15) = 0

x + 16 = 0

= 0

13 + x = 0

− x = −5

x −3 = 0 x = 3

83. x = −2, x = 6

x − 10 = 0

x = −2

x −3 = 0 x = 3

0 = x +3 −3 = x

Revenue = Cost R = C 140 x − x 2 = 2000 + 50 x 0 = x 2 − 90 x + 2000 0 = ( x − 40)( x − 50) x − 40 = 0

x − 50 = 0

x = 40

x = 50

Units

Units

40 or 50 systems can be produced and sold.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 5 91. The maximum number of solutions of an nth degree polynomial equation is n. The third-degree equation

(x

+ 1) = 0 has only one solution, x = −1. 3

93. When a quadratic equation has a repeated solution, the graph of the equation has one x-intercept, which is the vertex of the graph.

$2.13 = $0.071 per ounce 30 x +3 101. The domain of f ( x) = is all real numbers such x +1 that x ≠ −1 because x + 1 ≠ 0 means x ≠ −1.

99. Unit price =

103. The domain of g ( x) =

3 − x is all real numbers such

that x ≤ 3 because 3 − x ≥ 0 means − x ≥ −3 and x ≤ 3.

95. A solution to a polynomial expression is the value of x when y is zero. If a polynomial is not factorable, the equation can still have real number solutions for x when y is zero. 97. Unit price =

165

$0.75 = $0.0625 per ounce 12

Review Exercises for Chapter 5 1. x 4 ⋅ x5 = x 4 + 5 = x9

25.

3. (u 2 ) = u 2 ⋅ 3 = u 6 3

5. ( −2 z ) = ( −2) z 3 = −8 z 3 3

3

7. −(u 2v) ( −4u 3v) = −(u 4v 2 )(−4u 3v) 2

= 4u 4 + 3v 2 + 1

⎛ 3x −1 y 2 ⎞ 27. ⎜ 5 −3 ⎟ ⎝ 12 x y ⎠

= 4u 7v3 9.

11.

12 z 5 ⎛ 12 ⎞ = ⎜ ⎟ ⋅ z5 − 2 = 2z3 6z2 ⎝6⎠ 4

2

4

⎛3⎞ 17. ⎜ ⎟ ⎝4⎠

−3

= (8 ⋅ 9) 3

−1

= 72 −1 =

43 ⎛ 4⎞ = ⎜ ⎟ = 3 = 3 ⎝ 3⎠

12 x5 y −3 3 x −1 y 2

= 4 x 6 y −5

2

−1

=

2

25 g d 25 g d 5 5 2 = ⋅ ⋅ = ⋅ g 4 − 2d 2 − 2 = g 80 g 2 d 2 80 g 2 d 2 16 16

15. ( 23 ⋅ 32 )

−1

= 4 x5 − (−1) y (−3) − 2

2 2 ⎛ 72 x 4 ⎞ 13. ⎜ 2 ⎟ = (12 x 4 − 2 ) = (12 x 2 ) = 144 x 4 6 x ⎝ ⎠

=

64 27

4 x6 y5

29. u 3 (5u 0v −1 )(9u ) = u 3 (5v −1 )(81u 2 ) 2

= (5)(81)u 3 + 2v −1

1 72

19. (6 y 4 )( 2 y −3 ) = 12 y 4 + (−3) = 12 y1 = 12 y 21.

7 a 6b −2 7 a 6 b −2 = ⋅ ⋅ −1 4 14a b 14 a −1 b 4 1 = ⋅ a 6 − (−1) ⋅ b −2 − 4 2 1 7 −6 = ab 2 a7 = 2b6

=

405u 5 v

31. 0.0000319 = 3.19 × 10−5 33. 17,350,000 = 1.735 × 107 35. 1.95 × 106 = 1,950,000

−2

4x 2 = 2 x −2 −1 = 2 x −3 = 3 2x x

23. ( x3 y −4 ) = 1 0

37. 2.05 × 10−5 = 0.0000205 39. (6 × 103 ) = 62 × 106 = 36 × 106 = 36,000,000 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

166

41.

Chapter 5

P olynomials and Factoring

3.5 × 107 3.5 = × 107 − 4 7 × 104 7

45. x 4 + 3x5 − 4 − 6 x

Standard form: 3x5 + x 4 − 6 x − 4

= 0.5 × 103

Degree: 5

= 5 × 102

Leading coefficient: 3

= 500

47. (10 x + 8) + ( x 2 + 3 x) = x 2 + (10 x + 3 x) + 8

43. Standard form: − x 4 + 6 x3 + 5 x 2 − 4 x

= x 2 + 13x + 8

Leading coefficient: –1 Degree: 4

49. (5 x3 − 6 x + 11) + (5 + 6 x − x 2 − 8 x3 ) = (5 x3 − 8 x3 ) − x 2 + ( −6 x + 6 x) + (11 + 5) = −3x3 − x 2 + 16 51. (3 y − 4) − ( 2 y 2 + 1) = (3 y − 4) + (−2 y 2 − 1) = −2 y 2 + 3 y + (−4 − 1) = −2 y 2 + 3 y − 5 53. ( − x3 − 3 x) − 4( 2 x3 − 3x + 1) = − x3 − 3 x − 8 x3 + 12 x − 4 = (− x3 − 8 x3 ) + (−3 x + 12 x) + (−4) = −9 x3 + 9 x − 4 55. 3 y 2 − ⎡⎣2 y + 3( y 2 + 5)⎤⎦ = 3 y 2 − ⎡⎣2 y + 3 y 2 + 15⎤⎦ = 3 y 2 − 2 y − 3 y 2 − 15 = (3 y 2 − 3 y 2 ) − 2 y − 15 = −2 y − 15 57. (3 x5 + 4 x 2 − 8 x + 12) − ( 2 x5 + x) + (3x 2 − 4 x3 − 9) = (3x5 − 2 x5 ) − 4 x3 + ( 4 x 2 + 3 x 2 ) + (−8 x − x) + (12 − 9)

= x5 − 4 x 3 + 7 x 2 − 9 x + 3 59.

61.

67. − 2 x3 ( x + 4) = − 2 x 4 − 8 x3

3x 2 + 5 x −4 x

2

−

−x

2

+ 4x + 6

x + 6

69. 3x( 2 x 2 − 5 x + 3) = 6 x3 − 15 x 2 + 9 x

3t − 5

3t − 5

2 −(t 2 − t − 5) ⇒ −t + t + 5 −t 2 + 4t

63. P = (6 x) + (3 x + 5) + (3 x) + ( x) + (3x) + ( 2 x + 5) = (6 x + 3x + 3x + x + 3 x + 2 x) + (5 + 5)

71.

2: 3

65. Verbal Model:

Equation:

− 2)( x + 7) = x 2 + 7 x − 2 x − 14 = x 2 + 5 x − 14

73. (5 x + 3)(3 x − 4) = 15 x 2 − 20 x + 9 x − 12

= 15 x 2 − 11x − 12 75. ( 4 x 2 + 3)(6 x 2 + 1) = 24 x 4 + 4 x 2 + 18 x 2 + 3

= 18 + 10

When x =

(x

( 23 ) + 10 = 22 units

= 24 x 4 + 22 x 2 + 3

P = 18

Profit = Revenue − Cost P( x) = 35 x − (16 x + 3000) = 35 x − 16 x − 3000 = 19 x − 3000 P(1200) = 19(1200) − 3000 = 22,800 − 3000 = $19,800

77. ( 2 x 2 − 3 x + 2)( 2 x + 3) = 2 x 2 ( 2 x + 3) − 3 x( 2 x + 3) + 2( 2 x + 3)

= 4 x3 + 6 x 2 − 6 x 2 − 9 x + 4 x + 6 = 4 x3 + (6 x 2 − 6 x 2 ) + ( −9 x + 4 x) + 6 = 4 x3 − 5 x + 6

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 5

167

79. 2u (u − 7) − (u + 1)(u − 7) = 2u (u − 7) − u (u − 7) − 1(u − 7)

= 2u 2 − 14u − u 2 + 7u − u + 7 = ( 2u 2 − u 2 ) + ( −14u + 7u − u ) + 7 = u 2 − 8u + 7 81. ( 4 x − 7) = ( 4 x) − 2( 4 x)(7) + ( −7) 2

2

83. (6v + 9)(6v − 9) = (6v) − 92

2

2

= 16 x 2 − 56 x + 49

= 36v 2 − 81

85. ⎡⎣(u − 3) + v⎤⎦ ⎡⎣(u − 3) − v⎤⎦ = (u − 3) − v 2 2

= u 2 − 2(u )(3) + (−3) − v 2 2

= u 2 − 6u + 9 − v 2 Area of Area of Area of = − shaded region larger rectangle smaller rectangle

87. Verbal Model:

Width of larger rectangle = 2 x

Labels:

Length of larger rectangle = 2 x + 5 Width of smaller rectangle = 2 x − 3 Length of smaller rectangle = 2 x + 1 Equation:

Area = 2 x( 2 x + 5) − ( 2 x + 1)( 2 x − 3) = 4 x 2 + 10 x − ( 4 x 2 − 6 x + 2 x − 3) = 4 x 2 + 10 x − 4 x 2 + 4 x + 3 = 14 x + 3

89. Interest = 1000(1 + 0.06) = $1123.60

107. u 3 − 1 = (u − 1)(u 2 + u + 1)

91. 24 x 2 − 18 = 6( 4 x 2 − 3)

109. 8 x3 + 27 = ( 2 x) + (3)

2

95. 6 x 2 + 15 x3 − 3x = 3 x( 2 x + 5 x 2 − 1) 97. 28( x + 5) − 70( x + 5) = ( 28 − 70)( x + 5) = −42( x + 5)

99. v3 − 2v 2 − v + 2 = v 2 (v − 2) − 1(v − 2)

= (v − 2)(v 2 − 1) = (v − 2)(v − 1)(v + 1) 101. t + 3t + 3t + 9 = t (t + 3) + 3(t + 3) 2

2

= (t 2 + 3)(t + 3)

103. x − 36 = x − 6 = ( x − 6)( x + 6) 2

2

3

= ( 2 x + 3)( 4 x 2 − 6 x + 9)

93. −3b 2 + b = −b(3b − 1)

3

3

2

105. (u + 6) − 81 = (u + 6 − 9)(u + 6 + 9) 2

= (u − 3)(u + 15)

111. x3 − x = x( x 2 − 1)

= x( x − 1)( x + 1) 113. 24 + 3u 3 = 3u 3 + 24 = 3(u 3 + 8) = 3(u 3 + 23 ) = 3(u + 2)(u 2 − 2u + 4)

115. x 2 − 18 x + 81 = x 2 − 2(9) x + 92 = ( x − 9)

2

117. 4 s 2 + 40 st + 100t 2 = ( 2s ) + 2( 2 s)(10) + (10t ) 2

= ( 2s + 10t )

2

2

119. x 2 + 2 x − 35 = ( x + 7)( x − 5)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

168

Chapter 5

P olynomials and Factoring

121. 2 x 2 − 7 x + 6 = ( 2 x − 3)( x − 2)

3s 2 − 2 s − 8 = 0

145.

( 3s

123. 18 x 2 + 27 x + 10 = (3 x + 2)(6 x + 5)

+ 4)( s − 2) = 0

3s + 4 = 0 s =

125. 4 x 2 − 3 x − 1 = 4 x 2 − 4 x + x − 1

= 4 x( x − 1) + 1( x − 1)

s − 2 = 0 s = 2

− 43

147. m( 2m − 1) + 3( 2m − 1) = 0

= ( 4 x + 1)( x − 1)

(m

127. 5 x 2 − 12 x + 7 = 5 x 2 − 5 x − 7 x + 7

+ 3)( 2m − 1) = 0

m +3 = 0

= 5 x( x − 1) − 7( x − 1)

2m − 1 = 0

m = −3

2m = 1

= (5 x − 7)( x − 1)

m = 149. z (5 − z ) + 36 = 0

129. 7 s 2 + 10s − 8 = 7 s 2 + 14 s − 4s − 8

= 7 s( s + 2) − 4( s + 2)

5 z − z 2 + 36 = 0

= (7 s − 4)( s + 2)

z 2 − 5 z − 36 = 0

(z

131. 4a − 64a 3 = 4a(1 − 16a 2 )

= ( z 2 + 3)( z + 1)

( 12 x) + 2( 12 ) xy + = ( 12 x + y ) 2

+ xy + y 2 =

z + 4 = 0

z = 9

133. z 3 + z 2 + 3 z + 3 = z 2 ( z + 1) + 3( z + 1)

1 x2 4

− 9)( z + 4) = 0

z −9 = 0

= 4a(1 − 4a)(1 + 4a )

135.

1 2

z = −4

v 2 − 100 = 0

151.

(v

− 10)(v + 10) = 0

v − 10 = 0

y2

v + 10 = 0

v = 10

v = −10

2

153. 2 y 4 + 2 y 3 − 24 y 2 = 0

2 y 2 ( y 2 + y − 12) = 0

137. x 2 − 10 x + 25 − y 2 = ( x − 5) − y 2 2

2 y 2 ( y + 4)( y − 3) = 0

= ⎡⎣( x − 5) − y⎤⎡ ⎦⎣( x − 5) + y⎤⎦ = ( x − 5 − y )( x − 5 + y )

2 y2 = 0

= ( x − y − 5)( x + y − 5)

y = 0

4x = 0

x − 2 = 0

x = 0

x = 2

143.

(x

x( x − 9)( x − 2) = 0

x = 0

x = −10

x −9 = 0

x − 2 = 0

x = 9

x = 2

x −3 = 0

157.

x = 3

4x − 1 = 0 x =

1 4

b3 − 6b 2 − b + 6 = 0 b 2 (b − 6) − (b − 6) = 0

+ 10)( 4 x − 1)(5 x + 9) = 0

x + 10 = 0

y = 3

x( x 2 − 11x + 18) = 0

141. ( 2 x + 1)( x − 3) = 0 x = − 12

y = −4

y −3 = 0

155. x3 − 11x 2 + 18 x = 0

139. 4 x( x − 2) = 0

2x + 1 = 0

y + 4 = 0

5x + 9 = 0 x = − 95

(b − 6)(b 2 (b − 6)(b − 1)(b

− 1) = 0 + 1) = 0

b −6 = 0

b −1 = 0

b = 6

b =1

b +1= 0 b = −1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test for Chapter 5 159. x 4 − 5 x3 − 9 x 2 + 45 x = 0

163. Verbal Model:

x ( x − 5) − 9 x( x − 5) = 0 3

( x − 9 x)( x − 5) x( x 2 − 9)( x − 5) 3

Area = Length ⋅ Width Length = 2 14 x

Labels:

= 0

Width = x

= 0

900 = 2 14 x ⋅ x

Equation:

x( x − 3)( x + 3)( x − 5) = 0 x = 0

x −3 = 0

900 =

x +3 = 0

x = 3

169

3600 = 9 x 2

x −5 = 0

x = −3

9 2 x 4

400 = x 2

x = 5

20 = x First odd Second odd ⋅ = 99 integer integer

161. Verbal Model:

Labels:

45 = 2 14 x 45 inches × 20 inches

First odd integer = 2n + 1 Second odd integer = 2n + 3

( 2n

Equation:

165. h = −16t 2 + 3600

+ 1)( 2n + 3) = 99

0 = −16t 2 + 3600 0 = −16(t 2 − 225)

2

4n + 8n + 3 = 99 2

4n + 8n − 96 = 0

0 = −16(t + 15)(t − 15)

4( n 2 + 2n − 24) = 0

t + 15 = 0

4( n + 6)( n − 4) = 0 n + 6 = 0 n = −6 Reject

n − 4 = 0 n = 4

t − 15 = 0

t = −15

2n + 1 = 9 2n + 3 = 11

t = 15

Reject

The object reaches the ground after 15 seconds.

Chapter Test for Chapter 5 1. 3 − 4.5 x + 8.2 x3 = 8.2 x3 − 4.5 x + 3

Degree: 3

3. (a)

2−1 x5 y −3 1 = ⋅ x5 − (−2) ⋅ y −3 − 2 4 x −2 y 2 2⋅4

Leading coefficient: 8.2

=

2. (a) 690,000,000 = 6.9 × 108

⎛ −2 x 2 y ⎞ (b) ⎜ −3 ⎟ ⎝ z ⎠

(b) 4.72 × 10 − 5 = 0.0000472

−2

1 x7 ⋅ x 7 y −5 = 8 8 y5 2

⎛ z −3 ⎞ z −6 1 = ⎜ = = 2 ⎟ 4 x4 y 2 4 x4 y2 z 6 ⎝ −2 x y ⎠

3

⎛ −2u 2 ⎞ ⎛ 3v 2 ⎞ ⎛ 8u 6 ⎞⎛ 3v 2 ⎞ 4. (a) ⎜ −1 ⎟ ⎜ −3 ⎟ = ⎜ − −3 ⎟⎜ −3 ⎟ = −24u 6 − (−3)v 2 −(−3) = −24u 9v5 ⎝ v ⎠ ⎝u ⎠ ⎝ v ⎠⎝ u ⎠ (b) 5. (a)

(b)

(−3x2 y −1 ) 2 0

6x y

4

=

81x8 y −4 27 x8 − 2 27 x 6 = = 2 4 6x 2y 2 y4

(5a 2 − 3a + 4) + (a 2 − 4) = 6a 2 − 3a (16 − y 2 ) − (16 + 2 y + y 2 ) = 16 − y 2 − 16 − 2 y −

y 2 = −2 y 2 − 2 y

6. (a) −2( 2 x 4 − 5) + 4 x( x3 + 2 x − 1) = − 4 x 4 + 10 + 4 x 4 + 8 x 2 − 4 x = 8 x 2 − 4 x + 10

(b) 4t − ⎡⎣3t − (10t + 7)⎤⎦ = 4t − [3t − 10t − 7] = 4t − 3t + 10t + 7 = 11t + 7 7. (a) −3 x( x − 4) = −3x 2 + 12 x

(b)

(2 x

− 3 y )( x + 5 y ) = 2 x 2 + 7 xy − 15 y 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

170

Chapter 5

(x

8. (a)

− 1) ⎡⎣2 x + ( x − 3)⎤⎦ = ( x − 1)(3 x − 3) = 3 x 2 − 6 x + 3

(2s

(b)

− 3)(3s 2 − 4 s + 7) = 6s 3 − 8s 2 + 14s − 9 s 2 + 12 s − 21 = 6 s 3 − 17 s 2 + 26 s − 21

( 2 w − 7) 2

9. (a)

P olynomials and Factoring

= ( 2 w) − 2( 2 w)(7) + 7 2 = 4 w2 − 28w + 49 2

(4 − (a + b))(4 + (a + b))

(b)

= 42 − ( a + b) = 16 − ( a 2 + 2ab + b 2 ) = 16 − a 2 − 2ab − b 2 2

20. Area = 2 x( x + 15) − x( x + 4)

10. 18 y 2 − 12 y = 6 y(3 y − 2) 16 9

11. v 2 −

(

= v −

4 3

)(

4 3

v +

Shaded region = 2 x 2 + 30 x − x 2 − 4 x

)

= x 2 + 26 x

12. x3 − 3x 2 − 4 x + 12 = x 2 ( x − 3) − 4( x − 3)

= ( x − 3)( x 2 − 4) = ( x − 3)( x − 2)( x + 2) 13. 9u 2 − 6u + 1 = (3u − 1)(3u − 1) or (3u − 1)

2

21. Verbal Model:

Labels:

Area rectangle = Length ⋅ Width

Length =

Width = w Equation:

54 =

2

6 centimeters = width 9 centimeters = length

15. x3 + 27 = ( x + 3)( x 2 − 3 x + 9)

(x

22. Verbal Model:

− 3)( x + 2) = 14

Labels:

x 2 + 2 x − 3 x − 6 = 14 x − x − 20 = 0 + 4)( x − 5) = 0

x + 4 = 0

Equation:

x −5 = 0

x = −4

( y + 2) 2) − 3⎤⎡ ⎦⎣( y + 2)

⎣⎡( y +

y −1 = 0

20 =

x + 5 = 0

+ 3⎤⎦ = 0

− 43 = y

x = 4 feet 2 x + 2 = 10 feet

Reject

The base is 4 feet and the height is 10 feet. 23. Verbal Model:

+ 3 y) = 0

3 = y

x − 4 = 0

x = −5

y = −5

4 + 3y = 0

⋅ x ⋅ ( 2 x + 2)

0 = ( x + 5)( x − 4)

−9 = 0

3− y = 0

1 2

0 = x 2 + x − 20

18. 12 + 5 y − 3 y 2 = 0

(3 − y )(4

⋅ Base ⋅ Height

Base = x

y +5 = 0

y =1

1 2

20 = x 2 + x

x = 5 2

17.

Area of = a triangle

Height = 2 x + 2

2

(x

⋅w

36 = w2

= 2(3 x + 2)( x − 5)

16.

3 w 2

108 = 3w2

14. 6 x − 26 x − 20 = 2(3x − 13 x − 10) 2

3 w 2

Labels:

Revenue = Cost Revenue = R Cost = C

Equation:

R = C 2

x − 35 x = 150 + 12 x

19. 2 x3 + 10 x 2 + 8 x = 0

2

x − 47 x − 150 = 0

2 x ( x 2 + 5 x + 4) = 0

(x

2 x( x + 4)( x + 1) = 0

2x = 0 x = 0

x + 4 = 0 x = −4

− 50)( x + 3) = 0

x − 50 = 0 x +1= 0 x = −1

x = 50

x +3 = 0 x = −3

50 computer desks

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

C H A P T E R 6 Rational Expressions, Equations, and Functions Section 6.1

Rational Expressions and Functions..................................................172

Section 6.2

Multiplying and Dividing Rational Expressions...............................173

Section 6.3

Adding and Subtracting Rational Expressions..................................176

Section 6.4

Complex Fractions..............................................................................180

Mid-Chapter Quiz......................................................................................................184 Section 6.5

Dividing Polynomials and Synthetic Division..................................185

Section 6.6

Solving Rational Equations................................................................189

Section 6.7

Variation..............................................................................................197

Review Exercises ........................................................................................................199 Chapter Test .............................................................................................................205

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

C H A P T E R 6 Rational Expressions, Equations, and Functions Section 6.1 Rational Expressions and Functions 1. x − 3 ≠ 0

19.

x ≠ 3

D = ( −∞, 3) ∪ (3, ∞)

z ≠ 0

D = ( −∞, 0) ∪ (0, 4) ∪ ( 4, ∞)

=

− 4)(t + 4) ≠ 0

t ≠ 4

t ≠ −4

D = ( −∞, − 4) ∪ ( −4, 4) ∪ ( 4, ∞) 500 must x be defined, x ≠ 0. Therefore, the domain is x > 0 or (0, ∞).

7. Since length must be positive, x ≥ 0. Since

9. x = units of a product

D = {1, 2, 3, 4, …} 3 x( x − 3) ( x − 3) 3x 2 − 9 x = = 12 x 2 12 x 2 4x x 2 ( x − 8) x( x − 8)

=

15.

17.

( y − 8 x)( y + 8 x) y 2 − 64 x 2 = 5(3 y + 24 x) 15( y + 8 x) =

=

y + 6

y ≠ −8 x

29.

(u − 2v)(u + 2v) u 2 − 4v 2 = 2 u + uv − 2v (u − v)(u + 2v)

31.

u − 2v , u −v

− 2)

3( m 2 − 4n 2 ) 3m 2 − 12n 2 = 2 m + 4mn + 4n ( m + 2n)( m + 2n) 3( m − 2n)( m + 2n)

(m + 2n)(m 3( m − 2n)

+ 2n)

m + 2n

x( x + 1) Area of shaded portion = Area of total figure ( x + 1)( x + 3) =

y ≠ 2 35.

u ≠ −2v

2

=

33.

x ≠ 0

2

=

y( y − 2)( y + 2)

x ≠ 0

xy(5 + 3 xy ) 5 xy + 3 x 2 y 2 5 + 3 xy = = , 3 xy xy ⋅ y 2 y2

y( y 2 − 4)

( y + 6)( y y( y + 2) ,

y − 8x , 15

3 2

27.

u 2 − 12u + 36 (u − 6)(u − 6) = u −6 u −6 = u − 6, u ≠ 6

=

x ≠ −

25.

=

y − 4y = y 2 + 4 y − 12 ( y + 6)( y − 2)

x +8 , x −3

3xy 2 3 xy 2 3y2 = = , 2 xy + x y2 + 1 x( y 2 + 1)

x( x − 8)

3

x ≠ 4

23.

x ⋅ x( x − 8)

= x, x ≠ 0, x ≠ 8

3x + 5 , x +3

(2 x + 3)( x + 8) 2 x 2 + 19 x + 24 = 2 2 x − 3x − 9 (2 x + 3)( x − 3)

t 2 − 16 ≠ 0

(t

13.

21.

+ 5)( x − 4)

−1( x − 4)( x + 3)

= −

z − 4 ≠ 0 z ≠ 4

5.

(3 x

=

z ( z − 4) ≠ 0

3.

11.

3x 2 − 7 x − 20 (3x + 5)( x − 4) = 12 + x − x 2 −1( x 2 − x − 12)

Area of shaded portion = Area of total figure =

x , x +3 1 2

x > 0

( x + 1) ( 2 x)( 2 x + 2)

1 , 4

1x 2

x > 0

172

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.2 37. The rational expression is in simplified form if the numerator and denominator have no factors in common (other than ±1 ). 39. A change in signs is one additional step sometimes needed in order to find common factors. 41.

( x + 5)( x( x − 2)) x +5 = , 3x 3x 2 ( x − 2)

(x

53. The student incorrectly divided out; the denominator may not be split up.

Correct solution: x( x + 7) x2 + 7 x = = x, x + 7 x + 7

8x = , x ≠ −3 x−5

(x

Total Number Cost per Initial = ⋅ + cost of units unit cost Number of units = x

− 2) and divide by ( x − 2).

g ( x) =

Total cost = C

(b) Verbal Model:

x ≠ −7

55. To write the polynomial g ( x), multiply f ( x) by

45. (a) Verbal Model:

Equation:

1 x2 + 1

There are many correct answers.

8 x( x + 3) − 5)( x + 3)

Labels:

173

49. A rational expression is undefined if the denominator is equal to zero. To find the implied domain restriction of a rational function, find the values that make the denominator equal to zero (by setting the denominator equal to zero), and exclude those values from the domain. 51.

x ≠ 2

8 x( ) 8 x( ) 43. 2 = x − 2 x − 15 ( x − 5)( x + 3) =

Multiplying and Dividing Rational Expressions

f ( x)( x − 2)

(x

− 2)

, x ≠ 2

C = 6.50 x + 60,000

1⋅3 3 ⎛ 1 ⎞⎛ 3 ⎞ = 57. ⎜ ⎟⎜ ⎟ = 4⋅4 16 ⎝ 4 ⎠⎝ 4 ⎠

Average Total Number = ÷ cost cost of units

59.

Label:

Average cost = C

Equation:

C =

C 6.50 x + 60,000 ;C = x x

(c) D = {1, 2, 3, 4, …}

1⎛ 3 ⎞ 1⋅3⋅5 =1 ⎜ ⎟(5) = 3⎝ 5 ⎠ 3⋅ 5⋅1

61. ( −2a 3 )( −2a) = −2 ⋅ −2 ⋅ a 3 ⋅ a = 4a 4 63. ( −3b)(b 2 − 3b + 5) = −3b(b 2 ) − 3b( −3b) − 3b(5)

= −3b3 + 9b 2 − 15b

6.50(11,000) + 60,000 (d) C (11,000) = ≈ $11.95 11,000

π (3d ) ( d + 2) Circular pool volume = Rectangular pool volume d (3d )(3d + 6) 2

47.

π (3d ) ( d + 2) 2

=

3d ⋅ 3( d + 2) 2

π (3d ) ( d + 2) 2

=

(3d ) (d 2

+ 2)

= π, d > 0

Section 6.2 Multiplying and Dividing Rational Expressions 1.

7 7 x2 , = 3y 3 y( x 2 )

5. 4 x ⋅

x ≠ 0

3 x( x + 2) 3x = , 3. 2 x −4 ( x − 4)( x + 2) 2

x ≠ −2

7.

4 x(7) 7 7 = = , 12 x 12 x 3

x ≠ 0

8s 3 6 s 2 8s 3 ⋅ 3 ⋅ 2 s ⋅ s s3 ⋅ = = , 9s 32 s 3⋅3⋅s⋅8⋅2⋅2⋅s 6

s ≠ 0

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

174

Chapter 6

9. 16u 4 ⋅

R ational Expressions, Equations, and Functions

12 8 ⋅ 2 ⋅ u 2 ⋅ u 2 ⋅ 12 = = 24u 2 , 2 8u 8 ⋅ u2

u ≠ 0

11.

8 ⋅ 3(3 + 4 x) 8 3 ⋅ (9 + 12 x) = = 24, x ≠ − 3 + 4x 3 + 4x 4

13.

4 ⋅ 2 ⋅ u ⋅ u ⋅ v (u + v ) u + v 8u 2v ⋅ = 3u + v 12u (3u + v) ⋅ 4 ⋅ 3 ⋅ u 2uv(u + v)

=

3(3u + v)

,

−1( r − 12) ⋅ 3 12 − r 3 ⋅ = = −1, 3 3( r − 12) r − 12

15.

(2 x

17.

− 3)( x + 8) x3

⋅

(2 x − 3)( x + 8) x x = 3 − 2x x ⋅ x 2 ⋅ −1( 2 x − 3) = −

u ≠ 0

(2t + 5)(t − 3)(t − 3)(t + 2) = 2t + 5, 2t 2 − t − 15 t 2 − t − 6 ⋅ 2 = t + 2 t − 6t + 9 (t + 2)(t − 3)(t − 3)

23. ( 4 y 2 − x 2 ) ⋅

25.

27.

xy

(x

x 2 + 2 xy − 3 y 2

(x

+ y)

2

− 2 y)

2

=

−1( x 2 − 4 y 2 )( xy )

(x

− 2 y)

2

3 2

r ≠ 3, r ≠ 2

t ≠ 3, t ≠ −2

xy ( x + 2 y ) −( x − 2 y )( x + 2 y )( xy ) = − ( x − 2 y )( x − 2 y ) ( x − 2 y)

x2 − y 2 ( x + 3 y )( x − y ) ⋅ ( x − y )( x + y ) = ( x − y ) , = 2 x + 3y x + 3y x+ y ( x + y)

x ≠ −3 y

x + 5 2x2 − 9x − 5 x2 − 1 x + 5 ( 2 x + 1)( x − 5) ( x − 1)( x + 1) ⋅ ⋅ 2 = ⋅ ⋅ 2 x − 5 3x + x − 2 x + 7 x + 10 x − 5 (3 x − 2)( x + 1) ( x + 5)( x + 2)

=

( x + 5)(2 x + 1)( x − 5)( x − 1)( x + 1) ( x − 5)(3x − 2)( x + 1)( x + 5)( x + 2) (2 x + 1)( x − 1) , x ≠ ±5, −1 (3x − 2)( x + 2)

9 − x 2 4 x 2 + 8 x − 5 6 x 4 − 2 x3 (3 − x)(3 + x) ⋅ (2 x + 5)(2 x − 1) ⋅ 2 x3 (3x − 1) ⋅ 2 ⋅ = 2 2 x + 3 4x − 8x + 3 8x + 4x 2x + 3 (2 x − 3)(2 x − 1) 4 x(2 x + 1) = =

31.

x ≠

2

⋅

=

29.

= −

x +8 , x2

4( r − 3)( r − 2)( r + 2) 4r − 12 r 2 − 4 ⋅ = r −2 r −3 (r − 2) ⋅ (r − 3)

19.

= 4( r + 2), 21.

r ≠ 12

−1( x − 3)( x + 3)( 2 x + 5) x 2 (3 x − 1) (2 x + 3)(2 x − 3)2(2 x + 1) x 2 ( x 2 − 9)( 2 x + 5)(3x − 1) 2( 2 x + 3)(3 − 2 x )( 2 x + 1)

x( x + 1) x x x +1 3 ÷ = ⋅ = , x + 2 x +1 x+ 2 3 3( x + 2)

33. x 2 ÷

3x 4 4x = x2 ⋅ = , 4 3x 3

x ≠ 0,

,

x ≠ −1

x ≠ 0

35.

2(3)(5) x x2 2x 2 x 15 6 ÷ = ⋅ = = x 5 15 5 x2 5x2

37.

( x + 2)( x − 1) = 4 x = 4 , x2 + 2x 4x 4x ÷ 2 = ⋅ 3x − 3 x + x − 2 3( x − 1) 3x 3 x ( x + 2)

39.

7 xy 2 21x3 7 xy 2 45 xy 7 ⋅ x ⋅ y2 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ x ⋅ y 3 y2 ÷ = ⋅ = = , 2 2 3 2 3 10 x y 45 xy 10 x y 21x 2⋅5⋅ x ⋅ y⋅3⋅7⋅ x 2 x3

41.

3( a + b) ( a + b) ÷ 4 2

2

=

1 2

x ≠ −2, 0, 1

y ≠ 0

3( a + b) 3( a + b) ⋅ 2 2 3 ⋅ = = 2 4 2 ⋅ 2 ⋅ a + b a + b 2 a ( + b) ( )( ) ( a + b)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.2

43.

45.

175

2( x + y ) x2 − y2 x− y 2x + 2 y ÷ = ⋅ 3 3 x − y ( x − y )( x + y )

( x3 y ) (x

2

+ 2 y)

2

÷

=

2( x + y )( x − y ) 3( x − y )( x + y )

=

2 , 3

x2 y

(x

+ 2 y)

3

=

= =

47.

Multiplying and Dividing Rational Expressions

x ≠ y,

( x3 y ) (x

x ≠ −y

2

+ 2 y)

2

⋅

(x

+ 2 y) x2 y

3

( x3 y)( x3 y)( x + 2 y)2 ( x + 2 y) (x

+ 2 y) x2 y 2

( x3 y)( x2 ⋅ xy)( x + 2 y) x2 y

= x 4 y( x + 2 y ),

x ≠ 0, y ≠ 0, x ≠ −2 y

x 2 + 2 x − 15 x 2 − 8 x + 15 ( x + 5)( x − 3) ⋅ ( x + 6)( x − 4) ÷ 2 = 2 x + 11x + 30 x + 2 x − 24 ( x + 5)( x + 6) ( x − 3)( x − 5) =

( x + 5)( x − 3)( x + 6)( x − 4) ( x + 5)( x + 6)( x − 3)( x − 5)

=

x −4 , x −5

x ≠ −6, x ≠ −5, x ≠ 3

49. To find a model Y for the annual per capita income, divide the total personal income by the population. 0.184t + 4.3 0.029t + 1 0.029t + 1 = (6591.43t + 102,139.0) ⋅ 0.184t + 4.3 + t t + 1) 6591.43 102,139.0 0.029 ( )( = , 4 ≤ t ≤ 9 0.184t + 4.3

Y = (6591.43t + 102,139.0) ÷

51. To multiply two rational expressions, let a, b, c, and d represent real numbers, variables, or algebraic expressions such that b ≠ 0 and d ≠ 0. Then the

product of

a c a c ac and is ⋅ = . b d b d bd

53. To divide rational expressions, multiply the first expression by the reciprocal of the second expression.

⎡ x 2 3( x + 4) ⎤ x x 2 3( x + 4) x + 2 = ⋅ ⋅ 55. ⎢ ⋅ 2 ⎥ ÷ 9 x( x + 2) x ⎣ 9 x + 2x ⎦ x + 2 x + 4 , x ≠ −2, 0 = 3 y ( x + 1) 1 3x y ⎡ xy + y ⎤ ÷ (3 x + 3)⎥ ÷ = ⋅ ⋅ 57. ⎢ 4x 3( x + 1) y ⎣ 4x ⎦ 3x =

59.

1 , 4

x ≠ −1, 0, y ≠ 0

(2 x − 5)( x + 5) ⋅ x(3x + 2) 2 x 2 + 5 x − 25 3x 2 + 2 x ⎛ x ⎞ ⋅ ÷⎜ ⎟ = 3x 2 + 5 x + 2 x +5 x+5 (3x + 2)( x + 1) ⎝ x + 1⎠ 2

= =

⎛ x + 1⎞ ⋅⎜ ⎟ ⎝ x ⎠

2

(2 x

− 5)( x + 5) x(3x + 2)( x + 1)( x + 1) (3x + 2)( x + 1)( x + 5) x ⋅ x

(2 x

− 5)( x + 1) , x

x ≠ −1, − 5, −

2 3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

176

Chapter 6

61. x3 ⋅

R ational Expressions, Equations, and Functions

( x n − 3)( x n + 3) ⋅ x2n − 9 x2n − 2 xn − 3 x 3 x ÷ = ⋅ n n n x + 4xn + 3 x ( x + 3)( x + 1) ( x − 3)( x n + 1) 2n

=

( xn

x4

+ 1)

2

, x n ≠ −3, 3, 0

63.

Unshaded Area x⋅x x x = = = , Total Area 2 x( 4 x + 2) 2( 2)( 2 x + 1) 4( 2 x + 1)

x > 0

65.

Unshaded Area π x2 πx πx , = = = Total Area 2 x( 4 x + 2) 2( 2)( 2 x + 1) 4( 2 x + 1)

x > 0

67. In simplifying a product of national expressions, you divide the common factors out of the numerator and denominator. 69. The domain needs to be restricted, x ≠ a, x ≠ b. 71. The first expression needs to be multiplied by the reciprocal of the second expression ( not the second by the reciprocal of the

first ), and the domain needs to be restricted. x2 − 4 x+ 2 x2 − 4 x − 2 ( x − 2) ( x + 2) = ( x − 2) , x ≠ ±2 ÷ = ⋅ = 5x x −2 5x x+ 2 5 x ( x + 2) 5x 2

73.

1 3 5 1+3+5 9 + + = = 8 8 8 8 8

75.

3 4 9 4 9+ 4 13 + = + = = 5 15 15 15 15 15

2

77. x 2 + 3 x = 0 x( x + 3) = 0

x = 0

x +3 = 0 x = −3

Section 6.3 Adding and Subtracting Rational Expressions 1.

5x 4x 5x + 4x 9x 3x + = = = 6 6 6 6 2

3.

2 11 2 − 11 −9 3 − = = = − 3a 3a 3a 3a a

11. 5 x 2 = 5 ⋅ x ⋅ x

20 x3 = 5 ⋅ 2 ⋅ 2 ⋅ x ⋅ x ⋅ x LCM = 20 x3

5.

x − ( x + 2) x x + 2 x − x − 2 2 − = = = − 9 9 9 9 9

7.

2x − 1 1− x 2x − 1 + 1 − x + = x( x − 3) x( x − 3) x( x − 3) =

x x( x − 3)

=

1 , x −3

12 y = 2 ⋅ 2 ⋅ 3 ⋅ y LCM = 3 ⋅ 3 ⋅ 2 ⋅ 2 ⋅ y ⋅ y ⋅ y = 36 y 3 15. 15 x 2 = 5 ⋅ 3 ⋅ x ⋅ x

3( x + 5) = 3 ⋅ ( x + 5) LCM = 15 x 2 ( x + 5) 17. 63z 2 ( z + 1) = 7 ⋅ 9 ⋅ z ⋅ z ( z + 1)

x ≠ 0

14( z + 1) = 7 ⋅ 2 ⋅ ( z + 1) 4

c c −1 1 − = 9. 2 c + 3c − 4 c 2 + 3c − 4 (c + 4)(c − 1) =

13. 9 y 3 = 3 ⋅ 3 ⋅ y ⋅ y ⋅ y

1 , c + 4

c ≠ 1

LCM = 126 z 2 ( z + 1)

4

4

19. 8t (t + 2) = 2 ⋅ 2 ⋅ 2 ⋅ t ⋅ (t + 2) 14(t 2 − 4) = 2 ⋅ 7 ⋅ (t + 2)(t − 2) LCM = 2 ⋅ 2 ⋅ 2 ⋅ 7 ⋅ t ⋅ (t + 2)(t − 2) = 56t (t 2 − 4)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.3 21. 2 y 2 + y − 1 = ( 2 y − 1)( y + 1)

4 y 2 − 2 y = 2 y ( 2 y − 1)

Adding and Subtracting Rational Expressions

33.

20(1) 20( −1) 20 20 + = + x − 4 4− x ( x − 4)(1) (4 − x)(−1)

LCM = 2 y( 2 y − 1)( y + 1)

=

20 20 − ( x − 4) x − 4

23.

5(5) 3( 4 x) 5 3 25 12 x 25 − 12 x − = − = − = 4x 5 4 x(5) 5( 4 x) 20 x 20 x 20 x

=

20 − 20 = 0, x − 4

25.

7( a) 14(1) 7 14 7 a 14 7 a + 14 + 2 = + 2 = 2 + 2 = a a a(a) a (1) a a a2

27. 25 +

25( x + 4)

10 x + 4 x + 4 25 x + 100 + 10 25 x + 110 = = x + 4 x + 4

=

31.

3x 6 + x −8 x −8 3x + 6 = x −8 =

+

37.

3 x(1) 2( −1) 3x 2 + = + 3x − 2 2 − 3x 3 x − 2(1) 2 − ( 3x)(−1) 3x −2 + 3x − 2 3x − 2 3x − 2 2 = = 1, x ≠ 3x − 2 3 =

5( x + 3) x( x − 2) 5 x − = − x +3 x −2 ( x + 3)( x − 2) ( x − 2)( x + 3) =

x 2 − 2 x − 5 x − 15 ( x + 3)( x − 2) 2

x − 7 x − 15 + 3)( x − 2)

39.

3( x + 5) 2( x − 5) 3 2 + = + x −5 x+5 ( x − 5)( x + 5) ( x + 5)( x − 5)

(x

12 2 12 2 − = − − + −3 x2 − 9 x −3 x x x 3 3 ( )( ) = = = = =

2( x + 3) 12 − ( x − 3)( x + 3) ( x − 3)( x + 3) 12 − 2( x + 3)

(x

41.

=

3 x + 15 + 2 x − 10 ( x − 5)( x + 5)

=

5x + 5 − x ( 5)( x + 5)

9( x − 1) 3(5 x) 9 3 + = + 5x x −1 5 x( x − 1) 5 x( x − 1)

− 3)( x + 3)

=

12 − 2 x − 6 − 3)( x + 3)

9 x − 9 + 15 x 5 x( x − 1)

=

6 − 2x − 3)( x + 3)

24 x − 9 5 x( x − 1)

=

(x (x

−2( −3 + x)

(x

x ≠ 4

3 x(1) 6( −1) 3x 6 − = − x −8 8− x x 8 1 8 − − ( )( ) ( x)(−1)

25( x + 4) 10(1) 10 = + x + 4 1( x + 4) ( x + 4)(1) =

29.

35.

177

3(8 x − 3) 5 x( x − 1)

− 3)( x + 3)

−2 x +3 2 = − , x +3 =

x ≠ 3

43.

( x − 6) = x − ( x − 6) = x − x + 6 = x 1 x 6 − = − x − x − 30 x + 5 ( x + 5)( x − 6) ( x + 5)( x − 6) ( x + 5)( x − 6) ( x + 5)( x − 6) ( x + 5)( x − 6)

45.

4( x − 4) 16(1) 4 16 4 x − 16 16 4 x − 16 + 16 4x + = + = + = = 2 2 x − 4 ( x − 4) 2 ( x − 4)( x − 4) ( x − 4)2 (1) ( x − 4)2 ( x − 4)2 x − 4 x ( ) ( − 4)

2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

178

47.

Chapter 6

R ational Expressions, Equations, and Functions

y x y x − = − x 2 + xy xy + y 2 x( x + y ) y( x + y) =

x( x + y )( y )

x( x )

−

4( x 2 ) 4 x( x + 3) 2( x + 3) 4 2 4 − 2 + = − 2 + x x x + 3 x( x)( x + 3) x ( x + 3) ( x + 3) x 2

y ( x + y )( x)

=

4 x 2 + 12 x 2x + 6 4 x2 − 2 + 2 2 x ( x + 3) x ( x + 3) x ( x + 3)

=

y2 x2 − xy( x + y ) xy ( x + y )

=

4 x 2 + 12 x − 2 x − 6 + 4 x 2 x 2 ( x + 3)

=

y 2 − x2 xy( x + y )

=

8 x 2 + 10 x − 6 x 2 ( x + 3)

= =

51.

y( y)

49.

(y

− x)( y + x)

xy ( x + y )

y − x , xy

2( 4 x 2 + 5 x − 3)

=

x 2 ( x + 3)

x ≠ −y

u 3u 2 3u 2−u + − = + 2 u 2 − 2uv + v 2 u −v u −v u −v (u − v ) = = = =

3u (1)

(u

− v) (1) 2

3u

(u

− v)

+

2

+

(2 (u

− u )(u − v) − v)(u − v)

2u − 2v − u 2 + uv

(u

− v)

2

3u + 2u − 2v − u 2 + uv

(u

− v)

2

5u − 2v − u 2 + uv

= −

(u

− v)

2

u 2 − uv − 5u + 2v

(u

− v)

2

53. To find the model T, find the sum of M and F. T = = =

0.150t + 5.53 − 0.313t + 7.71 + − 0.049t + 1 − 0.055t + 1

(− 0.150t

+ 5.53)( − 0.055t + 1) + (− 0.313t + 7.71)( − 0.049t + 1)

(− 0.049t

2

+ 1)( − 0.055t + 1)

0.023587t − 1.14494t + 13.24 , (− 0.049t + 1)(− 0.055t + 1)

5 ≤ t ≤ 10

55. Two rational expressions with like denominators have a common denominator is a true statement. 57. To add or subtract rational expressions with like denominators, simply add (or subtract) the terms in the numerator and keep the common denominator.

59.

4( x 2 + 1) 4 4 4x2 − 2 = 2 2 − 2 2 x x +1 x ( x + 1) ( x + 1) x 2 =

4( x 2 + 1)

x ( x + 1) 2

2

−

4x2

x ( x 2 + 1)

=

4x2 + 4 − 4x2 x 2 ( x 2 + 1)

=

4 x 2 ( x 2 + 1)

2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.3

61.

=

x 2 + 8 x + 12 2x − 2 14 − − ( x − 1)( x + 6) ( x + 6)( x − 1) ( x + 6)( x − 1)

=

x 2 + 8 x + 12 − 2 x + 2 − 14 ( x − 1)( x + 6)

=

x2 + 6 x ( x − 1)( x + 6)

=

65.

179

14(1) x + 2 2 14 ( x + 2)( x + 6) − 2( x − 1) − − − 2 = x −1 x +6 x + 5x − 6 ( x − 1)( x + 6) ( x + 6)( x − 1) ( x + 6)( x − 1)(1)

=

63.

Adding and Subtracting Rational Expressions

(x

x ( x + 6) − 1)( x + 6)

x , x −1

x ≠ −6

t (3) t ( 2) t t 3t 2t 5t + = + = + = 4 6 4(3) 6( 2) 12 12 12

A +

4 2 2 −4 = + + x3 − x x x +1 x −1 2( x)( x − 1) 2( x)( x + 1) − 4( x + 1)( x − 1) = + + x( x + 1)( x − 1) ( x + 1)( x)( x − 1) ( x − 1)( x)( x + 1)

B + C = 0 −B + C = 0

−A

= 4

A = −4

=

Substitute into first equation. −4 + B + C = 0

Solve first and second equations. −B +

x( x 2 − 1)

− 4 x2 + 4 + 2 x2 − 2 x + 2 x2 + 2 x x3 − x 4 = 3 x − x =

B +C = 4

B +

− 4( x 2 − 1) + 2 x( x − 1) + 2 x( x + 1)

C = 4 C = 0 2C = 4 C = 2

B +

2 = 4 B = 2

67.

( x − 1) − (4 x − 11) = x − 1 − 4 x + 11 = −3x + 10 x −1 4 x − 11 − = x + 4 x + 4 x+ 4 x+ 4 x+ 4 The subtraction must be distributed to both terms of the numerator of the second fraction.

69. Yes.

75. x 2 − 7 x + 12 = ( x − 3)( x − 4)

3 x ( x + 2) + x + 2 2

71. 5v + ( 4 − 3v) = (5v − 3v) + 4 = 2v + 4 73.

( x2 − 4 x + 3) − (6 − 2 x) =

77. 2a 2 − 9a − 18 = ( a − 6)( 2a + 3)

x2 − 4x + 3 − 6 + 2 x

= x2 − 2x − 3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

180

Chapter 6

R ational Expressions, Equations, and Functions

Section 6.4 Complex Fractions ⎛3⎞ ⎜ ⎟ 3 9 3 12 3⋅2⋅2⋅3 1 16 ÷ = ⋅ = = 1. ⎝ ⎠ = ⎛ 9 ⎞ 16 12 16 9 2⋅2⋅2⋅2⋅3⋅3 4 ⎜ ⎟ 12 ⎝ ⎠

⎛ 25 x 2 ⎞ ⎜ ⎟ 25 x 2 10 x ⎝ x − 5⎠ 9. = ÷ 10 x x − 5 5 + 4x − x2 ⎛ ⎞ ⎜ 2⎟ ⎝ 5 + 4x − x ⎠ 25 x 2 5 + 4 x − x 2 ⋅ x −5 10 x

=

⎛ 8x2 y ⎞ ⎜ 2 ⎟ 3z ⎠ 8x2 y 4 xy 3. ⎝ = ÷ 5 3z 2 9z ⎛ 4 xy ⎞ ⎜ 5⎟ ⎝ 9z ⎠

5 ⋅ 5 ⋅ x ⋅ x ⋅ ( −1)( x 2 − 4 x − 5)

=

=

8x2 y 9 z5 ⋅ 3 z 2 4 xy

=

=

4 ⋅ 2 ⋅ 3 ⋅ 3x 2 ⋅ y ⋅ z 5 3 ⋅ 4 ⋅ x ⋅ y ⋅ z2

=

( x − 5) ⋅ 5 ⋅ 2 ⋅ x 5 ⋅ x ⋅ ( −1)( x − 5)( x + 1) ( x − 5)2 5 x( x + 1) − , x ≠ 0, 5, −1 2

3

= 6 xz , x, y, z ≠ 0

⎛ 6 x3 ⎞ ⎜ ⎟ ⎜ (5 y ) 2 ⎟ 3 2 ⎝ ⎠ = 6x ÷ 9x 5. 2 2 ⎛ (3 x ) ⎞ 25 y 15 y 4 ⎜ ⎟ ⎜ 15 y 4 ⎟ ⎝ ⎠

⎛ x 2 + 3x − 10 ⎞ ⎜ ⎟ 2 x + 4 ⎠ = x + 3 x − 10 ÷ 3x − 6 11. ⎝ 3x − 6 x + 4 1 + − x 5 x 2 ( )( )⋅ 1 = x + 4 3( x − 2)

6 x3 15 y 4 = ⋅ 25 y 2 9 x 2 3 ⋅ 2 ⋅ 5 ⋅ 3 ⋅ x3 ⋅ y 4 = 5 ⋅ 5 ⋅ 3 ⋅ 3 ⋅ x2 ⋅ y 2 =

2 xy 2 , 5

x ≠ 0, y ≠ 0

⎛ y ⎞ ⎜ ⎟ 3− y y y2 ÷ 7. ⎝ 2 ⎠ = ⎛ y ⎞ y −3 3− y ⎜ ⎟ ⎝ y − 3⎠ =

y y −3 ⋅ −1( y − 3) y2

=

y ( y − 3) −1y 2 ( y − 3)

1 = − , y

y ≠ 3

13.

=

( x + 5)( x − 2) ( x + 4)3( x − 2)

=

x +5 , 3( x + 4)

x ≠ 2

2 x − 14 2 x − 14 x 2 − 9 x + 14 = ÷ 1 x+3 ⎛ x − 9 x + 14 ⎞ ⎜ ⎟ x +3 ⎝ ⎠ 2

=

2( x − 7) x+3 ⋅ 1 ( x − 7)( x − 2)

=

2( x − 7)( x + 3) ( x − 7)( x − 2)

=

2( x + 3) , x −2

x ≠ −3, 7

⎛ 6 x 2 − 17 x + 5 ⎞ ⎜ ⎟ 3x 2 + 3x ⎠ 6 x 2 − 17 x + 5 3 x − 1 = ÷ 15. ⎝ 3x − 1 3x 2 + 3x 3x + 1 3x + 1 (3x − 1)(2 x − 5) ⋅ 3x + 1 = 3 x( x + 1) 3x − 1 = =

(3 x (2 x

− 1)( 2 x − 5)(3 x + 1)

3 x( x + 1)(3 x − 1) − 5)(3 x + 1)

3 x( x + 1)

,

x ≠ ±

1 3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.4

17.

181

16 x 2 + 8 x + 1 4 x 2 − 3 x − 1 16 x 2 + 8 x + 1 x 2 + 6 x + 9 ÷ 2 = ⋅ 3x 2 + 8 x − 3 x + 6x + 9 3x 2 + 8 x − 3 4 x 2 − 3x − 1 (4 x + 1)(4 x + 1) ⋅ ( x + 3)( x + 3) = (3x − 1)( x + 3) (4 x + 1)( x − 1) = =

19.

Complex Fractions

(4 x + 1)(4 x + 1)( x + 3)( x + 3) (3x − 1)( x + 3)(4 x + 1)( x − 1) (4 x + 1)( x + 3) , x ≠ −3, − 1 4 (3x − 1)( x − 1)

x( x + 3) − 2( x + 3) x 2 + 4 x + 4 x 2 + 3x − 2 x − 6 x +3 ÷ 2 = ⋅ 2 x − 4 x + 4x + 4 x2 − 4 x +3 x + 3)( x − 2) ( x + 2)( x + 2) ( = ⋅ x +3 ( x − 2)( x + 2)

(x

=

+ 3)( x − 2)( x + 2)( x + 2)

(x

− 2)( x + 2)( x + 3)

= ( x + 2),

x ≠ ±2, − 3

⎛ x 2 − 3x − 10 ⎞ ⎜ 2 ⎟ x − 4x + 4 ⎠ x 2 − 3 x − 10 21 + 4 x − x 2 = 2 ÷ 2 21. ⎝ 2 x − 4x + 4 x − 5 x − 14 ⎛ 21 + 4 x − x ⎞ ⎜ 2 ⎟ ⎝ x − 5 x − 14 ⎠ =

x 2 − 3 x − 10 x 2 − 5 x − 14 ⋅ 2 x − 4 x + 4 −1( x 2 − 4 x − 21)

=

( x − 5)( x + 2) ( x − 2)( x − 2)

= − = − = −

⎛ ⎜1 + 23. ⎝ y

4⎞ ⎟ y⎠

⎛ ⎜1 + = ⎝ y

4⎞ ⎟ y⎠

( x − 7)( x + 2) −1( x − 7)( x + 3)

⋅

( x 2 − 3x − 10)( x + 2) ( x 2 − 4 x + 4)( x + 3) (x

− 5)( x + 2)( x + 2)

(x (x (x

− 2) ( x + 3) 2

+ 2) ( x − 5) 2

− 2) ( x + 3) 2

,

x ≠ − 2, 7

y y + 4 ⋅ = y y2

⎛4 ⎞ ⎛4 ⎞ ⎜ + 3⎟ ⎜ + 3⎟ x x x ⎠ = ⎝ ⎠ ⋅ = 4 + 3x , 25. ⎝ ⎛4 ⎞ ⎛4 ⎞ x 4 − 3x ⎜ − 3⎟ ⎜ − 3⎟ x x ⎝ ⎠ ⎝ ⎠

x ≠ 0

⎛ x⎞ x ⎜ ⎟ 2x 2⎠ ⎝ 2 = ⋅ 27. 3 2x 3⎞ ⎛ 2+ ⎜2 + ⎟ x x⎠ ⎝ =

x2 x2 , = 4x + 6 2( 2 x + 3)

9 ⎞ 9 ⎞ ⎛ ⎛ ⎜3 + ⎟ ⎜3 + ⎟ x − 3⎠ x − 3⎠ x − 3 ⎝ ⎝ = ⋅ 29. 12 ⎞ 12 ⎞ x − 3 ⎛ ⎛ + + 4 4 ⎜ ⎟ ⎜ ⎟ x − 3⎠ x − 3⎠ ⎝ ⎝ 9 3( x − 3) + ( x − 3) −3 x = 12 4( x − 3) + ( x − 3) x −3 3x − 9 + 9 = 4 x − 12 + 12 3x 3 = = , x ≠ 0, 3 4x 4

x ≠ 0

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

182

Chapter 6

R ational Expressions, Equations, and Functions

1 ⎞ ⎛1 1 1 − ⎜ − ⎟ x x + 1⎠ ⎝ x x + 1 ⋅ x( x + 1) 31. = 1 1 x( x + 1) ⎛ ⎞ ⎜ ⎟ x +1 ⎝ x + 1⎠ =

1 x +1− x = , x x

⎛y ⎛y x⎞ x⎞ ⎜ − ⎟ ⎜ − ⎟ x y⎠ x y ⎠ xy ⎝ ⎝ = ⋅ 43. ⎛x + y⎞ ⎛ x + y ⎞ xy ⎜ ⎟ ⎜ ⎟ ⎝ xy ⎠ ⎝ xy ⎠ y x ( xy ) − ( xy ) x y = ⎛x + y⎞ ⎜ ⎟ xy ⎝ xy ⎠

x ≠ −1

1 2y − 2 y − y −1 y y2 = ⋅ 2 33. −2 1 10 − y 10 − 2 y y =

35.

7 x 2 + 2 x −1 5 x −3 + x

y( 2 y 2 − 1) 2 y3 − y = , 2 10 y − 1 10 y 2 − 1

37.

1 ⎞ ⎛ 1 ⎜R + R ⎟ 2⎠ ⎝ 1

y ≠ 0

x 2 (7 x3 + 2) 7 x5 + 2 x 2 , = x4 + 5 5 + x4

(y

− x)( y + x) x+ y

x ≠ 0

R1 ⎞ ⎛ R2 + ⎜ ⎟ R1 R 2 ⎠ ⎝ R1 R 2 1 = ⎛ R 2 + R1 ⎞ ⎜ ⎟ ⎝ R1 R 2 ⎠ R1 R 2 1 = ⋅ 1 R1 + R 2 R1 R 2

=

x⎞ ⎛ 2x − ⎟ ⎜ 4⎠ ⎝ 3 ⎛ x + 1⎞ ⎜ ⎟ ⎝ 6 ⎠

41. To simplify a complex fraction, multiply the numerator and denominator by the least common denominator of all of the fractions in the numerator and denominator.

10( x − 3) + 3 x3

3x 2 − 2 x 2 + 6 x 10 x − 30 + 3 x3

=

x2 + 6x , 3 x + 10 x − 30 3

x ≠ 0, x ≠ 3

⎛ 10 ⎞ ⎛ 10 ⎞ ⎜⎜ ( x + 1) ⎟⎟ ⎜ ⎟ 2( x + 1) x + 1⎠ ⎝ ⎠ ⎝ 47. = ⋅ 1 3 1 3 ⎛ ⎞ ⎛ ⎞ 2( x + 1) + + ⎜ ⎟ ⎜ ⎟ x + 1⎠ x + 1⎠ ⎝ 2x + 2 ⎝ 2x + 2

R1 + R 2

x +5 , 6 ⎞ ⎛ ⎜ ⎟ ⎝ 3 x + 15 ⎠

3 x 2 − 2 x( x − 3)

=

39. A complex fraction is a fraction that has a fraction in its numerator or denominator, or both. Examples of complex fractions: ⎛ x − 3⎞ ⎜ ⎟ ⎝ x ⎠ , ⎛ 3x − 9 ⎞ ⎜ ⎟ ⎝ 4 ⎠

x ≠ 0, y ≠ 0, x ≠ − y

2⎞ 2⎞ ⎛ x ⎛ x − ⎟ − ⎟ ⎜ ⎜ 3 3 3 3 x x − − ⎠ = ⎝ ⎠ ⋅ 3 x( x − 3) 45. ⎝ 2 2 ⎛ 10 ⎞ ⎛ 10 x x ⎞ 3 x( x − 3) + + ⎜ ⎟ ⎜ ⎟ x − 3⎠ x − 3⎠ ⎝ 3x ⎝ 3x

1

=

=

=

y 2 − x2 x+ y

= y − x,

2 7x + 3 x ⋅ x = 3 5 + x x 3 x 2

= 1

=

1 + x −1 + y −1 x = 49. −1 1 x − y −1 − x =

=

10( 2) 1 + 3( 2)

=

20 , 7

x ≠ −1

1 y xy ⋅ 1 xy y

y + x , y − x

x ≠ 0, y ≠ 0

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.4

51.

x −2 − y −2

(x

+ y)

2

=

1 1 − 2 x2 y

=

2

(y

8 x + 3x 24 11x = 24 =

− x)( y + x)

x 2 y 2 ( x + y )( x + y ) y − x x2 y 2 ( x + y)

1 1 − f ( 2 + h ) − f ( 2) 2 2 h + 53. = h h 1 1 − 2 2 ⋅ 2( 2 + h) h + = 2( 2 + h) h

b +5 b +5 2 2 + + b b 4 4 59. = ⋅ 2 2 b +5 ( 4b) + 4 = 2( 4b) =

2 − ( 2 + h) = 2 h( 2 + h ) =

=

2−2−h 2 h( 2 + h )

−h = 2 h( 2 + h ) =

−1 , 2( 2 + h)

h ≠ 0

1 ⎛ 1 1 1 ⎞ + + ⎜⎜ ⎟⎟ R R R 2 3⎠ ⎝ 1

=

67. In the second step, the set of parentheses cannot be moved because division is not associative. ⎛a⎞ ⎜ ⎟ ⎝b⎠ = a ⋅ 1 = a b b b b2

(x

+ 1) x +1 ( x + 1) ( x + 2) = x + 1 x + 2 2 , ÷ = ( )( ) 3 x + 2 ( x + 2)( x + 1) ( x + 2) 2

+ 5)b + 8 8b

2

b + 5b + 8 8b

x x 8x + = + 9 72 72 9x x x x2 = + = + 8 72 72 5x 10 x x + = x3 = 36 72 72 x1 =

x 9x x = = 72 72 8 10 x 5x x = = 72 72 36 11x x + = 72 72

R1 R 2 R 3 1 1 = = ⎛ R2 R3 ⎛ R1 R 2 + R1 R 3 + R 2 R 3 ⎞ R1 R 2 + R1 R 3 + R 2 R 3 R1 R 3 R1 R 2 ⎞ + + ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ R R R R R R R R R R R R 1 2 3 1 2 3⎠ 1 2 3 ⎝ 1 2 3 ⎝ ⎠

65. No. A complex fraction can be written as the division of two rational expressions, so the simplified form will be a rational expression.

75.

(b

4b 4b 2 ( 4b) b

x x x x − − 18 3x − 2 x x 6 9 6 9 = ⋅ = = 61. 4 4 18 72 72 1

x x x x + + 30 6 x + 5x 11x 5 6 5 6 = ⋅ = = 55. 2 2 30 60 60

63.

183

2x x 2x x + + 12 3 4 3 4 = ⋅ 57. 2 2 12 2x x (12) + (12) 3 4 = 2(12)

x2 y2 ⋅ 2 2 x y

+ y) 1 y 2 − x2 = 2 2 x y ( x + y )( x + y ) =

(x

Complex Fractions

2

69. ( 2 y ) (3 y ) = 8 y 3 ⋅ 9 y 2 = 72 y 5 3

2

71. 3x 2 + 5 x − 2 = (3 x − 1)( x + 2) 73.

x2 x2 1 x2 x ÷ 4x = ⋅ = = , 2 2 4x 8x 8

x ≠ 0

3

x ≠ −2,

x ≠ −1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

184

Chapter 6

R ational Expressions, Equations, and Functions

Mid-Chapter Quiz for Chapter 6 1. Domain: All real numbers x such that x ≠ −1 and

x ≠ 0.

10.

( x2

+ 2 x) ⋅

x( x + 2)5 5 = x2 − 4 x ( − 2)( x + 2)

2. Domain: All real numbers x such that x ≠ − 2 and

=

x ≠ 2. 3.

9 y2 3y , = 6y 2

11.

y ≠ 0

6u 4v3 3 ⋅ 2 ⋅ u 4 ⋅ v3 2u 3 4. = = , 3 3 15uv 3⋅5⋅u⋅v 5

5.

2

4x − 1 = x − 2 x2

(2 x − 1)(2 x + 1) x(1 − 2 x)

2x + 1 = , −x 6.

( z + 3)

2

2z 2 + 5z − 3

= =

7. 8.

=

u ≠ 0, v ≠ 0

− x( 2 x − 1)

=

12.

− 1)( z + 3)

z +3 , 2z − 1

13.

n 2 ( 2m − n) 2mn 2 − n3 = 2 2 2m + mn − n (2m − n)(m + n) n2 , m + n

11t 2 (9) t 11t 2 9 ⋅ = = , 9. 6 33t 6(33t ) 2

8x 3( x − 1) ( x + 3) 2

8x

3( x − 1)( x 2 + 2 x − 3) 8x 3( x − 1)( x + 3)( x − 1)

32 z 4 80 z 5 32 z 4 ⋅ 25 x8 y 6 ÷ = 5 5 8 6 5x y 25 x y 5 x5 y 5 ⋅ 80 z 5

z ≠ −3

ab(5a + 3b 2 ) 5a 2b + 3ab3 5a + 3b 2 = = 2 2 2 2 ab ab ab

=

=

( 2 x − 1)(2 x + 1)

( z + 3)( z + 3)

x ≠ −2

4(12 x) 4 12 x ⋅ = 2 3( x − 1) 6( x + 2 x − 3) 3( x − 1)6( x + 3)( x − 1) =

1 x ≠ 2

(2 z

5x , x − 2

=

16 ⋅ 2 ⋅ 5 ⋅ 5 ⋅ x8 ⋅ y 6 ⋅ z 4 16 ⋅ 5 ⋅ 5 ⋅ x5 ⋅ y 5 ⋅ z 5

=

2 x3 y , x ≠ 0, y ≠ 0 z

(a + 1)(a + 1) a −b a 2 − b2 a −b ÷ 2 = ⋅ 9a + 9b a + 2a + 1 9( a + b) ( a − b)( a + b) =

(a − b)( a + 1) 2 9( a + b) ( a − b)

=

(a + 1)2 , 2 9( a + b)

2

2m − n ≠ 0

t ≠ 0

14.

2 2(u 2 − v 2 ) 5u ⋅ 2(u − v)(u + v) ⋅ 18(u − v) 4(u − v ) 5u 25u 2 ⋅ ÷ = = , 3(u + v) 3v 18(u − v) 5uv 3(u + v)(3v)( 25u 2 )

15.

5x − 6 2x − 5 5x − 6 + 2 x − 5 7 x − 11 + = = x − 2 x − 2 x − 2 x − 2

16.

4( x − 3) 4( x − 3) x x − = − 2 x −9 x+3 ( x − 3)( x + 3) ( x − 3)( x + 3)

a ≠ b

u ≠ ±v

2

=

17.

2 x − 4( x 2 − 6 x + 9) x − 4( x − 3) x − 4 x 2 + 24 x − 36 − 4 x 2 + 25 x − 36 4 x 2 − 25 x + 36 = = = = − ( x − 3)( x + 3) ( x − 3)( x + 3) ( x − 3)( x + 3) ( x − 3)( x + 3) ( x − 3)( x + 3)

1( x − 2) x( x + 1) x2 + 2 1 x x2 + 2 + − = + − x − x − 2 x +1 x − 2 ( x − 2)( x + 1) ( x − 2)( x + 1) ( x − 2)( x + 1) 2

= =

x2 + 2 + x − 2 − x2 − x ( x − 2)( x + 1)

(x

0 = 0, − 2)( x + 1)

x ≠ 2, x ≠ −1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.5 9t 2 t −3 −9t 2 3t 18. 3 − t ⋅ = = − , 6t t −3 6t 2 t −3 10 x + 2x = 19. 15 x 2 + 3x + 2 2

= =

t ≠ 3

Dividing Polynomials and Synthetic Division

20.

3x −1 − y −1

( x − y)

−1

⎛3 1⎞ ⎛3 1⎞ ⎜ − ⎟ ⎜ − ⎟ xy x − y x y⎠ x y⎠ ( ) ⎝ = = ⎝ ⋅ ⎛ 1 ⎞ ⎛ 1 ⎞ xy( x − y ) ⎜ ⎟ ⎜ ⎟ ⎝ x − y⎠ ⎝ x − y⎠

10 x( x + 2) x( x + 2)( x + 1) ⋅ 15 x( x + 2)( x + 1) ( x + 2)( x + 1)

=

10( x + 1) 15 x 2( x + 1) 3x

,

x ≠ −2, x ≠ −1

3 y ( x − y ) − x( x − y ) xy

=

3xy − 3 y − x 2 + xy xy

=

−3 y 2 + 4 xy − x 2 xy

= = =

21. (a) Verbal Model:

185

2

−1(3 y 2 − 4 xy + x 2 ) xy −1(3 y − x)( y − x) xy

(3 y − x)( x − y) xy

,

x ≠ y

Total Set up Number of = + 144 ⋅ cost cost arrangements

Total cost = C

Labels:

Number of arrangements = x

C = 25,000 + 144 x

Equation: (b) Verbal Model:

Labels:

Average Total Number of = ÷ cost cost arrangements

Average cost = C Total cost = C Number of arrangements = x

Equation: (c) C (500) =

22.

x +

C =

25,000 + 144x x

25,000 + 144(500) 97000 = = $194 500 500

2x 2x x x + x+ + 2 3 = 2 3 ⋅ 6 = 6 x + 3 x + 4 x = 13x 3 3 6 18 18

Section 6.5 Dividing Polynomials and Synthetic Division 1. (7 x3 − 2 x 2 ) ÷ x =

7 x3 − 2 x 2 x

5. ( 4 x 2 − 2 x) ÷ ( − x) =

=

7 x3 2x2 − x x

=

= 7 x 2 − 2 x, 3.

4 x2 2x − −x −x = −4 x + 2, x ≠ 0

x ≠ 0

m 4 + 2m 2 − 7 m4 2m 2 7 = + − m m m m 7 = m3 + 2m − m

4 x2 − 2 x −x

7.

50 z 3 + 30 z 50 z 3 30 z = + −5 z −5 z −5 z = −10 z 2 − 6,

z ≠ 0

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

186

9.

Chapter 6

R ational Expressions, Equations, and Functions

4v 4 + 10v3 − 8v 2 4v 4 10v3 8v 2 5 = + − 2 = v 2 + v − 2, 2 2 2 4v 4v 4v 4v 2

11. (5 x 2 y − 8 xy + 7 xy 2 ) ÷ 2 xy =

5 x 2 y − 8 xy + 7 xy 2 5 x 2 y 8 xy 7 xy 2 5x 7 = − + = − 4 + y, 2 xy 2 xy 2 xy 2 xy 2 2

112 13. 9 1013 9 11 9 23 18 5

3t − 4, t ≠ 23. 2t − 3 6t − 17t + 12 6t 2 − 9t −8t + 12 −8t + 12

3 2

3x 2 − 3x + 1 + 5 9

25. 3x + 2 9 x3 − 3 x 2 − 3x + 4 9 x3 + 6 x 2 −9 x 2 − 3 x −9 x 2 − 6 x 3x + 4 3x + 2 2

215 15. 15 3235 30 23 15 85 75 10

x − 4+

So 3235 ÷ 15 = 215 +

10 15

= 215 +

2 3

242 17. 25 6055 50 105 100 55 50 5 6055 25

x ≠ 0, y ≠ 0

2

So 1013 ÷ 9 = 112 +

So,

v ≠ 0

= 242 +

5 25

= 242 +

1 5

x − 5, x ≠ 3 2 19. x − 8 x + 15 = x − 3 x 2 − 8 x + 15 x −3 x 2 − 3x −5 x + 15 −5 x + 15 x 21. − x + 3 − x 2 − 4 x − x 2 + 3x −7 x −7 x

+ 7, x ≠ 3 + 21

27. x + 4 x 2 + 0 x + 16 x2 + 4x −4 x + 16 −4 x − 16 32

x2 29. x + 5 x3 + 0 x 2 x3 + 5 x 2 −5 x 2 −5 x 2

2 3x + 2

32 x + 4

− 5 x + 25, x ≠ −5 + 0 x + 125 + 0x − 25 x 25 x + 125 25 x + 125

x + 2+

x2 31. 2 3 x + 2x + 3 x + 4 x2 x3 + 2 x 2 2 x2 2x2

+ + + + +

1 2x 7x 3x 4x 4x

+3 +7 +7 +6 1

+ 21 + 21

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.5

4 x 2 + 12 x + 25 + 33. x 2 − 3 x + 2 4 x 4 + 0 x3 4 x 4 − 12 x3 12 x3 12 x3

35.

5

45.

1

2

0

x3 + 3x 2 − 1 x + 4

1 1

–1 4

44

6

0

–1

–4

4

−16

−1

4

−17

1

4 x 3 − 3 x 2 − 1 = ( x − 1)( 2 x + 1)

47.

3

1 0

4 x 2 + 4 x + 1 = ( 2 x + 1)( 2 x + 1)

+ x − 6) ÷ ( x − 2) = x + 3, x ≠ 2

−4

4 0 –3 4

13

(x

52 x − 55 x 2 − 3x + 2

1 1 –6 2

–4

1

7

3

–63 36

–108 108

3 –9

–27

0

1

x3 + 3 x 2 − 9 x − 27 = x 2 ( x + 3) − 9( x + 3) = ( x 2 − 9)( x + 3) = ( x + 3)( x − 3)( x + 3) x + 7 x + 3 x − 63 x − 108 = ( x + 3) ( x − 3)( x + 4) 4

5 x3 − 6 x 2 + 8 39. x − 4

5 –6

0

49.

8

20 56 224 5 14 56 232 3

2

0.1

0.8

1

0.02

0.164

0.82

1.164

0.1x 2 + 0.8 x + 1 1.164 = 0.1x + 0.82 + x − 0.2 x − 0.2

43.

3

1 –1 –14 3

24 6

1

2 –8

15 x 2 − 2 x − 8 4 x − 5 15

–2 –8 12

8 0

4⎞ ⎛ 15 x 2 − 2 x − 8 = (15 x + 10)⎜ x − ⎟ 5⎠ ⎝

0.1x + 0.8 x + 1 x − 0.2

0.1

2

2

15 10

2

0.2

3

4 5

5x − 6x + 8 232 = 5 x 2 + 14 x + 56 + x − 4 x − 4

41.

2

–4 –12

x3 + 3x 2 − 1 −17 = x2 − x + 4 + x + 4 x + 4

4

187

5 50 55

( x2 + x − 6) ÷ ( x − 2) 2

37.

x − − 3x 2 + + 8x2 x − 11x 2 + − 36 x 2 + 24 x 25 x 2 − 23x − 25 x 2 − 75 x + 52 x −

Dividing Polynomials and Synthetic Division

–24 0

4⎞ ⎛ = 5(3 x + 2)⎜ x − ⎟ 5⎠ ⎝ 51.

x3 + 2 x 2 − 4 x + c x − 2

2

c

1 2 –4 2 14

8 4

8 0

c +8 = 0 c = −8

x + 2 x − 8 = ( x + 4)( x − 2) 2

x3 − x 2 − 14 x + 24 = ( x − 3)( x + 4)( x − 2)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

188

Chapter 6

R ational Expressions, Equations, and Functions

53. The equation 1253 ÷ 12 = 104 +

5 12

has a dividend of

69. Volume = Area of triangle ⋅ Height (of prism)

1253, a divisor of 12, a quotient of 104 and a remainder of 5.

Area of triangle =

55. A divisor divides evenly into a dividend when the remainder is zero.

=

3(uv) 8u 2v + 2u uv

2

Area of triangle = Height = =

73.

x + 3x + 2 ( x + 2)( x + 1) + 2 x + 3 + ( 2 x + 3) = ( ) x+2 x+2 = ( x + 1) + ( 2 x + 3) = ( x + 2 x) + (1 + 3) = 3x + 4,

x ≠ −2

x2 + 4 5 = x −1+ x +1 x +1

Dividend: x 2 + 4 Quotient: x − 1 Remainder: 5 75.

7 − 3x > 4 − x −3 x + x > 4 − 7

+ 4, x n ≠ −2 +8

−2 x > −3 x
0

19.

=

5b − 15 5(b − 3) 11. = 30b − 120 30(b − 4) 5(b − 3) = 5 ⋅ 6(b − 4) =

x 1 = , 2x 2

8− x 5 8− x + 5 13 − x x − 13 + = = = − 4x 4x 4x 4x 4x

39.

x ≠ − 2, x ≠ −1

2(3 y + 4) 2(3 y + 4) + 3 − y 3− y + = 2y + 1 2y + 1 2y + 1 =

6y + 8 + 3 − y 2y + 1

=

5 y + 11 2y + 1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

200

41.

43.

Chapter 6

4x 3x − 7 9 4 x + 3x − 7 − 9 + − = x + 2 x + 2 x + 2 x + 2 7 x − 16 = x + 2

1 3 1 ⎛ x − 12 ⎞ 3 ⎛ x + 5⎞ + = ⎜ ⎟+ ⎜ ⎟ x + 5 x − 12 x + 5 ⎝ x − 12 ⎠ x − 12 ⎝ x + 5 ⎠ 3( x + 5) x − 12 = + ( x + 5)( x − 12) ( x − 12)( x + 5) =

x − 12 + 3x + 15 ( x + 5)( x − 12)

=

4x + 3 ( x + 5)( x − 12)

2 3 5 x( x − 3)( x + 2) 2 ⎛ x + 2⎞ 3 ⎛ x − 3⎞ − = + ⎜ ⎟ ⎜ ⎟ − ( ( x − 3)( x + 2) − + + x −3 x + 2 x 3 x 2 x 2) ⎝ x − 3 ⎠ ( )⎝ ⎠ =

5 x 3 − 5 x 2 − 30 x + 2 x + 4 − 3x + 9 ( x − 3)( x + 2)

=

5 x 3 − 5 x 2 − 31x + 13 ( x − 3)( x + 2)

6 4x + 7 6 4x + 7 − 2 = − x −5 x − x − 20 x − 5 ( x − 5)( x + 4) 6( x + 4) 4x + 7 = − ( x − 5)( x + 4) ( x − 5)( x + 4) 6 x + 24 − 4 x − 7 = ( x − 5)( x + 4) =

51.

45.

3 4 6 4x 6 + 4x + = + = 5x2 10 x 10 x 2 10 x 2 10 x 2 2(3 + 2 x) 2x + 3 = = 2 10 x 5x2

47. 5 x +

49.

R ational Expressions, Equations, and Functions

2 x + 17 ( x − 5)( x + 4)

2 5 4x 1 5 ⎛ ( x + 3)( x − 3) ⎞ 4x ⎛ x − 3 ⎞ 1 ⎛ ( x + 3) ⎞ − − = − − ⎜⎜ ⎟ ⎜ ⎟ ⎜( ⎟ 2 2 2 x + 3 ( x + 3) x −3 x + 3 ⎝ x + 3)( x − 3) ⎠ ( x + 3) ⎝ x − 3 ⎠ x − 3 ⎝ ( x + 3) ⎟⎠

5 x 2 − 45 − 4 x 2 + 12 x − x 2 − 6 x − 9 ( x + 3)2 ( x − 3) 6 x − 54 = ( x + 3)2 ( x − 3) =

⎛6⎞ ⎜ ⎟ 6 2 3 ⋅ 2 x ⋅ x2 ÷ 3 = ⋅ = 3x 2 , 53. ⎝ x ⎠ = 2 x x x 2 ⎛ ⎞ ⎜ 3⎟ ⎝x ⎠

x ≠ 0

⎛ x ⎞ ⎜ ⎟ x 2− x 55. ⎝ x − 2 ⎠ = ⋅ 2 x x−2 2x ⎛ ⎞ ⎜ ⎟ ⎝2 − x⎠

=

x( −1)( x − 2) ( x − 2)( 2 x)

1 = − , 2

x ≠ 0, x ≠ 2

⎛ ⎞ 6x2 6 x2 ⎜ 2 ⎟ ( )( ) ( x + 7)( x − 5)( x + 5) 57. ⎝ x + 2 3x − 35 ⎠ = x + 7 3 x − 5 ⋅ ( x + 7)( x − 5)( x + 5) x ⎛ x ⎞ ⎜ 2 ⎟ ( )( ) x−5 x+5 ⎝ x − 25 ⎠ 6 x 2 ( x + 5) 6( x + 5) = 3 = , x ≠ 5, x ≠ −5 ( ) x x +7 x( x + 7) 59.

3t 2⎞ ⎛ ⎜5 − ⎟ t⎠ ⎝

⋅

3t 2 t , = t 5t − 2

t ≠ 0

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 6

201

2⎞ 2⎞ ⎛ ⎛ ⎜x − 3 + ⎟ ⎜x − 3 + ⎟ x x⎠ = ⎝ x⎠ ⋅ 61. ⎝ 2⎞ 2⎞ ⎛ ⎛ x ⎜1 − ⎟ ⎜1 − ⎟ x⎠ x⎠ ⎝ ⎝ x 2 − 3x + 2 , x ≠ 0 x − 2 ( x − 2)( x − 1) = , x ≠ 0 x − 2 = x − 1, x ≠ 0, x ≠ 2 =

⎛ ⎜ 2 63. ⎝ a ⎛ ⎜ 2 ⎝a

1 1⎞ − ⎟ a − ( a − 4)( a + 4) − 16 a ⎠ ⋅ a( a − 4)( a + 4) = ( )( ) 1 ⎞ aa −4 a + 4 a − 4 + 4a( a − 4)( a + 4) + 4⎟ + 4a ⎠ =

a − ( a 2 − 16) a − 4 + 4a( a 2 − 16)

=

a − a 2 + 16 a − 4 + 4a 3 − 64a

=

− a 2 + a + 16 , 4a 3 − 63a − 4

a ≠ 0, a ≠ −4

4 x3 − x 65. ( 4 x 3 − x) ÷ 2 x = 2x

5x − 8 +

69. x + 2 5 x 2 + 2 x + 3 5 x 2 + 10 x −8 x + 3 −8 x − 16 19

4 x3 x − 2x 2x 1 = 2x2 − , x ≠ 0 2 =

67.

3x3 y 2 − x 2 y 2 + x 2 y 3x3 y 2 x2 y 2 x2 y = − 2 + 2 2 2 x y x y x y x y = 3xy − y + 1,

3 2 5 4 3 2 73. x − 2 x + x − 1 x − 3 x + 0 x + x + 0 x + 6 5 4 3 2 x − 2x + x − x − x 4 − x3 + 2 x 2 + 0 x − x 4 + 2 x3 − x 2 + x − 3 x3 + 3x 2 − x + 6 − 3 x3 + 6 x 2 − 3x + 3 − 3x 2 + 2 x + 3

–1

1

35

4

23

x 2 + 3x + 5 3 = x + 2+ x +1 x +1

x 2 − 2, x ≠ ±1 3x 2 + 2 x2 + 2 + 2

−3 x 2 + 2 x + 3 x − 2 x2 + x − 1 3

77.

–2

1

–1 –2 1

3

71. x − 1 x + 0 x − − x4 −2 x 2 −2 x 2

x ≠ 0, y ≠ 0

x2 − x − 3 +

75.

2

19 x + 2

1

7

3

−14

–2 –10

14

5 –7

0

x 3 − 7 x 2 + 3 x − 14 = x 2 + 5 x − 7, x ≠ −2 x + 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

202 79.

Chapter 6 3

R ational Expressions, Equations, and Functions

1 0 –3 3 13

0

−25

9

18

54

6

18

29

( x 4 − 3x 2 − 25) ÷ ( x − 3) = x3 + 3x 2 + 6 x + 18 + 29

x −3

81.

2

1 2 –5

−6

2 14

8

6

3

0

12 ⎞ 1 ⎛ 89. (3t )⎜ 8 − ⎟ = (3t ) 3 t ⎠ ⎝ 24t − 36 = t 23t = 36

x3 + 2 x 2 − 5 x − 6 = ( x − 2)( x + 1)( x + 3)

t =

x2 + 4 x + 3

(x 83.

+ 3)( x + 1)

12 ⎛ 36 ⎞ ⎜ ⎟ ⎝ 23 ⎠ 23 8− 3 24 23 − 3 3 1 3

Check: 8 −

x 3 + =1 15 5 3⎞ ⎛x 15⎜ + ⎟ = (1)15 ⎝ 15 5 ⎠ x + 9 = 15 x = 6

85.

3x x = −15 + 8 4 ⎛3 ⎞ ⎛ x⎞ 8⎜ x ⎟ = ( −15)8 + ⎜ ⎟8 ⎝ 4⎠ ⎝8 ⎠ 3x = −120 + 2 x x = −120 Check:

3( −120) ? −120 = −15 + 8 4 −360 ? = −15 + −30 8 −45 = −45

1 x2 x − = 6 12 2

87.

36 23

⎛ x2 x ⎞ ⎛1⎞ 12⎜ − ⎟ = ⎜ ⎟12 12 ⎠ ⎝ 2 ⎠ ⎝6 2x2 − x = 6 2

2x − x − 6 = 0

91.

?

=

1 3

1 3 ? 1 = 3 1 = 3 ?

=

2 1 1 − = 3y 3 y 1 ⎞ ⎛1⎞ ⎛2 3 y⎜ − = ⎜ ⎟3 y 3 y y ⎟⎠ ⎝ 3 ⎠ ⎝ 6 −1 = y 5 = y Check:

2 1 − 5 3(5) 2 1 − 5 15 6 1 − 15 15 5 15 1 3

?

= ?

= ?

= ?

= =

1 3 1 3 1 3 1 3 1 3

(2 x + 3)( x − 2) = 0 2x + 3 = 0 3 x = − 2

x−2 = 0 x = 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 6 24 r 24 ⎞ ⎛ r (r ) = ⎜ 2 + ⎟r r ⎠ ⎝

93.

r = 2+

97.

r 2 = 2r + 24

3 y − 8( y + 1) = y( y + 1)

r − 2r − 24 = 0

3y − 8 y − 8 = y2 + y

( r − 6)( r + 4) = 0

−5 y − 8 = y 2 + y

r = 6, r = −4 Check: 6 = 2 +

24 6

?

Check: − 4 = 2 +

?

0 = y2 + 6 y + 8

24 −4

0 = ( y + 4)( y + 2) y + 4 = 0

?

6 = 2+ 4

−4 = 2 − 6

6 = 6

−4 = −4

y = −4 99.

4 t 95. = 4 t ⎛ t ⎞ ⎛ 4⎞ 4t ⎜ ⎟ = ⎜ ⎟ 4t ⎝ 4⎠ ⎝ t ⎠ t 2 = 16

y + 2 = 0 y = −2

2x 3 − = 0 x −3 x 3⎞ ⎛ 2x − ⎟ = (0) x( x − 3) x( x − 3)⎜ x⎠ ⎝x − 3 2 x( x) − 3( x − 3) = 0 2 x 2 − 3x + 9 = 0

t = ±4

No real solution

4 ? 4 = Check: 4 4 1 = 1 101.

3 8 − =1 y +1 y 8⎞ ⎛ 3 y( y + 1)⎜ − ⎟ = (1) y( y + 1) 1 y y⎠ + ⎝

2

?

203

−4 4 Check: = −4 4 −1 = −1

12 1 − = −1 x 2 + x − 12 x −3 1 ⎞ ( ) ( x − 3)( x + 4)⎛⎜ 2 12 − ⎟ = −1 ( x − 3)( x + 4) x − 3⎠ ⎝ x + x − 12 12 − ( x + 4) = −1( x 2 + x − 12) 12 − x − 4 = − x 2 − x + 12

( x 2 − 4) = 0 ( x − 2)( x + 2) = 0 x − 2 = 0 x = 2 Check:

? 12 1 − = −1 2 + 2 − 12 2−3 12 1 ? − = −1 −6 −1 2

?

−2 + 1 = −1 −1 = −1

x + 2 = 0 x = −2 Check:

12 1 − ( −2) + ( −2) − 12 ( −2) − 3 12 1 − −10 −5 6 1 − + 5 5 −1 2

?

= −1 ?

= −1 ?

= −1 = −1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

204

103.

Chapter 6

R ational Expressions, Equations, and Functions

5 6 − = −5 x2 − 4 x − 2 6 ⎞ ( x − 2)( x + 2)⎛⎜ 2 5 − ⎟ = (−5)( x − 2)( x + 2) x − 2⎠ ⎝x − 4 5 − 6( x + 2) = −5( x 2 − 4) 5 − 6 x − 12 = −5 x 2 + 20 5 x 2 − 6 x − 27 = 0

(5 x + 9)( x − 3) = 0 5x + 9 = 0

x −3 = 0

9 x = − 5 Check:

5 2

⎛ 9⎞ ⎜− ⎟ − 4 ⎝ 5⎠

−

? 6 = −5 9 − − 2 5

5 6 − 19 19 − − 25 5 125 30 − + 19 19 95 − 19 −5 105. Verbal Model:

?

= −5

x = 3 Check:

? 5 6 − = −5 3 − 4 3− 2 5 6 ? − = −5 5 1 −5 = −5 2

?

= −5 ?

= −5 = −5

Your time = Your friend’s time Your distance Friend’s distance = Your rate Friend’s rate

Labels:

Your distance = 24 Your rate = r Friend’s distance = 15 Friend’s rate = r − 6

Equation:

24 15 = r r −6 ⎛ 24 ⎞ ⎛ 15 ⎞ ( r ( r − 6)⎜ ⎟ = ⎜ ⎟r r − 6) ⎝ r ⎠ ⎝ r − 6⎠ 24( r − 6) = 15r 24r − 144 = 15r 9r = 144 r = 16 miles per hour

107.

d = kF 4 = k (100) 4 = k 100 1 = k 25 1 F 6 = 25

r − 6 = 10 miles per hour

109.

k r k 3 = 65 3(65) = k t =

t =

195 80

t ≈ 2.44 hours

195 = k

150 pounds = F

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test for Chapter 6

111.

ka p2

D =

113.

C = khw2

28.80 = k (16)(6)

k ( 20,000) 900 = 552

205

28.80 = k (576)

2

2 C = 0.05(14)(8)

C = $44.80

28.80 = k 576 0.05 = k

900(552 ) = k 20,000 136.125 = k 900 =

136.125( 25,000) p2

p2 =

136.125( 25,000) 900

p 2 = 3781.25 p = $61.49

Chapter Test for Chapter 6 1. f ( x) =

x +1 x2 − 6 x + 5

2 4. Least common multiple of x 2 , 3x3 , and ( x + 4) :

2 3x3 ( x + 4)

x2 − 6 x + 5 ≠ 0

( x − 1)( x − 5) ≠ 0 x −1 ≠ 0

x −5 ≠ 0

x ≠1

x ≠ 5

5.

Domain: ( −∞, 1) ∪ (1, 5) ∪ (5, ∞)

4z3 25 4 ⋅ z2 ⋅ z ⋅ 5 ⋅ 5 ⋅ = 2 5 12 z 5 ⋅ 4 ⋅ 3 ⋅ z2 5z = , z ≠ 0 3 2

4 − 2x −2( x − 2) = = −2, 2. x − 2 x − 2

3.

7.

=

( 2a + 3)( a − 4) 2a − 5a − 12 = 5a − 20 5( a − 4) 2a + 3 , a ≠ 4 = 5

(2 xy 2 ) 15

3

(2 xy 2 ) ⋅ 21 = 8 x3 y 6 ⋅ 7 ⋅ 3 = 14 y 6 , 12 x3 = 21 15 12 x3 5 ⋅ 3 ⋅ 4 ⋅ 3x3 15

4 , y + 4

y ≠ 2

3

÷

x ≠ 0

( 2 x − 3)( x + 1) 2x + 3 3 2 = ( 2 x − 3)( 2 x + 3) ⋅ = ( 2 x − 3) ( x + 1), x ≠ − , x ≠ −1 2x − x − 3 2x + 3 2 2

3 x − 2 3+ x − 2 x +1 + = = x −3 x −3 x −3 x −3

10. 2 x +

11.

y 2 + 8 y + 16 8 y − 16 ( y + 4) ⋅ 8( y − 2) ⋅ = 3 2 2( y − 2) 2( y − 2)( y + 4) ( y + 4) ( y + 4)

2

8. ( 4 x 2 − 9) ÷ 9.

x ≠ 2

6.

1 − 4 x2 2 x2 + 2 x 1 − 4 x2 −2 x 2 + 2 x + 1 ⎛ x + 1⎞ 1 − 4 x2 = 2 x⎜ = + = ⎟+ x +1 x +1 x +1 x +1 x+1 ⎝ x + 1⎠

5x 2 5x 2 5x ⎛ x − 3 ⎞ 2 5 x 2 − 15 x − 2 − 2 = − = − = ⎜ ⎟ ( x + 2)( x − 3) x + 2 x − x−6 x + 2 ( x − 3)( x + 2) x + 2 ⎝ x − 3 ⎠ ( x − 3)( x + 2)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

206

12.

Chapter 6

R ational Expressions, Equations, and Functions

3 5 2x 3 5 2x − 2 + 2 = − 2 + x x x + 2x + 1 x x ( x + 1)2 =

2 2 3 ⎡ x( x + 1) ⎤ 5 ⎡ ( x + 1) ⎤ 2x ⎛ x2 ⎞ − 2⎢ + ⎢ ⎜ ⎟ 2⎥ 2⎥ x ⎢⎣ x( x + 1) ⎥⎦ x ⎢⎣ ( x + 1) ⎥⎦ ( x + 1)2 ⎝ x 2 ⎠

=

3 x( x 2 + 2 x + 1) − 5( x 2 + 2 x + 1) + 2 x3 2 x 2 ( x + 1)

=

3 x3 + 6 x 2 + 3x − 5 x 2 − 10 x − 5 + 2 x3 2 x 2 ( x + 1)

=

5 x3 + x 2 − 7 x − 5 2 x 2 ( x + 1)

⎛ 3x ⎞ ⎜ ⎟ 3x 12 3x x 2 ( x + 2) x3 13. ⎝ x + 2 ⎠ = , ÷ 3 = ⋅ = 2 12 x + 2 x + 2x x+ 2 12 4 ⎛ ⎞ ⎜ 3 ⎟ ⎝ x + 2x2 ⎠ 1⎞ 1⎞ ⎛ ⎛ ⎜9x − ⎟ ⎜9x − ⎟ x x x⎠ ⋅ ⎝ ⎠ ⎝ 14. = ⎛1 ⎞ ⎛1 ⎞ x ⎜ − 3⎟ ⎜ − 3⎟ ⎝x ⎠ ⎝x ⎠ 1 9 x( x ) − ( x ) x = 1( ) x − 3( x) x

⎛3 ⎛3 1⎞ 1⎞ ⎜ 2 + ⎟ ⎜ 2 + ⎟ x2 y x + y x y⎠ x y⎠ ( ) 15. ⎝ = ⎝ ⋅ ⎛ 1 ⎞ ⎛ 1 ⎞ x 2 y( x + y ) ⎜ ⎟ ⎜ ⎟ ⎝x + y⎠ ⎝ x + y⎠ = =

9 x2 − 1 = 1 − 3x (3 x − 1)(3 x + 1) = −1( −1 + 3 x) = −(3 x + 1),

x ≠ 0,

x ≠ 0, − 2

1 3

3 y( x + y) + x 2 ( x + y) x2 y

(3 y + x 2 )( x + y) , x2 y

x ≠ −y

16.

6x2 − 4x + 8 6x2 4x 8 4 = − + = 3x − 2 + 2x 2x 2x 2x x

17.

2 x 4 − 15 x 2 − 7 x −3

3

20 6 26

−15

0

−7

18

9

27

3

9

20

2 x 4 − 15 x 2 − 7 20 = 2 x3 + 6 x 2 + 3x + 9 + x −3 x −3 t2

18.

+ 3−

t 4 + t 2 − 6t = t 2 − 2 t 4 + 0t 3 + t 2 − 6t + 0 t2 − 2 − 2t 2 t4 3t 2 − 6t − 6 3t 2 −6t + 6

6t − 6 t2 − 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test for Chapter 6

19.

3 1 = h + 2 6 ⎛ 3 ⎞ ⎛1⎞ 6( h + 2)⎜ ⎟ = ⎜ ⎟ 6( h + 2) ⎝ h + 2⎠ ⎝6⎠ 18 = h + 2

21.

1 1 2 + = 2 x +1 x −1 x −1 x −1+ x +1 = 2 2x = 2 x =1

16 = h 20.

1 1 2 Check: + ≠ 1+1 1−1 1−1

2 3 1 − = x +5 x +3 x 2 x( x + 3) − 3 x( x + 5) = ( x + 5)( x + 3) 2 x 2 + 6 x − 3 x 2 − 15 x = x 2 + 3 x + 5 x + 15

Division by zero is undefined. Solution is extraneous, so equation has no solution. 22.

−2 x 2 − 17 x − 15 = 0 2 x 2 + 17 x + 15 = 0

( 2 x + 15)( x + 1) = 0 2 x + 15 = 0

x = −

Check:

Check:

x = −1

15 2

?

= −

⋅

v = k

u

3 2

= k

36

3 2

= k ⋅6

3 2

= k

1 4

= k

v =

? 2 3 1 − = 15 15 15 − +5 − + 3 − 2 2 2 ? 12 10 2 − + = − 15 15 15 2 2 − = − 15 15

2 3 − −1 + 5 −1 + 3 2 3 − 4 2 1 3 − 2 2 −1

1 6

x +1 = 0

2 x = 15

1 1

?

= 1

207

23.

1 4

u

k V k 1= 180 180 = k P =

180 V 180 V = 0.75 V = 240 cubic meters

0.75 =

?

= −1 = −1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

C H A P T E R 7 Radicals and Complex Numbers Section 7.1

Radicals and Rational Exponents ......................................................209

Section 7.2

Simplifying Radical Expressions.......................................................211

Section 7.3

Adding and Subtracting Radical Expressions ...................................214

Mid-Chapter Quiz......................................................................................................217 Section 7.4

Multiplying and Dividing Radical Expressions ................................218

Section 7.5

Radical Equations and Applications..................................................223

Section 7.6

Complex Numbers..............................................................................229

Review Exercises .......................................................................................................232 Chapter Test .............................................................................................................237 Cumulative Test for Chapters 5–7..........................................................................238

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

C H A P T E R 7 Radicals and Complex Numbers Section 7.1 Radicals and Rational Exponents 1.

64 = 8 because 8 ⋅ 8 = 64.

39.

3. − 100 = −10 because 10 ⋅ 10 = 100. 5.

−25 is not a real number because no real number multiplied by itself yields −25.

7. −

−1 is not a real number because no real number

multiplied by itself yields − 4.

(103 )

3

27 = 3 because 33 = 27.

11.

3

−27 = −3 because −3 ⋅ −3 ⋅ −3 = −27. 3

13. − 1 = −1 because 1 = 1.

41. − 3

( 15 )

43. − 4 24 = −2

45. − 5 75 = − 7

(inverse property of powers and roots)

( 14 )

2

=

1. 16

23. 1728: Perfect cube because 123 = 1728. 25. 96: Neither because there is no integer that can be squared or cubed and yield 96.

29. l = 6 ft because

3

3

(5)

3

= 5

(index is odd) 35.

3

( − 7)

3

= −7

(index is odd) 37.

4

( −10)2 = −10 = 10 (index is even)

Rational Exponent Form

3

2563 4 = 64

256 = 64

51. x1 3 =

3

53. y 2 5 =

5

57. (33 )

23

x

(

55. 27 2 3 =

( y)

y 2 or 3

27

= ( 27)

59. 32−2 5 =

8 = 8 = 8 (index is even)

33.

49. Radical Form

2

31.

361 2 = 6

121 = 11. 216 = 6.

Rational Exponent Form

36 = 6

21. 49: Perfect square because 7 2 = 49.

27. l = 11 in. because

= − 15

47. Radical Form

17. 25 is a perfect square because 52 = 25.

is a perfect square because

3

(index is odd)

3

1 16

is not a real number

(even root of a negative number)

15. − 3 − 27 = 3 because ( − 3) = 27.

19.

2

(inverse property of powers and roots)

9.

3

−

(

)

2

32

)

2

= 32 = 9

23

1 5

5

=

2

(

=

3

27

)

2

= (3) = 9 2

1 1 = 2 2 4

Root is 5. Power is 2. 61.

( 278 )

23

=

( ) 3

8 27

2

=

( 23 )

2

=

4 9

Root is 3. Power is 2. 63. −361 2 = − 36 = −6

Root is 2. Power is 1. 65. ( − 27)

−2 3

=

1

(− 27)2 3

=

(

1 3

− 27

)

2

=

1 9

Root is 3. Power is 2.

209

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210

Chapter 7

R adicals and Complex Numbers 91. The square roots of 0.16 are −0.4 and 0.4 because (−0.4)(−0.4) = 0.16 and (0.4)(0.4) = 0.16.

67. x 3 x 6 = x ⋅ x 6 3 = x ⋅ x 2 = x3

Root is 3. Power is 6. 69. 71.

4

t5

x y = ( x y) 4

(x 4

14

3

y = (y

14

+ y)

93. The cube root of

t1 4 1 = t1 4 − 5 2 = t1 4 −10 4 = t −9 4 = 9 4 t5 2 t

=

3

4

73.

75.

t

12

y

1 4 ⋅1 2

= y

=

= ( x + y)

3 4 −1 4

= ( x + y)

24

= y

=

8

y

Root is 3. Power is 7. x

99. 101.

x 4

103. z 2

x + y

=

y3 ⋅

x1 2 1 = x1 2 − 3 2 = x −1 = 32 x x

3

y5 z 4 = z 2 ⋅ ( y5 z 4 )

12

2x + 9

= z 2 + 2 y5 2 = z 4 y5 2

2(0) + 9 =

9 = 3

(b) f (8) =

2(8) + 9 =

25 = 5

(c) f ( −6) =

2( −6) + 9 =

(d) f (36) = 5

y = y 3 4 ⋅ y1 3 = y 3 4 + 1 3

= z 2 y5 2 z 2

(a) f (0) =

79. f ( x ) =

3

= y 9 12 + 4 12 = y13 12

12

77. f ( x) =

2(36) + 9 =

105.

−3 = not a real number 107.

81 = 9

4

x3 =

4

x3 2 = ( x3 2 )

(3u − 2v)2 3 3 (3u − 2v)

x

Domain: ( − ∞, ∞) 81. f ( x ) = 3 x, x ≥ 0

13

87. The rule of exponents used to write x is the product rule of exponents.

( )( ) =

9 16

and

− 2v)

= (3u − 2v)

4 6−9 6

= (3u − 2v)

−5 6

1

(3u − 2v)

56

18

⎛1⎞ =1−⎜ ⎟ ⎝ 3⎠

≈ 0.128 ≈ 12.8%

Area = Side ⋅ Side Area = 529 Side = x

85. The principal cube root of x represented in radical form is 3 x and rational exponent form is x1 3.

89. The square roots of

23 32

2 3−3 2

Labels:

)

− 2v)

= x3 2 ⋅1 4 = x3 8

= (3u − 2v)

111. Verbal Model:

x ≥ − 92

Domain: ⎡⎣− 92 , ∞

(3u (3u

⎛ 25,000 ⎞ 109. r = 1 − ⎜ ⎟ ⎝ 75,000 ⎠

2 x ≥ −9

− 34

=

18

2x + 9

2x + 9 ≥ 0

− 34

14

=

Domain: [0, ∞) 83. h( x) =

1 . (101 )(101 )(101 ) = 1000

97. u 2 3 u = u 2 ⋅ u1 3 = u 2 + 1 3 = u 7 3

18

= ( x + y) =

1 because is 10

95. The cube root of 0.001 is 0.1 because (0.1)(0.1)(0.1) = 0.001.

34 14

( x + y )3 4 ( x + y )1 4

34

x + y

)

= x

1 1000

9 16

are − 34 and

( )( ) = 3 4

3 4

9 . 16

3 4

⋅ x

34

Equation:

529 = x ⋅ x 529 = x 2 529 = x

1 3+ 3 4

as x

23 = x 23 feet × 23 feet

because

113. If a and b are real numbers, n is an integer greater than or equal to 2, and a = b n , then b is the nth root of a.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 7.2 115. No. 2 is an irrational number. Its decimal representation is a nonterminating, nonrepeating decimal.

121.

117. (a) “Last digits:”

Simplifying Radical Expressions

2 5 = u + 4 8 ⎛ 2 ⎞ ⎛5⎞ 8(u + 4)⎜ ⎟ = ⎜ ⎟ 8(u + 4) ⎝u + 4⎠ ⎝8⎠ 16 = 5(u + 4)

1 (Perfect square 81)

16 = 5u + 20

4 (Perfect square 64) 5 (Perfect square 25)

−4 = 5u

6 (Perfect square 36)

−

9 (Perfect square 49) 0 (Perfect square 100)

4 = u 5

123. s = kt 2

(b) Yes, 4,322,788,986 ends in a 6, but it is not a perfect square. 119.

211

125. a = kbc

a a −3 = 5 2 10

⎛a⎞ ⎛ a − 3⎞ 10⎜ ⎟ = ⎜ ⎟ 5 ⎝ ⎠ ⎝ 2 ⎠

2a = 5( a − 3) 2a = 5a − 15 −3a = −15 a = 5

Section 7.2 Simplifying Radical Expressions 1.

28 =

4⋅7 =

4 ⋅

7 = 2 7

23.

120 x 2 y 3 =

4 ⋅ 30 ⋅ x 2 ⋅ y 2 ⋅ y = 2 x y 30 y

25.

192a 5b 7 =

64 ⋅ 3 ⋅ a 4 ⋅ a ⋅ b 6 ⋅ b = 8a 2b3

The figure supports the answer. For a square, A = s 2 , so

A = s.

A =

So, the figure shows

28 = s = 2 7.

8 =

5.

18 =

9⋅2 =

32 ⋅ 2 = 3 2

7.

45 =

9⋅5 =

32 ⋅ 5 = 3 5

9.

96 =

16 ⋅ 6 =

11.

153 =

3

48 =

29.

3

112 =

31.

4

x7 =

4

x 4 x3 =

4

x4 ⋅

4

x3 = x 4 x3

33.

4

x6 =

4

x4 x2 =

4

x4 ⋅

4

x2 = x

35.

3

40 x5 =

37.

4

324 y 6 =

16 ⋅ 3 =

24 ⋅ 3 = 2 3 3 ⋅ 2 = 2 3 6

2

3.

4⋅2 =

3

3

27.

3ab

2 ⋅2 = 2 2

9 ⋅ 17 =

42 ⋅ 6 = 4 6

3

8 ⋅ 14 =

3

3

8⋅

3

14 = 2 3 14

4

x2

8 ⋅ 5 ⋅ x3 ⋅ x 2 = 2 x 3 5 x 2

2

3 ⋅ 17 = 3 17 2

13.

1183 =

169 ⋅ 7 =

15.

4 y2 =

22 y 2 =

17.

9 x5 =

32 x 4 ⋅ x = 3 ⋅ x 2 ⋅

19.

48 y 4 =

21.

117 y 6 =

13 ⋅ 7 = 13 7 22 ⋅

y2 = 2 y

16 ⋅ 3 ⋅ y 4 = 4 y 2

9 ⋅ 13 ⋅ y 6 = 3 y 3

x = 3x 2 3

13

81 ⋅ 4 ⋅ y 4 ⋅ y 2

4

= 3y

4

4 y2

= 3y

4

22 y 2

= 3y

x 39.

3

x4 y3 =

41.

4

4x4 y6 =

43.

5

32 x5 y 6 =

3

2y

x3 ⋅ x ⋅ y 3 = xy 3 x

4 x 4 ⋅ y 4 ⋅ y 2 = xy

4

5

4

4 y2

25 ⋅ x5 ⋅ y 5 ⋅ y = 2 xy 5 y

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

212

Chapter 7 16 = 9

45.

47.

3

49.

3

39 2 y = 3

32a 4 = b2

5

13 y

16 ⋅ 2 ⋅ a 4 b2

32 x 2 = y5

55.

3 3

57.

1 = 7

1 ⋅ 7

7 = 7

7 7

4 4

6 = 61. 3 32 =

1

63.

y

67.

1

x 2 6

69.

71.

73.

3b3 3

3

⋅

22

4

22

4

22

3

3

2 ⋅2 6

36 + 9

=

45

=

9⋅5

= y

13

= 3 5

77. c =

a 2 + b2

=

(4 2 )

=

32 + 49

=

81

2

+ 72

79. Verbal Model:

Hypotenuse

Labels:

3

5 ⋅ 22 4

=

24

4

3

3 3

2x ⋅ 3y

24 x3 y 4 = 25 z = =

3

y2

2

Hypotenuse = c

c 2 = 262 + 102

Equation:

c 2 = 676 + 100 c =

If a and b are real numbers, variables, or algebraic expressions, and if the nth roots of a and b are real, then n ab = n a ⋅ n b .

y y

=

Example:

3

3 y

3

2

2

3 y

=

3

2

2x ⋅ 3 y 3

25 z

2 xy 3 3 y ⋅ 3

3

53 z 3

3

27 ⋅

3

4 = 33 4

• No radical contains a fraction.

5z 2

3

3 y

(23 )(3)( x3 )( y3 )( y) 3

108 =

• All possible nth-powered factors have been removed from each radical.

3b 6 3b 2 3b = = 2 b2 3b 3b 2

3

83. Three conditions that must be true for a radical expression to be in simplest form:

2 2 = 2 x 2

2

776 ≈ 27.86 feet

81. Product Rule for Radicals:

x 2 x = x x

2 ⋅ x ⋅

x 2

y

=

y

1

2

= Leg 1 + Leg 2

Leg 2 = 10

20 2

2 63 2 63 2 33 2 = = = 3 3 4 2 2 2 2

y

⋅

2

Leg 1 = 26

4

=

6

c 2 = 776

6 = ⋅ b 3b

2x = 3y

=

3

2

⋅

2 3 22

4 = x

=

62 + 32

6

y

4 = x

65.

5

1

=

=

= 9

3 = 3

59.

a 2 + b2

25 x 2 2 = 5 x2 y5 y

5

1 ⋅ 3

5 = 4

2

4a 2 2 b

=

1 = 3

4

75. c =

35 4

=

3

51.

53.

16 4 = 3 9

35 = 64 39 y 2

R adicals and Complex Numbers

⋅

3

2

=

3

5z 2

3

5z 2

• No denominator of a fraction contains a radical.

3

18 xy 3y

2

85.

0.04 =

87.

0.0072 =

4 ⋅ 0.01 =

=

4

0.01 = 2 ⋅ 0.1 = 0.2

36 ⋅ 2 ⋅ 0.0001 36 ⋅

2 ⋅

0.0001

= 6 ⋅ 0.01 2 = 0.06 2 89.

60 = 3

20 =

4⋅5 =

22 ⋅ 5 = 2 5

2 xy 3 15 yz 2 5z

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 7.2 13 = 25

91.

93.

3

103.

13 5

54a 4 = b9

4 × 10− 4 =

95.

107. Because

4 ⋅

=

97.

3

0.0001

3

2.4 × 10 = =

3

u ⋅

24 × 10

5

(23 )(3)(103 )(102 )

u and

4

u = u1 3 ⋅ u1 4 = u 4 12 + 3 12 = u 7 12 =

12

u7 .

3

2x + 3y = 12 2

3 ⋅ 102

−2 −2

400 × 10 5

6

6

10 12

4x − y = 10

−8 −10

≈ 8.9443 × 101

111. ⎧ y = x + 2 ⎨ ⎩y − x = 8

≈ 89.443 ≈ 89.44 cycles per second 101. (a) Verbal Model: Hypotenuse

( x + 2) − x = 8 2

30 2

2

= Leg 1 + Leg 2

Hypotenuse = c Leg 1 =

x 4

−4 −6

Labels:

3

y

= 20 3 300

1 100

u = u1 4 , when

Solution: (3, 2)

= 2 ⋅ 10 ⋅

99. f =

4

y = − 23 x + 4 109. ⎪⎧2 x + 3 y = 12 3 y = −2 x + 12 ⇒ ⇒ ⎨ − y = −4 x + 10 y = 4 x − 10 ⎪⎩4 x − y = 10

1 100

1 50 3

6

u = u1 3 and

u are multiplied, the rational exponents need to be added together. Therefore,

= 2 ⋅ 0.01 = 2⋅

3

4

4 ⋅ 0.0001

=

8 is not in simplest form because the perfect square factor 4 needs to be removed from the radical.

105. The display appears to be approaching 1.

33 ⋅ 2 ⋅ a 3 ⋅ a 3a = 3 3 2a b9 b

3

213

Simplifying Radical Expressions

2

2 ≠ 8 No solution

= 15

Leg 2 = 8 Equation:

c 2 = 152 + 82 c 2 = 225 + 64 c 2 = 289 c =

289

c = 17 ft (b) Area = 2 ⋅ 17 ⋅ 40 = 1360 sq ft

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

214

Chapter 7

R adicals and Complex Numbers

113. ⎧ x + 4 y + 3 z = 2 ⎪ ⎨ 2 x + y + z = 10 ⎪ ⎩− x + y + 2 z = 8

x + 4 y + 3z = 2 −7 y − 5 z = 6 5 y + 5 z = 10 x + 4 y + 3z = 2

−7( −8) − 5 z = 6

x + 4(−8) + 3(10) = 2

−7 y − 5 z = 6

56 − 5 z = 6

x − 32 + 30 = 2

−2 y

= 16

−5 z = −50

y

= −8

z = 10

x − 2 = 2 x = 4

(4, − 8, 10)

Section 7.3 Adding and Subtracting Radical Expressions 1. 3 2 −

3. 4 3 y + 9 3 y = 13 3 y 9. 2 2 + 5 2 − 11.

4

15. 7

7. 8 2 + 6 2 − 5 2 = (8 + 6 − 5) 2 = 9 2

2 + 3 2 = (2 + 5 − 1 + 3) 2 = 9 2

5 − 6 4 13 + 3 4 5 −

13. 9 3 7 −

7 + 3 7 − 2 7 = (1 + 3 − 2) 7 = 2 7

5.

2 = 2 2

4

13 = (1 + 3) 4 5 + (−6 − 1) 4 13 = 4 4 5 − 7 4 13

3 + 4 3 7 + 2 3 = (9 3 7 + 4 3 7 ) + ( −

3 + 2 3 ) = 13 3 7 +

x + 5 3 9 − 3 x + 3 3 9 = (5 + 3) 3 9 + (7 − 3)

x = 83 9 + 4

3

x

17. 3 4 x + 5 4 x − 3 x − 11 4 x = (3 + 5 − 11) 4 x − 3 x = − 3 4 x − 3 x 23.

19. 8 27 − 3 3 = 8 9 ⋅ 3 − 3 3

= 8(3) 3 − 3 3

16 x −

9x = 4

x −3 x =

25. 5 9 x − 3 x = 15

= 24 3 − 3 3

21. 3 45 + 7 20 = 3 9 ⋅ 5 + 7 4 ⋅ 5

29. 6

x3 − x 9 x = 6 x

31. 9

40 − 2 x4

= 3(3) 5 + 7( 2) 5 = 9 5 + 14 5

x − 3 x = 12

x

18 y + 4 72 y = 3 2 y + 24 2 y = 27 2 y

27.

= 21 3

x

x − 3x

x = 3x

x

90 18 10 6 10 12 10 = − = x4 x2 x2 x2

= 23 5 33.

48 x5 + 2 x 27 x3 − 3 x 2

3x = 4 x 2

3x + 6 x 2

35. 2 3 54 + 12 3 16 = 2 3 27 ⋅ 2 + 12 3 8 ⋅ 2

3x − 3x 2

3x = 7 x 2 37.

3

6 x4 +

3x 3

48 x =

3

6 ⋅ x3 ⋅ x +

3

= 6 3 2 + 24 3 2

= x 6x + 2 6x

= 30 3 2

= ( x + 2) 3 6 x

3

8⋅6⋅ x

3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 7.3

Adding and Subtracting Radical Expressions

215

39. 5 3 24u 2 + 2 3 81u 5 = 5 3 23 ⋅ 3u 2 + 2 3 33 ⋅ 3u 3 ⋅ u 2 = 5( 2) 3 3u 2 + 2(3u ) 3 3u 2 = 10 3 3u 2 + 6u 3 3u 2 = (10 + 6u ) 3 3u 2

41.

3

3x 4 +

43.

3

6x 1 + 4 3 2x = 24 2

45. h = 47.

3

3

81x +

x +

5 −

3

3

24 x 4 = x 3 3 x + 3 3 3 x + 2 x 3 3 x = (3 x + 3) 3 3 x

8x =

3

6x 1 + 4 3 2x = 3 2

3

x + 2 3 x = 3 3 x ft

⎛ 3 5 −⎜ ⋅ ⎝ 5

3 = 5

5⎞ ⎟ = 5⎠

3

5 −

2x + 4 3 2x =

3 5 5

9 2

3

2x

55.

7 y3 −

9 = 7 y3

3⎞ ⎛ = ⎜1 − ⎟ 5 5⎠ ⎝ 2 = 5 5 49.

32 +

1 = 2

= 4 2 +

51.

18 y −

7 y2 ⋅ y

= y 7y −

3 ⋅ y 7y

= y 7y −

3 7y y ⋅ 7y

= y 7y −

3 7y 7 y2

7y 7y

=

7 y2 ⎛ y 7 y ⎞ 3 7 y ⎜ ⎟ − 7 y2 ⎝ 1 ⎠ 7 y2

=

8 2 1 2 + 2 2

=

=

9 2 2

7 y3 7 y 3 7y − 7 y2 7 y2

=

7 y (7 y 3 − 3) 7 y2

y = 3 2y − 2y

= =

2 + 3x

2 2

2 2

= 3 2y −

53.

1 ⋅ 2

16 ⋅ 2 +

3

7 y2 ⋅ y −

3x =

5y

y

2y

2y 2y

=

54 x + 9 ⋅ 6x +

96 x +

150 x

16 ⋅ 6 x +

25 ⋅ 6 x

= 12 6 x 59. P =

5 2y 2

2 ⋅ 3x

57. P =

= 3 6x + 4 6x + 5 6x

2y 2y

175 x +

63 x +

252 x +

112 x

= 5 7x + 3 7x + 6 7x + 4 7x = 18 7 x

3x + 3x

2 3x 3x 3x + 3x 3x (3 x + 2) 3 x = 3x =

2y

y ⋅ 2y

3x

61. P = x 128 +

162 x +

200 x

= 8 x 2 + 9 2 x + 10 2 x = 19 2 x + 8 x 2

63. P = =

12 x + 4 ⋅ 3x +

48 x +

27 x +

16 ⋅ 3 x +

75 x

9 ⋅ 3x +

25 ⋅ 3 x

= 2 3x + 4 3 x + 3 3x + 5 3 x = ( 2 3x + 3 3x ) + (4 3 + 5 3 ) x = 5 3 x + (9 3 ) x = 9 x 3 + 5 3x

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

216

Chapter 7

R adicals and Complex Numbers

65. (a) T = G + S

) (

(

= − 3856 + 102,201 t − 58,994t + 8854 t 3 + 2,249,527 − 2,230,479 t + 742,197t − 82,167 t 3

)

= 2,245,671 − 2,127,778 t + 683,203t − 73,313 t 3

(b) When t = 9: T = 2,245,671 − 2,127,778 9 + 683,203(9) − 73,313 93 ≈ 31,713 people

67. The expression that was written in simplest form is 6 6 x + 6 x + 2 x 6 − 3 6. 69. The like radicals in 6 6 x +

are 6 6x and 71. 3 x + 1 + 10 73.

3

16t 4 −

3

77.

9x − 9 +

= 3 x −1+ = 4

6x + 2x 6 − 3 6

6 x , and 2 x 6 and − 3 6.

x3 − x 2 +

79.

3

23 ⋅ 2t 3 ⋅ t −

= x 3

33 ⋅ 2t 3 ⋅ t

x 2 ( x − 1) + x −1 + 2

4( x − 1) x −1

81. 2 3 a 4b 2 + 3a 3 ab 2 = 2 3 a 3 ⋅ a ⋅ b 2 + 3a 3 ab 2

= −t 3 2t 5a + 2 45a 3 =

x −1

= ( x + 2) x − 1

= 2t 3 2t − 3t 3 2t

75.

x −1

x −1

4x − 4 =

x + 1 = 13 x + 1

54t 4 =

9( x − 1) +

x −1 =

= 2a 3 ab 2 + 3a 3 ab 2 = 5a 3 ab 2

5a + 2 9 ⋅ 5 ⋅ a 2 ⋅ a

=

5a + 2(3)( a ) 5a

=

5a + 6a 5a

= (1 + 6a ) 5a = (6a + 1) 5a 4r 7 s 5 + 3r 2

83.

r 3s 5 − 2rs

r 5s3 =

4r 6 ⋅ r ⋅ s 4 ⋅ s + 3r 2

= 2r 3 s 2 3 2

= 3r s 85.

3

128 x9 y10 − 2 x 2 y 3 16 x3 y 7 =

3

rs + 3r 3 s 2

r 2 ⋅ r ⋅ s 4 ⋅ s − 2rs

rs − 2r 3s 2

r 4 ⋅ r ⋅ s2 ⋅ s

rs

rs

64 ⋅ 2 ⋅ x9 ⋅ y 9 ⋅ y − 2 x 2 y 3 8 ⋅ 2 ⋅ x3 ⋅ y 6 ⋅ y

= 4 x3 y 3 3 2 y − 4 x3 y 3 3 2 y = 0 87.

7 +

18 >

7 + 18

89. 5
4 105.

x +

y −

z =

1

2

3

4

5

6

7

⎡ 1 1 −1 # 4⎤ ⎡ 1 1 −1 # 4⎤ ⎢ ⎥ −2 R1 + R2 ⎢ ⎥ 2 x + y + 2 z = 10 ⇒ ⎢2 1 2 # 10⎥ ⎢0 −1 4 # 2⎥ − R + R 1 3⎢ ⎢ 1 −3 − 4 # − 7 ⎥ ⎥ x − 3 y − 4 z = −7 ⎣ ⎦ ⎣0 − 4 −3 # −11⎦ 4

4⎤ 3 # 6⎤ ⎡ 1 1 −1 # ⎡1 0 ⎢ ⎥ − R2 + R1 ⎢ ⎥ 1 1 −4 # −2⎥ − R2 ⎢0 ⎢0 1 − 4 # −2⎥ − 19 R3 4 R + R 2 3 ⎢ ⎥ ⎢ ⎥ ⎣0 −4 −3 # −11⎦ ⎣0 0 −19 # −19⎦ 3 # 6⎤ ⎡1 0 ⎡ 1 0 0 # 3⎤ ⎢ ⎥ −3R3 + R1 ⎢ ⎥ − − # 0 1 4 2 ⎢ ⎥ 4 R + R ⎢0 1 0 # 2⎥ 3 2 ⎢0 0 ⎢0 0 1 # 1⎥ 1 # 1⎥⎦ ⎣ ⎣ ⎦

x = 3 y = 2

(3, 2, 1)

z =1

107. 5 3 − 2 3 = (5 − 2) 3 = 3 3 109. 16 3 y − 9 3 x = 16 3 y − 9 3 x

Radical expressions are not alike so cannot be combined. 111.

16m 4 n3 + m m 2 n = 4m 2 n n + m 2 n = ( 4m 2 n + m 2 ) n = ( 4n + 1)m 2

n

Section 8.2 Completing the Square 1. x 2 + 8 x + 16 ⎡16 = ⎢⎣

( 82 ) ⎤⎥⎦ 2

3. y 2 − 20 y + 100

( )

⎡100 = − 20 2 ⎤ 2 ⎥ ⎢⎣ ⎦

5. x 2 + 14 x + 49 ⎡49 = ⎢⎣

(142 ) ⎤⎥⎦ 2

25 4

7. t 2 + 5t + ⎡ 25 = ⎢⎣ 4

( 52 ) ⎤⎥⎦ 2

9. x 2 − 9 x +

( )

11. a 2 − 13 a +

1 36

( )( 12 ) ⎤⎦⎥

⎡ 1 = −1 3 ⎣⎢ 36 13. y 2 +

⎡16 = ⎣⎢ 25

8y 5

+

2

16 25

( 12 ⋅ 85 ) ⎤⎦⎥ 2

15. r 2 − 0.4r + 0.04 2 ⎡ ⎛ 0.4 ⎞ ⎤ ⎢0.04 = ⎜ − ⎟ ⎥ ⎝ 2 ⎠ ⎥⎦ ⎢⎣

17. (a) x 2 − 20 x + 100 = 100

(x

− 10) = 100 2

x − 10 = ±10

81 4

⎡ 81 = − 9 ⎤ 2 ⎥ ⎢⎣ 4 ⎦

x = 10 ± 10 x = 20, 0

2

(b) x 2 − 20 x = 0 x( x − 20) = 0 x = 0, 20

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

250

Chapter 8

Quadratic Equations, Functions, and Inequalities

19. (a) x 2 + 6 x + 9 = 0 + 9

(x

27. (a) x 2 − 3 x +

+ 3) = 9 2

( x − 32 )

x + 3 = ±3

x −

x = −3 ± 3 x = −6, 0

(b) x 2 + 6 x = 0 (b)

x +6 = 0

2

3 2

25 4

( y − 52 )

2

5 2

y −

3 2

x =

12 , 2

29. (a)

25 4

=

25 4

=

± 52 5 2

x2 + 8x + 7 = 0

(x

2

x + 4 = ±3

5 2

±

+ 4) = 9 x = −4 ± 3 x = −1, − 7

(b)

x2 + 8x + 7 = 0

(x

y = 0,

x +1 = 0

y −5 = 0

+ 1)( x + 7) = 0

− 4) = 9 2

x 2 − 10 x + 25 = −21 + 25

(x

t − 4 = ±3 t = 4± 3

− 5) = 4 2

x − 5 = ±2

t = 7, 1

x = 5± 2 x = 7, 3

t 2 − 8t + 7 = 0 − 7)(t − 1) = 0

(b) x 2 − 10 x + 21 = 0

(x

t =1 49 4

(x + ) 7 2

2

x +

7 2

49 4

= −12 + =

1 4

− 72

±

1 2

x = − 62 , − 82 x = −3, − 4 (b)

2

x + 7 x + 12 = 0

(x

− 7)( x − 3) = 0

x −7 = 0

x −3 = 0

x = 7

x = 3

33. x 2 − 4 x − 3 = 0

= ± 12

x =

x = −7

31. (a) x 2 − 10 x + 21 = 0

23. (a) t 2 − 8t + 16 = −7 + 16

25. (a) x 2 + 7 x +

x + 7 = 0

x = −1

y = 5

t = 7,

− 62

− 6)( x + 3) = 0

y( y − 5) = 0

(t

9 2

x 2 + 8 x + 16 = −7 + 16

y2 − 5 y = 0

(b)

±

x − 3x − 18 = 0

= 0, 5

(t

= ± 92

x = 6, −3

=

y =

(b)

81 4

x =

y2 − 5 y = 0 y2 − 5 y +

=

2

(x

x = −6, 0 21. (a)

9 4

= 18 +

x = 6, − 3

x ( x + 6) = 0

x = 0,

9 4

x2 − 4x + 4 = 3 + 4

(x

− 2) = 7 2

x − 2 =

7

x = 2±

7

x ≈ 4.65, − 0.65

+ 4)( x + 3) = 0

x = − 4,

x = −3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 8.2 35. x 2 + 4 x − 3 = 0

43.

x2 + 4 x + 4 = 3 + 4

(x

+ 2) = 7 2

x + 2 = ± 7 x = −2 ±

2

1⎞ 127 ⎛ ⎜x − ⎟ = 3⎠ 9 ⎝

7

x −

37. 3x 2 + 9 x + 5 = 0

1 127 = ± 3 9 x =

9 5 9 x 2 + 3x + = − + 4 3 4

x =

2

3⎞ −20 + 27 ⎛ ⎜x + ⎟ = 2⎠ 12 ⎝

x =

2

3⎞ 7 ⎛ ⎜x + ⎟ = 2⎠ 12 ⎝ 7 ⋅ 12

3 x = − ± 2

3 3

2

2

5⎞ 17 ⎛ ⎜x + ⎟ = 2⎠ 4 ⎝ 5 17 = ± 2 4 5 17 ± 2 2 −5 ± 17 x = 2 x ≈ −0.44, − 4.56 x = −

5 ⋅ 2

2 2

10 2 x ≈ −0.42, − 3.58

47. z 2 + 4 z + 13 = 0

z 2 + 4 z + 4 = −13 + 4

(z

41. 4 y 2 + 4 y − 9 = 0 1 9 1 = + 4 4 4

+ 2) = −9 2

z + 2 = ± −9 z = −2 ± 3i

2

1⎞ 10 ⎛ ⎜y + ⎟ = 2⎠ 4 ⎝

49.

1 ± 2

x 2 − 4 x = −9 x 2 − 4 x + 4 = −9 + 4

1 10 = ± 2 4 10 2

−1 ± 10 2 y ≈ 1.08, − 2.08 y =

≈ −3.42

2

5 2

y = −

127 3

−8 + 25 5⎞ ⎛ ⎜x + ⎟ = 2 4 ⎝ ⎠

x +

x = −2 ±

y +

1−

≈ 4.09

x2 + 5x + 2 = 0 25 25 = −2 + x2 + 5x + 4 4

21 6

3 + 4 2

x + 2 = ±

y2 + y +

3

0.1x + 0.5 x + 0.2 = 0

39. 2 x 2 + 8 x + 3 = 0

+ 2) =

127 3 127

2

−9 ± 21 6 x ≈ − 0.74, − 2.26

(x

1 ± 3 1+

0.1x 2 + 0.5 x = −0.2

45.

x =

x2 + 4x + 4 = −

251

2⎞ ⎛ x⎜ x − ⎟ = 14 3⎠ ⎝ 2 1 1 2 = 14 + x − x + 3 9 9

x ≈ 0.65, − 4.65

3 = ± x + 2

Completing the Square

(x

− 2) = −5 2

x − 2 = ± −5 x = 2±

5i

x ≈ 2 + 2.24i, 2 − 2.24i

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

252

Chapter 8

Quadratic Equations, Functions, and Inequalities 57. Verbal Model:

51. − x 2 + x − 1 = 0

Volume = Length ⋅ Width ⋅ Height

x2 − x + 1 = 0 x2 − x +

Labels:

1 1 = −1 + 4 4

Length = x + 4 Width = x

2

Height = 6

1⎞ 3 ⎛ ⎜x − ⎟ = − 2 4 ⎝ ⎠ 1 x − = ± 2

840 = ( x + 4)( x )(6)

Equation:

3 − 4

140 = x( x + 4) 140 = x 2 + 4 x

1 3 x = i ± 2 2 x =

1±

3

0 = x 2 + 4 x − 140 0 = ( x + 14)( x − 10)

i

2 x ≈ 0.5 + 0.87i

x + 14 = 0

x ≈ 0.5 − 0.87i

Thus, the dimensions are: length = 10 + 4 = 14 inches, width = x = 10 inches, height = 6 inches

3 9 −1 9 = + z + 4 64 2 64 2

3⎞ 23 ⎛ ⎜z − ⎟ = − 8 64 ⎝ ⎠ z −

59. A perfect square trinomial is one that can be written as

(x

3 23 = ± − 8 64 z =

3 ± 8

23 i 8

z =

3 + 8

23 i ≈ 0.38 + 0.60i 8

3 − 8

23 i ≈ 0.38 − 0.60i 8

z =

+ k) . 2

61. Each side of the equation must be divided by 2 to obtain a leading coefficient of 1. The resulting equation is the same by the Multiplication Property of Equality. 63.

0.2 x 2 + 0.1x = −0.5 x2 +

1 1 −5 1 = + x + 2 16 2 16 2

1⎞ 39 ⎛ ⎜x + ⎟ = − 4⎠ 16 ⎝

55. (a) Area of square = x ⋅ x = x 2

x +

Area of vertical rectangle = 4 ⋅ x = 4 x Area of horizontal rectangle = 4 ⋅ x = 4 x Total area = x 2 + 4 x + 4 x = x 2 + 8 x (b) Area of small square = 4 ⋅ 4 = 16 2

Total area = x + 8 x + 16 (c)

(x

x = 10

Not a solution

53. 4 z 2 − 3z + 2 = 0

z2 −

x − 10 = 0

x = −14

+ 4)( x + 4) = x 2 + 8 x + 16

65.

1 39 = ± − 4 16 x = −

1 ± 4

39 i 4

x = −

1 + 4

39 i ≈ −0.25 + 1.56i 4

x = −

1 − 4

39 i ≈ −0.25 − 1.56i 4

x 1 − =1 x 2 1⎞ ⎛x 2 x⎜ − ⎟ = (1)2 x 2 x⎠ ⎝ x2 − 2 = 2 x x2 − 2x + 1 = 2 + 1

(x

− 1) = 3 2

x −1= ± 3 x =1±

3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 8.2 x2 x +3 = 8 2

67.

(

73.

R = x 80 − 2750 = 80 x −

⎛ x2 ⎞ ⎛ x + 3 ⎞ 8⎜ ⎟ = ⎜ ⎟8 ⎝8⎠ ⎝ 2 ⎠

0 = x − 50

2

x = 6, − 2

75. Use the method of completing the square to write the quadratic equation in the form u 2 = d . Then use the Square Root Property to simplify.

2x + 1 = x − 3 2

= ( x − 3)

x = 110

50 pairs, 110 pairs

x − 2 = ±4

)

2

77. (a) d = 0

2x + 1 = x2 − 6x + 9

(b) d > 0, and d is a perfect square.

2

0 = x − 8x + 8

(c) d > 0, and d is not a perfect square.

2

16 − 8 = x − 8 x + 16 8 = ( x − 4)

(d) d < 0

2

79. 3 5

± 8 = x−4

500 = 3 ⋅

4±2 2 = x 71. Verbal Model:

(

2 3−

(

2

81. 3 + Area = Length ⋅ Width

Labels:

83. 3 +

Length = x

)( )

2

)

2 = 32 −

( 2)

2

= 9− 2 = 7

( 2) + ( 2)

= 32 + 2(3)

2

= 11 + 6 2

⎡ ⎛ 200 − 4 x ⎞⎤ 1400 = 2 ⎢ x ⋅ ⎜ ⎟⎥ 3 ⎠⎦ ⎣ ⎝

85.

⎡ 200 4 x2 ⎤ x − 1400 = 2 ⎢ ⎥ 3 ⎦ ⎣ 3

87. Product Rule:

1400 =

25 ⋅ 100 = 3 ⋅ 5 ⋅ 10 = 150

= 9+ 6 2 + 2

200 − 4 x Width = 3 Equation:

5 ⋅ 500

= 3⋅

8 = x

4±

x − 110 = 0

50 = x

− 2) = 16

2x + 1

1 x2 2

0 = ( x − 50)( x − 110)

x 2 − 4 x + 4 = 12 + 4

(

)

0 = x 2 − 160 x + 5500

x 2 = 4 x + 12

69.

1x 2

253

0 = − 12 x 2 + 80 x − 2750

x 2 = 4( x + 3)

(x

Completing the Square

8 = 10

8 ⋅ 10

10 8 10 4 10 = = 10 5 10 ab =

a ⋅

b

400 x 8 x 2 − 3 3

4200 = 400 x − 8 x 2 2

8 x − 400 x + 4200 = 0 x 2 − 50 x + 525 = 0

(x

− 35)( x − 15) = 0

x − 35 = 0 x = 35 meters 200 − 4 x = 20 meters 3

x − 15 = 0 x = 15 meters 200 − 4 x 2 = 46 meters 3 3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

254

Chapter 8

Quadratic Equations, Functions, and Inequalities

Section 8.3 The Quadratic Formula 9. x 2 − 2 x − 4 = 0

2x2 = 7 − 2x

1.

2 x2 + 2x − 7 = 0

x =

x(10 − x) = 5

3.

10 x − x 2 = 5

x =

−( −2) ±

x =

x 2 − 10 x + 5 = 0 2

5. (a) x − 11x + 28 = 0

x = x =

(−11)2

11 ±

− 4(1)( 28)

2(1)

2±

4 + 16 2

2±

20

11 ±

121 − 112 2

11 ± 9 2 11 ± 3 x = 2 x = 7, 4

(x

t =

x −7 = 0

x − 4 = 0

x = 7

x = 4

t = t =

7. (a) x 2 + 6 x + 8 = 0

x = x =

6±

62 − 4(1)(8) 2(1) 36 − 32 2

−6 ± 4 2 −6 ± 2 x = 2 x = −2, − 4

(x

x = −4

t =

5

42 − 4(1)(1) 2(1)

−4 ± −4 ±

16 − 4 2

4±

12 2

−4 ± 2 3 2

(

2 −2 ±

3

)

2

t = −2 ±

x = x =

+ 4)( x + 2) = 0

x + 4 = 0

)

2

3

13. − x 2 + 10 x − 23 = 0

x =

(b)

5

x =1±

t =

− 7)( x − 4) = 0

−6 ±

(

21 ±

11. t 2 + 4t + 1 = 0

x =

(b)

− 4(1)( −4)

2

2± 2 5 x = 2 x =

2

2(1)

2

− x + 10 x − 5 = 0

( − 2)

102 − 4(−1)( − 23) 2( −1)

−10 ±

100 − 92 −2

x =

−10 ± −2

x =

−10 ± 2 2 −2

x + 2 = 0 x = −2

−10 ±

x =

(

−2 5 ±

x = 5±

8

2

)

−2 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 8.3

x =

82 − 4(16)(1) 2(16)

−8 ± −8 ±

64 − 64 32

5y = 0

2

(x

4x + 1 = 0

4 x = −1

4 x = −1

1 x = − 4

1 x = − 4

−3 ±

9 − 24 4

−3 ±

−15

(y

x =

21 9 , 5 5

6 5

±

− 18)( 2 y + 3) = 0 2y + 3 = 0 2 y = −3 y = − 32

35. x 2 + 8 x + 25 = 0 + 4) = − 9 2

x + 4 = ±

−9

x = − 4 ± 3i

19. b − 4ac = 1 − 4(1)(1) = 1 − 4 = −3

37. 3x 2 − 13 x − 169 = 0

2

2 distinct complex solutions 21. b 2 − 4ac = ( −2) − 4(8)( −5) = 4 + 160 = 164 2

Two distinct rational solutions 23. b 2 − 4ac = (−24) − 4(9)(16) = 576 − 576 = 0 2

1 repeated rational solution 25. b 2 − 4ac = ( −1) − 4(3)( 2) = 1 − 24 = −23 2

2 distinct complex solutions

z 2 = 169

15 5

y = 18

(x

15 i 4

27. z 2 − 169 = 0

x =

6 5

y − 18 = 0

−3 ± i 15 x = 4

2

36 25

x 2 + 8 x + 16 = −25 + 16

4

3 ± 4

36 25

33. 2 y ( y − 18) + 3( y − 18) = 0

32 − 4( 2)(3) 2( 2)

−3 ±

x = −

2

x = 3±

17. 2 x + 3 x + 3 = 0

x =

− 3) =

x −3 = ±

2

x =

y = −3

+ 1)( 4 x + 1) = 0

4x + 1 = 0

x =

y +3 = 0

31. 25( x − 3) − 36 = 0

−8 ± 0 32 8 1 x = − = − 32 4

(4 x

5 y ( y + 3) = 0

y = 0

x =

(b)

255

29. 5 y 2 + 15 y = 0

15. (a) 16 x 2 + 8 x + 1 = 0

x =

The Quadratic Formula

x = x = x = x =

−( −13) ±

(−13)

2

− 4(3)(169)

2(3) 13 ±

169 − 2028 6

13 ±

−1859 6

13 13 11 ± i 6 6

39. 25 x 2 + 80 x + 61 = 0

x = x =

z = ±13

−80 ±

802 − 4( 25)(61) 2( 25)

−80 ±

6400 − 6100 50

x =

−80 ± 300 50

x =

−80 ± 10 3 50

x = −

8 3 ± 5 5

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

256

Chapter 8

Quadratic Equations, Functions, and Inequalities

41. 7 x( x + 2) + 5 = 3 x( x + 1) 2

51.

2

7 x + 14 x + 5 = 3 x + 3 x

x = x = x =

)

2 ⎤⎦ = 0

(

2) = 0

− 1) − 2

x =1−

)

2

2 = 0

2 )⎤⎦ = 0

2 ⎤⎦ ⎡⎣( x − 1) +

(x

121 − 80 8

−11 ± 8

(

x − 1−

2 )⎤⎦ ⎡⎣ x − (1 −

⎡⎣( x − 1) −

2( 4)

2

2 = 0

⎡⎣ x − (1 +

112 − 4( 4)(5)

−11 ±

(

x − 1+

4 x 2 + 11x + 5 = 0 −11 ±

x =1+

2

x2 − 2 x + 1 − 2 = 0

41

x2 − 2 x − 1 = 0

x = 5i

53.

43. y = x 2 − 4 x + 3

x = −5i

x − 5i = 0

Keystrokes:

(x

Y= X,T,θ x 2 − 4 X,T,θ + 3 GRAPH

x + 5i = 0

− 5i )( x + 5i ) = 0 x 2 − 25i 2 = 0

0 = x2 − 4x + 3

x 2 + 25 = 0

0 = ( x − 3)( x − 1) x −3 = 0

x −1 = 0

x = 3

x =1

x = 12

55.

x = 12

x − 12 = 0

(x

(3, 0), (1, 0)

x − 12 = 0

− 12)( x − 12) = 0

2

x − 24 x + 144 = 0

6

x =

57.

1 2

x − −4

1 2

x =

x2 − x +

−2

2

45. −0.03 x + 2 x − 0.4 = 0

Keystrokes:

Y= (−) 0.03 X,T,θ x 2 + 2 X,T,θ − 0.4 GRAPH

x =

−2 ±

22 − 4( −0.03)( −0.4) 4 − 0.048 −0.06

− 2 ± 3.952 −0.06 x ≈ 0.20, 66.47

50

x = 5

47.

x −5 = 0

(x

− 5)( x + 2) = 0

x 2 − 3 x − 10 = 0

x =1

x = 7 x −7 = 0

(x

− 1)( x − 7) = 0 2

x − 8x + 7 = 0

3x 2 − 8 x − 12 = 0 x = x = x =

x −1 = 0

49.

90 −10

x + 2 = 0

2x x2 − =1 4 3 ⎛ x2 2x ⎞ 12⎜ − ⎟ = (1)12 4 3⎠ ⎝

−10

x = −2

= 0

59. The Quadratic Formula states: The opposite of b, plus or minus the square root of b squared minus 4ac, all divided by 2a.

63.

x =

(0.20, 0), (66.47, 0)

1 4

61. The equation in general form is 3x 2 + x − 3 = 0, so a = 3, b = 1, and c = − 3.

2( −0.03)

−2 ±

= 0

( x − 12 )( x − 12 ) = 0

8

x =

1 2

x −

= 0

1 2

−( −8) ± 8±

(−8) − 4(3)(−12) 2(3) 2

64 + 144 6

8±

208 6

x =

8 ± 4 13 6

x =

4 2 13 ± 3 3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 8.3 x + 3 = x −1

65.

(

x +3

)

2

2

Labels:

Length = x + 6.3 Width = x

x + 3 = x2 − 2x + 1 0 = x − 3x − 2

(−3) 2(1)

2

58.14 = x 2 + 6.3 x

− 4(1)(−2)

0 = x 2 + 6.3 x − 58.14

9+8 2 3 ± 17 x = 2 3 + 17 x = 2 3 − 17 x = does not check. 2 x =

3±

Check:

3+

17 2

?

+3 =

x = x =

3+

17 2

−1

17

17

3−

17 2

?

=

1+ 2 1+

t =

17

17 2

−1

3−

17 + 6 ? 3 − = 2

17 − 2 2

9−

17

17

2 18 − 2 17 2 1.5616

?

=

1−

0 ≤ t ≤ 7

0 = t 2 − 6.8t + 10

t =

=

272.25

0 = −0.25t 2 + 1.7t − 2.5

3−

?

−6.3 ±

6 = −0.25t + 1.7t + 3.5

t =

+3 =

39.69 + 232.56 2

69. d = −0.25t 2 + 1.7t + 3.5,

2 = 2.5616 ?

−6.3 ±

2

9+

=

2(1)

x + 6.3 ≈ 11.4 inches

17 − 2 2

?

6.32 − 4(1)( −58.14)

−6.3 ±

2 x ≈ 5.1 inches

17 + 6 ? 3 + = 2

18 + 2 17 2 2.5616 Check:

x =

3+

2

58.14 = ( x + 6.3) ⋅ x

Equation:

2

x =

257

67. Verbal Model: Area = Length ⋅ Width

= ( x − 1)

−( −3) ±

The Quadratic Formula

t = t =

−( −6.8) ± 6.8 ±

6.82 − 4(1)(10)

2(1)

46.24 − 40 2

6.8 ±

6.24 2

6.8 +

6.24 2

6.8 −

6.24 2

≈ 4.65 hours after a heavy rain ≈ 2.15 hours after a heavy rain

2 1−

17

2 ≠ −1.5616

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

−4

258

Chapter 8

Quadratic Equations, Functions, and Inequalities

71. (a) Keystrokes: Y= − 0.013 X,T,θ x 2 + 1.25 X,T,θ + 5.6 GRAPH 45

5

75 0

(b) 32 = −0.013x 2 + 1.25 x + 5.6 0 = −0.013x 2 + 1.25 x − 26.4 0 = 1.3x 2 − 125 x + 2640 x = x =

−( −125) ± 125 ±

(−125) − 4(1.3)(2640) 2(1.3) 2

15,625 − 13,728 2.6

125 + 1897 2.6 x ≈ 64.8 miles per hour x =

125 − 1897 2.6 x ≈ 31.3 miles per hour x =

Using the graph, you would have to travel approximately 65 miles per hour or 31 miles per hour to obtain a fuel economy of 32 miles per gallon. 73. The Square Root Property would be convenient because the equation is of the form u 2 = d .

83. f ( x) = ( x − 1) y

75. The Quadratic Formula would be convenient because the equation is already in general form, the expression cannot be factored, and the leading coefficient is not equal to 1. 77. When the Quadratic Formula is applied to ax 2 + bx + c = 0, the square root of the discriminant is evaluated. When the discriminant is positive, the square root of the discriminant is positive and will yield two real solutions (or x-intercepts). When the discriminant is zero, the equation has one real solution (or x-intercept). When the discriminant is negative, the square root of the discriminant is negative and will yield two complex solutions (or no x-intercepts). 79.

(−1 − 2)

2

+ (11 − 2)

2

=

9 + 81

=

90

(− 6 − (− 3))

2

+ (− 2 − ( − 4))

2

5 4

1 x

−3 − 2 − 1 −1

1

2

3

4

5

−2 −3

85. f ( x) = ( x − 2) + 4 2

y 8 6 4 2 −4

= 3 10 81.

2

x

−2

2

4

6

−2

=

9+ 4

=

13

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Mid-Chapter Quiz for Chapter 8

259

Mid-Chapter Quiz for Chapter 8 2 x 2 − 72 = 0

1.

7. x 2 + 4 x − 6 = 0

2( x 2 − 36) = 0

x =

2( x − 6)( x + 6) = 0

x −6 = 0

x + 6 = 0

x = 6 2.

x =

x = −6

2

2 x + 3 x − 20 = 0

(2 x

− 5)( x + 4) = 0

2x − 5 = 0 x =

x = −4

16 + 24 2

− 4 ± 40 2

x =

− 4 ± 2 10 = −2 ± 2

10

8. 6v 2 − 3v − 4 = 0

−( −3) ±

v =

2

3. 3x = 36

x 2 = 12 x = ±2 3

(−3)2

− 4(6)(− 4)

2(6) 3±

v =

x = ± 12

9 + 96 12

3±

v =

4. (u − 3) − 16 = 0

−4 ±

x =

x + 4 = 0

5 2

42 − 4(1)(−6) 2(1)

−4 ±

105 12

2

(u

9. x 2 + 5 x + 7 = 0

− 3) = 16 2

u − 3 = ±4

x =

u = 3 ± 4 = 7, −1 5.

x =

2

m + 7m + 2 = 0 m2 + 7m +

49 49 = −2 + 4 4 2

7⎞ 41 ⎛ ⎜m + ⎟ = 2⎠ 4 ⎝ 7 = ± 2

m +

41 4

25 − 28 2

−5 ± −3 2

x =

−5 ± 3 i 5 3 = − ± i 2 2 2 36 = (t − 4)

2

4± 6 = t 10, − 2 = t

2

11.

5 2 9 5 9 2 y + 3y + = + 4 2 4

−5 ±

±6 = t − 4

7 41 m = − ± 2 2

y2 + 3y =

52 − 4(1)(7) 2(1)

x =

10.

6. 2 y + 6 y − 5 = 0

−5 ±

(x

− 10)( x + 3) = 0

(x

− 10) = 0

x +3 = 0 x = −3

x = 10

2

12.

3⎞ 10 9 ⎛ + ⎜y + ⎟ = 2⎠ 4 4 ⎝

x 2 − 3 x − 10 = 0

(x

2

3⎞ 19 ⎛ ⎜y + ⎟ = 2 4 ⎝ ⎠

− 5)( x + 2) = 0

x −5 = 0 x = 5

x + 2 = 0 x = −2

3 19 y + = ± 2 2 y = −

3 ± 2

19 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

260 13.

Chapter 8

Quadratic Equations, Functions, and Inequalities

4b 2 − 12b + 9 = 0

17. x − 4

(2b − 3)( 2b − 3) = 0 2b − 3 = 0

b =

u 2 − 4u + 3 = 0

(u

3 2

14. 3m 2 + 10m + 5 = 0

m = m = m = m =

100 − 60 6

u = 3

u =1

10 ±

x = 3

x =1

x = 9

x =1 ?

Check: 9 − 4 9 + 3 = 0 ?

40

9 − 12 + 3 = 0

6

0 = 0

−10 ± 2 10 6

Solution ?

Check: 1 − 4 1 + 3 = 0

−5 ± 10 m = 3 15.

u −1 = 0

2(3)

−10 ±

− 3)(u − 1) = 0

u −3 = 0

102 − 4(3)(5)

−10 ±

x − 4

x , u 2 = x.

Let u =

2b − 3 = 0

3 2

b =

x +3 = 0

?

1 − 4 + 3=0 0 = 0

x − 21 = 0 x − 21 = 4

(x

(

− 21) = 4 2

Solution

x x

)

2

18. x 4 − 14 x 2 + 24 = 0

x 2 − 42 x + 441 = 16 x

Let u = x 2 , u 2 = x 4 .

x 2 − 58 x + 441 = 0

u 2 − 14u + 24 = 0

(x

− 9)( x − 49) = 0

x −9 = 0

u 2 − 14u + 49 = −24 + 49

(u − 7)2 = 25

x − 49 = 0

x = 9

x = 49

u − 7 = ±5 u = 7 ±5

?

Check: 9 − 4 9 − 21 = 0

u = 12

?

9 − 12 − 21 = 0

2

x = 12

− 24 ≠ 0

x = ± 12

Not a solution

?

49 − 28 − 21 = 0 0 = 0 Solution 16.

+ 4)( x 2 + 3) = 0

x2 = − 4

x = ±

2

19.

R = x(180 − 1.5 x) 5400 = 180 x − 1.5 x 2 0 = 1.5 x 2 − 180 x + 5400 0 = x 2 − 120 x + 3600

x 4 + 7 x 2 + 12 = 0

( x2

x2 = 2

x = ±2 3

?

Check: 49 − 4 49 − 21 = 0

u = 2

x 2 = −3

x = ± −4

x = ± −3

x = ± 2i

x = ± 3i

0 = ( x − 60)( x − 60) 0 = x − 60 60 = x

x − 60 = 0 x = 60

60 video games

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 8.4

10 = −0.003 x 2 + 1.19 x + 5.2

2275 = x ⋅ (100 − x)

0 = −0.003 x 2 + 1.19 x − 4.8

2275 = 100 x − x 2

0 = 3 x 2 − 1190 x + 4800

0 = x 2 − 100 x + 2275 0 = ( x − 35)( x − 65)

x − 35 = 0

x =

−( −1190) ±

x − 65 = 0

x = 35 meters

261

21. h = −0.003 x 2 + 1.19 x + 5.2

20. Verbal Model: Area = Length ⋅ Width

Equation:

Graphs of Quadratic Functions

x = 65 meters

x =

35 meters × 65 meters x = x =

1190 ±

(−1190) 2(3)

2

− 4(3)( 4800)

1,416,100 − 57600 6

1190 +

1,358,500 ≈ 392.6 feet 6

1190 −

1,358,500 ≈ 4.07; not reasonable 6

Section 8.4 Graphs of Quadratic Functions

(

)

1. y = x 2 − 2 x = x 2 − 2 x + 1 − 1 = ( x − 1) − 1

vertex = (1, −1) 3. y = x 2 − 4 x + 7

= ( x 2 − 4 x + 4) + 7 − 4 = ( x − 2) + 3 2

2

11. f ( x) = x 2 − 8 x + 15 a = 1, x =

b = −8

−( −8) −b = = 4 2a 2(1)

⎛ −b ⎞ f ⎜ ⎟ = 42 − 8( 4) + 15 = 16 − 32 + 15 = −1 ⎝ 2a ⎠

vertex = ( 2, 3)

vertex = ( 4, −1)

5. y = x 2 + 6 x + 5

13. g ( x) = − x 2 − 2 x + 1

y = ( x 2 + 6 x + 9) + 5 − 9

a = −1, b = −2

y = ( x + 3) − 4

x =

2

vertex = ( −3, − 4) 2

7. y = − x + 6 x − 10 y = −1( x 2 − 6 x) − 10 y = −1( x 2 − 6 x + 9) − 10 + 9 y = −1( x − 3) − 1 2

vertex = (3, −1) 9. y = 2 x 2 + 6 x + 2 9 + 2 − 92 ( 4) = 2( x + 32 ) − 52 vertex = ( − 32 , − 52 )

= 2 x 2 + 3x + 2

−( −2) −b = = −1 2a 2( −1)

2 ⎛ −b ⎞ g ⎜ ⎟ = −( −1) − 2(−1) + 1 ⎝ 2a ⎠ = −1 + 2 + 1

= 2 vertex = ( −1, 2) 15. y = 4 x 2 + 4 x + 4

a = 4,

b = 4

1 −b −4 x = = = − 2a 2( 4) 2 2

⎛ 1⎞ ⎛ 1⎞ y = 4⎜ − ⎟ + 4⎜ − ⎟ + 4 ⎝ 2⎠ ⎝ 2⎠ ⎛1⎞ = 4⎜ ⎟ − 2 + 4 ⎝ 4⎠ =1− 2+ 4 = 3 ⎛ 1 ⎞ vertex = ⎜ − , 3⎟ ⎝ 2 ⎠

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

262

Chapter 8

Quadratic Equations, Functions, and Inequalities

17. g ( x) = x 2 − 4

23. y = x 2 − 9 x − 18 y

x-intercepts: 0 = x −4

(− 2, 0) −3

0 = ( x − 2)( x + 2) x = 2,

x-intercepts:

1

2

−1

x = −2

(2, 0) 1

x

0 = x 2 − 9 x − 18

3

−1

x =

−2 −3

vertex: g ( x) = ( x − 0) − 4 2

x =

(0, − 4)

(0, − 4)

−5

2

− 4(1)( −18)

81 + 72 9 ± 153 = 2 2

9 ± 3 17 9 3 17 = ± 2 2 2 ≈ (10.68, 0), ( −1.68, 0)

y

x-intercepts:

vertex:

5

0 = − x2 + 4

(0, 4)

x2 = 4

3

x = ±2

2

81 ⎞ 81 ⎛ y = ⎜ x2 − 9x + ⎟ − 18 − 4⎠ 4 ⎝ 2

9⎞ 72 81 ⎛ = ⎜x − ⎟ − − 2 4 4 ⎝ ⎠

1

vertex:

(2, 0) x

−3

f ( x) = −( x − 0) + 4 2

(− 2, 0)

1

3

2

9⎞ 153 ⎛ = ⎜x − ⎟ − 2⎠ 4 ⎝

(0, 4)

⎛ 9 153 ⎞ ⎜ ,− ⎟ = ( 4.5, − 38.25) 4 ⎠ ⎝2

2 y

x-intercepts: 0 = ( x − 4)

9±

(−9) 2(1)

=

19. f ( x ) = − x 2 + 4

21. y = ( x − 4)

−( −9) ±

y 20

2

16

0 = x − 4

x

(

12

4 = x

8

vertex:

4

y = ( x − 4) + 0

−4

5

(

(

9 3 17 , 0 + 2 2

− 15

(4, 0) 2

(

9 3 17 , 0 − 2 2

4

8

x

12

16

20

−4

(4, 0)

− 25 − 35

(

9 153 ,− 4 2

(

25. vertex = ( 2, 0), point = (0, 4)

4 = a ( 0 − 2) + 0

y = 1( x − 2) + 0

4 = a( 4)

y = ( x − 2)

1 = a

y = x2 − 4x + 4

2

2

2

27. vertex = ( 2, 6), point = (0, 4) y = a ( x − 2) + 6 2

y = − 12 ( x − 2) + 6 2

4 = a ( 0 − 2) + 6

y = − 12 ( x 2 − 4 x + 4) + 6

4 = a ( 4) + 6

y = − 12 x 2 + 2 x − 2 + 6

2

−2 = a( 4)

y = − 12 x 2 + 2 x + 4

− 24 = a

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 8.4 29. vertex = ( −1, − 3), point = (0, − 2)

Graphs of Quadratic Functions

39. vertex = ( 2, − 2), point = (7, 8)

y = a( x − ( −1)) − 3

y = 1( x + 1) − 3

y = a( x − 2) − 2

y = a( x + 1) − 3

y = ( x + 1) − 3

8 = a(7 − 2) − 2

2

2

2

2

2 5

(x

− 2) − 2 2

10 = a( 25)

2

2 5

2

1 = a 31. vertex = ( 2, 1), a = 1

= a

y = a( x − 0) + 0 2

41.

15 = a(30 − 0)

y = 1( x − 2) + 1 = x 2 − 4 x + 5 2

33. vertex = ( 2, − 4), point = (0, 0) 0 = a ( 0 − 2) − 4 2

= a

1 60

= a

y =

1 = a y = 1( x − 2) − 4 = x − 4 x 2

35. vertex = ( −2, −1), point = (1, 8) y = a( x − ( −2)) − 1 2

2

15 = a(900) 15 900

4 = a ( 4) 2

y =

2

2

− 2 = a(0 + 1) − 3 1 = a(0 + 1)

263

1 x2 60

43. The graph is the shape of a parabola that opens upward when a > 0 and downward when a < 0. 45. To find any x-intercepts, set y = 0 and solve the resulting equation for x. To find the y-intercept, set x = 0 and solve the resulting equation for y. 47. h( x) = x 2 − 1

8 = a(1 + 2) − 1 2

9 = a ( 9)

Vertical shift 1 unit down y

1 = a y = 1( x + 2) − 1 2

5 4

y = ( x + 2) − 1 2

3 2

37. vertex = ( −1, 1), point = (− 4, 7)

y = a( x + 1) + 1 2

7 = a( − 4 + 1) + 1

1 −3

−2

x

−1

1

2

3

2

6 = a ( 9) 2 3

=

6 9

y =

(x

2

Horizontal shift 2 units left

= a 2 3

49. h( x) = ( x + 2)

y

+ 1) + 1 2

5 4

1

−5

−4

−3

−2

−1

x 1

−1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

264

Chapter 8

Quadratic Equations, Functions, and Inequalities

51. h( x) = −( x + 5)

2

59. y = − 94 x 2 +

y 2 −10 −8

−2

x 2

y =

−2 −4

y =

−6

53. h( x) = −( x − 2) − 3 2

Vertical shift 3 units down

2 x

−2

Reflection in the x-axis

2 −2 −4 −6 −8

1 2 x + 2x + 4 12

1 2 (0) + 2(0) + 4 12 y = 4 feet

(a) y = −

1 2 x + 2x + 4 12 1 y = − ( x 2 − 24 x + 144) + 4 + 12 12 1 2 y = − ( x − 12) + 16 12

(b) y = −

Maximum height = 16 feet

1 (c) 0 = − x 2 + 2 x + 4 12 0 = x 2 − 24 x − 48

1x 2 2

(b) y =

1 − 480

(x

2

1 2

(0)

= 0 yards

− 240 x + 14,400) + 30

1 x − 120 = − 480 ( ) + 30 2

Maximum height = 30 yards 1 x2 + (c) 0 = − 480

1x 2

0 = x 2 − 240 x 0 = x( x − 240) 0 = x

− 6 x + 9) + 4 + 10

4

6

8

63. Find the y-coordinate of the vertex of the graph of the function. 65. (0, 0), ( 4, − 2)

−2 − 0 2 1 = − = − 4−0 4 2 y = mx + b

m =

1 y = − x + 0 2 1 y = − x 2 67. ( −1, − 2), (3, 6) m =

6 − ( −2) 8 = = 2 3 − ( −1) 4

y − y1 = m( x − x1 ) y − 6 = 2( x − 3) y − 6 = 2x − 6 y = 2x

576 + 192 ≈ 25.9 feet 2

1 0 (a) y = − 480 () +

− 6 x + 9 − 9) + 10

61. If the discriminant is positive, the parabola has two x-intercepts; if it is zero, the parabola has one x-intercept; and if it is negative, the parabola has no x-intercepts.

y

Horizontal shift 2 units right

1 x2 + 57. y = − 480

− 6 x) + 10

The maximum height of the diver is 14 feet.

Reflection in the x-axis

x =

+ 10

2

Horizontal shift 5 units left

24 ±

(x − 94 ( x 2 − 94 ( x 2 2

y = − 94 ( x − 3) + 14

−8

55. y = −

y =

− 94

24 x 9

x − 240 = 0

⎛ 3 ⎞ ⎛ 11 5 ⎞ 69. ⎜ , 8 ⎟, ⎜ , ⎟ ⎝ 2 ⎠ ⎝ 2 2⎠ 5 5 16 11 −8 − − 11 2 2 2 m = = = 2 = − 11 3 8 4 8 − 2 2 2

y − y1 = m( x − x1 ) y −8 = −

11⎛ 3⎞ ⎜x − ⎟ 8⎝ 2⎠

11 33 x+ +8 8 16 11 33 128 y = − x+ + 8 16 16 11 161 y = − x+ 8 16 y = −

x = 240 240 yards

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 8.5 71. (0, 8), (5, 8)

8−8 0 = = 0 5−0 5 y = 8

m =

Applications of Quadratic Equations

73.

−64 =

75.

−0.0081 =

265

−1 ⋅ 64 = 8i −1 ⋅ 0.0081 = 0.09i

Horizontal line

Section 8.5 Applications of Quadratic Equations 1. Verbal Model:

Selling price Cost per Profit per = + per dozen eggs dozen eggs dozen eggs

Equation:

21.60 21.60 = + 0.30 x x + 6

Labels:

Number of eggs sold = x Number of eggs purchased = x + 6 21.60( x + 6) = 21.60 x + 0.30 x( x + 6) 21.6 x + 129.6 = 21.6 x + 0.3x 2 + 1.8 x 0 = 0.3 x 2 + 1.8 x − 129.6 0 = 3 x 2 + 18 x − 1296 0 = x 2 + 6 x − 432 0 = ( x + 24)( x − 18) x = −24, Selling price =

3. Verbal Model:

Labels:

x = 18 dozen

21.60 = $1.20 per dozen 18

Area = Length ⋅ Width Length = x + 4 Width = x

Equation:

7.

A = P(1 + r )

572.45 = 500(1 + r )

192 = ( x + 4) x

572.45 2 = (1 + r ) 500

192 = x 2 + 4 x

1.1449 = (1 + r )

2

2

0 = x 2 + 4 x − 192

1.07 = 1 + r

0 = ( x + 16)( x − 12)

0.07 = r or 7%

x = −16,

2

x = 12 inches

x + 4 = 16 inches 5.

A = P(1 + r )

2

11,990.25 = 10,000(1 + r ) 1.199025 = (1 + r )

2

2

1.095 = 1 + r 0.095 = r or 9.5%

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

266

Chapter 8

Quadratic Equations, Functions, and Inequalities

9. Verbal Model:

Equation:

Cost per Number of ⋅ = 210 ticket people going

⎛ 210 ⎞ − 3.50 ⎟ ⋅ ( x + 3) = 210 ⎜ ⎝ x ⎠ 630 210 + − 3.5 x − 10.50 = 210 x 210 x + 630 − 3.5 x 2 − 10.5 x = 210 x −3.5 x 2 − 10.5 x + 630 = 0 0 = 3.5 x 2 + 10.5 x − 630 0 = 3.5( x 2 + 3 x − 180) 0 = 3.5( x − 12)( x + 15)

x − 12 = 0

x + 15 = 0

x = 12

x = −15

There are 12 + 3 = 15 people going to the game. 11. Verbal Model:

Equation:

Work done by Work done by One complete + = Machine A Machine B job 1 1 (6) + (6) = 1 x x +3 1 ⎡1 x( x + 3) ⎢ (6) + (6)⎤ = (1) x( x + 3) x + 3 ⎥⎦ ⎣x 6( x + 3) + 6 x = x( x + 3) 6 x + 18 + 6 x = x 2 + 3 x − x 2 + 9 x + 18 = 0 x 2 − 9 x − 18 = 0

x =

9±

(−9)2 − 4(1)(−18) 2(1)

=

9±

x = 10.684658, −1.6846584 x + 3 = 13.684658 13. Common formula:

81 + 72 9 ± 153 = 2 2 x ≈ 10.7 minutes

x + 3 ≈ 13.7 minutes

a 2 + b2 + c2 x 2 + (100 − x) = 802 2

Equation:

x 2 + 10,000 − 200 x + x 2 = 6400 2 x 2 − 200 x + 3600 = 0 x 2 − 100 x + 1800 = 0 x = x =

−( −100) ± 100 ±

(−100) 2(1)

2

− 4(1)(1800)

10,000 − 7200 100 ± 2800 = ≈ 76.5 yards, 23.5 yards 2 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 8.5 3.5 = 5 + 18t − 16t 2

15. 2

16t − 18t − 1.5 = 0

t = t =

−( −18) ± 18 ±

(−18) − 4(16)(−1.5) 2(16)

23. Verbal Model:

Length = 2.5w

Equation:

250 = 2.5w ⋅ w

Width = w 250 = 2.5w2

324 + 96 32

100 = w2 10 = w

18 ±

25 = 2.5w

1.2 seconds will pass before you hit the ball. 17. The Pythagorean Theorem can be used to set the sum of the square of a missing length, and the square of the difference of the sum of the lengths and the missing length, equal to the square of the length of the hypotenuse. 19. The quotient of the investment amount and the number of units is equal to the quotient of the investment amount and the number of units sold, plus the profit per unit sold. 21. Verbal Model:

Labels: Equation:

267

Area = Length ⋅ Width

Labels: 2

400 32 t ≈ 1.2, − 0.1 t =

Applications of Quadratic Equations

Verbal Model:

2 Length + 2 Width = Perimeter

Equation:

2( 25) + 2(10) = P 70 feet = P

25. Verbal Model:

Labels:

2 Length + 2 Width = Perimeter Length = w + 3 Width = w

Equation:

2( w + 3) + 2w = 54 2 w + 6 + 2w = 54 4w = 48 w = 12 km

2 Length + 2 Width = Perimeter

l = w + 3 = 15 km

Length = l Width = 1.4l

Verbal Model:

2l + 2(1.4l ) = 54

Equation:

Length ⋅ Width = Area

15 ⋅ 12 = 180 km 2 = A

2l + 2.8l = 54 4.8l = 54 l = 11.25 inches w = 1.4l = 15.75 inches

Verbal Model: Equation:

Length ⋅ Width = Area 11.25 ⋅ 15.75 = A 177.1875 in.2 = A

27. Verbal Model:

Labels:

Length ⋅ Width = Area Length = l Width = l − 20

Equation:

l ⋅ (l − 20) = 12,000 l 2 − 20l = 12,000 l 2 − 20l + 100 = 12,000 + 100

(l

− 10) = 12,100 2

l − 10 = ± 12,100 l = 10 + 110 = 120 meters w = l − 20 = 100 meters

Verbal Model:

2 Length + 2 Width = Perimeter

Equation:

2(120) + 2(100) = 440 meters = P

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

268

Chapter 8

Quadratic Equations, Functions, and Inequalities

29. Verbal Model:

Selling price Cost per Profit per = + per DVD DVD DVD

Number of DVDs sold = x

Labels:

Number of DVDs purchased = x + 15 50 50 = +3 x x + 15 50( x + 15) = 50 x + 3x( x + 15)

Equation:

31. Verbal Model:

Second integer = n + 1 2

n + n − 182 = 0

(n

+ 14)( n − 13) = 0

n + 14 = 0 n = −14 ⎧ reject ⎨ ⎩n + 1 = −13

0 = x 2 + 15 x − 250

50 Selling price = = $5 10

n ⋅ ( n + 1) = 182

Equation:

0 = 3 x 2 + 45 x − 750

x = −25, x = 10 DVDs

First integer = n

Labels:

50 x + 750 = 50 x + 3x 2 + 45 x

0 = ( x + 25)( x − 10)

Integer ⋅ Integer = Product

33. Verbal Model:

Labels:

n − 13 = 0 n = 13 n + 1 = 14

Height ⋅ Width = Area Height = x Width = 48 − 2 x

x ⋅ ( 48 − 2 x) = 288

Equation: 2

2 x − 48 x + 288 = 0 x 2 − 24 x + 144 = 0

(x

− 12)( x − 12) = 0 x = 12

height = 12 inches

width = 48 − 2(12) = 48 − 24 = 24 inches 35. Verbal Model:

Equation:

Time for Time for + = 5 part 1 part 2

100 135 + =5 r r−5 135 ⎤ ⎡100 r ( r − 5) ⎢ + = 5r ( r − 5) r − 5 ⎥⎦ ⎣ r 100( r − 5) + 135r = 5r 2 − 25r 100r − 500 + 135r = 5r 2 − 25r 0 = 5r 2 − 260r + 500 0 = 5( r 2 − 52r + 100) 0 = 5( r − 50)( r − 2)

r − 50 = 0 r = 50

r −2 = 0 r = 2, reject

Thus, the average speed for the first part of the trip was 50 miles per hour.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 8.6 37.

d =

( x1

− x2 ) + ( y1 − y2 )

10 =

( x1

− 2) + (9 − 3)

2

2

41. No. For each additional person, the cost-per-person decrease gets smaller because the discount is distributed to more people.

2

2

2

43. 5 − 3 x > 17

2

−3 x > 12

100 = ( x1 − 2) + 36 64 = ( x1 − 2)

269

Quadratic and Rational Inequalities

x −6 −5 −4 −3 −2 −1

0

1

x < −4

±8 = x1 − 2

x2 − 8x = 0

45.

2 ± 8 = x1 x1 = 10

x1 = −6

(10, 9)

(−6, 9)

x 2 − 8 x + 16 = 16

(x

− 4) = 16 2

x − 4 = ±4

39. To solve a rational equation, each side of the equation is multiplied by the LCD. The resulting equations in this section are quadratic equations.

x = 4± 4 x = 8, 0

Section 8.6 Quadratic and Rational Inequalities 5. 4 x( x − 5)

1. x − 4

Negative: ( −∞, 4) Positive: ( 4, ∞)

−

x −5 = 0

x = 0

x = 5

0

Positive: ( −∞, 0) ∪ (5, ∞) Choose a test value from each interval.

(−∞, 0) ⇒ x = −1 ⇒ 4(−1)(−1 − 5) = 24 (0, 5) ⇒ x = 1 ⇒ 4(1)(1 − 5) = −16 < 0 (5, ∞) ⇒ x = 6 ⇒ 4(6)(6 − 5) = 24 > 0

+ x 4

3. 3 − 12 x

Negative: (6, ∞)

x

⇒ x = 0 ⇒ 3 − 12 (0) = 3 > 0

⇒ x = 8 ⇒ 3−

1 2

− x 6

5

7. 4 − x 2 = ( 2 − x)( 2 + x)

Choose a test value from each interval.

+

> 0

+

0

Positive: ( −∞, 6)

(6, ∞)

−

+

(−∞, 6)

Critical numbers

Negative: (0, 5)

Choose a test value from each interval.

(−∞, 4) ⇒ x = 0 ⇒ 0 − 4 = − 4 < (4, ∞) ⇒ x = 5 ⇒ 5 − 4 = 1 > 0

4x = 0

(8)

= −1 < 0

Negative: ( −∞, − 2) ∪ ( 2, ∞) Positive: ( −2, 2) Choose a test value from each interval.

(−∞, − 2) ⇒ x = −3 ⇒ ( 2 − (−3))(2 + (−3)) = (−2, 2) ⇒ x = 0 ⇒ (2 − 0)(2 + 0) = 4 > 0 (2, ∞) ⇒ x = 3 ⇒ (2 − 3)(2 + 3) = −5 < 0 −

−5 < 0

−

+

x −2

2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

270

Chapter 8

Quadratic Equations, Functions, and Inequalities

9. 3x( x − 2) < 0

x2 > 4

17.

x2 − 4 > 0

Critical numbers: x = 0, 2 Test intervals:

(x

Positive: ( −∞, 0)

Critical numbers: x = 2, − 2

Negative: (0, 2)

Test intervals: Positive: ( −∞, 2)

Positive: ( 2, ∞)

Negative: ( −2, 2)

Solution: (0, 2)

Positive: ( 2, ∞)

x −1

0

1

− 2)( x + 2) > 0

2

3

Solution: ( −∞, − 2) ∪ ( 2, ∞)

11. 3x( 2 − x) ≥ 0

x

−4

−2

0

2

4

Critical numbers: x = 0, 2 Test intervals: Negative: ( −∞, 0]

x 2 + 5 x − 36 ≤ 0

Positive: [0, 2]

(x

Negative: [2, ∞)

Critical numbers: −9, 4

Solution: [0, 2]

Test intervals:

0

1

2

+ 9)( x − 4) ≤ 0

Positive: ( −∞, − 9)

x −1

3

Negative: ( −9, 4)

13. x 2 + 4 x > 0

Positive: ( 4, ∞)

x ( x + 4) > 0

Solution: [−9, 4]

Critical numbers: x = −4, 0

−9 x

Test intervals:

− 10 − 8 − 6 − 4 − 2

Positive: ( −∞, − 4)

21.

Negative: ( −4, 0)

0

2

4

6

u 2 + 2u − 2 > 1 u 2 + 2u − 3 > 0

Positive: (0, ∞)

(u

Solution: ( −∞, − 4) ∪ (0, ∞)

Critical numbers: u = −3, 1

x

−8 −6 −4 −2

15.

x 2 + 5 x ≤ 36

19.

0

2

4

6

Test intervals: Positive: ( −∞, − 3)

2

x − 3 x − 10 ≥ 0

(x

+ 3)(u − 1) > 0

Negative: ( −3, 1)

− 5)( x + 2) ≥ 0

Critical numbers: x = −2, 5 Test intervals:

Positive: (1, ∞) Solution: ( −∞, − 3) ∪ (1, ∞ ) u

Positive: ( −∞, − 2)

−4

−3

−2

−1

0

1

2

Negative: ( −2, 5) Positive: (5, ∞) Solution: ( −∞, − 2] ∪ [5, ∞) 5 x −6 −4 −2

0

2

4

6

8

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 8.6 23. x 2 + 4 x + 5 < 0 x =

−4 ±

31.

16 − 20 2

Positive: ( −∞, 3)

Solution: 25. x + 6 x + 10 > 0

Negative: (3, ∞)

none

2

x =

Solution: ( −∞, 3)

36 − 4(10)

x

2 −6 ±

0

−4

33.

2 −6 ± 2i x = 2 x = −3 ± i

1

2

3

4

x + 4 > 0 x − 2

Positive: ( −∞, − 4)

x + 6 x + 10 is greater than 0 for all values of x.

Negative: ( − 4, 2)

Solution: ( −∞, ∞)

Positive: ( 2, ∞) x

−2

−1

0

1

2

Solution: ( −∞, − 4) ∪ ( 2, ∞)

3

x

27. y 2 + 16 y + 64 ≤ 0

(y

−6

+ 8) ≤ 0 2

Critical number: y = −8 Test intervals: Positive: ( −∞, − 8)

y −4

−2

−2

0

2

0

2

4

3 ≤ −1 y −1 3 +1≤ 0 y −1 y −1

Solution: −8 −6

35.

−4

3 + ( y − 1)

Positive: ( − 8, ∞)

− 10 − 8

6

Test intervals:

2

−3

5

Critical numbers: x = − 4, 2

No critical numbers

29.

−5 > 0 x −3

Test intervals:

x 2 + 4 x + 5 is not less than zero for any value of x.

x =

271

Critical number: x = 3

No critical numbers

−6 ±

Quadratic and Rational Inequalities

≤ 0

y + 2 ≤ 0 y −1 Critical numbers: x = −2, 1 Test intervals:

5 > 0 x −3

Positive: ( −∞, − 2)

Critical number: x = 3

Negative: ( −2, 1)

Test intervals:

Positive: (1, ∞)

Negative: ( −∞, 3)

Solution: [−2, 1)

Positive: (3, ∞)

y −3

−2

−1

0

1

2

Solution: (3, ∞) x 0

1

2

3

4

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

272

Chapter 8

Quadratic Equations, Functions, and Inequalities

37. h = −16t 2 + 128t

6 − ( x − 2) < 0 2

47.

height > 240

⎡⎣ 6 − ( x − 2)⎤⎦ ⎡⎣ 6 + ( x − 2)⎤⎦ < 0

2

−16t + 128t > 240

(

− 3)(t − 5) < 0

(

Critical numbers: x = 3, 5

Negative: −∞, 2 −

Test intervals:

Positive: 2 −

(

6 < 0

6, 2 +

Test intervals:

Positive: ( −∞, 3)

)

6 x−2+

Critical numbers: 2 −

t 2 − 8t + 15 < 0

(t

)(

− x−2−

−16t 2 + 128t − 240 > 0

6

6

)

6, 2 +

6

)

6, ∞) ( Solution: ( −∞, 2 − 6 ) ∪ ( 2 +

Negative: 2 +

Negative: (3, 5) Positive: (5, ∞)

2−

Solution: (3, 5)

2+

6

6, ∞

)

6 x

−2

0

2

4

6

8

39. The critical numbers are –1 and 3. 41. x = 4 is not a solution of the inequality x( x − 4) < 0.

16 ≤ (u + 5)

49.

(u

It does not satisfy the inequality:

2

u 2 + 10u + 9 ≥ 0

?

4(0) < 0

(u

0 ⬍ 0

+ 9)(u + 1) ≥ 0

Critical numbers: u = −9, −1

43. x( 2 x − 5) = 0

Test intervals: Positive: ( −∞, −9]

2x − 5 = 0 x =

Critical numbers: 0,

Negative: ( −9, −1]

5 2

Positive: [−1, ∞)

5 2

Solution: ( −∞, −9] ∪ [−1, ∞ )

45. 4 x 2 − 81 = 0

x2 =

+ 5) ≥ 16

u 2 + 10u + 25 − 16 ≥ 0

?

4( 4 − 4) < 0

x = 0,

2

−9

x = ± 92 Critical numbers: 92 , − 92

−1 u

81 4

−10

51.

−8

−6

−4

−2

0

u −6 ≤ 0 3u − 5

Critical numbers: u = 6,

5 3

Test intervals: 5⎞ ⎛ Positive: ⎜ −∞, ⎟ 3⎠ ⎝ ⎛5 ⎤ Negative: ⎜ , 6⎥ ⎝3 ⎦ Positive: [6, ∞) ⎛5 ⎤ Solution: ⎜ , 6⎥ ⎝3 ⎦ 5 3

u 1

2

3

4

5

6

7

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 8.6

53.

2( 4 − t ) 4+ t

6x −5 < x − 4 6 x − 5( x − 4) < x − 4 6 x − 5 x + 20 < x − 4 x + 20 < x − 4

Test intervals: Negative: ( −∞, 4) Positive: ( − 4, 4) Negative: ( 4, ∞) Solution: ( − 4, 4) −2

0

2

4

6

1 +3 > x + 2 1 + 3( x + 2) > x + 2 1 + 3x + 6 > x + 2 3x + 7 > x + 2

0

Positive: ( 4, ∞)

0

Solution: ( −20, 4) − 20

0

x

0

59.

0

8

1000(1 + r ) > 1150 2

1000(1 + 2r + r 2 ) > 1150 1000 + 2000r + 1000r 2 > 1150 20r 2 + 40r − 3 > 0 Critical numbers: r =

− 40 + 1840 − 40 − 1840 , 40 40

r cannot be negative.

Positive: ( − 2, ∞) 7⎞ ⎛ Solution: ⎜ −∞, − ⎟ ∪ ( − 2, ∞) 3⎠ ⎝ x −2

−8

1000r 2 + 2000r − 150 > 0

⎛ 7 ⎞ Negative: ⎜ − , − 2 ⎟ 3 ⎝ ⎠

3

4

− 16

7⎞ ⎛ Positive: ⎜ −∞, − ⎟ 3⎠ ⎝

−7

0

Negative: ( −20, 4)

Test intervals:

3

0

Positive: ( −∞, − 20)

7 Critical numbers: x = − , − 2 3

−8

0

Test intervals:

1 > −3 x + 2

55.

0

Critical numbers: x = −20, 4

t −4

273

6x < 5 x − 4

57.

> 0

Critical numbers: t = − 4, 4

−6

Quadratic and Rational Inequalities

−5 3

Test intervals: ⎛ − 40 + 1840 ⎞ Negative: ⎜⎜ 0, ⎟⎟ 40 ⎝ ⎠ ⎛ − 40 + 1840 ⎞ , ∞ ⎟⎟ Positive: ⎜⎜ 40 ⎝ ⎠ ⎛ − 40 + 1840 ⎞ Solution: ⎜⎜ , ∞ ⎟⎟ 40 ⎝ ⎠

(0.0724, ∞),

r > 7.24%

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

274

Chapter 8

61.

Quadratic Equations, Functions, and Inequalities Area > 240

l (32 − l ) > 240 32l − l 2 > 240 −l 2 + 32l − 240 > 0

l 2 − 32l + 240 < 0

(l

− 20)(l − 12) < 0

Critical numbers: l = 20, 12 Test intervals: Positive: ( −∞, 12) Negative: (12, 20) Positive: ( 20, ∞) Solution: (12, 20) 63. Verbal Model: Profit = Revenue − Cost P( x) = x(125 − 0.0005 x) − (3.5 x + 185,000) = 125 x − 0.0005 x 2 − 3.5 x − 185,000 = −0.0005 x 2 + 121.5 x − 185,000

Inequality:

−0.0005 x 2 + 121.5 x − 185,000 ≥ 6,000,000 −0.0005 x 2 + 121.5 x − 6,185,000 ≥ 0 x =

−121.5 ±

(121.5)2 − 4(−0.0005)(−6,185,000) 2( −0.0005)

1476.25 − 12,370 121.5 ± 2392.25 = 0.001 0.001 x ≈ 72,589; 170,411 x =

121.5 ±

Critical numbers: x ≈ 72,589; 170,411 Test intervals: Positive: ( −∞, 72,589) Negative: (72,589, 170,411) Positive: (170,411, ∞) Solution: (72,589, 170,411) 65. The critical numbers of a polynomial are its zeros, so the value of the polynomial is zero at its critical numbers.

67. No solution. The value of the polynomial is positive for every real value of x, so there are no values that would make the polynomial negative. 69.

4 xy 3 y 4 xy 4 y3 ⋅ = = , y ≠ 0 2 3 x y 8x 8x y 2x2

71.

( x − 3)( x + 2)( x + 1) = ( x − 3)( x + 1) , x2 − x − 6 x+1 ⋅ 2 = x + 5x + 6 4 x3 4 x 3 ( x + 3)( x + 2) 4 x 3 ( x + 3)

73.

( x + 4) ( x + 4)( x + 4) ⋅ 1 x 2 + 8 x + 16 ÷ (3 x − 24) = = 3( x − 8) 3 x( x − 6)( x − 8) x2 − 6 x x ( x − 6)

x ≠ −2 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 8 75.

77.

x = − 13

( )

x 2 = − 13

2

=

275

x = 1.06 100 100 = = 79.21 x4 (1.06)4

1 9

Review Exercises for Chapter 8 1. x 2 + 12 x = 0

17. z 2 = −121

x( x + 12) = 0

z = ±

x = 0

z = ±11i

x + 12 = 0

x = 0

x = −12

19. y 2 + 50 = 0

y 2 = −50

3 y 2 − 27 = 0

3.

3( y 2 − 9) = 0

y = ± −50

3( y − 3)( y + 3) = 0

y = ±5 2 i

y −3 = 0

y +3 = 0

y = 3

y = −3

21.

(y

5. 4 y + 20 y + 25 = 0

(2 y

2

2 y = −5

y = − 52

y = − 52

x4 − 4 x2 − 5 = 0

23.

( x2

x2 − 5 = 0

x 2 = −1

2

x = ±

x = ± 5

x = ±i

x = 5

2( x − 10)( x + 9) = 0 x+9 = 0 x = −9

2 x 2 − 9 x − 18 = 0 + 3)( x − 6) = 0

2x + 3 = 0

x −6 = 0

− 32

2

11. z = 144

− 5)( x 2 + 1) = 0 x2 + 1 = 0

2( x 2 − x − 90) = 0

x = 10

−18

y = −4 ± 3 2i

7. 2 x 2 − 2 x − 180 = 0

x =

2

2y + 5 = 0

2 y = −5

x − 10 = 0

+ 4) = −18

y + 4 = ±

+ 5)( 2 y + 5) = 0

2y + 5 = 0

(2 x

+ 4) + 18 = 0

(y

2

9.

−121

x = 6

25.

( (

x − 4

−1

x +3 = 0

)( x − 1) = 0 x − 3) = 0 ( x − 1) = 0

x −3

x = 3

( x)

2

= 32

x =1

( x)

x = 9

2

x =1

z = ± 144 z = ±12

Check: 9 − 4 9 + 3 = 0

13. y 2 − 12 = 0

9 − 12 + 3 = 0

y 2 = 12 y = ± 12 y = ±2 3

15.

(x

− 16) = 400 2

= 12

? ?

0 = 0 ?

Check: 1 − 4 1 + 3 = 0 ?

1− 4+3 = 0 0 = 0

x − 16 = ± 400 x = 16 ± 20 x = 36, − 4

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

276

Chapter 8

( x2

27.

Quadratic Equations, Functions, and Inequalities

− 2 x) − 4( x 2 − 2 x) − 5 = 0 2

⎡( x 2 − 2 x) − 5⎤⎡( x 2 − 2 x) + 1⎤ = 0 ⎣ ⎦⎣ ⎦

( x2

2±

x =

2±

x =

v 2 + 5v +

(−2) 2(1)

− 4(1)(−5)

2

v +

= −7

3

x = −7 = ( −7)

(x

− 1) = 0 2

x =1

= 0

= ± 32

2 y + 2 = 0 3 2 y 2 − y = −2 3 2 1 1 = −2 + y2 − y + 3 9 9 1⎞ −17 ⎛ ⎜y − ⎟ = 3⎠ 9 ⎝

x1 3 = 4 3 3

3

y −

x = 4

( x)

(182 ) ⎤⎦⎥

3

225 4

33. x − 15 x +

( )

⎡ 225 = − 15 2 ⎤ 2 ⎥ ⎣⎢ 4 ⎦ 2 1 y + 5 25

2 ⎡1 ⎛ 2 5⎞ ⎤ = ⎜ ⎢ ⎟ ⎥ ⎝ 2 ⎠ ⎥⎦ ⎢⎣ 25 2 ⎡1 ⎛1⎞ ⎤ = ⎜ ⎟ ⎥ ⎢ ⎝ 5 ⎠ ⎦⎥ ⎣⎢ 25

1 = ± 3

−17 9

1 17 i ± 3 3 y ≈ 0.33 + 1.37i, 0.33 − 1.37i

y =

= 43

x = 64

43. v 2 + v − 42 = 0

2

2

3 2

41. y 2 −

x1 3 − 4 = 0

31. z 2 + 18 z + 81

35. y 2 +

9 4

2

x = −343

⎡81 = ⎣⎢

25 4

v = − 22 , − 82

6

x

3

=

v = −1, − 4

x1 3 + 7 = 0

3

5 2

= − 16 + 4

24

( x1 3 + 7)( x1 3 − 4)

( x)

2

25 4

v = − 52 ±

x 2 3 + 3 x1 3 − 28 = 0

13

2

= −4 +

4 + 20 2

2± 2 6 2

x = 1±

25 4

(v + 52 ) (v + 52 )

2

x =

29.

v 2 + 5v + 4 = 0

− 2 x − 5)( x 2 − 2 x + 1) = 0

−( −2) ±

x =

39.

v = v =

−1 ±

12 − 4(1)( − 42) 2(1)

−1 ±

1 + 168 2

169 2 −1 ± 13 v = 2 12 −14 v = , 2 2 v = 6, − 7 v =

−1 ±

37. x 2 − 6 x − 3 = 0

x2 − 6 x + 9 = 3 + 9

(x

− 3) = 12 2

x − 3 = ± 12 x = 3± 2 3 x ≈ 6.46, − 0.46

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 8 57.

45. 5 x 2 − 16 x + 2 = 0 x = x = x = x =

−( −16) ±

(−16)2 2(5)

16 ±

256 − 40 10

16 ±

216

x = 3

x = −7

x −3 = 0

− 4(5)( 2)

(x

x + 7 = 0

− 3)( x + 7) = 0 2

x + 4 x − 21 = 0

59.

x = 5+

(

(x

16 ± 6 6 10

− 5) −

x = x =

−( −3) ±

9 − 16 8

3±

−7 8

− 5) − 2

7 = 0 7 = 0

7 ⎤⎦ = 0

( 7)

2

= 0

x − 10 x + 25 − 7 = 0 x 2 − 10 x + 18 = 0

2

− 4( 4)(1)

3 = ± 8

7 i 8

61.

x = 6 + 2i

x − (6 − 2i ) = 0

(x

(x

− 6) − 2i = 0

− 6) + 2i = 0

⎣⎡( x − 6) − 2i⎤⎡ ⎦⎣( x − 6) + 2i⎤⎦ = 0

(x

− 6) − ( 2i ) = 0 2

2

x 2 − 12 x + 36 + 4 = 0

49. x + 4 x + 4 = 0 b − 4ac = 4 − 4(1)( 4)

x = 6 − 2i

x − (6 + 2i ) = 0

2

2

− 5) +

)

7

2

(Divide by 2.)

(−3) 2( 4)

3±

(x

7 ⎤⎦ ⎡⎣( x − 5) +

4 x2 − 3x + 1 = 0

x =

(

7 = 0

(x

x = 5−

x − 5−

7 = 0

⎡( x − 5) − ⎣

8± 3 6 x = 5

7

)

x − 5+

10

47. 8 x 2 − 6 x + 2 = 0

277

x 2 − 12 x + 40 = 0

2

= 16 − 16 = 0

One repeated rational solution

63. y = x 2 − 8 x + 3

= ( x 2 − 8 x + 16) + 3 − 16 = ( x − 4) − 13 2

Vertex: ( 4, −13)

51. s 2 − s − 20 = 0 b 2 − 4ac = ( −1) − 4(1)( −20) 2

= 1 + 80 = 81

Two distinct rational solutions 53. 4t 2 + 16t + 10 = 0 b 2 − 4ac = 162 − 4( 4)(10) = 256 − 160

65. y = 2 x 2 − x + 3

( 12 x) + 3 1 + 3 − 1 = 2( x − 12 x + 16 ) 8 23 1 = 2( x − 4 ) + 8 Vertex: ( 14 , 23 8) = 2 x2 − 2

2

= 96

Two distinct irrational solutions 55. v 2 − 6v + 21 = 0

b 2 − 4ac = ( −6) − 4(1)( 21) 2

= 36 − 84 = − 48 Two distinct complex solutions

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

278

Chapter 8

Quadratic Equations, Functions, and Inequalities 73. Vertex: (5, 0); passes through the point (1, 1)

67. y = x 2 + 8 x

x-intercepts:

y = a ( x − h) + k 2

0 = x2 + 8x

1 = a(1 − 5) + 0 2

0 = x( x + 8) x = 0,

1 = a(16)

x = −8

1 16

Vertex:

y =

y = x 2 + 8 x + 16 − 16 y = ( x + 4) − 16 2

(0, 0) −2

2

− 16

x-intercepts:

(c) 0 = −

0 = x2 + 2x − 4 −2 ±

4 + 16

y

2

25 16

⎛ 1⎞ 32 − 4⎜ − ⎟(6) ⎝ 10 ⎠ ⎛ 1⎞ 2⎜ − ⎟ ⎝ 10 ⎠ 32

−3 ± x =

9

3

(−1 − 5, 0)

−2 ± 2 5 2

x = −1 ±

− 85 x +

1 2 x + 3x + 6 10

(−1, 5) 6

−2 ± 20 x = 2 x =

1 x2 16

b −3 3 = − = = 15 1 2a ⎛ 1⎞ − 2⎜ − ⎟ 5 ⎝ 10 ⎠ 1 y = (15)2 + 3(15) + 6 10 1 = − ( 225) + 45 + 6 10 = −22.5 + 45 + 6 = 28.5 feet

x

69. f ( x) = − x 2 − 2 x + 4

x =

2

(b) x = −

4

(−8, 0)

(−4, −16)

− 5) + 0 or y =

y = 6 feet

8

−4

(x

1 2 (0) + 3(0) + 6 10 y = 0 + 0+ 6

y

−6

1 16

75. (a) y = −

(− 4, −16)

− 10

= a

−9

(−1 + 5, 0)

−6

3

6

x

x =

−3 ±

−3

5

−6

x =

Vertex:

f ( x) = −1( x + 2 x) + 4

−3 ±

2

(

9 + 2.4 1 − 5

1 − 5

11.4 0

32 0

)

= −( x 2 + 2 x + 1) + 4 + 1

x = −5 −3 ±

= −( x + 1) + 5

The ball is 31.9 feet from the child when it hits the ground.

2

(−1, 5)

11.4 = 15 ± 5 11.4 = 31.9

71. Vertex: ( 2, − 5); y-intercept: (0, 3)

y = a( x − h) + k 2

y = a( x − 2) − 5 2

3 = a(0 − 2) − 5 2

3 = a( 4) − 5 8 = a( 4) 2 = a y = 2( x − 2) − 5 or y = 2 x 2 − 8 x + 3 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 8

279

77. Verbal Model: Selling price per car = Cost per car + Profit per car

Number of cars sold = x Number of cars bought = x + 4 Selling price per car = 80,000 x

Labels:

Cost per car: 80,000 ( x + 4) Profit per car: 1,000 80,000 80,000 = + 1000 x x + 4 80,000( x + 4) = 80,000 x + 1000 x( x + 4), x ≠ 0, x ≠ − 4

Equation:

80,000 x + 320,000 = 80,000 x + 1000 x 2 + 4000 x 0 = 1000 x 2 + 4000 x − 320,000 0 = x 2 + 4 x − 320 0 = ( x − 16)( x + 20) x − 16 = 0 x = 16

x + 20 = 0 x = − 20

By choosing the positive value, it follows that the dealer sold 16 cars at an average price of 80,000 16 = $5000 per car. 79. Verbal Model:

Labels:

Area = Length ⋅ Width

Width = x Length = x + 12

Equation:

85 = ( x + 12) x 0 = x 2 + 12 x − 85 0 = ( x + 17)( x − 5)

reject x = −17,

x = 5 inches x + 12 = 17 inches

81. Verbal Model:

Labels:

Cost per ticket ⋅ Number of tickets = $96

Number in team = x Number going to game = x + 3

Equation:

⎛ 96 ⎞ − 1.60 ⎟( x + 3) = 96 ⎜ x ⎝ ⎠ ⎛ 96 − 1.60 x ⎞ ⎜ ⎟( x + 3) = 96 x ⎝ ⎠

(96

− 1.6 x )( x + 3) = 96 x

2

96 x − 1.6 x − 4.8 x + 288 = 96 x 1.6 x 2 + 4.8 x − 288 = 0 x 2 + 3x − 180 = 0

(x

− 12)( x + 15) = 0

x − 12 = 0

x + 15 = 0

x = 12

x = −15 reject

x + 3 = 15 15 people are going to the game.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

280

Chapter 8

Quadratic Equations, Functions, and Inequalities c2 = a2 + b2

83. Formula:

Labels:

c = 16

a + b = 20

a = x

x + b = 20 b = 20 − x

Equation:

162 = x 2 + ( 20 − x)

2

256 = x 2 + 400 − 40 x + x 2 0 = 2 x 2 − 40 x + 144 0 = x 2 − 20 x + 72 x =

(− 20) − 4(1)(72) 2(1) 2

20 ±

x =

20 ± 112 2

x =

20 ± 4 7 2

x = 10 ± 2 7 x ≈ 15.3, 4.7 So, the possible distances from campus to the secretary’s location are about 4.7 miles or 15.3 miles. Work done Work done + = One complete job by Person 1 by Person 2

85. Verbal Model:

Labels:

Time Person 1 = x Time Person 2 = x + 2

1 1 (10) + (10) = 1 x x + 2

Equation:

⎡ ⎛1 1 ⎞⎤ x( x + 2) ⎢10⎜ + ⎟ = [1] x( x + 2) x x 2 ⎠⎦⎥ + ⎝ ⎣ 10( x + 2) + 10 x = x( x + 2) 10 x + 20 + 10 x = x 2 + 2 x 0 = x 2 − 18 x − 20 x = x = x =

−( −18) ±

(−18)2 2(1)

18 ±

324 + 80 2

18 ±

404

− 4(1)(−20)

2

18 ± 2 101 x = 2 x = 9±

101

x ≈ 19, x = −1, reject x + 2 ≈ 21 19 hours, 21 hours

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 8 87. 2 x( x + 7)

2x = 0

95.

x = 0

x = −7

5 2

Test intervals: Positive: ( −∞, − 4)

2

89. x − 6 x − 27 − 9)( x + 3)

x −9 = 0

− 5)( x + 4) < 0

Critical numbers: x = − 4,

Critical numbers: x = 0, − 7

(x

2 x 2 + 3x − 20 < 0

(2 x

x +7 = 0

281

(

Negative: − 4, x +3 = 0

x = 9

5 2

)

( 52 , ∞) Solution: ( − 4, 52 )

Positive:

x = −3

Critical numbers: x = 9, − 3

5 2

91. 5 x(7 − x) > 0

x −6

Critical numbers: x = 0, 7 Test intervals: 97.

Negative: ( −∞, 0) Positive: (0, 7)

−4

−2

0

2

4

x +3 ≥ 0 2x − 7 Critical numbers: x = −3,

Negative: (7, ∞)

7 2

Test intervals:

Solution: (0, 7)

Positive: ( −∞, − 3) 7

7⎤ ⎡ Negative: ⎢−3, ⎥ 2⎦ ⎣

x −2

0

2

4

6

8

16 − ( x − 2) ≤ 0 2

93.

(4

⎛7 ⎞ Positive: ⎜ , ∞ ⎟ ⎝2 ⎠

− x + 2)( 4 + x − 2) ≤ 0

(6

⎛7 ⎞ Solution: ( −∞, − 3] ∪ ⎜ , ∞ ⎟ ⎝2 ⎠

− x )( 2 + x ) ≤ 0

Critical numbers: x = −2, 6

7 2

Test intervals:

x

Negative: ( −∞, − 2]

−4 −3 −2 −1

Positive: [−2, 6]

99.

Negative: [6, ∞) Solution: ( −∞, − 2] ∪ [6, ∞) x −4 −2

0

2

4

6

8

0

1

2

3

4

x + 4 < 0 x −1 Critical numbers: x = − 4, 1 Test intervals: Positive: ( −∞, − 4) Negative: ( − 4, 1) Positive: (1, ∞) Solution: ( − 4, 1) 1 x −6

−4

−2

0

2

4

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

282

Chapter 8

Quadratic Equations, Functions, and Inequalities

101. h = −16t 2 + 312t

−16t 2 + 312t > 1200

(Divide by

−16t 2 + 312t − 1200 > 0

−16.)

2

t − 19.5t + 75 < 0 t = t =

−( −19.5) ± 19.5 ±

(−19.5) 2(1)

2

− 4(1)(75)

80.25

2 t ≈ 14.2, 5.3 Critical numbers: t = 14.2, 5.3 Test intervals: Positive: ( −∞, 5.3) Negative: (5.3, 14.2) Positive: (14.2, ∞) Solution: (5.3, 14.2) 5.3 < t < 14.2

Chapter Test for Chapter 8 1. x( x − 3) − 10( x − 3) = 0

(x

− 3)( x − 10) = 0

x −3 = 0

x 2 − 3x +

x − 10 = 0

x = 3 2.

5. 2 x 2 − 6 x + 3 = 0

2

3⎞ −6 + 9 ⎛ ⎜x − ⎟ = 2⎠ 4 ⎝

x = 10

2

6 x 2 − 34 x − 12 = 0

3⎞ 3 ⎛ ⎜x − ⎟ = 2⎠ 4 ⎝

2(3 x 2 − 17 x − 6) = 0 2(3 x + 1)( x − 6) = 0

3x + 1 = 0 x = 3.

(x

− 2) = 0.09 2

x = 2 ± 0.3 x = 2.3, 1.7

(x

+ 4) + 100 = 0 2

(x

+ 4) = −100 x = − 4 ± 10i

3 4

3 3 ± 2 2

2 y( y − 2) = 7

6. 2

2y − 4y − 7 = 0 y = y =

2

x + 4 = ± −100

3 = ± 2 x =

x = 6

x − 2 = ± 0.3

4.

x −

x −6 = 0

− 13

9 3 9 = − + 4 2 4

y =

−( − 4) ± 4±

(− 4)2 2( 2)

− 4( 2)( −7)

16 + 56 4

4±

72 4

y =

4±6 2 4

y =

2±3 2 ≈ 7.41 and − 0.41 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test for Chapter 8

7.

1 6 − + 4 = 0 x2 x

11. y = − x 2 + 2 x − 4

x-intercepts:

1 − 6x + 4x2 = 0

0 = x2 − 2x + 4

4 x2 − 6x + 1 = 0 x = x = x =

−( −6) ±

(−6) 2( 4)

6±

36 − 16 8

6±

20

2

− 4( 4)(1)

( − 2) 2 2(1)

− 4(1)( 4)

4 − 16 2

2 ±

−12 2

Not real No x-intercepts Vertex:

5

y = −1( x 2 − 2 x) − 4

4

= −( x 2 − 2 x + 1) − 4 + 1

x 2 3 − 9 x1 3 + 8 = 0

8.

2 ±

x =

8

3±

− ( − 2) ±

x = x =

6± 2 5 x = 8 x =

283

= −( x − 1) − 3 2

( x1 3 − 8)( x1 3 − 1) =

0

x1 3 − 8 = 0

x1 3 − 1 = 0

x1 3 = 8

x1 3 = 1

( x1 3 ) = 83

( x1 3 ) = 13

3

(1, − 3) y 1

3

x = 512

−1

x =1

x 1 −2 −3

9. b 2 − 4ac = ( −12) − 4(5)(10) 2

2

3

4

5

−1

(1, − 3)

−4

= 144 − 200 = − 56 A negative discriminant tells us the equation has two complex solutions. 10.

x = −7 x +7 = 0

(x

x = −3 x+3 = 0

+ 7)( x + 3) = 0

x 2 + 10 x + 21 = 0

12. y = x 2 − 2 x − 15

x-intercepts: 0 = x 2 − 2 x − 15 0 = ( x − 5)( x + 3) x = 5,

x = −3

Vertex: y = ( x 2 − 2 x + 1) − 15 − 1 y = ( x − 1) − 16 2

(1, −16) y −8

x

(−3, 0) −2

2

4

8

(5, 0)

−4

−14 −16

(1, −16)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

284

Chapter 8

Quadratic Equations, Functions, and Inequalities 16 ≤ ( x − 2)

13.

(x

14. 2 x( x − 3) < 0

2

Critical numbers: x = 0, 3

− 2) ≥ 16 2

Test intervals:

x 2 − 4 x + 4 ≥ 16

Positive: ( −∞, 0)

x 2 − 4 x − 12 ≥ 0

(x

− 6)( x + 2) ≥ 0

Negative: (0, 3)

Critical numbers: x = −2, 6

Positive: (3, ∞)

Test intervals:

Solution: (0, 3)

Positive: ( −∞, − 2] Negative: [−2, 6]

15.

x −1

0

Critical numbers: x = −1, 5

Solution: ( −∞, − 2] ∪ [6, ∞)

Test intervals:

0

2

4

6

8

Solution: [−1, 5)

Labels:

4

Negative: ( −1, 5) Positive: (5, ∞)

16. Verbal Model:

3

Positive: ( −∞, −1)

x −2

2

x +1 ≤ 0 x −5

Positive: [6, ∞)

−4

1

−1

5 x

−4

−2

0

2

4

6

Area = Length ⋅ Width Length = x Width = x − 22

Equation:

240 = x( x − 22) 0 = x 2 − 22 x − 240 0 = ( x − 30)( x + 8)

0 = x − 30 30 feet = x

reject x = −8 x − 22 = 8 feet

8 feet × 30 feet

17. Verbal Model:

Labels:

Cost per person Cost per person − = 6.25 Current Group New Group

Number Current Group = x Number New Group = x + 10

Equation:

1250 1250 − = 6.25 x x + 10 1250 ⎞ ⎛ 1250 x( x + 10)⎜ − ⎟ = (6.25) x( x + 10) x + 10 ⎠ ⎝ x 1250( x + 10) − 1250 x = 6.25 x( x + 10) 1250 x + 12500 − 1250 x = 6.25 x 2 + 62.5 x 0 = 6.25 x 2 + 62.5 x − 12,500 0 = x 2 + 10 x − 2000 0 = ( x + 50)( x − 40) reject x = −50,

x = 40 club members

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test for Chapter 8 18.

35 = −16t 2 + 75

Labels:

16t 2 = 40 t2 = t =

40 5 = 16 2 5 2

10 ≈ 1.5811388 2 t ≈ 1.58 seconds t =

19. Formula:

c2 = a2 + b2

c = 125

a + b = 155

a = x

x + b = 155

b = 155 − x Equation:

285

b = 155 − x

125 = x + (155 − x) 2

2

2

15,625 = x 2 + 24025 − 310 x + x 2 0 = 2 x 2 − 310 x + 8400 0 = x 2 − 155 x + 4200 0 = ( x − 35)( x − 120)

0 = x − 35 35 feet = x

x − 120 = 0 x = 120 feet

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

C H A P T E R 9 Exponential and Logarithmic Functions Section 9.1

Exponential Functions........................................................................287

Section 9.2

Composite and Inverse Functions......................................................291

Section 9.3

Logarithmic Functions .......................................................................294

Mid-Chapter Quiz......................................................................................................298 Section 9.4

Properties of Logarithms....................................................................299

Section 9.5

Solving Exponential and Logarithmic Equations .............................302

Section 9.6

Applications ........................................................................................307

Review Exercises ........................................................................................................310 Chapter Test .............................................................................................................316

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

C H A P T E R 9 Exponential and Logarithmic Functions Section 9.1 Exponential Functions 13. f ( x) = 1000(1.05)

1. 2− 2.3 ≈ 0.203 Keystrokes: Scientific: 2 g x 2.3 + − =

Graphing: 2 ^ − 2.3 ENTER 3. 5

2

2x

(a) f (0) = 1000(1.05) (b) f (5) = 1000(1.05)

(2)(0)

2(5)

(c) f (10) = 1000(1.05)

≈ 9.739

Keystrokes:

15. h( x) =

Scientific: 5 g x 2

=

Graphing: 5 ^

2 ) ENTER

5. 61 3 ≈ 1.817

≈ 1628.895

2(10)

≈ 2653.298

5000 8x

(1.06)

5000

(a) h(5) =

(1.06)8(5)

(b) h(10) =

Keystrokes: Scientific: 6 g x ( 1 ÷ 3 ) =

= 1000

≈ 486.111

5000

≈ 47.261

(1.06)8(10)

(c) h( 20) =

Graphing: 6 ^ ( 1 3 ) Enter

5000

≈ 0.447

(1.06)8(20)

17. f ( x) = 2 x ; d

7. f ( x) = 3x

(a) f ( −2) = 3−2 =

18. g ( x) = 6 x ; c

1 9

(b) f (0) = 30 = 1

19. h( x) = 2− x ; a

(c) f (1) = 31 = 3

20. k ( x) = 6− x ; b

9. g ( x) = 2.2− x

(a) g (1) = 2 ⋅ 2−1 ≈ 0.455

21. f ( x) = 3x

y

4

(b) g (3) = 2 ⋅ 2−3 ≈ 0.094 (c) g

( 6) = 2 ⋅ 2

−

6

3

≈ 0.145

2 1

( 12 )

11. f (t ) = 500

t −2

( 12 )

(a) f (0) = 500

0

( 12 )

1

(b) f (1) = 500

= 250

()

(c) f (π ) = 500

1 2

= 500

π

≈ 56.657

−1

x 1

2

Domain: − ∞ < x < ∞ Range: y > 0

Table of values: x f ( x)

–2

–1

0

12

0.1

0.3

1

39

287

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

288

Chapter 9

E xponential and Logarithmic Functions

23. f ( x) = 3− x =

( 13 )

31. The function g is related to f ( x) = 2 x by

x y

g ( x) = f ( x − 4). To sketch the graph of g, shift the

5

graph of f four units to the right.

4

(

1 y-intercept: 0, 16

3

)

y

Asymptote: x-axis

1

8

x –3

–2

–1

Domain: − ∞ < x < ∞

1

−1

2

6

3

4

Range: y > 0

2

25. The function g is related to f ( x) = 3x by

x 2

g ( x) = f ( x) − 1. To sketch g, shift the graph of f one unit downward.

4

6

8

33. g ( x) = 2− x ; b

y

y-intercept: (0, 0)

34. g ( x) = 2 x − 1; d

4

Asymptote: y = −1

3

35. g ( x) = 2 x −1; a

2 1 x –2

–1

1

36. g ( x) = − 2 x ; c

2

37. The function g is related to f ( x) = 4 x by

g ( x) = − f ( x). To sketch the graph of g, reflect the

27. The function g is related to f ( x) = 5 x by

g ( x) = f ( x − 1). To sketch the graph of g, shift the

graph of f in the x-axis. y

graph of f one unit upward to the right.

( 15 )

y-intercept: 0,

1 x

y

Asymptote: x-axis

−3

−2

7

−1

6

−2

5

−3

4

1

2

3

−4

3

−5

2 1 x

−5 −4 −3 −2 −1

1

2

39. The function g is related to f ( x) = 5 x by

3

g ( x) = f ( − x). To sketch the graph of g, reflect the

29. The function g is related to f ( x) = 2 x by

g ( x) = f ( x) + 3. To sketch the graph of g, shift the

graph of f three units upward.

graph of f in the x-axis. y

y

y-intercept: (0, 4)

4

6

Asymptote: y = 3

3

5

1

2 x

1

–2

–1

1

2

x –3

–2

–1

1

2

3

41. e1 3 ≈ 1.396

Keystrokes: Scientific: ( 1 ÷ 3 ) INV ln x = Graphing: e x ( 1 ÷ 3 ) ENTER

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 9.1

Exponential Functions

289

49. The function g is related to f ( x) = e x by

43. 3( 2e1 2 ) = 3 ⋅ 8 ⋅ e3 2 = 24e1.5 ≈ 107.561 3

g ( x) = f ( x) + 1. To sketch the graph of g, shift the

Keystrokes:

graph of f one unit upward.

Scientific: 24 × 1.5 e x =

y

Graphing: 24 e x 1.5 ) ENTER

5 4

45. g ( x) = 10e

−0.5 x

3

(a) g ( −4) = 10e −0.5(−4) = 10e2 ≈ 73.891 1

(b) g ( 4) = 10e

−0.5( 4)

= 10e

−2

≈ 1.353

x –2

(c) g (8) = 10e −0.5(8) = 10e −4 ≈ 0.183

–1

1

( 12 )

51. y = 16

47. The function g is related to f ( x) = e by x

g ( x) = − f ( x). To sketch the graph of g, reflect the graph of f in the x-axis. y

80 30

2

≈ 2.520 grams

Keystrokes: Scientific: 16 × 0.5 y x ( 8 ÷ 3 ) = Graphing: 16 × 0.5 ^ ( 8 ÷ 3 ) ENTER

x –2

1

2

–1 –2 –3 –4

53.

n

1

4

12

365

Continuous

A

$275.90

$283.18

$284.89

$285.74

$285.77

1(15)

0.07 ⎞ ⎛ Compounded 1 time: A = 100⎜1 + ⎟ 1 ⎠ ⎝

0.07 ⎞ ⎛ Compounded 4 times: A = 100⎜1 + ⎟ 4 ⎠ ⎝

= $275.90

4(15)

= $283.18

12(15)

0.07 ⎞ ⎛ Compounded 12 times: A = 100⎜1 + ⎟ 12 ⎠ ⎝

0.07 ⎞ ⎛ Compounded 365 times: A = 100⎜1 + ⎟ 365 ⎠ ⎝

= $284.89

365(15)

= $285.74

Compounded continuously: A = Pe rt = 100e0.07(15) = $285.77

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290 55.

Chapter 9

E xponential and Logarithmic Functions

n

1

4

12

365

Continuous

A

$4956.46

$5114.30

$5152.11

$5170.78

$5171.42

1(10)

0.095 ⎞ ⎛ Compounded 1 time: A = 2000⎜1 + ⎟ 1 ⎠ ⎝

0.095 ⎞ ⎛ Compounded 4 times: A = 2000⎜1 + ⎟ 4 ⎠ ⎝

= $4956.46

4(10)

= $5114.30

12(10)

0.095 ⎞ ⎛ Compounded 12 times: A = 2000⎜1 + ⎟ 12 ⎠ ⎝

0.095 ⎞ ⎛ Compounded 365 times: A = 2000⎜1 + ⎟ 365 ⎠ ⎝

= $5152.11

365(10)

= $5170.78

Compounded continuously: A = 2000e0.095(10) = $5171.42 57. To obtain the graph of g ( x) = − 2 x from the graph of f ( x ) = 2 x , reflect the graph of f in the x-axis. 59. The behavior is reversed. When the graph of the original function is increasing, the reflection is decreasing. When the graph of the original function is decreasing, the reflected graph is increasing.

65. 67.

61. 3x ⋅ 3x + 2 = 3x + ( x + 2) = 32 x + 2 63. 3(e x )

69.

−2

= 3⋅

1

(e ) x

2

=

−8e3 x = −2e x because

3 e2 x

1

4

12

365

Continuous

P

$2541.75

$2498.00

$2487.98

$2483.09

$2482.93

1(10)

0.07 ⎞ ⎛ 5000 = P⎜1 + ⎟ 1 ⎠ ⎝ 5000 = P 10 (1.07)

Compounded 4 times:

$2541.75 = P Compounded 12 times:

3

−2 ⋅ −2 ⋅ −2 ⋅ e x ⋅ e x ⋅ e x = −8e3 x .

n

Compounded 1 time:

ex + 2 = e x + 2 − x = e2 ex

0.07 ⎞ ⎛ 5000 = P⎜1 + ⎟ 12 ⎠ ⎝ 5000 = P 120 1.00583

)

$2487.98 = P Compounded continuously:

4(10)

$2498.00 = P 12(10)

(

0.07 ⎞ ⎛ 5000 = ⎜1 + ⎟ 4 ⎠ ⎝ 5000 = P 40 (1.0175)

0.07 ⎞ ⎛ 5000 = P⎜1 + ⎟ 365 ⎠ ⎝

Compounded 365 times: 5000

(1.0001918)

3650

365(10)

= P

$2483.09 = P

5000 = Pe0.07(10) 5000 = P e0.7 $2482.93 = P

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Section 9.2 71. p = 25 − 0.40.02 x

(b) p = 25 − 0.4e0.02(125) = 25 − 0.4e 2.5 ≈ $20.13 t 15

(a) v(5) = 64,000( 2)

5 15

(b) v( 20) = 64,000( 2) = 64,000( 2)

= 64,000( 2)

13

≈ $80,634.95

2

291

< 22.

83. When k > 1, the values of f will increase. When 0 < k < 1, the values of f will decrease. When k = 1, the values of f remain constant. 85. g ( s ) =

s − 4,

Domain: [4, ∞)

s − 4 ≥ 0

20 15

s ≥ 4

43

87. y 2 = x − 1

≈ $161,269.89

75.

2 < 2 and 2 > 0, 21 < 2

81. Because 1
0, y > 0

67. log 4 ( x + 8) − 3 log 4 x = log 4 ( x + 8) − log 4 x 3 = log 4

(x

+ 8) x3

, x > 0

69. f (t ) = 80 − log10 (t + 1)

12

= 80 − 12 log10 (t + 1) f ( 2) = 80 − 12 log10 ( 2 + 1) ≈ 74.27 f (8) = 80 − 12 log10 (8 + 1) ≈ 68.55

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 9.4

Properties of Logarithms

75. The Power Property of Logarithms can be used to ⎛ 3x ⎞ condense 2⎜ log 2 ⎟. y⎠ ⎝

⎛ I ⎞ 71. B = 10 log10 ⎜ −12 ⎟ ⎝ 10 ⎠

= 10 log10 I − 10 log10 (10−12 ) = 10 log10 I + 120 log10 (10)

77. 3 ln x + ln y − 2 ln z = ln x 3 + ln y − ln z 2 = ln

= 10 log10 I + 120

B = 10 log10 (10−1 ) + 120

79. 2 ⎣⎡ln x − ln ( x + 1)⎤⎦ = 2 ln

= −10 log10 (10) + 120 = −10 + 120

= ln

73. The expression is obtained by condensing.

83. 5 log 6 (c + d ) −

85.

3

1 13 log 5 ( x + 3) − log 5 ( x − 6) = log 5 ( x + 3) − log 5 ( x − 6) = log 5 3

x3 y z2

x x +1

⎛ x ⎞ = ln ⎜ ⎟ ⎝ x + 1⎠

= 110 decibels

81.

301

2

x2

(x

+ 1)

2

, x > 0

x+3 x−6

1 (c + d ) 5 12 log 6 ( m − n) = log 6 (c + d ) − log 6 ( m − n) = log 6 2 m−n 5

x3 ⎞ 1 1 1⎛ (3 log 2 x − 4 log 2 y ) = (log 2 x3 − log 2 y 4 ) = ⎜ log 2 4 ⎟ = log 2 y ⎠ 5 5 5⎝

5

x3 , y > 0 y4

87. ln 3e2 = ln 3 + ln e 2 = ln 3 + 2 ln e = ln 3 + 2 89. log 5 91. log8

50 =

1 ⎡log 5 2⎣

(52 ⋅ 2)⎤⎦

=

1 2

[2 log5 5 + log5 2]

1 2

[2 +

log 5 2] = 1 +

1 2

log 5 2

97. False; log 3 (u + v) does not simplify.

8 = log8 8 − log8 x3 = 1 − 3 log8 x x3

93. E = 1.4(log10 C2 − log10 C1 )

=

99. True, f ( ax) = log a ax = log a a + log a x

⎛ C ⎞ = 1.4⎜ log10 2 ⎟ C1 ⎠ ⎝

= 1 + log a x = 1 + f ( x)

1.4

⎛C ⎞ = log10 ⎜ 2 ⎟ ⎝ C1 ⎠

101. False; 0 is not in the domain of f. 103. False; f ( x − 3) = ln ( x − 3) ≠ ln x − ln 3

95. True; log 2 8 x = log 2 8 + log 2 x = 3 + log 2 x

105. Choose two values for x and y, such as x = 3 and y = 5, and show the two expressions are not equal. ln 3 3 ≠ ln = ln 3 − ln 5 ln 5 5 0.6826062 ≠ −0.5108256 = −0.5108256

or

e ln e ≠ ln ln e e 1 ≠ ln 1 1 ≠ 0

107.

2x 3

+

2 3

= 4x − 6

2 x + 2 = 12 x − 18 −10 x = −20 x =

20 10

x = 2

109.

5 4 − = 3 x 2x 5 − 8 = 6x −3 = 6 x 3 = x 6 1 − = x 2 −

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

302

Chapter 9

E xponential and Logarithmic Functions 115. g ( x) = −2 x 2 + 4 x − 7

111. x − 4 = 3 x −4 = 3

x − 4 = −3

or

x = 7

x =1

113. g ( x) = −( x + 2)

2

= −2( x − 1) − 5 Vertex: (1, − 5)

1

x-intercepts: none

x −6 −5 −4

= −2( x 2 − 2 x + 1) − 7 + 2

2 y

(−2, 0)

g ( x) = −2( x 2 − 2 x) − 7

−2

1

y 1 x 1 2 3 4 5 6 7 8 9

−1 −2 −3 −4 −5 −6 −7 −8

−4 −5 −6

Vertex: ( −2, 0) x-intercept: ( −2, 0)

(1, −5)

117. f ( x) = 4 x + 9, g ( x) = x − 5

(a)

(f

D g )( x) = f ( x − 5) = 4( x − 5) + 9 = 4 x − 20 + 9 = 4 x − 11

Domain: ( −∞, ∞) (b)

(g

D f )( x) = g ( 4 x + 9) = ( 4 x + 9) − 5 = 4 x + 4

Domain: ( −∞, ∞) 1 , g ( x) = x + 2 x

119. f ( x) =

(a)

(f

D g )( x) = f ( x + 2) =

1 x + 2

Domain: ( −∞, − 2) ∪ ( −2, ∞) (b)

(g

1 1 + 2x ⎛1⎞ + 2 = D f )( x) = g ⎜ ⎟ = x x ⎝ x⎠

Domain: ( −∞, 0) ∪ (0, ∞)

Section 9.5 Solving Exponential and Logarithmic Equations 1. 7 x = 73

x = 3 3.

e1 − x = e 4 1− x = 4 −x = 3 x = −3

5.

32 − x = 81 32 − x = 34 2− x = 4 −x = 2 x = −2

7. 62 x = 36

62 x = 6 2 2x = 2 x =1 9. ln 5 x = ln 22

5 x = 22 x =

22 5

11. ln (3 − x) = ln 10

3 − x = 10 −x = 7 x = −7

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 9.5 13. log 3 ( 4 − 3 x) = log 3 ( 2 x + 9)

e 2 − x = 21

−5 = 5 x

ln e 2 − x = ln 21

−1 = x

2 − x = ln 21 − x = ln 21 − 2

3x = 91 log 3 3x = log 3 91 x = log 3 91 x =

log 91 log 3

x = log 5 8.2 log 8.2 log 5

x ≈ 1.31

2 − x = log 3 8 − x = log 3 8 − 2 log 8 log 3

x ≈ 0.11 10 x + 6 = 250 log 10 x + 6 = log 250 x + 6 = log 250 x = log 250 − 6 x ≈ −3.60 1 ex 4

= 5

e x = 20 ln e x = ln 20 x = ln 20 x ≈ 3.00 25.

4e− x = 24 e− x = 6 ln e− x = ln 6 − x = ln 6 x = −ln 6 x ≈ −1.79

31. 17 − e x 4 = 14

ex 4 = 3

log 3 32 − x = log 3 8

23.

ln 6 2 x ≈ 0.90 x =

−e x 4 = −3

32 − x = 8

x = 2−

29. 4 + e 2 x = 10

2 x = ln 6

log 5 5 x = log 5 8.2

19.

x ≈ −1.04

ln e 2 x = ln 6

5 x = 8.2

x =

x = 2 − ln 21

e2 x = 6

x ≈ 4.11 17.

303

27. 7 + e 2 − x = 28

4 − 3x = 2 x + 9

15.

21.

Solving Exponential and Logarithmic Equations

ln e x 4 = ln 3 x = ln 3 4 x = 4 ln 3 x ≈ 4.39 33. 8 − 12e − x = 7

−12e − x = −1 e− x =

1 12

ln e − x = ln

1 12

− x = ln

1 12

x = −ln

1 12

≈ 2.48

35. 4(1 + e x 3 ) = 84 1 + e x 3 = 21 e x 3 = 20 ln e x 3 = ln 20 x = ln 20 3 x = 3 ln 20 x ≈ 8.99

37. log10 x = −1

10log10 x = 10−1 x = 10−1 x =

1 10

= 0.1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

304

Chapter 9

E xponential and Logarithmic Functions

39. 4 log 3 x = 28 log 3 x = 7 log3 x

3

41.

1 6

A = Pe rt

53. Formula:

Labels:

Principal = P = $10,000

7

= 3

Amount = A = $11,051.71

x = 37

Time = t = 2 years

x = 2187

Annual interest rate = r 11,051.71 = 10,000e r(2)

Equation:

1 3

log 3 x =

11,051.71 = e2 r 10,000

log 3 x = 2 x = 32

1.105171 = e2 r

x = 9

ln 1.105171 = ln e 2 r ln 1.105171 = 2r

43. log10 4 x = 2

10log10 4 x

= 10

2

4 x = 10

2

102 4 100 x = = 25 4 x =

ln 1.105171 = r 2 0.05 ≈ r ≈ 5%

F = 200e −0.5πθ 180

55.

80 = 200e −0.5πθ 180 0.4 = e −0.5πθ 180

45. 2 log 4 ( x + 5) = 3

log 4 ( x + 5) =

ln 0.4 = ln e−0.5πθ 180

3 2

4log4 ( x + 5) = 41.5 x + 5 = 41.5

ln 0.4 =

180 ⎞ ⎟ =θ ⎝ 0.5π ⎠ 105° ≈ θ

(ln 0.4)⎛⎜ −

x = 41.5 − 5 x = 3 47. 2 log10 x = 10

log10 x = 5 x = 105 x = 100,000 49. 3 ln 0.1x = 4

ln 0.1x =

4 3

0.1x = e

x = 10e

43

x ≈ 37.94 51. log10 x + log10 ( x − 3) = 1

log10 x( x − 3) = 1 10log10 x( x − 3) = 101 x( x − 3) = 10 x 2 − 3 x − 10 = 0

(x

57. The one-to-one property can be applied to 3x − 2 = 33 because each side of the equation is in exponential form with the same base. 59. To apply an inverse property to the equation log 6 3 x = 2, exponentiate each side of the equation

using the base 6, and then rewrite the left side of the equation as 3x. 61. 5 x =

43

− 5)( x + 2) = 0

x = 5, x = −2 ( which is extraneous)

−0.5πθ 180

1 125

5 x = 5 −3 so x = −3. 63. 2 x + 2 =

1 16

2 x + 2 = 2− 4 so x + 2 = − 4 x = −6. 65. log 6 3 x = log 6 18 so 3x = 18 x = 6.

x = 5

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 9.5 67. log 4 ( x − 8) = log 4 ( − 4)

Solving Exponential and Logarithmic Equations 77. log10 4 x − log10 ( x − 2) = 1 ⎛ 4x ⎞ log10 ⎜ ⎟ =1 ⎝ x − 2⎠ 4x = 101 x − 2 4 x = 10( x − 2)

No solution log 4 ( − 4) does not exist. 69.

1 x+2 4 = 300 5 4 x + 2 = 1500 log 4 4

x+2

4 x = 10 x − 20 −6 x = −20

= log 4 1500

x + 2 = x =

20 6 10 x = 3 x ≈ 3.33

log 1500 log 4

x =

log 1500 − 2 log 4

x ≈ 3.28

79. log 2 x + log 2 ( x + 2) − log 2 3 = 4

71. 6 + 2 x −1 = 1

2 x −1 = −5

log 2

log 2 2 x −1 = log 2 ( −5)

2

No solution

(

)

= 24

x2 + 2x = 16 3 x 2 + 2 x = 48

73. log 5 ( x + 3) − log 5 x = 1

2

x + 2 x − 48 = 0

⎛ x + 3⎞ log 5 ⎜ ⎟ =1 ⎝ x ⎠ log ⎡⎣( x + 3) x⎦⎤ 5

x ( x + 2) = 4 3

log 2 ⎡ x 2 + 2 x 3⎤ ⎣⎢ ⎦⎥

log 2 ( −5) is not possible.

5

305

(x

+ 8)( x − 6) = 0

x = −8 (which is extraneous)

= 51

x = 6.00

x+3 = 5 x x + 3 = 5x

81.

5000 = 2500e0.09t 5000 = e0.09t 2500

3 = 4x 3 = x 4 0.75 = x

2 = e0.09t ln 2 = ln (e0.09t ) ln 2 = 0.09t

75. log 2 ( x − 1) + log 2 ( x + 3) = 3

ln 2 = t 0.09 7.70 years ≈ t

log 2 ( x − 1)( x + 3) = 3 x 2 + 2 x − 3 = 23 x 2 + 2 x − 11 = 0 x =

−2 ±

4 − 4(1)(−11) 2(1)

= =

−2 ±

4 + 44 2

−2 ± 48 2

x ≈ 2.46 and − 4.46 (which is extraneous)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

306

Chapter 9

E xponential and Logarithmic Functions

⎛ I ⎞ B = 10 log10 ⎜ −16 ⎟ ⎝ 10 ⎠

83.

⎛ I ⎞ 80 = 10 log10 ⎜ −16 ⎟ ⎝ 10 ⎠ ⎛ I ⎞ 8 = log10 ⎜ −16 ⎟ ⎝ 10 ⎠ I ⎞ ⎜ −16 ⎟ 10 ⎝ ⎠

10 =

⎛ 10log10

108 =

I 10−16

8

89. To solve an exponential equation, first isolate the exponential expression, then take the logarithms of both sides of the equation, and solve for the variable.

To solve a logarithmic equation, first isolate the logarithmic expression, then exponentiate both sides of the equation, and solve for the variable. 91. x 2 = −25

x = ± −25 x = ± 5i 93. 9n 2 − 16 = 0

108 ⋅ 10−16 = I

n2 =

10−8 = I

watts per square centimeter 85. (a)

kt = ln

16 9

T − S T0 − S

n = ± 43

78 − 65 85 − 65 1 13 k = ln 3 20 k ≈ −0.144

k (3) = ln

85 − 65 98.6 − 65 20 ln 33.6 t = −0.144 t ≈ 3.6 hours

(b) −0.144t = ln

95.

t 4 − 13t 2 + 36 = 0

(t 2

− 9)(t 2 − 4) = 0 t2 = 9

10 P.M. − 3 hr 36 min = 6:24 P.M.

T − 65 98.6 − 65 T − 65 −0.288 = ln 33.6 − 65 T e −0.288 = 33.6

−0.144( 2) = ln

33.6e −0.288 + 65 = T 90.2°F ≈ T 87. The equation 2 logarithms.

2 x −1 = 32 2 x − 1 = 25 x −1 = 5 x = 6

x −1

= 32 can be solved without

2 x −1 = 30 x − 1 = log 2 30 x = 1 + log 2 30

t2 = 4

t = ±3

97. Verbal Model:

Labels:

t = ±2

Area = Length ⋅ Width Length = x Width = 2.5 x

Equation:

2 x + 2( 2.5 x) = 42 7 x = 42

≈ 3 hours 36 minutes

(c)

16 9

n = ±

x = 6 A = 6 ⋅ 2.5(6) A = 90 in.2

99. Verbal Model:

Labels:

Perimeter = 2 ⋅ Length + 2 ⋅ Width Length = w + 4 Width = w

Equation:

w( w + 4) = 192 w2 + 4 w − 192 = 0

( w − 12)( w + 16)

= 0

w = 12, w = −16 P = 2(12 + 4) + 2(12) = 32 + 24 = 56 km

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 9.6

Applications

307

Section 9.6 Applications r⎞ ⎛ A = P⎜1 + ⎟ n⎠ ⎝

1.

nt

r⎞ ⎛ 11. A = P⎜1 + ⎟ n⎠ ⎝ 12(10)

r ⎞ ⎛ 1004.83 = 500⎜1 + ⎟ 12 ⎝ ⎠ 120

r ⎞ ⎛ 2.00966 = ⎜1 + ⎟ 12 ⎠ ⎝ r 1 120 (2.00966) = 1 + 12 r 1.0058333 ≈ 1 + 12 r 0.0058333 ≈ 12 0.07 ≈ r

12(1)

0.07 ⎞ ⎛ A = 1000⎜1 + ⎟ 12 ⎠ ⎝ A = $1072.29

72.29 1000 = 0.07229 ≈ 7.23%

Effective yield =

r⎞ ⎛ 13. A = P⎜1 + ⎟ n⎠ ⎝

nt

0.06 ⎞ ⎛ A = 1000⎜1 + ⎟ 4 ⎠ ⎝ A = $1061.36

7% ≈ r

3.

nt

A = Pe rt

Effective yield =

2 P = Pe0.08t 2 = e0.08t ln 2 = 0.08t

Effective yield =

365(1)

53.90 = 0.0539 = 5.39% 1000

17. No. The effective yield is the ratio of the year’s interest to the amount invested. This ratio will remain the same regardless of the amount invested.

A = Pe rt 2 P = Pe0.0675t 2 = e0.0675t

y = Cekt

19. (a)

ln 2 = 0.0675t

1 × 106 = 73e k (20)

ln 2 = t 0.0675 t ≈ 10.27 years

1 × 106 = e 20 k 73

0.06 ⎞ ⎛ 7. 8954.24 = 5000⎜1 + ⎟ n ⎠ ⎝

ln

n(10)

1(10)

Yearly compounding 9. A = Pert

A = 1000e0.08(1) A = $1083.29 83.29 1000 = 0.08329 ≈ 8.33%

1 × 106 = 20k 73

1 1 × 106 ln = k 20 73 0.4763 ≈ k

0.06 ⎞ ⎛ 8954.24 = 5000⎜1 + ⎟ 1 ⎠ ⎝ 8954.24 = 8954.24

Effective yield =

61.36 = 0.06136 ≈ 6.14% 1000

0.0525 ⎞ ⎛ 15. A = 1000⎜1 + ⎟ 365 ⎠ ⎝ A = $1053.90

ln 2 = t 0.08 t ≈ 8.66 years

5.

4(1)

y = 73e0.4763t (b)

5300 = 73e0.4763t 5300 73 5300 ln 73 5300 ln 73 0.4763 9 hours

= e0.4763t = 0.4763t

= t ≈ t

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

308

Chapter 9

E xponential and Logarithmic Functions

21. y = Ce kt 5 = Ce k (0) 5 = C

2.5 = 5e k (1620)

y = 5e−0.00043(1000)

0.5 = e1620 k

y ≈ 3.3 grams

1620 k

ln 0.5 = ln e

ln 0.5 = 1620k ln 0.5 = k 1620 −0.00043 ≈ k

23. May 22, 1960: R = log10 I

r⎞ ⎛ A = P⎜1 + ⎟ n⎠ ⎝

33.

9.5 = log10 I

I = 109.5 February 11, 2011:

0.0575 ⎞ ⎛ 1800 = 900⎜1 + ⎟ 4 ⎠ ⎝

R = log10 I 6.8 = log10 I

2 = (1.014375)

I = 106.8

4t

log 2 = t 4 log 1.014375

25. June 13, 2011: R = log10 I

t ≈ 12.14 years

6.0 = log10 I

I = 106.0

kt 35. y = Ce

3 = Ce

R = log10 I

k ( 0)

3 = C

7.6 = log10 I

I = 107.6 July 6 107.6 = 6.0 = 101.6 ≈ 40 June 13 10

The earthquake on July 6 was about 40 times as intense. 27. The formula A = Pe rt is for continuously compounded interest. 29. Substituting 0.06 for r is replacing the variable r, representing the interest rate, with the interest rate in decimal form. 31.

4t

log 2 = 4t log 1.014375

The earthquake of 1960 was about 501 times as intense.

Ratio:

4t

log1.014375 2 = log1.014375 (1.014375)

May 22 109.5 = 6.8 = 102.7 ≈ 501 Ratio: February 11 10

July 6, 2011:

nt

r⎞ ⎛ A = P⎜1 + ⎟ n⎠ ⎝

37.

y = Ce kt 400 = Ce k(0) 400 = C

nt

12 t

0.075 ⎞ ⎛ 5000 = 2500⎜1 + ⎟ 12 ⎠ ⎝ 12 t

2 = (1.00625)

12t

log1.00625 2 = log1.00625 (1.00625) log 2 = 12t log 1.00625

8 = 3e k(2) 8 3 8 ln 3 8 ln 3 8 ln 3 2

= e2k = ln e 2 k = 2k

= k ≈ 0.4904

200 = 400e k(3) 1 2 1 ln 2 1 ln 2 1 ln 2 3

= e3 k = ln e3k = 3k

= k ≈ −0.2310

39. pH = −log10 ⎡⎣H + ⎤⎦

pH = −log10 (9.2 × 10−8 ) ≈ 7.04

log 2 = t 12 log 1.00625 t ≈ 9.27 years

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 9.6 41. pH = −log10 ⎡⎣H + ⎤⎦

x =

2.5 = −log10 ⎡⎣H + ⎤⎦ −2.5 = −log10 ⎡⎣H + ⎤⎦

0.0031623 ≈ H

x =

log10 ⎢⎡H + ⎥⎤ ⎣ ⎦

x =

+

9.5 = −log10 ⎡⎣H + ⎤⎦ −9.5 = −log10 ⎡⎣H + ⎤⎦ log10 ⎢⎡H + ⎥⎤ ⎣ ⎦

x =

H + of fruit 3.1623 × 10−3 ≈ = 107 + H of tablet 3.1623 × 10−10

x =

The H + of fruit is 107 times as great.

x =

10.9 1 + 0.80e − 0.031t

10.9 1 + 0.80e − 0.031(22) ≈ 7.761 billion people

When t = 22: P =

45. (a) p(0) =

(b) p(9) = (c)

= 2.5

e

−t 6

= 0.375

t = ln 0.375 6 t = −6 ln 0.375 t ≈ 5.88 years

47. If k > 0, the model represents exponential growth and if k < 0, the model represents exponential decay. 49. When the investment is compounded more than once in a year (quarterly, monthly, daily, continuous), the effective yield is greater than the interest rate.

49 + 20 2

7 ±

69

− 4(1)( −5)

2 69 2

92 − 4(3)( 4)

−9 ±

2(3)

−9 ±

81 − 48 6

−9 ± 33 6 3 ± 2

33 6

4 > 0 x − 4 Critical number: 4 Test intervals: Negative: ( −∞, 4) Positive: ( 4, ∞) Solution: ( 4, ∞)

5000 1 + 4e − t 6

1 + 4e

−

55.

5000 ≈ 2642 rabbits 1 + 4e −9 6

−t 6

7 ±

x = −

5000 5000 = = 1000 rabbits 1 + 4e −0 6 5

2000 =

( −7 ) 2 2(1)

53. 3x 2 + 9 x + 4 = 0

3.1623 × 10−10 ≈ H +

43. P =

−( −7) ±

7 x = ± 2

Tablet:

10−9.5 = 10

309

51. x 2 − 7 x − 5 = 0

Fruit:

10−2.5 = 10

Applications

x 0

2

4

6

8

2x >1 x −3

57.

2x −1 > 0 x −3 2 x − ( x − 3) > 0 x −3 x +3 > 0 x −3 Critical numbers: −3, 3 Test intervals: Positive: ( −∞, − 3) Negative: ( −3, 3) Positive: (3, ∞) Solution: ( −∞, − 3) ∪ (3, ∞) −3

3 x

−6 −4 −2

0

2

4

6

8

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

310

Chapter 9

E xponential and Logarithmic Functions

Review Exercises for Chapter 9 9. f ( x) = 3( x + 1)

1. f ( x) = 4 x

y

1 1 = 43 64

(a) f ( −3) = 4−3 =

6 5

(b) f (1) = 41 = 4

4 3

(c) f ( 2) = 42 = 16

1

3. g (t ) = 5

−t 3

x

−4 −3 −2 −1 −1

(a) g ( −3) = 5−(−3) 3 = 51 = 5 (b) g (π ) = 5 (c) g (6) = 5

−π 3

−6 3

= 5

=

5. f ( x) = 3x

2

3

Table of values:

≈ 0.185 −2

1

1 25

x

–1

01

y

1

3

9

11. f ( x) = 3e −2 x y

(a) f (3) = 3e −2(3) = 3e −6 ≈ 0.007

6

(b) f (0) = 3e −2(0) = 3e0 = 3

5 4

(c) f ( −19) = 3e−2(−19) = 3e38 ≈ 9.56 × 1016

3 2 1 x

−4 −3 −2 −1 −1

1

2

3

–1

y

1 3

y-intercept: (0, 2)

01 1

y = f ( − x) + 1. To sketch y, reflect the graph of f in the y-axis, and shift the graph one unit upward.

Table of values: x

13. The function y is related to f ( x) = e x by

Asymptote: y = 1

3

y

7. f ( x) = 3x − 3

6 5

y

4

3

3

2

2

1 x

−3

−2

−1

2 −1

3

−3

−2

−1

x 1

2

3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 9 15. The function g is related to f ( x) = e x by

y = f ( x + 2). To sketch g, shift the graph of f two

( 12 ) = 21( 12 )

19. y = 21

y

units to the left.

t 25

,

311

t ≥ 0

58 25

y ≈ 4.21 grams

y-intercept: (0, e2 )

21. (a)

Asymptote: y = 0

(f

D g )( x) = x 2 + 2

so ( f D g )( 2) = 22 + 2 = 6.

y 10

(b)

8

(g

D f )( x) = ( x + 2) = x 2 + 4 x + 4 2

so ( g D f )( −1) = ( −1) + 4( −1) + 4 2

6

= 1 − 4 + 4 = 1. 2 −6

−4

23. (a)

x

−2

2

r⎞ ⎛ 17. (a) A = P⎜1 + ⎟ n⎠ ⎝

4

(b) 1( 40)

0.10 ⎞ ⎛ = 5000⎜1 + ⎟ 1 ⎠ ⎝

(a)

nt

x2 = x

D f )( x) =

(

x +1

)

2

−1 = x +1−1 = x

(f

x + 6, g ( x) = 2 x D g )( x) = f ( 2 x) =

2x + 6

Domain: [−3, ∞) 2x + 6 ≥ 0

4( 40)

x ≥ −3

(b)

= 5000(1.025)

160

(g

D f )( x) = g

(

)

x + 6 = 2

x + 6

Domain: [−6, ∞)

≈ $259,889.34

x +6 ≥ 0

nt

x ≥ −6 12( 40)

0.10 ⎞ ⎛ = 5000⎜1 + ⎟ 12 ⎠ ⎝

(

= 5000 1.0083

)

480

f ( x) = 3 x + 4 y = 3x + 4 x = 3y + 4

nt

0.10 ⎞ ⎛ = 5000⎜1 + ⎟ 365 ⎠ ⎝

27. No, f ( x) does not have an inverse. f is not one-to-one. 29.

≈ $268,503.32

r⎞ ⎛ (d) A = P⎜1 + ⎟ n⎠ ⎝

(g

25. f ( x) =

40

0.10 ⎞ ⎛ = 5000⎜1 + ⎟ 4 ⎠ ⎝

r⎞ ⎛ (c) A = P⎜1 + ⎟ n⎠ ⎝

x2 − 1 + 1 =

so ( g D f )( −1) = −1.

≈ $226,296.28 r⎞ ⎛ (b) A = P⎜1 + ⎟ n⎠ ⎝

D g )( x) =

so ( f D g )(5) = 5 = 5.

nt

= 5000(1.10)

(f

x − 4 = 3y 365( 40)

≈ 5000(1.00027397)

14600

x − 4 = y 3 1 x − 4 = f −1 ( x) = ( x − 4) 3 3

≈ $272,841.23 (e) A = Pe rt = 5000e0.10(40) = 5000e4 ≈ $272,990.75

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

312

Chapter 9

E xponential and Logarithmic Functions

31. h( x) =

5x

y =

5x

x =

5y

47. f ( x) = −1 + log 3 x y

1

x2 = 5 y

= y

1 2 x 5

= h −1 ( x ),

4

5

−3 −4

Vertical asymptote: x = 0

y = t3 + 4

49. f ( x) = log 2 ( x − 4)

t = y3 + 4 t − 4 = y 3

3

−2

x ≥ 0

f (t ) = t 3 + 4

3

1 −1

1 x2 5

33.

x

−1

3

y

t − 4 = y

8

−1

t − 4 = f

(t )

6 4 2

35.

y

x

f(x) = 3x + 4

8

(1, 7)

10 12 14

−6 3

Vertical asymptote: x = 4

(7, 1) (4, 0) x 4

6

8

Table of values:

(1, −1)

37. log10 1000 = 3 because 103 = 1000. 1 9

8

(0, 4) g(x) = 1 (x − 4)

(−1, 1)

39. log 3

6

−4

6 4

2

−2

= −2 because 3−2 = 19 .

41. log 2 64 = 6 because 26 = 64.

x

5

6

y

0

1

51. ln e7 = 7 53. y = ln ( x − 3) y

0

43. log 3 1 = 0 because 3 = 1.

6 4

45. f ( x) = log 3 x

2

y

x 2

3

−2

2

−4

4

6

8

10

1 x

−1

1

2

3

4

−1 −2

Vertical asymptote: x = 0

Table of values: x

4

5

y

0

0.7

Table of values: x

1

3

y

0

1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 9 55. y = 5 − ln x

⎛ 2x ⎞ 77. −2(ln 2 x − ln 3) = ln ⎜ ⎟ ⎝ 3⎠

y

313

−2

2

⎛ 3⎞ = ln ⎜ ⎟ ⎝ 2x ⎠ 9 = ln 2 , x > 0 4x

8 6 4 2

x −2

2

4

6

⎡ ⎛ k ⎞⎤ 79. 4 ⎡⎣log 2 k − log 2 ( k − t )⎤⎦ = 4 ⎢log 2 ⎜ ⎟⎥ ⎝ k − t ⎠⎦ ⎣

8

−2

4

⎛ k ⎞ = log 2 ⎜ ⎟ , t < k, k > 0 ⎝k − t⎠

Table of values: x

1

e

y

5

4

81. False log 2 4 x = log 2 4 + log 2 x = 2 + log 2 x

57. (a) log 4 9 =

(b) log 4 9 =

log 9 ≈ 1.5850 log 4

83. True

log10 102 x = 2 x log10 10 = 2 x

ln 9 ≈ 1.5850 ln 4

85. True

log 4

log 160 59. (a) log8160 = ≈ 2.4409 log 8

⎛ I ⎞ 87. B = 10 log10 ⎜ −12 ⎟ ⎝ 10 ⎠

ln 160 ≈ 2.4409 ln 8

(b) log8160 =

⎛ 10− 4 ⎞ = 10 log10 ⎜ −12 ⎟ ⎝ 10 ⎠

61. log 5 18 = log 5 32 + log 5 2

= 10 log10 (108 )

= 2 log 5 3 + log 2 ≈ 2(0.6826) + 0.4307

= 10 ⋅ 8

≈ 1.7959 63. log 5

1 2

= 80 decibels

= log 5 1 − log 5 2

89. 2 x = 64

≈ 0 − (0.4307)

2 x = 26

≈ −0.4307 65. log 5 (12)

23

= ≈

x = 6

[2 log5 2 + log5 3] 2 ⎡2 0.4307 + 0.6826⎤ ) ⎦ 3⎣ ( 2 3

91.

1 2

x =1 93. log 7 ( x + 6) = log 7 12

log 5 ( x + 2)

x + 6 = 12 x = 6

x + 2 71. ln = ln ( x + 2) − ln ( x + 3) x + 3

95.

75. log8 16 x + log8 2 x = log8 (16 x ⋅ 2 x

3x = 500 log 3 3x = log 3 500

73. 5 log 2 y = log 2 y 5 2

1 16

x − 3 = −2

67. log 4 6 x 4 = log 4 6 + 4 log 4 x

x + 2 =

4x − 3 =

4 x − 3 = 4−2

≈ 1.0293

69. log 5

16 = log 4 16 − log 4 x = 2 − log 4 x x

2

) = log8 (32 x ) 3

x =

log 500 log 3

x ≈ 5.66

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

314

Chapter 9

E xponential and Logarithmic Functions

2e0.5 x = 45

97.

111.

5751.37 = 5000e r(2)

e0.5 x = 22.5 ln e0.5 x = ln 22.5

5751.37 = e2r 5000

0.5 x = ln 22.5

ln 1.150274 = ln e 2 r

x = 2 ln 22.5

ln 1.150274 = 2r

x ≈ 6.23

ln 1.150274 = r ≈ 7% 2

99. 12(1 − 4 x ) = 18

1 − 4x =

18 12

x

−4 =

3 2

−4 x =

1 2

−1

113.

r⎞ ⎛ 1.6436 = ⎜1 + ⎟ 4 ⎝ ⎠ r 1 40 (1.6436) = 1 + 4 r 1.0124997 ≈ 1 + 4 r 0.0124997 ≈ 4 0.0499 ≈ r

No solution; there is no power that will raise 4 to − 12 . 101. ln x = 7.25

eln x = e7.25 x = e7.25 x ≈ 1408.10 103. log10 4 x = 2.1 4 x = 102.1

105. log 3 ( 2 x + 1) = 2

3log3(2 x +1) = 32 2x + 1 = 9 2x = 8

x = 4.00 1 3

log 2 x + 5 = 7 1 3

nt

4(10)

40

5% ≈ r

2.1

10 4 x ≈ 31.47

107.

r⎞ ⎛ A = P⎜1 + ⎟ n ⎝ ⎠

r⎞ ⎛ 410.90 = 250⎜1 + ⎟ 4⎠ ⎝

4 x = − 12

x =

A = Pe rt

log 2 x = 2 log 2 x = 6 2log2 x = 26 x = 26 = 64.00

115.

r⎞ ⎛ A = P⎜1 + ⎟ n ⎝ ⎠

nt

r ⎞ ⎛ 15,399.30 = 5000⎜1 + ⎟ 365 ⎠ ⎝ r ⎞ ⎛ 3.07986 = ⎜1 + ⎟ 365 ⎝ ⎠ r 1 5475 = 1+ (3.07986) 365 r 1.000205479 ≈ 1 + 365 r 0.000205479 ≈ 365 0.074999 ≈ r

365(15)

5475

7.5% ≈ r

109. log 3 x + log 3 7 = 4

log 3 7 x = 4 7 x = 34 34 7 x ≈ 11.57

x =

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 9 A = Pert

117.

r⎞ ⎛ 121. A = P⎜1 + ⎟ n⎠ ⎝

46,422.61 = 1800er(50) 46,422.61 = e50 r 1800 ln 25.790338 = 50r

Effective yield =

ln 25.790338 = r 50 0.065 ≈ r

A = $1077.88

nt

Effective yield =

Effective yield = y = Ce kt 3.5 = Ce k(0) 3.5 = C

77.14 = 0.07714 ≈ 7.71% 1000

A = 1000e0.075(1)

0.055 ⎞ ⎛ A = 1000⎜1 + ⎟ 365 ⎠ ⎝ A = $1056.54

125.

4(1)

123. A = Pe rt

6.5% ≈ r r⎞ ⎛ 119. A = P⎜1 + ⎟ n⎠ ⎝

nt

0.075 ⎞ ⎛ A = 1000⎜1 + ⎟ 4 ⎠ ⎝ A = $1077.14

25.790338 = e50 r

315

365(1)

77.88 = 0.07788 ≈ 7.79% 1000

56.54 = 0.05654 ≈ 5.65% 1000 1.75 = 3.5e k(1620) 0.5 = e1620 k

y = 3.5e−0.00043(1000) y ≈ 2.282 grams

1620 k

ln 0.5 = ln e

ln 0.5 = 1620k ln 0.5 = k 1620 −0.00043 ≈ k

127. April 18, 1906: R = log10 I

8.3 = log10 I I = 108.3 March 5, 2012:

R = log10 I 4.0 = log10 I I = 104.0

Ratio:

April 18 108.3 = 4.0 = 104.3 ≈ 19,953 March 5 10

The earthquake of 1906 was about 19,953 times as intense.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

316

Chapter 9

E xponential and Logarithmic Functions

Chapter Test for Chapter 9

( 23 ) = 54( 32 )

1. (a) f ( −1) = 54

f ( x) = 9 x − 4

4.

−1

y = 9x − 4 x = 9y − 4

= 81

( 23 )

(b) f (0) = 54

x + 4 = 9y 0

x + 4 = y 9

= 54 (c) f

1 ( x + 4) = f −1( x) 9

( ) = 54( ) 1 2

2 3

12

≈ 44.09

() = 54( 74 )

(d) f ( 2) = 54

2 3

5. f ( g ( x)) = f ( −2 x + 6)

2

= − 12 ( −2 x + 6) + 3 = x −3+ 3 = x

= 24

(

)

g ( f ( x)) = g − 12 x + 3

2. f ( x) = 2 x − 5

(

= −2

y

− 12 x

)

+3 + 6

= x −6+6

18

= x

15 12

6. log 4

9 6

x 3

6

9

= −4 because 4−4 =

1 . 256

7. The graph of g is a reflection in the line y = x of the graph of f.

3 −6 −3 −3

1 256

12 15

y

Horizontal asymptote: y = 0

12 10

3. (a) f D g = f ( g ( x)) = f (5 − 3 x) = 2(5 − 3 x) + (5 − 3 x) 2

= 2( 25 − 30 x + 9 x 2 ) + 5 − 3 x 2

= 50 − 60 x + 18 x + 5 − 3x = 18 x 2 − 63x + 55

Domain: ( −∞, ∞) (b) g D f = g ( f ( x)) = g ( 2 x 2 + x)

g

8 6 4

f

2

x 2

4

10 12

x − log8 y 4

= log8 4 + log8 x1 2 − log8 y 4 =

= 5 − 6 x 2 − 3x Domain: ( −∞, ∞)

8

⎛4 x ⎞ 8. log8 ⎜⎜ 4 ⎟⎟ = log8 4 ⎝ y ⎠

= 5 − 3( 2 x 2 + x) = −6 x 2 − 3 x + 5

6

2 1 + log8 x − 4 log8 y 3 2

9. ln x − 4 ln y = ln

x , y > 0 y4

10. log 2 x = 5

2log2 x = 25 x = 32

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test for Chapter 9 11.

17. 30(e x + 9) = 300

92 x = 182 log 9 92 x = log 9 182 2x =

e x + 9 = 10

log 182 log 9

ex = 1 e x = e0

log 182 x = 2 log 9 x ≈ 1.18

so x = 0. 0.07 ⎞ ⎛ 18. (a) A = 2000⎜1 + ⎟ 4 ⎠ ⎝

12. 400e0.08t = 1200

ln e0.08t = ln 3

19.

0.08t = ln 3 ln 3 0.08 t ≈ 13.73 t =

13. 3 ln ( 2 x − 3) = 10

e

ln( 2 x − 3)

10 3

20.

e10 3 + 3 x = 2 x ≈ 15.52

7 − 2 = −25

ln

−2 x = −32

3 4

= ek

3 4

= k

y ≈ $4746.09

2 x = 25

22. p(0) =

2400 = 600 foxes 1 + 3e −0 4

23. p( 4) =

2400 ≈ 1141 foxes 1 + 3e −4 4

15. log 2 x + log 2 4 = 5

4 x = 32 x = 8 16. ln x − ln 2 = 4

e

x = 4 2

ln( x 2)

= e

4

x = e4 2 x = 2e 4 x ≈ 109.20

≈ r ≈ r

ln 3 4 5 y = 20,000e ⎡⎣ ( )⎤⎦( )

2 x = 32

2log2 4 x = 25

= r

15,000 = 20,000e k(1)

x

log 2 x( 4) = 5

= ln e10 r = 10r

y = Ce kt

21.

300 12

x = 5

4( 25)

1006.88 = 500e r(10) ln 2.01376 ln 2.01376 ln 2.01376 10 0.07 7%

= e

14. 12(7 − 2 x ) = −300

0.09 ⎞ ⎛ 100,000 = P⎜1 + ⎟ 4 ⎠ ⎝ 100,000 = P 100 (1.0225)

2.01376 = e10 r

2 x − 3 = e10 3

ln

≈ $8012.78

$10,806.08 ≈ P

10 3

7 − 2x = −

4( 20)

(b) A = 2000e0.07(20) ≈ $8110.40

e0.08t = 3

ln ( 2 x − 3) =

317

24.

1200 = 1 + 3e − t 4 = 3e − t 4 = e−t 4 =

ln e − t 4 = −

t = 4 t =

2400 1 + 3e − t 4 2400 1200 1 1 3 1 ln 3 1 ln 3 1 −4 ln ≈ 4.4 years 3

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C H A P T E R Conics

1 0

Section 10.1

Circles and Parabolas .........................................................................319

Section 10.2

Ellipses ................................................................................................322

Mid-Chapter Quiz......................................................................................................325 Section 10.3

Hyperbolas ..........................................................................................327

Section 10.4

Solving Nonlinear Systems of Equations..........................................331

Review Exercises ........................................................................................................336 Chapter Test .............................................................................................................342 Cumulative Test for Chapters 8–10........................................................................345

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

C H A P T E R Conics

1 0

Section 10.1 Circles and Parabolas 15. Center: (6, −5); radius: 3

1. The graph is a parabola.

(x

3. The graph is a circle.

2

x + y = r

2

x 2 + y 2 = 52 x 2 + y 2 = 25

2

x + y = r

2

2

2

2

4 9

x + y =

r =

⎡0 − (−2)2 ⎤ + (1 − 1)2 ⎣ ⎦

r =

4+0

2 3

(x

2

( 23 )

x2 + y2 =

− h) + ( y − k ) = r 2 2

2

2

⎡⎣ x − ( −2)⎤⎦ + ( y − 1) = 22

2 2

(x

2

or 9 x + 9 y = 4

2

+ 2) + ( y − 1) = 4 2

2

x2 + y2 + 2 x + 6 y + 6 = 0

19.

9. x + y = 16

x 2 + 2 x + y 2 + 6 y = −6

( x2 + 2 x + 1) + ( y 2 + 6 y + 9)

Center: (0, 0); radius: 4

(x

y 5

1 2 3

5

1 −6 −5 − 4 − 3 −2 −1

−2 −3

−4 −6

Center: (0, 0); radius: 6

−7

y

21. The center of the circle is (0, 0) and the radius of the

8

circle r = 40 feet. This implies

4

(x

2

− 0) + ( y − 0) = 402 2

x 2

4

2

x 2 + y 2 = 1600

8

−4

y = ± 1600 − x 2 . Finally, take the positive square root to obtain the

−8

13. Center: ( 4, 3); radius: 10

(x

− h) + ( y − k ) = r 2

(x

− 4) + ( y − 3) = 102

(x

− 4) + ( y − 3) = 100

2

2 2

2

−3

11. x 2 + y 2 = 36

−4 −2

x 1

−2

−5

−8

2

y x

−3 −2 −1

= −6 + 1 + 9

+ 1) + ( y + 3) = 4 2

Center: ( −1, −3); radius: 2

3 2 1 −5

2

r = 2

7. Center: (0, 0); radius: 2

2

17. Center: ( −2, 1); point: (0, 1)

5. Center: (0, 0); radius: 5 2

− 6) + ( y + 5) = 9

2

2 2

equation for the semicircle, y =

1600 − x 2 .

23. Vertex: (0, 0), focus: (0, − 32 ), vertical parabola x 2 = 4 py p = − 32

x 2 = 4( − 32 ) y x 2 = −6 y

319

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

320

Chapter 10

Conics

25. Vertex: (0, 0), focus: ( −2, 0), horizontal parabola

35.

(x

− 1) + 8( y + 2) = 0 2

(x

y 2 = 4 px p = −2

− 1) = −8( y + 2) 2

Vertical parabola

y = 4( −2) x 2

4 3

4 p = −8

y 2 = −8 x

2

p = −2

− k ) = 4 p ( x − h)

(y

− 2) = 4(−2)( x − 3)

(y

− 2) = −8( x − 3)

x

−3 −2 −1

Focus: (1, − 4)

p = −2

(y

1

Vertex: (1, −2)

27. Vertex: (3, 2), focus: (1, 2), horizontal parabola

1

2

3

4

5

−3 −4

2

37.

2 2

( y + 12 )

2

= 2( x − 5)

y

Horizontal parabola

4

4p = 2 29.

y

y =

1 x2 2

p = y

2 y = x2

p =

2 4

3

−

1 2

2 1

Vertex: (0, 0)

−3

( )

−2

−1

Focus: 0, 12

x 1

2

3

x 2

39.

y =

4

6

8

2

4

6

−2 −4

1 3

( x2

− 2 x + 10)

2

x − 2 x + 10 = 3 y

−1

x 2 − 2 x + 1 = 3 y − 10 + 1

31. y 2 = −10 x

y

Horizontal parabola

4

4 p = −10

(x

− 1) = 3 y − 9

(x

− 1) = 3( y − 3)

2

y

2

6

Vertical parabola

p = − 52 −6

Vertex: (0, 0)

(

=

( ) Focus: ( 11 , − 12 ) 2

4

4p = 2

2

1 2

Vertex: 5, − 12

5

Vertical parabola

2 4

−4

)

Focus: − 52 , 0

Vertex: (0, 0) Focus: (0, − 2)

−1

−2

( )

41. y 2 + 6 y + 8 x + 25 = 0 x 3

4

( y 2 + 6 y + 9)

−2 −3 −4 −5 −6

x

−2

Focus: 1, 15 4

1 −4 −3

2

3 4

Vertex: (1, 3)

2

Vertical parabola

4p = 3 p =

y

x 2 = −8 y

p = − 84 = −2

2

−4

33. x 2 + 8 y = 0

4 p = −8

x

−2

= −8 x − 25 + 9

(y

+ 3) = −8 x − 16

(y

+ 3) = −8( x + 2)

2 2

y

Horizontal parabola −2

4 p = −8 p = −2 Vertex: ( −2, −3) Focus: ( − 4, −3)

− 10

−8

−6

x

−4 −2 −4 −6 −8

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Section 10.1 43. First write the equation in standard form: x 2 + y 2 = 62. Because h = 0 and k = 0, the center

Vertical parabola x 2 = 4 py 602 = 4 p( 20)

45. The equation ( x − h) = 4 p( y − k ), p ≠ 0, should be 2

3600 = 4 p( 20)

used because the vertex and the focus lie on the same vertical axis. 47.

( x + 94 ) Center:

(

180 = 4 p x 2 = 180 y

+ ( y − 4) = 16 2

x2 = y 180

); radius: 4

− 94 , 4

(b) Let x = 0.

y

y = 6

2 −4

y = x

−2

2

4

2

y =

2

x + y + 10 x − 4 y − 7 = 0 x 2 + y 2 + 10 x − 4 y = 7

(x

2

+ 10 x + 25) + ( y − 4 y + 4) = 7 + 25 + 4 2

402 1600 80 8 = = = 8 180 180 9 9

Let x = 60.

2

(x

202 400 20 2 = = = 2 180 180 9 9

Let x = 40.

−2

49.

02 = 0 180

Let x = 20.

4

−6

321

53. (a) Vertex: (0, 0), point: (60, 20)

is the origin. Because r 2 = 62 , the radius is six units.

2

Circles and Parabolas

y =

2

+ 5) + ( y − 2) = 36

Center: ( −5, 2); radius: 6 y

602 3600 = = 20 180 180

x

0

20

40

60

y

0

2 92

8 89

20

10 8

55. No. The equation of the circle is

(x

4

−8 −6 −4 −2 −4

2

2

51. x + y = r

2

x 2 + y 2 = 752

x 2

2

2

satisfy the equation.

2 −12

+ 1) + ( y − 1) = 13, and the point (3, 2) does not

57. No. For each x > 0, there correspond two values of y. 59. Yes. The directrix of a parabola is perpendicular to the line through the vertex and focus. 61.

x2 + 6x = − 4 x2 + 6 x + 9 = − 4 + 9

(x

+ 3) = 5 2

x +3 = ± 5 x = −3 ±

5

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

322

Chapter 10

Conics

63. 4 x 2 − 12 x − 10 = 0

67. log10

2

4 x − 12 x = 10 5 x − 3x = 2 9 5 9 = + x 2 − 3x + 4 2 4 2

69. ln

2

3⎞ 19 ⎛ ⎜x − ⎟ = 2⎠ 4 ⎝ x −

3 ± 2

65.

1 2

log10 ( xy 3 )

=

1 ⎡log x 10 2⎣

+ log10 y 3 ⎤⎦

=

1 2

(log10 x

+ 3 log10 y )

x = ln x − ln y 4 = ln x − 4 ln y y4

71. 2 log 3 x − log 3 y = log 3 x 2 − log 3 y = log 3

3 19 = ± 2 4 x =

xy 3 =

x2 y

19 2

9 x 2 − 12 x = 14 x 2 − 43 x = x2 −

4x 3

+

4 9

( x − 23 ) x −

2

2 3

14 9

=

14 9

=

18 9

+

2

= ±

x =

2 3

4 9

±

2

73. 4(ln x + ln y ) − ln ( x 4 + y 4 ) = 4 ln x + 4 ln y − ln ( x 4 + y 4 ) = ln x 4 + ln y 4 − ln ( x 4 + y 4 ) = ln

x4 y 4 x + y4 4

Section 10.2 Ellipses 1. Because the major axis is vertical, the standard form of x2 y2 the equation of the ellipse is 2 + 2 = 1. b a 3. Because the major axis is horizontal, the standard form x2 y2 of the equation of the ellipse is 2 + 2 = 1. a b 5. Center: (0, 0)

Vertices: ( − 4, 0), ( 4, 0) Co-vertices: (0, −3), (0, 3) x2 y2 + 2 =1 2 a b

7. Center: (0, 0)

Vertices: (0, −6), (0, 6) Co-vertices: ( −3, 0), (3, 0) x2 y2 + 2 =1 2 b a

Major axis is y-axis so b = 3. Minor axis is x-axis so a = 6.

x2 y2 + 2 =1 2 3 6 x2 y2 + =1 9 36

Major axis is x-axis so a = 4. Minor axis is y-axis so b = 3. x2 y2 + 2 =1 2 4 3 x2 y2 + =1 16 9

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 10.2 9. Vertices: ( − 4, 0), ( 4, 0)

323

19. Center: ( 4, 0)

Co-vertices: (0, 2), (0, − 2)

Vertices: ( 4, 4), ( 4, − 4) Co-vertices: (1, 0), (7, 0)

y 5 4 3

Major axis is vertical so a = 4. Minor axis is horizontal so b = 3.

1 −5

Ellipses

x

−2 −1

1 2 3

(x

5

−3 −4 −5

− 4) ( y − 0) + 32 42 2

(x

11. 4 x 2 + y 2 − 4 = 0 2

2

=1

− 4) y2 + =1 9 16 2

21. 4( x − 2) + 9( y + 2) = 36 2

2

x y + =1 1 4

(x

Vertices: (0, 2), (0, − 2)

2

− 2) ( y + 2) + 9 4 2

2

=1

Center: ( 2, − 2)

Co-vertices: (1, 0), ( −1, 0)

Vertices: ( −1, − 2), (5, − 2)

y 3

a2 = 9

b2 = 4

a = 3

b = 2

1 −3

y

x

−2

3

2

2 1

13. Because the major axis is vertical, the standard form of

the equation of the ellipse is

(x

− b) b2

2

+

(y

− k) a2

− b)

2

a2

+

(y

− k)

2

b2

= 1.

1

5

6

−6

23.

9 x 2 + 4 y 2 + 36 x − 24 y + 36 = 0

(9 x 2 + 36 x) + (4 y 2 − 24 y) = 9( x 2 + 4 x + 4) + 4( y 2 − 6 y + 9) =

−36 −36 + 36 + 36

9( x + 2) + 4( y − 3) = 36 2

(x

Vertices: ( −1, 2), (5, 2) Co-vertices: ( 2, 0), ( 2, 4) Major axis is horizontal so a = 3. Minor axis is vertical so b = 2.

(x

− 2) ( y − 2) + 2 3 22

2

(x

− 2) ( y − 2) + 9 4

2

2

3

−5

= 1.

17. Center: ( 2, 2)

2

2

−4

2

15. Because the major axis is horizontal, the standard form of the equation of the ellipse is

(x

x

−2 −1

−3

=1 =1

2

+ 2) ( y − 3) + 4 9 2

2

=1

Center: ( −2, 3)

y 6

Vertices: ( −2, 0), ( −2, 6)

4

a2 = 9

b2 = 4

a = 3

b = 2

2

−6

−4

x

−2

2 −2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

324

Chapter 10

Conics

25 x 2 + 9 y 2 − 200 x + 54 y + 256 = 0

25.

35. 16 x 2 + 25 y 2 − 9 = 0

(25x 2 − 200 x) + (9 y 2 + 54 y) = −256 25( x 2 − 8 x + 16) + 9( y 2 + 6 y + 9) = −256 + 400 + 81

16 x 2 25 y 2 + =1 9 9

( x − 4)2 9

2

+

( y + 3)2 25

a 2 = 25

b2 = 9

a = 5

b = 3

4 2 −4 −2

x 2

4

6

8 10 12

−4 −6 −8 −10

Length of major axis: 1230 Length of minor axis: 580 2a = 1230

2b = 580

a = 615

b = 290

x2 y2 + =1 2 615 2902

or x2 y2 + =1 2 290 6152

29. An ellipse is the set of all points ( x, y ) such that the sum

of the distances between ( x, y ) and two distinct fixed

37.

x2 y2 x2 y2 + 2 = 1 or 2 + 2 = 1 2 a b b a

31. The length of the major axis is 2a and the length of the minor axis is 2b. 33. Center: (0, 0)

Major axis (vertical) 10 units Minor axis 6 units

2

2

x y + 2 =1 32 5 x2 y2 + =1 9 25

1

2

−1 −2

x2 y2 + =1 324 196

a 2 = 324

b 2 = 196

a = 18

b = 14

2a = 36

2b = 28

24 feet = shortest distance 39. (a) Every point on the ellipse represents the maximum distance (800 miles) that the plane can safely fly

with enough fuel to get from airport A to airport B.

(b) Center: (0, 0)

Airport A: ( −250, 0) Airport B: ( 250, 0)

2c = 500

c = 250 (c) Airplane flies maximum of 800 miles without refueling, so 800 − 500 = 300. 300 = twice the distance from the airport to the vertex. 150 = the distance from the airport to the vertex.

Vertices: ( ±400, 0)

points is a constant.

b = 3, a = 5

x

−1

36 feet = longest distance

27. Center: (0, 0)

x2 y2 + 2 =1 2 b a

−2

3⎞ ⎛ Co-vertices: ⎜ 0, ± ⎟ 5 ⎝ ⎠

6

Vertices: ( 4, 2), ( 4, −8)

1

⎛ 3 ⎞ Vertices: ⎜ ± , 0 ⎟ ⎝ 4 ⎠

=1 y

Center: ( 4, −3)

2

x2 y2 + =1 9 16 9 25

25( x − 4) + 9( y + 3) = 225 2

y

−250 − 150 = − 400

250 + 150 = 400

(d)

c2 = a 2 − b2 2502 = 4002 − b 2

x2 y2 + 2 =1 2 a b

4002 − 2502 = b 2 97,500 = b 2

x2 y2 + 2 400 50 39

(

97,500 = b 50 39 = 6 (e) A = π ab

(

A = π ( 400) 50 39

)

2

=1

)

= 20,000 39π ≈ 392,385 square miles

41. A circle is an ellipse in which the major axis and the minor axis have the same length. Both circles and ellipses have foci; however, in a circle the foci are both at the same point, whereas in an ellipse they are not.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Mid-Chapter Quiz for Chapter 10 43. The sum of the distances between each point on the ellipse and the two foci is a constant. 45. The graph of an ellipse written in the standard form

(x

− h) ( y − k ) = 1 intersects the y-axis if + a2 b2 h > a and intersects the x-axis if k > b. Similarly, 2

2

(x

− h) ( y − k ) = 1 intersects the + b2 a2 y-axis if h > b and intersects the x-axis if k > a.

the graph of

2

325

51. h( x) = log16 4 x

(a) h( 4) = log16 4( 4) = log16 16 = 1 (b) h(64) = log16 4(64) = log16 256 = log16 162 = 2 y 4

2

2 x

−2

2

4

6

−2

47. f ( x) = 3− x

−4

53. f ( x) = log 4 ( x − 3)

(a) f ( −2) = 3−(−2) = 32 = 9 (b) f ( 2) = 3−2 =

(a) f (3) = log 4 (3 − 3) = log 4 0 = does not exist

1 1 = 2 3 9

(b) f (35) = log4(35 − 3)

y

= log 4 32 = log 4 25

4

= 5 log 4 2 = 5 ⋅

1 2

=

5 2

y x

−2

2

4

4 2

−2

x

−2

4

6

8

−2

49. g ( x) = 6e0.5 x

−4

(a) g ( −1) = 6e0.5(−1) (b) g ( 2) = 6e

2

0.5( 2)

−6

≈ 16.310 y

8 6

2

−8

−6

−4

−2

x

Mid-Chapter Quiz for Chapter 10 1. Center: (0, 0); radius: 5

(x

− h) + ( y − k ) = r 2

(x

− 0) + ( y − 0) = 52

2

2

2

2

x 2 + y 2 = 25

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

326

Chapter 10

Conics

2. Center: (3, − 5); passes through the point: (0, −1)

(3 − 0)2

r =

+ ( − 5 − ( −1))

=

32 + ( − 4)

=

9 + 16

=

25

x2 + y2 + 6 y + 9 = 7 + 9

2

x 2 + ( y + 3) = 16 2

2

Center: (0, −3) Radius: 4 x2 + y 2 + 2x − 4 y + 4 = 0

8.

= 5

(x

7. x 2 + y 2 + 6 y − 7 = 0

( x 2 + 2 x + 1) + ( y 2 − 4 y + 4)

− h) + ( y − k ) = r 2 2

2

(x

− 3) + ( y − (− 5)) = 52

(x

2

2

(x

− 3) + ( y + 5) = 25 2

2

− k ) = 4 p( x − h) ⇒ p = 2

(y

− 1) = 4( 2)( x − ( − 2))

(y

− 1) = 8( x + 2)

2

2

x + 7 = ( y 2 − 6 y + 9) − 9

(x

− h) = 4 p( y − k ) ⇒ p = − 2

(x

− 2) = 4( − 2)( y − 3)

(x

− 2) = − 8( y − 3)

2 2

4p = 1 p =

1 4

Co-vertices: ( −2, 1), ( −2, −3) Major axis is horizontal so a = 4. Minor axis is vertical so b = 2.

Co-vertices: ( −6, 0), (6, 0) Vertical ellipse 2

2

x y + 2 =1 b2 a x2 y2 + 2 =1 2 6 10 x2 y2 + =1 36 100

)

x 2 − 8 x + 16 = − y − 12 + 16

Vertices: ( −6, −1), ( 2, −1)

6. Vertices: (0, −10), (0, 10)

) (

10. x 2 − 8 x + y + 12 = 0

5. Center: ( −2, −1)

2

) (

(

Focus: −16 + 14 , 3 = − 64 + 14 , 3 = − 63 ,3 4 4

2

+ 2) ( y + 1) + 16 4

2

Vertex: ( −16, 3)

4. Vertex: ( 2, 3); focus: ( 2, 1)

(x

2

x = y2 − 6 y − 7

9.

x + 16 = ( y − 3)

2

2

2

Radius: 1

(y

⎡⎣ x − ( −2)⎤⎦ ⎡ y − ( −1)⎤⎦ + ⎣ 42 22

+ 1) + ( y − 2) = 1

Center: ( −1, 2)

3. Vertex: ( − 2, 1); focus: (0, 1)

2

= −4 + 1 + 4

2

=1 2

=1

(x

− 4) = − y + 4

(x

− 4) = −1( y − 4)

2 2

Vertex: ( 4, 4) 4 p = −1 p = − 14

(

Focus: 4, 4 −

1 4

) = (4, 164 − 14 ) = (4, 154 )

11. 4 x 2 + y 2 − 16 x − 20 = 0

4 x 2 − 16 x + y 2 = 20 4( x 2 − 4 x) + y 2 = 20 4( x 2 − 4 x + 4) + y 2 = 20 + 16 4( x − 2) + y 2 = 36 2

(x

− 2) y2 + =1 9 36 2

Center: ( 2, 0) Vertices: ( 2, −6), ( 2, 6) a 2 = 36 a = 6

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 10.3

(4 x 2 − 48x) + (9 y 2 + 36 y) = −144 4( x 2 − 12 x + 36) + 9( y 2 + 4 y + 4) = −144 + 144 + 36 2

( x − 6) 2

+

9

( y + 2) 2 4

(x

Circle: Center: ( −5, 1) −6 −5 −4 −3

14. 9 x 2 + y 2 = 81

b2 = 9

x

−1

1

b = ±3

6 4 2

−8 −6 −4 −2 −4

a = ±9

− 10

4

5

2

4( x + 3) ( y − 2) + 16 16

(0, 9)

(x

(0, 0) (3, 0)

x

2 4 6 8 10

+ 3) ( y − 2) + 4 16 2

2

(− 3, 6) 6

2

5

=1

4

(− 5, 2)

Ellipse

(0, − 9)

y

=1

(− 3, 2)

3

(− 1, 2)

2 1

a 2 = 16

b2 = 4

a = 4

b = 2

Center: ( −3, 2)

−7 −6 −5

(− 3, −2)

−3 −2

x −1

1

−2

Vertices: ( −3, 6), ( −3, − 2) y

2

Co-vertices: ( −5, 2), ( −1, 2)

2

2

1

−2

Vertex: ( 4, − 2)

x 3

( 154, −2(

4

5

(4, −2)

−3

4 p = −1

−5

p = − 14

(

3

x 2

(1, 0)

−2

) = (1, 0.25)

2

−3 −2 −1

Parabola

1 4

2

x = −1( y 2 + 4 y + 4) + 4 −1( x − 4) = ( y + 2)

−3 −2 −1

1 4

18. 4( x + 3) + ( y − 2) = 16

x = −y − 4y x − 4 = −1( y + 2)

p =

(1, 0.25)

1

(

2

15.

4

Focus: 1, 0 +

y 10

a 2 = 81

4

5

4p = 1

−2 −3 −4 −5

(− 3, 0)

3

6

Vertex: (1, 0)

5 4 3 2 1

x2 y2 + =1 9 81

2

y

2

Parabola

y

−9 −8

x 1

−2

y = ( x − 1)

+ 5) + ( y − 1) = 9

Vertical ellipse

Center: (0, − 4)

17. y = x 2 − 2 x + 1

2

Radius: 3

1 −4 −3 −2 −1

−6

Vertices: (3, − 2), (9, − 2) 13.

Circle

=1

Center: (6, − 2)

2

2

Radius: 1

4( x − 6) + 9( y + 2) = 36 2

y

16. x 2 + ( y + 4) = 1 2

4 x 2 + 9 y 2 − 48 x + 36 y + 144 = 0

12.

327

Hyperbolas

−6

) (164 − 14 , −2) = (154 , −2)

Focus: 4 − 14 , − 2 =

Section 10.3 Hyperbolas 1. Because the transverse axis is vertical, the standard form y2 x2 of the equation of the hyperbola is 2 − 2 = 1. a b

3. Because the transverse axis is horizontal, the standard x2 y2 form of the equation of the hyperbola is 2 − 2 = 1. a b

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

328

5.

Chapter 10

Conics

x2 y2 − =1 9 25 5 x 3 5 y = − x 3

Asymptotes: y =

1 Asymptotes: y = ± x 2

6

Vertices: (3, 0), ( −3, 0)

7.

13. Vertices: (0, ± 4)

y

4

−6

2

4

6

−4 −6

y2 x2 − =1 9 25

y2 x2 − 2 =1 2 a b

y 6

Vertices: (0, ±3) 3 Asymptotes: y = ± x 5

y2 x2 − 2 =1 2 4 8

4 2 −6

y2 x2 − =1 16 64

x

−4

4

6

−2 −4

15. Vertices: ( ±9, 0)

−6

9.

y2 − x2 = 9

2 Asymptotes: y = ± x 3

y

y2 x2 − =1 9 9

6

Vertices: (0, ±3)

2 x −6

−4

−2

2

4

6

−4

x2 y2 − 2 =1 2 a b

−6

x2 y2 − =1 92 62

11. Vertices: ( ±4, 0)

x2 y2 − =1 81 36

Asymptotes: y = ± 2 x a = 4

b = ±2 a b = ±2 4 b = ±8

17.

(x

− 1) ( y + 2) − 4 1 2

2

y

=1

Vertices: ( −1, − 2), (3, − 2) 2

x2 y2 − 2 =1 2 4 8 19.

4 2

Center: (1, − 2)

x2 y2 − 2 =1 2 a b

x2 y2 − =1 16 64

b 2 = ± a 3 b 2 = ± 9 3 b = 6

a = 9

4

Asymptotes: y = ± x

1 a = ± 2 b 4 1 = ± 2 b b = 8

a = 4

x

−4

x

−6 −4

8 −4 −6

2

a = 4

b =1

a = 2

b =1

−8 −10

(y

+ 4) − ( x − 3) = 25

(y

+ 4) ( x − 3) − 25 25

2

2

2

2

=1

y 4

Center: (3, − 4) Vertices: (3, 1), (3, −9) a = 5

b = 5

−6 −4 −2

x 2

4

8 10

−4 −6

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Section 10.3

(9 x2 − 36 x) − ( y 2 + 6 y) = 9( x 2 − 4 x + 4) − ( y 2 + 6 y + 9) =

−18

draw and extend the rectangles of the central rectangle.

−18 + 36 − 9

29. Central rectangle dimensions: 2a × 2b; Center: ( h, k )

9( x − 2) − ( y + 3) = 9 2

(x

2

− 2) ( y + 3) − 1 9 2

=1

Point: ( −2, 5) Center: (0, 0)

2

b2 = 9

a =1

Vertical axis

y

Vertices: (3, −3), (1, −3) a2 = 1

31. Vertices: (0, ±3)

2

Center: ( 2, −3)

−4

b = 3

−2

y2 x2 − 2 =1 2 a b

x 4

6

8

−2 −4

(5)2

−6

32

−8

4 x 2 − y 2 + 24 x + 4 y + 28 = 0

23.

(4 x 2 + 24 x) − ( y 2 − 4 y) = 4( x 2 + 6 x + 9) − ( y 2 − 4 y + 4) =

−28 −28 + 36 − 4

4( x + 3) − ( y − 2) = 4 2

(x

2

+ 3) ( y − 2) − 1 4 2

2

=1

( −2)2

=1 b2 25 4 − 2 =1 9 b 25 9 4 − = 2 9 9 b 16 4 = 2 9 b 36 9 = b2 = 16 4 −

y2 x2 − =1 9 94

Center: ( −3, 2)

y

Vertices: ( − 4, 2), ( −2, 2)

33. Vertices: (1, 2), (5, 2)

6 5

a2 = 1

b2 = 4

4

Horizontal axis

a =1

b = 2

3 2

Point: (0, 0)

1 −7 −6 −5

−3 −2

x −1

Center: (3, 2)

1

(x 2

2

25. 89 x − 55 y − 4272 x + 31,684 = 0

89( x 2 − 48 x + 576) − 55 y 2 = −31,684 + 51,264

− h) ( y − k) − a2 b2 2

(0 − 3)2 22

2

(x

− 24) y2 − =1 220 356 2

a 2 = 220 220

Center: ( 24, 0)

(

Vertex: 24 +

)

220, 0 ≈ (38.8, 0)

−

2

=1

( 0 − 2) 2 b2

=1

9 4 − =1 4 b2 9 4 4 − = 2 4 4 b 5 4 = 2 4 b 16 b2 = 5

89( x − 24) − 55 y 2 = 19,580

a =

329

27. The sides of the central rectangle will pass through the points ( ± a, 0) and (0, ±b). To sketch the asymptotes,

9 x 2 − y 2 − 36 x − 6 y + 18 = 0

21.

Hyperbolas

(x

− 3) ( y − 2) − 4 16 5 2

2

=1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

330

35.

Chapter 10

Conics

x2 y2 − =1 1 94

41. x 2 − y 2 = 1

Hyperbola of the form

Vertices: ( ±1, 0)

center and a 2 = b 2 .

32 x 1 3 y = ± x 2

Asymptotes: y = ±

y 2 − x 2 − 2 y + 8 x − 19 = 0

43.

( y 2 − 2 y) − ( x2 − 8x) = 19 ( y 2 − 2 y + 1) − ( x 2 − 8x + 16) = 19 + 1 − 16

y 3 2

x

−3 −2

2

(y

− 1) − ( x − 4) = 4

(y

− 1) ( x − 4) − 4 4

3

−2

(1, 4) 37. 4 y 2 − x 2 + 16 = 0 45.

−16 x2 4 y2 − = −16 −16 −16

x2 y2 − =1 16 4 2 1 x = x 4 2 2 1 y = − x = − x 4 2

− k) ( x − h) − 2 a b2 2

2

= 1 where

is the center and a 2 = b 2 .

x2 y2 − =1 100 4

−3 y = − x + 5

6

−6 y = −2 x − 5

y = 1x − 5 3 3

4

y = 1x + 5 3 6

m1 = m2 and b1 ≠ b2 ; inconsistent

2 x 6

53. ⎧4 x + 3 y = 3 ⎨ ⎩ x − 2y = 9

−2 −4 −6

4x + 3y = − 3) ( y − 4) + 42 62

Ellipse of the form

=1

51. ⎧ x − 3 y = 5 ⎨ ⎩2 x − 6 y = −5

y

2

(y

2

49. Answers will vary.

Asymptotes: y =

(x

2

47. Infinitely many. The constant difference of the distances can be different for an infinite number of hyperbolas that have the same set of foci.

Vertices: ( 4, 0), ( − 4, 0)

4

2

The shortest horizontal distance would be the distance between the center and a vertex of the hyperbola, that is, 10 miles.

− y2 x2 + =1 4 16

39.

2

Hyperbola of the form

−3

−6

x2 y2 − 2 = 1 where (0, 0) is the 2 a b

2

− 4 x + 8 y = −36

=1

(x

a 2 = 62 and b 2 = 42

11 y = −33

− h) ( y − k) + b2 a2

where (3, 4) is the center.

3

2

y = −3

2

=1

x − 2( −3) = 9 x+6 = 9 x = 3

(3, −3)

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Section 10.4

Solving Non-linear Systems of Equations

331

Section 10.4 Solving Non-linear Systems of Equations 1. ⎧⎪ x + y = 2 ⎨ 2 ⎪⎩x − y = 0

5. ⎧⎪x 2 + y 2 = 100 ⎨ ⎪⎩ x + y = 2

Solve for y.

Solve for y.

y = −x + 2 y = x

y = ± 100 − x 2

2

y = 2− x

Solutions: ( −2, 4), (1, 1) Check:

Solutions: ( −6, 8), (8, −6) Check:

?

?

−2 + 4 = 2

1 + 1= 2

2 = 2

2 = 2

(−2)

2

?

1 − 1= 0

0 = 0

0 = 0

x2

(−6)2

Check: ?

?

+ 82 = 100

82 + ( −6) = 100

100 = 100

100 = 100

?

2

− 4=0

y

Check:

2

?

?

−6 + 8 = 2

8 + ( −6) = 2

2 = 2

2 = 2

y

−y=0 15

5

x 2 + y 2 = 100

4 3 2 1 −3

−2

−15

x+y=2

x 15 −5

x

−1 −1

1

2

3

x+y=2

−15

3. ⎪⎧x 2 + y = 9 ⎨ ⎪⎩ x − y = −3

7. ⎪⎧x 2 + y 2 = 25 ⎨ ⎪⎩ 2 x − y = −5

Solve for y.

Solve for y.

2

y = −x + 9

y = ± 25 − x 2

y = x +3

y = 2x + 5

Solutions: ( 2, 5), ( −3, 0) Check:

Solutions: (0, 5), ( − 4, −3)

Check: ?

2

−5

(−3)

2 + 5=9

2

?

Check:

Check: ?

+ 0=9

9 = 9

02 + 52 = 25

9 = 9

25 = 25

?

?

?

2 − 5 = −3

−3 − 0 = − 3

−3 = −3

− 3 = −3

2

?

+ ( −3) = 25 2

25 = 25 ?

2(0) − 5 = −5

2( − 4) − ( −3) = −5

−5 = −5

−5 = −5

y

x2 + y = 9

( − 4)

y

9

2x − y = −5

6

6 3 −9

−6

x − y = −3 2 x 6

−3 −6 −9

9

−6

−2

x 2

4

6

−2 −4 −6

x 2 + y 2 = 25

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

332

Chapter 10

Conics

9. ⎧⎪x − 2 y = 4 ⎨ 2 ⎪⎩x − y = 0

y 3

Solve for y.

y = 2− x

2 1

x − 4 = 2y 1x 2

−3

−2 = y

−2

−1

x 1

2 x( x − 2) = 0 2x = 0 x = 0

x = 2

y = 2−0

y = 2−2

4

y = 2

y = 0

3x − y = 12

2

x y − =1 16 16

2 −6

−2

Solve for y.

x 2 −2 −4

y = 3 x − 12

−6

Solutions: (5, 3), ( 4, 0)

(0, 2)

2

2

?

5 − 3 = 16

4 − 0 = 16

16 = 16

16 = 16

?

?

3(5) − 3 = 12

3( 4) − 0 = 12

12 = 12

12 = 12

17. ⎪⎧x 2 − 4 y 2 = 16 ⎨ 2 ⎪⎩ x + y 2 = 1 y 2 = 1 − x2 x 2 − 4(1 − x 2 ) = 16 x 2 − 4 + 4 x 2 = 16 5 x 2 = 20 x2 = 4 x = ±2 2

13. ⎪⎧ y = 2 x ⎨ ⎪⎩ y = 6 x − 4 2

y = 1 − 4 = −3

No real solution

2x2 = 6x − 4

19. ⎧⎪ y = x2 − 3 ⎨ 2 ⎪⎩x + y 2 = 9

2

2x − 6x + 4 = 0 2( x 2 − 3 x + 2) = 0

x2 = y + 3

2( x − 2)( x − 1) = 0

y + 3 + y2 = 9 x =1

y2 + y − 6 = 0

y = 2( 2) = 8

y = 2(1) = 1

(y

(2, 8)

(1, 2)

y +3 = 0

2

(2, 0)

6

x 2 − y 2 = 16

Check:

x = 2

x −2 = 0

6

y

Rewrite.

?

2x2 − 4x = 0

−3

11. ⎪⎧x 2 − y 2 = 16 ⎨ ⎪⎩ 3x − y = 12

2

2

x2 + 4 − 4x + x2 = 4

x − 2y = 4

Solutions: none

2

2

−1

x = y

Check:

x + ( 2 − x) = 4 2

2

2

15. ⎧⎪x 2 + y 2 = 4 ⎨ ⎪⎩ x + y = 2

x2 − y = 0

2

+ 3)( y − 2) = 0 y −2 = 0

y = −3

y = 2

2

x = −3 + 3

x2 = 2 + 3

x2 = 0

x2 = 5

x = 0

(0, −3)

x = ± 5

(±

5, 2

)

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Section 10.4

4 x = 12 − 3 y 12 − 3 y x = 4 2

⎛ 12 − 3 y ⎞ 2 16⎜ ⎟ + 9 y = 144 4 ⎝ ⎠ ⎛ 144 − 72 y + 9 y 2 ⎞ 2 16⎜ ⎟ + 9 y = 144 16 ⎝ ⎠ 144 − 72 y + 9 y 2 + 9 y 2 = 144 18 y 2 − 72 y = 0 18 y( y − 4) = 0

18 y = 0

y −4 = 0

y = 0

y = 4

12 − 3(0) 4 x = 3

12 − 3( 4) 4 x = 0

x =

x =

(0, 4)

23. ⎧⎪ x 2 − y 2 = 9 ⎨ 2 ⎪⎩x + y 2 = 1 2

x =1− y 2

x2 −x

2

− y2

+ 2y = −

y

2

(y

− 3)( y + 1) = 0

y −3 = 0 2

x +3 = 4

y = −1 x + ( −1) = 4 2

2

x 2 = −5

x2 = 3 x = ± 3

No real solution

(±

)

3, −1

2 27. ⎧− ⎪ x + y = 10 ⎨ 2 ⎪⎩ x − y 2 = −8

− x + x2 = 2 x2 − x − 2 = 0

(x 2

− 2)( x + 1) = 0 x−2 = 0 2

No real solution

y +1 = 0

y = 3 2

x = 2

y2 = − 4

−3

y2 − 2 y − 3 = 0

2

−2 y = 8

1

= −4

+ 2y =

2

1− y − y = 9

333

25. ⎪⎧x 2 + 2 y = 1 ⎨ 2 ⎪⎩x + y 2 = 4

21. ⎪⎧16 x 2 + 9 y 2 = 144 ⎨ ⎪⎩ 4 x + 3 y = 12

(3, 0)

Solving Non-linear Systems of Equations

−2 + y = 10

x +1 = 0 x = −1 −1( −1) + y 2 = 10

y 2 = 12

y2 = 9 y = ±3

y = ± 12 y = ±2 3

(2, ±2 3 )

(−1, ± 3)

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334

Chapter 10

Conics

29. Equation of circle: ( x − 0) + ( y − 0) = 12 2

2

x2 + y2 = 1 Equation of Clarke Street (line): points: ( −2, −1) and (5, 0); slope: m =

0 − ( −1) 1 = ; 5 − ( −2) 7

1 ( x − 5) 7 1 5 y = x − 7 7 7y = x − 5

Equation: y − 0 =

Find the intersection of the circle and the line by solving the system: x2 + y 2 = 1 7y = x − 5 x = 7y + 5

(7 y

+ 5) + y 2 = 1 2

49 y 2 + 70 y + 25 + y 2 = 1 50 y 2 + 70 y + 24 = 0 25 y 2 + 35 y + 12 = 0

(5 y

+ 4)(5 y + 3) = 0

4 5 28 25 ⎛ 4⎞ x = 7⎜ − ⎟ + 5 = − + 5 5 ⎝ 5⎠ 3 x = − 5 3 4⎞ ⎛ ⎜− , − ⎟ ⎝ 5 5⎠

3 5 21 25 ⎛ 3⎞ x = 7⎜ − ⎟ + 5 = − + 5 5 ⎝ 5⎠ 4 x = 5 4 3⎞ ⎛ ⎜ ,− ⎟ ⎝ 5 5⎠

y = −

y = −

31. In addition to zero, one, or infinitely many solutions, a system of nonlinear equations can have two or more solutions.

37. ⎧⎪ y = 4− x ⎨ + x 3 y = 6 ⎪⎩ x +3 4− x = 6

33. Substitution

3 4− x = 6− x

y = x2 − 5 35. ⎧⎪ ⎨ ⎪⎩3 x + 2 y = 10

9( 4 − x) = 36 − 12 x + x 2 36 − 9 x = 36 − 12 x + x 2

3x + 2( x 2 − 5) = 10

0 = x 2 − 3x

3 x + 2 x 2 − 10 = 10

0 = x( x − 3)

2 x 2 + 3 x − 20 = 0

(2 x

0 = x

− 5)( x + 4) = 0

x + 4 = 0

2x − 5 = 0

( 52 , 54 )

y =

x = y =

( 52 )

2

=

25 4

−

=

5 4

−5 20 4

4−0 = 2

x = 3 y =

x = −4

5 2

x −3 = 0

y = ( − 4) − 5 2

(0, 2)

4−3 =1

(3, 1)

= 16 − 5 = 11

(− 4, 11)

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Section 10.4

Solving Non-linear Systems of Equations

43. Verbal Model: 15 + Base + Height = Perimeter

⎧ x2 39. ⎪ + y 2 = 1 ⎪4 ⎨ 2 ⎪x 2 + y = 1 ⎪⎩ 4

Base

15 + x + y = 36

System:

2

12 15 4 x2 = 5

(x

x = ±

2 ⋅ 5

5 5

2 5 x = ± 5 2

⎛ 2 5⎞ 2 ⎜⎜ ± 5 ⎟⎟ + 4 y = 4 ⎝ ⎠ 4 + 4 y2 = 4 5 16 4 y2 = 5 4 2 y = 5 y = ±

− x 2 = −3 x2 = 3 x = ± 3

(±

3, ± 13

x + y 2 = 225

− 9)( x − 12) = 0

x = 9

x = 12

y = 12

y = 9

9 meters × 12 meters 45. Find the equation of the line with points (0, 10) and

(5, 0). m =

0 − 10 = −2, y = −2 x + 10 5−0

Find the point of intersection of the line and the hyperbola. y = −2 x + 10 y = −2 x + 10 ⎧ ⇒ ⎪ 2 2 2 16 x − 9 y 2 = 144 ⎨x y =1 ⎪ − 16 ⎩9

2 5 5

16 x 2 − 9( −2 x + 10) = 144 2

16 x 2 − 9( 4 x 2 − 40 x + 100) = 144 16 x 2 − 36 x 2 + 360 x − 900 − 144 = 0 −20 x 2 + 360 x − 1044 = 0

41. ⎧⎪ y 2 − x 2 = 10 ⎨ 2 ⎪⎩x + y 2 = 16

13 − x 2 = 10

x + y = 21 2

x 2 − 21x + 108 = 0 4 5

y = ± 13

⇒

2 x 2 − 42 x + 216 = 0

x = ±

y 2 = 13

2

x 2 + 441 − 42 x + x 2 = 225

x2 =

2 y 2 = 26

2

2

2

−15 x = −12

x 2 + y 2 = 16

= Hypotenuse

x + ( 21 − x) = 225

2

− x + y = 10

2

y = 21 − x

−16 x 2 − 4 y 2 = −16

2

2

x + y = 15

2

x + 4y = 4

2

+ Height

Height = y

4x2 + y 2 = 4

⎛ 2 5 2 5⎞ ,± ⎜⎜ ± ⎟ 5 5 ⎟⎠ ⎝

2

Base = x

Labels:

x2 + 4 y 2 = 4 2

335

5 x 2 − 90 x + 261 = 0 x =

90 ±

902 − 4(5)( 261) 10

≈ 3.633

y = −2(3.633) + 10 y ≈ 2.733

(3.633, 2.733) 47. Solve one of the equations for one variable in terms of the other. Substitute that expression into the other equation. Solve the equation. Back-substitute the solution into the first equation to find the value of the other variable. Check the solution to see that it satisfies both of the original equations.

)

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336

Chapter 10

Conics

49. Two. The line can intersect a branch of the hyperbola at most twice, and it can intersect only one point on each branch at the same time. 51.

(

6 − 2x

)

6 − 2( −5) = 4

Check:

6 − 2x = 4 2

?

log 5 x −1 = log 310

(x

− 1) log 5 = log 310

x log 5 − log 5 = log 310 x log 5 = log 310 + log 5

?

= 42

6 + 10 = 4

6 − 2 x = 16

x =

?

16 = 4

−2 x = 10

?

Check: 54.564 −1 = 310

x = x −6

( x)

2

= ( x − 6)

log 310 + log 5 log 5

x ≈ 4.564

4 = 4

x = −5 53.

5 x −1 = 310

57.

?

53.564 = 310

2

?

310 = 310

x = x 2 − 12 x + 36 0 = ( x − 9)( x − 4)

9 = x ?

9 =9 − 6

Check:

3 = 3

3x = 35

0.01 = 0.01

1.023 ≈ x ?

4=4 − 6 2 ≠ −2

61. 2 ln ( x + 1) = −2

ln ( x + 1) = −1 e −1 = x + 1

Solution: x = 9 55. 3x = 243

Check: log10 1.023 = 0.01

100.01 = x x = 4

Check:

?

59. log10 x = 0.01

0 = x 2 − 13 x + 36

e −1 − 1 = x Check:

−0.632 ≈ x

?

35 = 243 243 = 243

Check: 2 ln ( −0.632 + 1) = −2 ?

2 ln (0.368) = − 2

x = 5

− 2 = −2

Review Exercises for Chapter 10 1. Ellipse

y

13. x 2 + y 2 = 64

3. Circle

Center: (0, 0)

5. Hyperbola

Radius: 8

10 6 4 2 − 10

7. Circle

Radius: 6

x 2 4 6

10

−4 −6

9. Parabola 11. Center: (0, 0)

−6 −4 −2

− 10

15. Center: ( 2, 6); Radius: 3

x 2 + y 2 = 62

(x

− h) + ( y − k ) = r 2

x 2 + y 2 = 36

(x

− 2) + ( y − 6) = 9

2

2

2

2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

337

Review Exercises for Chapter 10 25. Vertices: (0, −5), (0, 5)

x 2 + y 2 + 6 x + 8 y + 21 = 0

17.

( x2 + 6 x + 9) + ( y 2 + 8 y + 16) = (x

Co-vertices: ( −2, 0), ( 2, 0)

−21 + 9 + 16

+ 3) + ( y + 4) = 4 2

2

Center: ( −3, − 4)

x2 y2 + 2 =1 2 a b

y

x2 y2 + 2 =1 2 2 5

1

Radius: 2

x −6 −5 −4 −3 −2 −1

1

x2 y2 + =1 4 25

−2 −3

27. Major axis (vertical) 6 units

−5

Minor axis 4 units

−6

19. Vertex: (0, 0); focus: (6, 0)

x2 y2 + 2 =1 2 a b

y

2b = 6

20

(y

− k ) = 4 p( x − h) 2

15

p = 6

(y

5

− 0) = 4(6)( x − 0) 2

−2

2

4

6

8

10

x2 y2 + =1 4 9

− 15 − 20

29.

21. Vertex: (0, 5); focus: ( 2, 5)

(y

y

2

18

p = 2

12

− k ) = 4 p( x − h)

15

(y

− 5) = 4( 2)( x − 0)

(y

− 5) = 8 x

6 3

2

−3 −3

x 3

9 12 15 18 21

y =

1 x2 2

a 2 = 64

b 2 = 16

a = 8

b = 4

Co-vertices: (0, ± 4)

2 y = x 2 − 16 x + 14

x2 +

2 y − 14 = x 2 − 16 x a2 = 4

2 y − 14 + 64 = x 2 − 16 x + 64 2

2( y + 25) = ( x − 8)

2

a = ±2

p =

1 2 1 ( 2) 50 1 (8, − 2 + 2 ) (8, − 492 )

Focus: 8, − 25 +

6

2 x

−6 −4 −2

2

6

−8

y

y2 =1 4 b2 = 1 b = ±1

Co-vertices: ( ±1, 0)

y

4

−6

Vertices: (0, ± 2)

Vertex: (8, − 25) 4p = 2

8

31. 36 x 2 + 9 y 2 − 36 = 0

− 8x + 7

2 y + 50 = ( x − 8)

y

Horizontal axis

−6

23.

x2 y2 + =1 64 16

Vertices: ( ±8, 0)

9

2

a = 2 2

x y + 2 =1 22 3

x 2

−5

− 10

y 2 = 24 x

2a = 4

b = 3

10

3

1 −3

x

−2

2

3

−3

10 5 −10 −5 −5

x 5

10

20 25

−10 −15 −20 −25

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

338

Chapter 10

Conics

33. Vertices: ( −2, 4), (8, 4)

41.

Co-vertices: (3, 0), (3, 8)

x 2 − y 2 = 25 2

y

2

x y − =1 25 25

Center: (3, 4)

8 6 4

Center: (0, 0)

Major axis is horizontal so a = 5.

2

Minor axis is vertical so b = 4.

Vertices: ( ±5, 0)

(x

Asymptotes: y = ± x

− h) ( y − k) + a2 b2 2

2

=1

(x

− 3) ( y − 4) + 52 42

2

(x

− 3) ( y − 4) + 25 16

2

2

2

2

=1

43.

=1

4

6

8

−4 −6 −8

2

a = 25

b = 25

a = 5

b = 5

y2 x2 − =1 25 4

5 Asymptotes: y = ± x 2

Center: (0, 4)

a 2 = 25

Major axis is vertical so b = 4.

(x

− h) ( y − k) + a2 b2

2

(x

− 0) ( y − 4) + 32 42

2

x ( y − 4) + 9 16

2

b2 = 4

a = 5

Minor axis is horizontal so a = 3.

2

x 2

Vertices: (0, ±5)

Co-vertices: ( −3, 4), (3, 4)

2

−2

Center: (0, 0)

35. Vertices: (0, 0), (0, 8)

2

−8 −6

b = 2 y

=1

8

4

=1

x −8 −6 −4 −2

2

4

6

8

=1 −8

37. 9( x + 1) + 4( y − 2) = 144 2

(x

2

+ 1) ( y − 2) + 16 36 2

45. Vertices: ( ± 2, 0)

y

2

=1

3 Asymptotes: y = ± x 2

6

Center: ( −1, 2)

Center: (0, 0)

4

Vertices: ( −1, − 4), ( −1, 8) 2

−6

2

a = 36

b = 16

a = 6

b = 4

−2

x 2

−2

4

6

39. 16 x 2 + y 2 + 6 y − 7 = 0

x2 y2 − 2 =1 2 2 3

16 x 2 + ( y + 3) = 16 2

(y

+ 3) 16

2

Center: (0, −3) Vertices: (0, −7), (0, 1) a 2 = 16 a = 4 Vertical axis

b 3 = ± a 2 b 3 = ± 2 2 b = 3

x2 y2 − =1 a2 b2

16 x 2 + ( y 2 + 6 y + 9) = 7 + 9

x2 +

a = 2

=1

x2 y2 − =1 4 9

y 1 x

−4 −3 −2

2

3

4

b2 = 1 b =1 −7

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

339

Review Exercises for Chapter 10 47. Center: (0, 0)

53. Center: ( − 4, 6)

Vertices: (0, −8),

(0, 8)

Vertices: ( −6, 6), ( −2, 6)

Asymptotes: y =

4 4 x, y = − x 5 5

Point: (0, 12) Horizontal hyperbola

Vertical transverse axis

(x

y2 x2 − 2 =1 2 a b

− 3) ( y + 1) − 9 4 2

2

2

(x

=1

2

b2 = 4

a = 3

b = 2

2

=1

55. ⎧⎪ y = x 2 ⎨ ⎪⎩ y = 3 x

Vertices: (0, −1), (6, −1) a2 = 9

2

+ 4) ( y − 6) − 4 12

Center: (3, −1)

y

Solutions: (0, 0), (3, 9)

y 8 6 4

4 3

Check:

Check:

0 = 02

9 = 32

?

y = x2

0 = 3(0)

9 = 9

−4 −3 −2 −1

2 1 x 1

2

3

4

y = 3x

?

9 = 3(3)

x

−4 −2

−4

9 = 9

8 10 12

57. ⎧⎪x 2 + y 2 = 16 y = ± 16 − x 2 ⎨ ⇒ y = x+ 4 ⎪⎩ − x + y = 4

−6 −8

8 y 2 − 2 x 2 + 48 y + 16 x + 8 = 0

51.

=1

⎡⎣0 − ( − 4)⎤⎦ (12 − 6) = 1 − b2 4 16 36 − 2 =1 b 4 36 4 −1 = 2 b 36 3 = 2 b b 2 = 12

y2 x2 − =1 64 100

(x

2

a2 = 4

y2 x2 − 2 =1 2 8 10

49.

2

a = 2

4 a = 5 b 8 4 = 5 b b = 10

a = 8

− h) ( y − k) − a2 b2

Solutions: ( − 4, 0), (0, 4)

(8 y 2 + 48 y) − (2 x 2 − 16 x) = −8

y

8( y + 6 y + 9) − 2( x − 8 x + 16) = −8 + 72 − 32 2

2

6

8( y + 3) − 2( x − 4) = 32 2

(y

2

+ 3) ( x − 4) − 4 16 2

2

2

Center: ( 4, −3)

−6

=1

−x + y = 4

y

Vertices: ( 4, −1), ( 4, −5) a2 = 4

b 2 = 16

a = 2

b = 4

4 2 x −4 −8 −10

x 2

10 12

( − 4)

2

6

−2

x2 + y2 = 16

−6

Check:

6

−4 −2

−2

Check: ?

?

+ 02 = 16

02 + 42 = 16

16 = 16

16 = 16

?

?

− ( − 4) + 0 = 4

−0 + 4 = 4

4 = 4

4 = 4

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

340

Chapter 10

Conics

59. ⎧⎪ y = 5 x 2 ⎨ ⎪⎩ y = −15 x − 10

65. ⎧⎪x 2 + y 2 = 9 ⎨ ⎪⎩ x + 2 y = 3 x = 3 − 2y

5 x 2 = −15 x − 10

(3 − 2 y )

5 x 2 + 15 x + 10 = 0

5 y 2 − 12 y = 0

+ 2)( x + 1) = 0

x + 2 = 0

y(5 y − 12) = 0

x +1 = 0

x = −2

5 y − 12 = 0

x = −1

y = 5( −2) = 5( 4) = 20

y = 5( −1) = 5(1) = 5

(−2, 20)

(−1, 5)

2

2

2

y =

x = 3 − 2(0)

x = 3−2 =

x 2 + ( −1 − x) = 1 2

x = ±2

2 x( x + 1) = 0

(±2)

x +1 = 0

2

+ y 2 = 13 y2 = 9

x = −1

y = ±3

y = −1 − (−1) = 0

(±2, ±3)

(−1, 0)

69. ⎪⎧x 2 + y 2 = 7 ⎨ 2 ⎪⎩x − y 2 = 1

63. ⎪⎧4 x + y 2 = 2 ⎨ ⎪⎩ 2 x − y = −11 y = 2 x + 11

2 x2 = 8

4 x + ( 2 x + 11) = 2 2

x2 = 4 x = ±2

4 x + 4 x 2 + 44 x + 121 = 2

(± 2)2

2

4 x + 48 x + 119 = 0

(2 x

+ y2 = 7

+ 17)( 2 x + 7) = 0

2 x + 17 = 0 x =

( ) + 11

y = 2

(− 172 , − 6)

y2 = 3 y = ± 3

2x + 7 = 0

− 17 2

− 17 2

= − 95

x2 = 4

2x2 + 2x = 0

(0, −1)

24 5

7 x 2 = 28

x2 + 1 + 2 x + x2 − 1 = 0

y = −1 − 0 = −1

−

67. ⎧⎪6 x 2 − y 2 = 15 ⎨ 2 ⎪⎩ x + y 2 = 13

y = −1 − x

x = 0

15 5

(125 )

(− 95 , 125 )

(3, 0)

2x = 0

12 5

y = 0 = 3

61. ⎧⎪x + y = 1 ⎨ ⎪⎩ x + y = −1 2

+ y2 = 9

9 − 12 y + 4 y 2 + y 2 = 9

x 2 + 3x + 2 = 0

(x

2

x =

− 72

( ) + 11

y = 2

− 72

= −17 + 11

= −7 + 11

= −6

= 4

(− 72 , 4)

(±2, ± 3) 71. ⎪⎧ x 2 − y 2 = 4 ⎨ 2 ⎪⎩x + y 2 = 4

2 x2 = 8 x2 = 4 x = ±2

(± 2)2

− y2 = 4 − y2 = 0 y = 0

(± 2, 0)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

341

Review Exercises for Chapter 10 73. ⎧⎪ x 2 + y 2 = 13 ⎨ 2 ⎪⎩2 x + 3 y 2 = 30

79. Verbal Model:

Length × Width = Area Length

−2 x 2 − 2 y 2 = −26

2

+ Width

= Diagonal

Width = y

y2 = 4

System:

y = ±2

xy = 3000 2

x + y 2 = 852

x 2 + ( ± 2) = 13 2

y =

2

x = 9

2

(±3, ± 2) 75. ⎧⎪4 x 2 + 9 y 2 = 36 ⎨ 2 ⎪⎩2 x − 9 y 2 = 18

x 4 + 9,000,000 = 7225 x 2 x 4 − 7225 x 2 + 9,000,000 = 0

2

6 x = 54

( x 2 − 5625)( x 2 − 1600) =

2

x = 9 x = ±3

4( ±3) + 9 y 2 = 36 2

x 2 = 5625

x 2 = 1600

x = 75

x = 40

3000 y = 75 = 40

36 + 9 y 2 = 36 9 y2 = 0

0

3000 40 = 75

y =

40 feet × 75 feet

y2 = 0 y = 0

81. Verbal Model:

(±3, 0)

Length

2

+ Width

2

= Diagonal

2

Length of Length of + = 100 first piece second piece Area of Area of = + 144 Square 1 Square 2

77. Verbal Model: 2 ⋅ Length + 2 ⋅ Width = Perimeter

Labels: Length of first piece = x Length of second piece = y

Length = x Width = y

x + y = 100

System:

System: x 2 + y 2 = 102 x + y = 14 y = 14 − x x 2 + (14 − x) = 100 2

x 2 + 196 − 28 x + x 2 − 100 = 0 2 x 2 − 28 x + 96 = 0

x = 6 y = 8

2

2

10,000 − 200 y + y 2 y2 = + 144 16 16 10,000 − 200 y + y 2 = y 2 + 2304 −200 y = −7696 y ≈ 38.48

− 8)( x − 6) = 0

y = 6

2

⎛ 100 − y ⎞ ⎛ y⎞ ⎜ ⎟ = ⎜ ⎟ + 144 4 ⎝ ⎠ ⎝ 4⎠

x 2 − 14 x + 48 = 0 x = 8

2

⎛ x⎞ ⎛ y⎞ ⎜ ⎟ = ⎜ ⎟ + 144 ⎝ 4⎠ ⎝ 4⎠ x = 100 − y

2 x + 2 y = 28

(x

3000 x

⎛ 3000 ⎞ x2 + ⎜ ⎟ = 7225 ⎝ x ⎠ 9,000,000 = 7225 x2 + x2

x = ±3

Labels:

2

Length = x

Labels:

2 x 2 + 3 y 2 = 30

2

x ≈ 100 − 38.48 ≈ 61.52 38.48 inches; 61.52 inches

6 cm × 8 cm

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

342

Chapter 10

Conics

Chapter Test for Chapter 10 1. Center: ( −2, −3)

x = −3 y 2 + 12 y − 8

4.

x + 8 = −3( y 2 − 4 y )

Radius: 4

(x

− h) + ( y − k ) = r 2

(x

+ 2) + ( y + 3) = 16

2

x + 8 − 12 = −3( y 2 − 4 y + 4)

2

2

2

x − 4 = −3( y − 2)

x2 + y 2 − 2 x − 6 y + 1 = 0

2.

( x 2 − 2 x) + ( y 2 − 6 y ) = ( x2 − 2 x + 1) + ( y 2 − 6 y + 9) = (x

−1

2

Vertex: ( 4, 2) 4 p = − 13 1 p = − 12

−1 + 1 + 9

− 1) + ( y − 3) = 9 2

2

(

Focus: 4 −

Center: (1, 3)

1, 12

) ( 1248 − 121 , 2) = ( 1247 , 2)

2 =

y 5

Radius: 3

4 y

6

2

5 4 −1

3 2 1 1

2

3

2

3

4

5

5. Vertex: (7, − 2)

x −3 −2 −1

x 1 −1

4

Focus: (7, 0) 3.

x2 + y 2 + 4x − 6 y + 4 = 0

( x 2 + 4 x) + ( y 2 − 6 y ) = − 4

Vertical parabola

(x

− h) = 4 p ( y − k )

(x

− 7) = 4 p ⎡⎣ y − (−2)⎤⎦ p = 2

Center: ( −2, 3)

(x

− 7) = 4( 2)( y + 2)

Radius: 3

(x

− 7) = 8( y + 2)

( x 2 + 4 x + 4) + ( y 2 − 6 y + 9) (x

= −4 + 4 + 9

+ 2) + ( y − 3) = 9 2

2

2

2 2

6. Vertices: ( −3, 0), (7, 0)

y 6

Co-vertices: ( 2, 3), ( 2, −3)

5 4

Horizontal axis

3

Center: ( 2, 0)

2 1 x −6 −5 −4 −3 −2 −1

2

1

a = 5

b = 3

2

a = 25

(x

b2 = 9

− h) ( y − k) + a2 b2 2

(x

2

=1

− 2) y2 + =1 25 9 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test for Chapter 10 7. 16 x 2 + 4 y 2 = 64

10. Vertices: (0, −5), (0, 5)

y

x2 y2 + =1 4 16

5 Asymptotes: y = ± x 2

3 2

Center: (0, 0)

1

Vertical transverse axis

x −4 −3

Vertices: (0, ± 4)

−1

1

3

4

y2 x2 − 2 =1 2 a b

−2

a 2 = 16

b2 = 4

a = 4

b = 2

−3

a 5 = b 2 5 5 = 2 b b = 2

a = 5

25 x 2 + 4 y 2 − 50 x − 24 y − 39 = 0

8.

(25x 2 − 50 x) + (4 y 2 − 24 y) = 39 25( x 2 − 2 x + 1) + 4( y 2 − 6 y + 9) = 39 + 25 + 36

y2 x2 − =1 25 4

25( x − 1) + 4( y − 3) = 100 2

(x

2

− 1) ( y − 3) + 4 25 2

11. 4 x 2 − 2 y 2 − 24 x + 20 = 0

2

=1

Center: (1, 3)

(4 x 2 − 24 x) − 2 y 2 4( x 2 − 6 x + 9) − 2 y 2

y 8

2

a = 25

(x 6

a = 3

− 3) y2 − =1 4 8

x2 y2 − =1 9 4

−4

−2

x 2

4

6

8

−2 −4

a2 = 4 a = 2 12.

16 y 2 − 25 x 2 + 64 y + 200 x − 736 = 0

(16 y 2 + 64 y) − (25x 2 − 200 x) = 736 16( y 2 + 4 y + 4) − 25( x 2 − 8 x + 16) = 736 + 64 − 400 16( y + 2) − 25( x − 4) = 400 2

( y + 2)2

x2 y2 − 2 =1 2 a b x2 y2 − 2 =1 2 3 2

4 2

Vertices: (1, 0), (5, 0)

9. Vertices: ( ±3, 0)

b 2 = ± a 3 b 2 = ± 3 3 b = 2

y

2

Center: (3, 0)

x 4 −2

2 Asymptotes: y = ± x 3

= −20 + 36

4( x − 3) − 2 y 2 = 16

Vertices: (1, − 2), (1, 8) −2

= −20

2

a = 5

−4

343

25

2

−

( x − 4)2 16

=1

Center: ( 4, − 2) Vertices: ( 4, −7), ( 4, 3) a 2 = 25

b 2 = 16

a = 5

b = 4

y

2 x −4 −2

2

6

8 10 12

−4 −6

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

344

Chapter 10

Conics 15. ⎧⎪x 2 + y 2 = 10 ⎨ x2 = y 2 + 2 ⎪⎩

⎧ x2 y2 13. ⎪ + =1 9 ⎨16 ⎪ 3 x + 4 y = 12 ⎩

x2 + y2 =

Solve for y in second equation.

2

10

2

x − y =

4 y = 3x + 12

=

12

x2

=

6

2x

3 y = − x +3 4 Substitute into first equation after multiplying by 144. 9 x 2 + 16 y 2 = 144

x

= ± 6

(± 6 )

2

2

2

2

+ y 2 = 10

⎛ 3 ⎞ 9 x 2 + 16⎜ − x + 3⎟ = 144 ⎝ 4 ⎠

6 + y 2 = 10

9 ⎛9 ⎞ 9 x 2 + 16⎜ x 2 − x + 9 ⎟ = 144 2 ⎝ 16 ⎠

y = ±2

9 x 2 + 9 x 2 − 72 x + 144 = 144 2

18 x − 72 x = 0 18 x( x − 4) = 0 18 x = 0

x−4 = 0

x = 0

x = 4

y = −

3 ( 0) + 3 4

= 3

y = −

3 (4) + 3 4

= 0

(0, 3)

(4, 0)

14. ⎧ x 2 + y 2 = 16 ⎪ ⎨ x2 y2 =1 ⎪ − 9 ⎩16

y2 = 4

(

)(

6, 2 , −

)(

)(

6, 2 ,

6, − 2 , − 6, − 2

16. r = 5000

x2 + y2 = r 2 x 2 + y 2 = 50002 x 2 + y 2 = 25,000,000 17. Verbal Model: 2 ⋅ Length + 2 ⋅ Width = Perimeter Length

2

+ Width

System:

x 2 + y 2 = 202 x + y = 28

x + y = 16

y = 28 − x x + ( 28 − x) = 400

2

x 2 + 784 − 56 x + x 2 − 400 = 0

16 x + 16 y = 256

2 x 2 − 56 x + 384 = 0

9 x − 16 y = 144 2

= 400

x2

= 16

x

= ±4

25 x

(±4)2

+ y 2 = 16

16 + y 2 = 16

2

2

Multiply first equation by 16. 2

2

2 x + 2 y = 56

9 x 2 + 16 y 2 = 144 2

= Diagonal

Width = y

2

2

2

Length = x

Labels:

Multiply second equation by 144. 2

)

x 2 − 28 x + 192 = 0

(x

− 16)( x − 12) = 0

x = 16

x = 12

y = 28 − 16

y = 28 − 12

= 12

= 16

12 inches × 16 inches

y2 = 0 y = 0

(4, 0), (− 4, 0)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Cumulative Test for Chapters 8–10

345

Cumulative Test for Chapters 8–10 4x2 − 9x − 9 = 0

1.

(4 x

+ 3)( x − 3) = 0

4x + 3 = 0

3x 2 + 8 x − 3 ≤ 0

(x

(3 x

x −3 = 0 x = 3

x = − 34 2.

− 1)( x + 3) ≤ 0

Critical numbers: −3, 13 Test intervals:

− 5) − 64 = 0 2

(x

3x 2 + 8 x ≤ 3

6.

Positive: ( −∞, −3)

− 5) = 64 2

(

Negative: −3, 13

x − 5 = ±8 x = 5±8

Positive:

x = 13, −3

( 13 , ∞)

Solution: ⎡⎣−3, 13 ⎤⎦

2

3. x − 10 x − 25 = 0

1 3

2

x − 10 x = 25

x −3

2

x − 10 x + 25 = 25 + 25

(x

− 5) = 50 2

x − 5 = ± 50 x = 5±5 2

7.

x =

x =

−6 2 3 ± 6 6

−4

x −2

( x2 − 5)( x 2 − 3) = x = 5 x = ± 5

1 2

3

x 4 − 8 x 2 + 15 = 0

2

⎛1 ⎞ Positive: ⎜ , ∞ ⎟ ⎝2 ⎠ ⎛ 4 1⎞ Solution: ⎜ − , ⎟ ⎝ 3 2⎠

3 x = −1 ± 3 5.

⎛ 4 1⎞ Negative: ⎜ − , ⎟ ⎝ 3 2⎠

36 − 24 6

−6 ± 12 6

1

4⎞ ⎛ Positive: ⎜ −∞, − ⎟ 3⎠ ⎝

62 − 4(3)( 2)

x =

0

4 1 Critical numbers: x = − , 3 2

2(3)

−6 ±

−1

3x + 4 < 0 2x − 1

a = 3, b = 6, c = 2

−6 ±

−2

Test intervals:

4. 3x 2 + 6 x + 2 = 0

x =

)

0

2

x = 3

−1

0

1

8. x = −2 and x = 6

x + 2 = 0 and x − 6 = 0

(x

+ 2)( x − 6) = 0

x 2 − 4 x − 12 = 0

x = ± 3

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

346

Chapter 10

Conics

9. f ( x) = 2 x 2 − 3, g ( x) = 5 x − 1

(a)

(f

12. f ( x) = 4 x −1

Horizontal asymptote: y = 0

D g )( x) = f ⎡⎣ g ( x)⎤⎦

= f [5 x − 1]

y 7

= 2(5 x − 1) − 3 2

6

= 2( 25 x 2 − 10 x + 1) − 3

5 4

= 50 x − 20 x + 2 − 3

3

= 50 x 2 − 20 x − 1

1

2

2 x

Domain: ( −∞, ∞) (b)

(g

−2 −1

D f )( x) = g ⎡⎣ f ( x)⎤⎦

1

13.

= g ⎡⎣2 x 2 − 3⎤⎦

2

3

4

5

6

y 3

= 5( 2 x − 3) − 1

2

= 10 x 2 − 15 − 1

1

2

f

g

= 10 x 2 − 16

x −1

Domain: ( −∞, ∞) 10.

1

5 − 3x 4 5 − 3x y = 4 5 − 3y x = 4 4x = 5 − 3y

f and g are inverse functions, so the graphs are reflections in the line y = x. 14. g ( x) = log 3 ( x − 1)

Vertical asymptote: x = 1 y 4

4 x − 5 = −3 y

3 2

4 5 − x+ = y 3 3 4 5 − x + = f −1 ( x) 3 3

1 x −2 −1

2

3

4

5

6

−2 −3 −4

−x

(a) f (1) = 7 + 2−1 = 7 +

1 = −2 because 4 −2 = 15. log 4 16

1 15 = 2 2

3

1 14 2 14 + = + = 2 2 2 2

(c) f (3) = 7 + 2−3 = 7 +

1. 16

16. 3(log 2 x + log 2 y ) − log 2 z = log 2 ( xy ) − log 2 z

(b) f (0.5) = 7 + 2−0.5 = 7+

3

−1

f ( x) =

11. f ( x) = 7 + 2

2

1 56 1 57 = + = 8 8 8 8

2

= log 2

17. log10

x +1 = log10 x4

3

z

= log 2

x3 y 3 z

x + 1 − log10 x 4 12

= log10 ( x + 1) =

( xy )

− 4 log10 x

1 log10 ( x + 1) − 4 log10 x 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Cumulative Test for Chapters 8–10

⎛1⎞ 18. log x ⎜ ⎟ = −2 ⎝9⎠ 1 = x −2 9 1 1 = 2 x 9 9 = x

6000 = 1500e0.07t 4 = e 0.07t ln 4 = ln e0.07t ln 4 = 0.07t ln 4 = t 0.07 19.8 ≈ t years

2

19. 4 ln x = 10 10 4

ln x =

5 2

A = Pe rt

24.

3 = x

ln x =

25.

eln x = e5 2

x 2 + y 2 − 6 x + 14 y − 6 = 0

( x 2 − 6 x) + ( y 2 + 14 y) ( x 2 − 6 x + 9) + ( y 2 + 14 y + 49) (x

x = e5 2

2 x −6 −4

2

1+ e

2x

= 20 =

e2 x =

ln e 2 x = 2x = x = x ≈

20 3 17 3 17 ln 3 17 ln 3 ln (17 3) 2 0.867

22. C (t ) = P(1.028)

4

−8 −10

−16

y = 2 x 2 − 20 x + 5

26.

y − 5 = 2( x 2 − 10 x) y − 5 + 50 = 2( x 2 − 10 x + 25) y + 45 = 2( x − 5)

(y

1 2

+ 45) = ( x − 5)

2

2

Vertex: (5, − 45) 4p =

1 2

p =

1 8

Focus: 5, 45 +

C (5) = 29.95(1.028) ≈ $34.38 5

23. A = Pert

10 12

−6

(

t

6

−4

log1.08 1.08t = log1.08 4 log 4 t = log 1.08 t ≈ 18.013

)

2

y

2000 500 1.08t = 4 1.08t =

21. 3(1 + e

= 6 + 9 + 49

− 3) + ( y + 7) = 64 2

Radius: 8

t

500(1.08) = 2000

2x

= 6

Center: (3, −7)

x ≈ 12.182

20.

347

1 8

) = (5, − 3608 + 18 ) = (5, − 3598 )

y

10

0.08(1)

A = 1000e A = $1083.29

83.29 Effective yield = = 0.08329 ≈ 8.33% 1000

x −2

2

4

6

8

12

−20 −30 −40 −50

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

348

27.

Chapter 10

Conics

(x

− h) ( y − k) + b2 a2

(x

+ 3) ( y − 2) + 4 25

2

2

2

x 2 − (9 y 2 − 18 y ) = 153

2

x 2 − 9( y 2 − 2 y + 1) = 153 − 9

=1

b = 2

a = 5

b2 = 4

a 2 = 25

x 2 − 9( y − 1) = 144 2

( y − 1) x2 − 144 16

28. 4 x 2 + y 2 = 4

y

x2 y2 + =1 1 4 Vertical ellipse

x −2

2

3

a 2 = 144

b 2 = 16

a = 12

b = 4

Vertices: (0, ± 2) a2 = 4

b2 = 1

a = 2

b =1

29. Hyperbola ( Vertical)

y −3

8 6 4 x −16

−4

4

Vertices: (0, ±3)

−4

Asymptotes: y = ±3 x

−8

Center: (0, 0) a = ±3 b 3 = ±3 b b =1

a = 3 a2 = 9

8 12 16

−6

y = x2 − x − 1 31. ⎧⎪ ⎨ ⎪⎩3 x − y = 4 Substitute for y in second equation. 3x − ( x 2 − x − 1) = 4 3x − x 2 + x + 1 − 4 = 0 − x2 + 4 x − 3 = 0

y2 x2 − 2 =1 2 a b 2

=1

Vertices: ( ±12, 1)

1 −3

2

Center: (0, 1)

3

Center: (0, 0)

x 2 − 9 y 2 + 18 y = 153

30.

=1

x2 − 4x + 3 = 0

2

y x − =1 9 1

(x

− 3)( x − 1) = 0

x = 3

x =1

3(3) − y = 4

3(1) − y = 4

− y = −5

−y = 1

y = 5

(3, 5)

y = −1

(1, −1)

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Cumulative Test for Chapters 8–10 32. ⎧⎪x 2 + 5 y 2 = 21 ⎨ ⎪⎩ − x + y 2 = 5

34. Verbal Model: Length ⋅ Width = Area

2 ⋅ Length + 2 ⋅ Width = Perimeter 2

y = 5+ x x + 5(5 + x) = 21

Length = x

Labels:

2

2

x + 25 + 5 x − 21 = 0

+ 4)( x + 1) = 0

2 x + 2 y = 20

x = −4

x = −1

y = 5 + ( −4)

x + y = 10

y = 5 + ( −1)

2

y = 10 − x

2

2

y =1

2

x(10 − x) = 21

y = ±2

10 x − x 2 = 21

y = 4

y = ±1

(− 4, ±1)

Width = y System: ⎧ xy = 21 ⎨ 2 x + 2 y = 20 ⎩

x2 + 5x + 4 = 0

(x

(−1, ± 2)

33. Verbal Model: Length ⋅ Width = Area

2 ⋅ Length + 2 ⋅ Width = Perimeter Length = x

Labels:

349

0 = x 2 − 10 x + 21 0 = ( x − 7)( x − 3) x = 7

x = 3

y = 3

y = 7

Dimensions: 7 feet × 3 feet

Width = y xy = 32

System:

2 x + 2 y = 24 x + y = 12 y = 12 − x x(12 − x) = 32 12 x − x 2 = 32 0 = x 2 − 12 x + 32 0 = ( x − 8)( x − 4) x = 8

x = 4

y = 4

y = 8

4 feet × 8 feet

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C H A P T E R 1 1 Sequences, Series, and the Binomial Theorem Section 11.1

Sequences and Series..........................................................................351

Section 11.2

Arithmetic Sequences.........................................................................355

Mid-Chapter Quiz......................................................................................................358 Section 11.3

Geometric Sequences and Series .......................................................359

Section 11.4

The Binomial Theorem ......................................................................362

Review Exercises ........................................................................................................365 Chapter Test .............................................................................................................368

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

C H A P T E R 1 1 Sequences, Series, and the Binomial Theorem Section 11.1 Sequences and Series 1. a1 = 2(1) = 2

9. a1 =

a2 = 2( 2) = 4 a3 = 2(3) = 6 a4 = 2( 4) = 8 a5 = 2(5) = 10

a2 =

3( 2) 6 2 = = 5( 2) − 1 9 3

a3 =

3(3) 9 = 5(3) − 1 14

a4 =

3( 4) 12 = 5( 4) − 1 19

a5 =

3(5) 15 5 = = 5(5) − 1 24 8

2, 4, 6, 8, 10, … , 2n, …

( 14 ) = ( 14 ) = ( 14 ) = ( 14 ) = ( 14 )

3. a1 =

a2 a3 a4 a5

1 2

3 4

5

=

1 4

=

1 16

=

1 64

=

1 256

=

1 1024

3(1) 3 = 5(1) − 1 4

3 2 9 12 15 3n , , , , ,…, 4 3 14 19 24 5n − 1 11. a1 = ( −1) 2(1) = −2 1

a2 = ( −1) 2( 2) = 4 2

()

1, 1 , 1 , 1 , 1 ,… , 1 4 16 64 256 1024 4

5. a1 = 5(1) − 2 = 3

n

a3 = ( −1) 2(3) = −6 3

a4 = ( −1) 2( 4) = 8 4

a2 = 5( 2) − 2 = 8

a5 = ( −1) 2(5) = −10

a3 = 5(3) − 2 = 13

−2, 4, −6, 8, −10, …

a4 = 5( 4) − 2 = 18 a5 = 5(5) − 2 = 23

5

3, 8, 13, 18, 23, … , 5h − 2

a2

4 =1 1+3 4 4 a2 = = 2+3 5 4 4 2 a3 = = = 3+3 6 3 4 4 a4 = = 4+3 7 4 4 1 a5 = = = 5+3 8 2 4 2 4 1 4 1, , , , , … , 5 3 7 2 n+3

a3

7. a1 =

( ) = ( − 12 ) = ( − 12 ) = ( − 12 ) = ( − 12 )

13. a1 = − 12

a4 a5 1, 4

2 3

4 5

6

=

1 4

= − 18 =

1 16

1 = − 32

=

1,− 1, − 18 , 16 32

1 64 1 ,…, 64

(− 12 )

n +1

,…

351

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

352

Chapter 11

15. a1 =

a2 = a3 = a4 = a5 =

(1)1

( −1)2 22

=

( −1)4 2

(−1)5 52

= − =

=

1 9

= 53,130

1 16

= −

1 25 n

( −1) 1 1 1 1 −1, , − , , − , … , 2 , … n 4 9 16 25 17. 6! = 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 = 720 19. 9! = 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 = 362,880 21. a1 =

a2 = a3 = a4 = a5 =

1! =1 1 2! =1 2 3! = 2 3 4! = 6 4 5! = 24 5

1, 1, 2, 6, 24, … ,

23. a1 =

(1 + 1)! 1!

(2 + 1)!

=

n! ,… n

2! 2 ⋅1 = = 2 1! 1

3! 3 ⋅ 2! = = 2! 2! 2! (3 + 1)! = 4! = 4 ⋅ 3! a3 = 3! 3! 3! 4 + 1)! 5! 5 ⋅ 4! ( a4 = = = 4! 4! 4! (5 + 1)! = 6! = 6 ⋅ 5! a5 = 5! 5! 5! 1 ! + n ( ) 2, 3, 4, 5, 6 … , n! a2 =

25.

25! 25 ⋅ 24 ⋅ 23 ⋅ 22 ⋅ 21 ⋅ 20! = 20!5! 20!5! 25 ⋅ 24 ⋅ 23 ⋅ 22 ⋅ 21 5 ⋅ 4 ⋅3⋅ 2 ⋅1 = 5 ⋅ 6 ⋅ 23 ⋅ 11 ⋅ 7

1 4

3

32 4

27.

= −1

12

( −1)

Sequences, Series, and the Binomial Theorem

5! 5 ⋅ 4 ⋅ 3⋅ 2 ⋅1 = = 5 4! 4 ⋅ 3⋅ 2 ⋅1

= 3 = 4 = 5

29.

n! n ⋅1 1 = = (n + 1)! (n + 1)n ⋅ 1 n + 1

31.

(n + 1)! (n − 1)!

=

(n

+ 1)n( n − 1)! = ( n + 1)n (n − 1)!

6 n +1 6 a1 = = 3 2 6 a2 = = 2 3 6 3 a3 = = 4 2 6 a4 = 5 6 a5 = =1 6 6 a6 = 7 6 3 a7 = = 8 4 6 2 a8 = = 9 3 6 3 = a9 = 10 5 6 a10 = 11

33. (b); an =

35. (c); an = (0.6) n −1

a1 = (0.6) = 1 0

a2 = (0.6) = 0.6 1

= 6

a3 = (0.6) = 0.36 2

a4 = (0.6) ≈ 0.22 3

a5 = (0.6) ≈ 0.13 4

a6 = (0.6) ≈ 0.08 5

a7 = (0.6) ≈ 0.05 6

a8 = (0.6) ≈ 0.03 7

a9 = (0.6) ≈ 0.02 8

a10 = (0.8) ≈ 0.01 9

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 11.1

1 2 3 4 5

43. a1 = 2(1) + 5 = 7

Terms : 1 3 5 7 9

a2 = 2( 2) + 5 = 9

Apparent pattern: Each term is twice n minus 1.

a3 = 2(3) + 5 = 11

37. n:

an = 2n − 1 39. n:

1

2 3 4

a5 = 2(5) + 5 = 15

5

a6 = 2(6) + 5 = 17

Apparent pattern: Each term is the square of n minus 1. an = n − 1 1

2

−1 5

1 −1 25 125

3

⎛ 1⎞ an = ⎜ − ⎟ ⎝ 5⎠

45. a1 =

a2 = a3 = a4 = a5 = a6 = a7 = a8 = a9 =

47.

5

∑6

S6 = a1 + a2 + a3 + a4 + a5 + a6

4

5

1 625

−1 3125

Apparent pattern: The fraction − power.

S1 = a1 = 7 S 2 = a1 + a2 = 7 + 9 = 16

2

Terms :

353

a4 = 2( 4) + 5 = 13

Terms : 0 3 8 15 24

41. n:

Sequences and Series

= 7 + 9 + 11 + 13 + 15 + 17 = 72

1 is raised to the n 5

n

1 =1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9

1 3 = 2 2 1 1 6+3+ 2 11 S3 = a1 + a2 + a3 = 1 + + = = 2 3 6 6 S9 = a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 + a9

S 2 = a1 + a2 = 1 +

1 1 1 1 1 1 1 1 + + + + + + + 2 3 4 5 6 7 8 9 2520 + 1260 + 840 + 630 + 504 + 420 + 360 + 315 + 280 7129 = = 2520 2520 =1+

= 6 + 6 + 6 + 6 + 6 = 30

k =1

49.

6

∑ (2i + 5) i =0

= ⎡⎣2(0) + 5⎤⎦ + ⎡⎣2(1) + 5⎤⎦ + ⎡⎣2( 2) + 5⎤⎦ + ⎡⎣2(3) + 5⎤⎦ + ⎡⎣2( 4) + 5⎤⎦ + ⎡⎣2(5) + 5⎤⎦ + ⎡⎣2(6) + 5⎤⎦ = 5 + 7 + 9 + 11 + 13 + 15 + 17 = 77

51.

3

∑

j =0

1 1 1 1 1 = 2 + + + j2 + 1 0 + 1 12 + 1 22 + 1 32 + 1 =

1 1 1 1 1 1 1 1 10 5 2 1 18 9 + + + = + + + = + + + = = 0 + 1 1 + 1 4 + 1 9 + 1 1 2 5 10 10 10 10 10 10 5

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

354

53.

Chapter 11 5

⎛ 1⎞

∑ ⎜⎝ − 3 ⎟⎠

n

n=0

Sequences, Series, and the Binomial Theorem 0

1

2

3

4

⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ = ⎜− ⎟ + ⎜− ⎟ + ⎜− ⎟ + ⎜− ⎟ + ⎜− ⎟ + ⎜− ⎟ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠

5

1 243 − 81 + 27 − 9 + 3 − 1 182 ⎛ 1⎞ 1 ⎛ 1 ⎞ ⎛ 1 ⎞ = 1 + ⎜− ⎟ + + ⎜− ⎟ + + ⎜− = ⎟ = 243 243 ⎝ 3 ⎠ 9 ⎝ 27 ⎠ 81 ⎝ 243 ⎠ 55.

5

∑k k =1

57.

180(5 − 4) 180(1) 180 = = = 36° 5 5 5 180(6 − 4) 180( 2) 360 d6 = = = = 60° 6 6 6 180(7 − 4) 180(3) 540 d7 = − = ≈ 77.1° 7 7 7 180(8 − 4) 180( 4) 720 d8 = = = = 90° 8 8 8 180(9 − 4) 180(5) 900 d9 = − = = 100° 9 9 9 180(10 − 4) 180(6) 1080 d10 = = = = 108° 10 10 10

65. d5 =

11

∑k k =1

k +1

59. The third term in the sequence an =

2 2 is . 3n 9

61. To find the 6th term of the sequence an =

6 for n in an : a6 =

2 , substitute 3n

2 1 = . 3(6) 9

67. A sequence is a function because there is only one value for each term of the sequence.

63. (a) A1 = 500(1 + 0.07)1 = $535.00

A2 = 500(1 + 0.07) = $572.45 2

A3 = 500(1 + 0.07) = $612.52 3

A4 = 500(1 + 0.07) = $655.40 4

A5 = 500(1 + 0.07) = $701.28 5

A6 = 500(1 + 0.07) = $750.37 6

A7 = 500(1 + 0.07) = $802.89 7

A8 = 500(1 + 0.07) = $859.09 8

(b) A40 = 500(1 + 0.07)

40

= $7487.23

(c) Keystrokes (calculator in sequence and dot mode): Y = 500 ( 1 + 0.07 ) ^ n TRACE 8000

0

40 0

(d) No. Investment earning compound interest increases at an increasing rate. 69. an = 4n! = 4(1 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ … ⋅ ( n − 1) ⋅ n)

an = ( 4n)! = ( 4 ⋅ 1) ⋅ ( 4 ⋅ 2) ⋅ ( 4 ⋅ 3) ⋅ ( 4 ⋅ 4) ⋅ … ⋅ ( 4n − 1) ⋅ ( 4n) 73. −2n + 15 for n = 3: −2(3) + 15 = 9

71. True. 4

∑ 3k k =1

4

= 3 + 6 + 9 + 12 = 3(1 + 2 + 3 + 4) = 3∑ k k =1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 11.2 75. 25 − 3( n + 4) for n = 8:

6 4

p =

2

77. x + y = 36

Center: (0, 0) Radius: 6

4

=

−8 −6 −4 −2 −2

2

4

Vertex: (0, − 4) 4 p = −8

y

79. x 2 + y 2 + 4 x − 12 = 0

−4

−2

2 −8 −6 −4 −2

x 2

4

6

8

−6 −10 −12

Focus: (0, −6)

2 −8

y 4

−8

p = − 84 = −2

6

8

−6

x = −8( y + 4)

2

6

−4

2

+ 2) + y 2 = 16

4

x 2 + 8( y + 4) = 0

8

y 2 = 12 + 4

x 2

83. x 2 + 8 y + 32 = 0

x

−8

Center: ( −2, 0)

2

3 2

Focus: 0,

2 −4 −2

4

−4

(x

6

( 32 )

y 8

( x2 + 4 x + 4) +

8

4p = 6

= −11

−8

10

Vertex: (0, 0)

= 25 − 36

2

y

81. x 2 = 6 y

25 − 3(8 + 4) = 25 − 3(12)

355

Arithmetic Sequences

x −2

4

Radius: 4 −6

Section 11.2 Arithmetic Sequences 1. 2, 5, 8, 11, …

a1 = 5

d = 3

5 − 2 = 3, 8 − 5 = 3, 11 − 8 = 3 3. 100, 94, 88, 82, …

d = −6

5. 4, 9 2

a3 = −1 a4 = − 4

11 , 2

6, …

11. an =

1 2

a2 =

9 2

1 2

a3 = 2

−5 =

1 2

6−

11 2

1 2

d =

1 2

=

7. an = 4n + 5 a1 = 9 a2 = 13 a3 = 17

+ 1)

3 2

−4 = =

(n

a1 = 1

1 2

5− 11 2

5,

a2 = 2

d = −3

94 − 100 = −6, 88 − 94 = −6, 82 − 88 = −6 9, 2

9. an = 8 − 3n

a4 = d =

5 2 1 2

13. a1 = 4, d = 3 an = a1 + ( n − 1)d an = 4 + ( n − 1)3 an = 4 + 3n − 3 an = 3n + 1

a4 = 21 d = 4

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

356

Chapter 11

15. a1 =

1, d 2

=

Sequences, Series, and the Binomial Theorem 25. a1 = 14, ak +1 = ak + 6 so, d = 6.

3 2

a2 = a1+1 = a1 + 6 = 14 + 6 = 20

an = a1 + ( n − 1)d an =

1 2

+ ( n − 1) 32

an =

1 2

+ 32 n −

an =

3n 2

−

an =

3n 2

−1

a3 = a2 +1 = a2 + 6 = 20 + 6 = 26 a4 = a3 +1 = a3 + 6 = 26 + 6 = 32

3 2

2 2

a5 = a4 +1 = a4 + 6 = 32 + 6 = 38

27. a1 = 23, ak +1 = ak − 5 so, d = −5. a2 = a1+1 = a1 − 5 = 23 − 5 = 18 a3 = a2 +1 = a2 − 5 = 18 − 5 = 13

17. a1 = 100, d = −5

a4 = a3 +1 = a3 − 5 = 13 − 5 = 8

an = a1 + ( n − 1)d

a5 = a4 +1 = a4 − 5 = 8 − 5 = 3

an = 100 + ( n − 1)( −5) an = 100 − 5n + 5 an = −5n + 105

19. a3 = 6, d =

29. (a) a4 = a1 = ( n − 1)d

23 = a1 + ( 4 − 1) ⋅ 6 23 = a1 + 18

3 2

a1 = 5

an = a1 + ( n − 1)d

(b) an = a1 + ( n − 1)d

6 = a1 + (3 − 1) 32 6 = a1 + 3

a 5 = 5 + (5 − 1) ⋅ 6

3 = a1

a 5 = 29

an = 3 + ( n − 1) 32 = 3+ an =

3n 2

3n 2

+

−

3 2

3 2

21. an = a1 + ( n − 1)d

15 = 5 + (5 − 1)d 15 = 5 + 4d

31.

=

10 4

33.

35.

37.

an = 5 + ( n − 1) 52

23.

5n 2

+ 3) =

10

∑ (5k 500

n

∑2

=

n =1

5n 2

(1 +

20) = 210

50 2

k =1

So,

an =

50

∑ (k

20 2

− 2) =

−

39.

+ 52 .

41.

= 11 − 5

43.

= 6 an = a1 + ( n − 1)d a10 = 5 + (10 − 1)(6)

30

∑ ( 13 n − 4)

30 2

(− 113 + 6) = 35

12

∑ (7n − 2)

=

12 2

(5 + 82)

=

50 2

50

∑ (12n − 62) n =1

45.

12

∑ (3.5n − 2.5)

47.

10

∑ (0.4n + 0.1) n =1

49.

75

∑n n =1

=

75 2

(1 +

= 522

(−50 + 538)

=

12 2

(1 + 39.5)

=

10 2

(0.5 +

n =1

a10 = 59

48) = 255

=

n =1

d = a 2 − a1

(3 +

= 1425

500 ⎛ 1 ⎞ ⎜ + 250 ⎟ = 62,625 2 ⎝2 ⎠

n =1

5 2

( 4 + 53)

10 2

k =1

= d

an = 5 +

=

k =1

10 = 4d 5 2

20

∑k

= 12,200

= 243

4.1) = 23

75) = 2850

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 11.2 51. a1 = $54,000

65. a1 =

a2 = $57,000

a2 =

a3 = $60,000

a3 =

a4 = $63,000 a5 = $66,000

a4 =

a6 = $69,000

a5 =

Total salary = =

(54,000 + 69,000) 3(123,000) 6 2

= $369,000

53. Sequence = 93, 89, 85, 81, … 8

∑ (97

− 4 n) =

n =1

8 2

(93 +

65) = 632 bales

55. To find the next term in the sequence, add 3 to the value of the term.

(1) − 1 = 32 5 2 −1 = 4 2( ) 5 3 − 1 = 13 2( ) 2 5 4 −1 = 9 2( ) 5 5 − 1 = 23 2( ) 2

67. (d) 68. (c) 69. (a) 70. (b) 71. Sequence = 16, 48, 80, … , n = 8, d = 32

an = a1 + ( n − 1)d an = 16 + ( n − 1)32 = 16 + 32n − 32 an = 32n − 16

59. a1 = 7, d = 5

an = 16 + 224

a2 = 7 + 5 = 12 a3 = 12 + 5 = 17 a4 = 17 + 5 = 22 a5 = 22 + 5 = 27 61. a1 = 11, d = 4

a1 = 11 a2 = 11 + 4 = 15 a3 = 15 + 4 = 19 a4 = 19 + 4 = 23 a5 = 23 + 4 = 27 63. a1 = 3(1) + 4 = 7 a2 = 3( 2) + 4 = 10 a3 = 3(3) + 4 = 13 a4 = 3( 4) + 4 = 16 a5 = 3(5) + 4 = 19

357

5 2

57. The nth partial sum can be found by multiplying the number of terms by the average of the first term and the nth term.

a1 = 7

Arithmetic Sequences

an = 16 + (8 − 1)32 an = 240 8

∑ (32n − 16)

=

n =1

8 2

(16 +

240) = 1024 feet

73. Sequence = 1, 2, 3, 4,…

an = a1 + ( n − 1)d an = 1 + ( n − 1)(1) an = 1 + n − 1 an = n 12

∑n n =1

=

12 2

(1 + 12)

= 78 chimes

3 chimes each hour × 12 hours = 36 chimes Total chimes = 78 + 36 = 114 chimes 75. A recursion formula gives the relationship between the terms an +1 and an . 77. Yes. Because a2n is n terms away from an , add n times the difference d to an .

a2n = an + nd

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

358

Chapter 11

Sequences, Series, and the Binomial Theorem

79. (a) 1 + 3 = 4

1+3+5 = 9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 + 9 = 25 1 + 3 + 5 + 7 + 9 + 11 = 36 (b) No. There is no common difference between consecutive terms of the sequence. (c) 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 n

∑ ( 2k k =1

− 1) = n 2

(d) Looking at the sums in part (a), each sum is the square of a consecutive positive integer.

1 + 3 = 22 1 + 3 + 5 = 32 1 + 3 + 5 + 7 = 42 1 + 3 + 5 + 7 + 9 = 52 1 + 3 + 5 + 7 + 9 + 11 = 62 So the sum of n odd integers is n 2 . An odd integer is represented by 2k − 1 which gives the formula n

∑ ( 2k k =1

81.

(x

− 1) = n 2 .

+ 2) 2 + ( y − 8) = 1 4 2

83.

x 2 + 4 y 2 − 8 x + 12 = 0

( x2 − 8x + 16) + 4 y 2

Center: ( −2, 8)

(x

2

a = 4

− 4) + 4 y 2 = 4

(x

a = 2, horizontal axis Vertices: ( −2 + 2, 8), ( −2 − 2, 8)

(0, 8), ( − 4, 8)

= −12 + 16

2

− 4) + y2 = 1 4 2

Center: ( 4, 0)

a2 = 4 a = 2, horizontal axis Vertices: ( 4 + 2, 0), ( 4 − 2, 0)

(6, 0), (2, 0) 85. 3 + 6 + 9 + 12 + 15 =

5

∑3k k =1

87. 2 + 22 + 23 + 24 + 25 =

5

∑2k k =1

Mid-Chapter Quiz for Chapter 11 1. an = 4n

2. an = 2n + 5

a1 = 4(1) = 4

a1 = 2(1) + 5 = 7

a2 = 4( 2) = 8

a2 = 2( 2) + 5 = 9

a3 = 4(3) = 12

a3 = 2(3) + 5 = 11

a4 = 4( 4) = 16

a4 = 2( 4) + 5 = 13

a5 = 4(5) = 20

a5 = 2(5) + 5 = 15

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 11.3

( 14 ) = 32( 14 ) = 32( 14 ) = 32( 14 ) = 32( 14 ) = 32( 14 )

3. an = 32

n −1

1 −1

a1 a2

10.

3 −1

a3 a4

4 −1

5 −1

a5

11.

= 8

1 2

=

1 8

20

a1 = a2 = a3 = a4 =

(−3)

12.

13.

n n+ 4 1+ 4

14.

∑ 20

∑

4

⋅4 81 = 4+ 4 2

4 2

=

k =1

6.

10

∑4

=

i =1

7.

10 2

(10 +

( 4 + 4)

∑

j =1

11 = 20 + ( 4 − 1)d

an = 20 − 3n + 3

−9 = 3d

an = −3n + 23

an = a1 + ( n − 1)d an = 32 + ( n − 1)( − 4)

40) = 100

an = 32 − 4n + 4 an = − 4n + 36

= 40

60 60 60 60 60 60 = + + + + 2 3 4 5 6 j +1

200

∑ 2n

∑

n =1

9.

∑ (3n − 1)

=

n =1

(2 +

400) = 100( 402) = 40,200

= 0.50 + ( n − 1)(0.50) = 0.50 + 0.50n − 0.50 an = 0.50n

12 = 12 + 6 + 4 + 3 = 25 n

5

200 2

20. an = a1 + ( n − 1)d

= 87

8.

=

n =1

= 30 + 20 + 15 + 12 + 10 4

an = 20 + ( n − 1)(−3)

18. a1 = 32, d = − 4

19.

5

1 2

−3 = d

5

∑ 10k

k2 2

17. an = a1 + ( n − 1)d

3

(−3) ⋅ 5 = −135 a5 = 5+ 4 5.

k −1 k

16. d = −6

⋅3 81 = − 3+ 4 7

4

10

∑

k +1

k3

15. d =

⋅2 = 3 2+ 4

( −3)

( −1)

25

k =1

3 = − 5

( −3)2 ( −3)

2

k =1

n

( −3)1 ⋅ 1

− 1) = 0 + 3 + 8 + 15 = 26

∑ 3k

k =1

4. an =

359

k =1

= 2 =

4

∑ (k 2 k =1

= 32

2 −1

Geometric Sequences and Series

5 2

(2 + 14)

365

∑ 0.50n

=

n =1

= 40

365 2

(0.50 + 182.5)

= $33,397.50

Section 11.3 Geometric Sequences and Series 1. 3, 6, 12, 24, …

5. The sequence is not geometric, because

r = 2 since 6 3

15 10

24 = 2. = 2, 12 = 2, 12 6 3

r = π since

π 1

= π,

π π = π, = π. π π 2

3

and

20 15

=

4. 3

7. The sequence is geometric. r =

3. 1, π , π , π , … 2

3 2

=

32 64

=

1 2

since 1 , 16 2 32

=

1, 8 2 16

=

1. 2

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

360

Chapter 11

Sequences, Series, and the Binomial Theorem

9. The sequence is geometric: r = −

( ) ( ) − 12

4 an +1 = an 4 − 12

n +1 n

1 since 2

1 = − . 2

17. an = a1r n −1 ⇒ r =

( 14 )

an = 8 19.

n −1

21.

( 23 )

an = 9

15. an = a1r

25.

29.

31.

33.

⇒ r =

a7 = 2( 2)

7 −1

i =1

27.

n −1

an = 2( 2)

∑

i =1

12

i −1

⎛ 3⎞ ∑ 3⎝⎜ 2 ⎟⎠ i =1

∞

⎛ 3⎞ ∑ 8⎜⎝ 4 ⎟⎠ n =1

⎛ ⎛ 3 ⎞12 ⎞ ⎜ ⎜ ⎟ − 1⎟ 2 ⎟ = 3⎛ 128.74634 ⎞ ≈ 772.48 = 3⎜ ⎝ ⎠ ⎜ ⎟ ⎜ 3 ⎟ 0.5 ⎝ ⎠ −1 ⎟ ⎜⎜ 2 ⎟ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎜ 1 ⎟ ⎜1⎟ = 8⎜ = 8 ⎟ ⎜ ⎟ = 8( 4) = 32 3 ⎜⎜ 1 − ⎟⎟ ⎜⎜ 1 ⎟⎟ 4⎠ ⎝ ⎝4⎠

n −1

⎛ 2⎞

∑ 2⎜⎝ 3 ⎟⎠

=

⎛ 3⎞

n

=

⎛1⎞

∑ ⎜⎝ 2 ⎟⎠ ∞

⎛ 1⎞

∑ ⎜⎝ − 2 ⎟⎠ n =1

2

2 1− 3

=

=

2 = 6 1 3

1 1 7 = = 10 ⎛ 3⎞ 10 1 − ⎜− ⎟ 7 ⎝ 7⎠

n −1

n =1

39.

i =1

41.

∞

⎛1⎞

∑ ⎜⎝ 10 ⎟⎠

n

1

=

1 1− 10

n=0

43. Total salary =

=

40

1 10 = 9 9 10

∑ 30,000(1.05)

n

n =1

⎛ 1.0540 − 1 ⎞ = 30,000⎜ ⎟ ≈ $3,623,993.23 ⎝ 1.05 − 1 ⎠ 45. an = 0.01( 2)

n −1

(a) Total income =

29

∑ 0.01(2)

n −1

n =1

n

∑ ⎜⎝ − 7 ⎟⎠ ∞

⎛1⎞

∑ 8⎜⎝ 2 ⎟⎠

= 128

⎛ 210 − 1 ⎞ 1024 − 1 = 1⎜ = 1023 ⎟ = 2 1 1 − ⎝ ⎠

n=0

37.

= 2

23.

⎛ ⎛ 1 ⎞15 ⎞ ⎜ ⎜ ⎟ − 1⎟ 2 ⎟ ≈ 16.00 = 8⎜ ⎝ ⎠ ⎜ 1 ⎟ −1 ⎟ ⎜⎜ ⎟ 2 ⎝ ⎠

⎛ ⎛ 3 ⎞6 ⎞ ⎛ − 3367 ⎞ ⎜ ⎜ ⎟ − 1⎟ 3⎜ ⎝ 4 ⎠ 3 ⎛3⎞ ⎟ = ⎜⎜ 4096 ⎟⎟ = 3 ⎛ 3367 ⎞ ≈ 2.47 ⎜ ⎟ = ⎜ 3 ⎜ ⎟ ⎟ 4 ⎛ ⎞ 4⎜ − 1 ⎟ 4 ⎝ 1024 ⎠ ⎝ 4⎠ ⎜ ⎟ ⎜⎜ ⎜ ⎟ − 1 ⎟⎟ 4 ⎠ ⎝ ⎝ ⎝ 4⎠ ⎠

10

∞

15

i −1

⎛ ( −3)10 − 1 ⎞ ⎟ = −14,762 = ⎜ ⎜ −3 − 1 ⎟ ⎝ ⎠

n

n=0

35.

4 2

∑ 2i −1

∞

i −1

n −1

n −1

6

∑ 1(−3) i =1

13. an = a1r n −1

1 4

⎛ 38 − 1 ⎞ ⎛ 6560 ⎞ = 4⎜ ⎟ = 4⎜ ⎟ = 13,120 ⎝ 2 ⎠ ⎝ 3 −1⎠

i −1

10

=

n −1

i =1

11. an = a1r n −1

an = 1( 2)

8

∑ 4(3)

2 8

1

1 1− 2

n −1

=

=

1 = 2 1 2

⎡ 229 − 1⎤ = 0.01⎢ ⎥ ⎣ 2 −1 ⎦ ≈ $5,368,709.11

(b) Total income =

30

∑ 0.01(2)

n −1

n =1

⎡ 230 − 1⎤ = 0.01⎢ ⎥ ⎣ 2 −1 ⎦ ≈ $10,737,418.23

1 1 2 = = 3 1 ⎛ ⎞ 3 1−⎜ ⎟ 2 ⎝ 2⎠

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 11.3 r⎞ ⎛ 47. A = P⎜1 + ⎟ n⎠ ⎝

12(10)

= 50(1.0075)

120

a2

⎡1.0075120 − 1⎤ A = ⎡⎣50(1.0075)⎤⎦ ⎢ ⎥ ⎣ 1.0075 − 1 ⎦ ≈ $9748.28

a4

0.08 ⎞ ⎛ = 30⎜1 + ⎟ 12 ⎠ ⎝

⎛ 151 ⎞ = 30⎜ ⎟ ⎝ 150 ⎠

61.

480

1

nt

= 100(1.005)

a1 = 100(1.005)

6

∑ 3(2)

53. The partial sum

i −1

is the sum of the first

i =1

6 terms of the geometric sequence whose nth term .

55. The common ratio of the geometric series with the sum

∑ 5( 54 )

3 1− 4

=

8 = 32 1 4

1 3 1 9 1 16 12 9 37 + ⋅ + ⋅ = + + = 4 4 4 16 4 64 64 64 64

or Total area of shaded region 3

1⎛ 3 ⎞

∑ 4 ⎜⎝ 4 ⎟⎠

n −1

⎡ ⎛ 3 ⎞3 ⎤ ⎢ ⎜ ⎟ − 1⎥ 1 4 ⎥ = 37 ≈ 0.578 square unit = ⎢⎝ ⎠ 4 ⎢ 3 − 1 ⎥ 64 ⎢ 4 ⎥ ⎣⎢ ⎦⎥

65. an = a1r n

Tn = 70(0.8)

⎡1.005360 − 1⎤ A = ⎡⎣100(1.005)⎤⎦ ⎢ ⎥ ≈ $100,953.76 ⎣ 1.005 − 1 ⎦

n=0

8

=

( ) = ( − 4)( 14 ) = −1 = (− 4)( − 18 ) = 12 1 = −1 = ( − 4)( 16 ) 4 = (− 4) − 12 = 2

360

1

n

n

i =1

12(30)

∞

=

=

0.06 ⎞ ⎛ a360 = 100⎜1 + ⎟ 12 ⎠ ⎝

is an = 3( 2)

⎛ 3⎞

= −4

63. Total area of shaded region

⎡ ⎛ 151 ⎞ 480 ⎤ − 1⎥ ⎢⎜ ⎟ ⎡ ⎛ 151 ⎞⎤ ⎢ ⎝ 150 ⎠ ⎥ ≈ $105,428.44 Balance = ⎢30⎜ ⎟⎥ ⎣ ⎝ 150 ⎠⎦ ⎢ ⎛ 151 ⎞ − 1 ⎥ ⎢ ⎜ ⎟ ⎥ ⎢⎣ ⎝ 150 ⎠ ⎥⎦

i −1

∞

∑ 8⎜⎝ 4 ⎟⎠

n=0

⎛ 151 ⎞ a1 = 30⎜ ⎟ ⎝ 150 ⎠

4 −1

5 −1

a5

12( 40)

2 −1

3 −1

a3

nt

n −1

1 −1

a1

1

r⎞ ⎛ 51. A = P⎜1 + ⎟ n⎠ ⎝

( ) = ( − 4)( − 12 ) = ( − 4)( − 12 ) = ( − 4)( − 12 ) = ( − 4)( − 12 ) = ( − 4)( − 12 )

an = ( − 4) − 12

a1 = 50(1.0075)

a480

361

59. an = a1r n −1

nt

0.09 ⎞ ⎛ a120 = 50⎜1 + ⎟ 12 ⎠ ⎝

r⎞ ⎛ 49. A = P⎜1 + ⎟ n ⎝ ⎠

Geometric Sequences and Series

is 54 .

n

T6 = 70(0.8) ≈ 18.4° 6

67. The general formula for the nth term of a geometric sequence is an = a1r n −1. 69. An arithmetic sequence has a common difference between consecutive terms and a geometric sequence has a common ratio between terms. 71. The terms of a geometric sequence decrease when a1 > 0 and a < r < 1 because raising a real number between 0 and 1 to higher powers yields smaller numbers.

57. an = a1r n −1 an = 5( −2)

n −1

a1 = 5( −2)

1 −1

a2 = 5( −2) a3 = 5( −2)

= 5( −2) = 5(1) = 5 0

2 −1

= 5( −2) = 5(−2) = −10

3 −1

= 5(−2) = 5( 4) = 20

a4 = 5( −2) a5 = 5( −2)

1

2

4 −1

= 5( −2) = 5(−8) = − 40

5 −1

= 5( −2) = 5(16) = 80

3

4

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

362

Chapter 11

Sequences, Series, and the Binomial Theorem

73. ⎧⎪ y = 2 x 2 ⎨ ⎪⎩ y = 2 x + 4

r⎞ ⎛ 77. A = P⎜1 + ⎟ n⎠ ⎝

1( 20)

r⎞ ⎛ 10,619.63 = 2500⎜1 + ⎟ 1⎠ ⎝

2x2 = 2x + 4 2 x2 − 2 x − 4 = 0

4.247852 = (1 + r )

x2 − x − 2 = 0

(x

− 2)( x + 1) = 0

x = 2

4.2478521 20 − 1 = r 0.075 ≈ r

y = 2( −1) = 2

2

2

7.5% ≈ r

(2, 8), (−1, 2) r⎞ ⎛ 75. A = P⎜1 + ⎟ n ⎝ ⎠

20

4.2478521 20 = 1 + r x = −1

y = 2( 2) = 8

nt

79.

nt

x2 y2 − =1 16 9

y 16 12

Horizontal axis

8

Vertices: ( ±4, 0)

12(10)

r ⎞ ⎛ 2219.64 = 1000⎜1 + ⎟ 12 ⎝ ⎠

3 Asymptotes: y = ± x 4

120

r ⎞ ⎛ 2.21964 = ⎜1 + ⎟ 12 ⎠ ⎝ r 2.219641 120 = 1 + 12

4 − 12 − 8

x −4

8 12 16

−8 − 12 − 16

12( 2.219641 120 − 1) = r 0.08 ≈ r 8% ≈ r

Section 11.4 The Binomial Theorem 13.

6⋅5 = 15 2 ⋅1

1.

6 C4

3.

10 C5

5.

12 C12

=

12! =1 0!12!

7.

20 C6

=

20! 20 ⋅ 19 ⋅ 18 ⋅ 17 ⋅ 16 ⋅ 15 = = 38,760 14!6! 6 ⋅ 5 ⋅ 4 ⋅ 3⋅ 2 ⋅1

9.

= 6C2 =

20 C14

=

1

1

1 8

1 7

15.

1 6 28

1 5 21

1 4 15 56

1 3 10 35

1 2 6 20 70

1 3

17.

10 35

4 15 56

1 5 21

1 6 28

1 7

= 35

8 C4

= 70

Row 8: 1 8 28 56 70 56 28 8 1 ↑ entry 4 19.

1

7 C3

Row 7: 1 7 21 35 35 21 7 1 ↑ entry 3

20! 20 ⋅ 19 ⋅ 18 ⋅ 17 ⋅ 16 ⋅ 15 = = 38,760 6!14! 6 ⋅ 5 ⋅ 4 ⋅ 3⋅ 2 ⋅1

11.

= 15

Row 6: 1 6 15 20 15 6 1 ↑ entry 2

10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 = 252 5 ⋅ 4 ⋅ 3⋅ 2 ⋅1

=

6 C2

5 C3

Row 5: 1 5 10 10 5 1 ↑ entry 3 1 8

5 C3

1

= 10

1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 11.4 21.

7 C4

363

23. (t + 5) = (1)t 3 + (3)t 2 (5) + (3)t (5) + (1)53 3

Row 7: 1 7 21 35 35 21 7 1 ↑ entry 4 7 C4

The Binomial Theorem 2

= t 3 + 15t 2 + 75t + 125

= 35

25. ( m − n) = (1)m5 + (5)m 4 ( − n) + (10)m3 ( − n) + 10m 2 ( − n) + 5m( − n) + 1(− n) 5

2

3

4

5

= m5 − 5m 4 n + 10m3n 2 − 10m 2 n3 + 5mn 4 − n5 27.

(x

+ 3) = 1x 6 + 6 x5 (3) + 15 x 4 (3) + 20 x3 (3) + 15 x 2 (3) + 6 x(3) + 1(3) 6

2

3

4

5

6

= x 6 + 18 x5 + 135 x 4 + 540 x3 + 1215 x 2 + 1458 x + 729 29. (u − v) = 1(u 3 ) + 3(u 2 )( − v) + 3(u )( − v) + 1( − v) 3

2

3

= u 3 − 3u 2v + 3uv 2 − v3

31. (3a − 1) = (1)(3a) + (5)(3a ) (−1) + (10)(3a) ( −1) + (10)(3a) ( −1) + (5)(3a )(−1) + (1)(−1) 5

5

4

3

2

2

3

4

5

= 243a 5 − 405a 4 + 270a 3 − 90a 2 + 15a − 1 33. ( 2 y + z ) = (1)( 2 y ) + 6( 2 y ) z + 15( y ) z 2 + 20( 2 y ) z 3 + 15( 2 y ) z 4 + 6( 2 y ) z 5 + 1z 6 6

6

5

4

3

2

= 64 y 6 + 192 y 5 z + 240 y 4 z 2 + 160 y 3 z 3 + 60 y 2 z 4 + 12 yz 5 + z 6 35.

( x 2 + 2)

= 1( x 2 ) + 4( x 2 ) ( 2) + 6( x 2 ) ( 2) + 4( x 2 )( 2) + 1( 2)

4

4

3

2

2

3

4

= x8 + 8 x 6 + 24 x 4 + 32 x 2 + 16 37. (3a + 2b) = 1(3a ) + 4(3a) ( 2b) + 6(3a ) ( 2b) + 4(3a)( 2b) + 1( 2b) 4

4

3

2

2

3

4

= 81a 4 + 216a 3b + 216a 2b 2 + 96ab3 + 16b 4 4

2

3

⎛ ⎛ 2⎞ ⎛ 2⎞ 2⎞ 4 3⎛ 2 ⎞ 2⎛ 2 ⎞ 39. ⎜ x + ⎟ = 1( x ) + 4( x) ⎜ ⎟ + 6( x) ⎜ ⎟ + 4( x )⎜ ⎟ + 1⎜ ⎟ y⎠ ⎝ ⎝ y⎠ ⎝ y⎠ ⎝ y⎠ ⎝ y⎠ = x4 +

4

8 x3 24 x 2 32 x 16 + + 3 + 4 y y2 y y

41. ( 2 x 2 − y ) = 1( 2 x 2 ) + 5( 2 x 2 ) ( − y ) + 10( 2 x 2 ) ( − y ) + 10( 2 x 2 ) (− y ) + 5( 2 x 2 )( − y ) + 1(− y ) 5

5

4

3

2

2

3

4

5

= 32 x10 − 80 x8 y + 80 x 6 y 2 − 40 x 4 y 3 + 10 x 2 y 4 − y 5

43.

( x + y )10 , 4th term (r + 1)th term = nCr x n − r y r r +1 = 4

n = 10 x = x y = y

9

(r

+ 1) th term = nCr x n − r y r

r +1 = 5 n = 9

x = a

y = 6b

r = 4

r = 3

4th term =

45. ( a + 6b) , 5th term

10 − 3 3 y 10C3 x

= 120 x 7 y 3

5th term = 9C4 a 9 − 4 (6b)

4

= 126a 5 (1296b 4 ) = 163,296a 5b 4

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

364

Chapter 11

Sequences, Series, and the Binomial Theorem

47. ( 4 x + 3 y ) , 8th terms 9

(r

51.

+ 1) th term = nCr x n − r y r

Coefficient = (3) 12 C3 = 27( 220) = 5940

y = 3y

r = 7 2

53. Pascal’s Triangle is formed by making the first and last numbers in each row 1. Every other number in the row is formed by adding the two numbers immediately above the number.

7

= 36(16 x 2 )( 2187 y 7 ) = 1,259,712 x 2 y 7 49.

n Cr x

n−r

n

55. The signs in the expansion of ( x + y ) are all positive.

yr

n

The signs in the expansion of ( x − y ) alternate.

n = 10 n − r = 7 r = 3 x = x y = 1

59.

12 ⋅ 11 ⋅ 10 = 220 3⋅ 2 ⋅1

=

3

r + 1 = 8 n = 9 x = 4x

8th term = 9C7 ( 4 x) (3 y )

12 C3

(

57.

)

x + 5

3

=

( x)

3

( x ) (5) + 3( x )(5 ) + 5

+ 3

10 C3 x

7 3

10 C3

=

( x2 3

− y1 3 ) = ( x 2 3 ) − 3( x 2 3 ) ( y1 3 ) + 3( x 2 3 )( y1 3 ) − ( y1 3 ) = x 2 − 3x 4 3 y1 3 + 3 x 2 3 y 2 3 − y

1

10 ⋅ 9 ⋅ 8 = 120 3⋅ 2 ⋅1 3

(

4

61. 3 t +

t

)

(

= 3 t

2

)

2

3

= x 3 2 + 15 x + 75 x1 2 + 125

3

4

2

(

4

+ 43 t

2

3

) ( t ) + 6(3 t ) ( t ) 3

2

4

4

(

2

+ 43 t

)( t ) + ( t ) 4

3

4

4

= 81t 2 + 108t 7 4 + 54t 3 2 + 12t 5 4 + t

67. Keystrokes:

63. Keystrokes:

30 MATH PRB 3 6 ENTER

30C6

= 593,775

800 MATH PRB 3 797 ENTER 800 C797

65. Keystrokes:

52 MATH PRB 3 5 ENTER

52 C5

= 85,013,600

69. (1 + i ) = 14 + 4 ⋅ 13 i + 6 ⋅ 12 ⋅ i 2 + 4 ⋅ 1 ⋅ i 3 + i 4 4

= 2,598,960

= 1 + 4i − 6 − 4i + 1 = −4 71. (1.02) = (1 + 0.02) 8

8

= (1) + 8(1) (0.02) + 28(1) (0.02) + 56(1) (0.02) + … 8

7

6

2

5

3

≈ 1 + 0.16 + 0.0112 + 0.000448 ≈ 1.172

73. ( 2.99)

= (3 − 0.01)

12

12

= 1(3)

12

− 12(3) (0.01) + 66(3) 11

10

(0.01)

2

− 220(3) (0.01) + 495(3) (0.01) − 792(3) (0.01) + … 9

3

8

4

7

5

≈ 531,441 − 21,257.64 + 389.7234 − 4.33026 + 0.03247695 − 0.0001732104 ≈ 510,568.785 75.

( 12 + 12 )

5

( 12 )

=1 =

77.

( 14 + 34 )

4

1 32

5

+

( 14 )

=1

4

( 12 ) ( 12 ) + 10( 12 ) ( 12 ) 4

+5 5 32

+

10 32

3

+

10 32

+

5 32

+

3

2

( 12 ) ( 12 )

+ 10

2

3

( 12 )( 12 )

+5

4

( 12 )

+1

5

1 32

( 14 ) ( 43 ) + 6( 14 ) ( 43 )

+ 4

2

2

( 14 )( 43 )

+ 4

3

79. The sum of the numbers in each row is a power of 2. Because Row 2 is 1 + 2 + 1 = 4 = 22 , Row n is 2n.

( 43 )

+1

4

=

1 256

+

12 256

+

54 256

+

108 256

+

81 256

81. No. The coefficient of x 4 y 6 is r = 6.

10 C6

because n = 10 and

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 11 83. The value for r is 6 because r + 1 = 7. 85.

15

∑ (2 + 3i)

=

i =1

15 2

(5 +

47) =

15 2

(52)

87.

365

⎡ 58 − 1⎤ = 1⎢ ⎥ = 97,656 ⎣5 − 1⎦

8

∑ 5 k −1 k =1

= 390

a1 = 2 + 3(1) = 5, an = 2 + 3(15) = 47

Review Exercises for Chapter 11 1. an = 3n + 5

9. an = 3n + 1

a1 = 3(1) + 5 = 8 a2 = 3( 2) + 5 = 11 a3 = 3(3) + 5 = 14 a4 = 3( 4) + 5 = 17 a5 = 3(5) + 5 = 20

n 3n − 1 1 1 a1 = = 3(1) − 1 2

3. an =

2

2 a2 = = 3( 2) − 1 5

a3 =

3

3(3) − 1

=

4 4 = 3( 4) − 1 11

a5 =

5 5 = 3(5) − 1 14

5. an = ( n + 1)!

a1 = (1 + 1)! = 2! = 2 ⋅ 1 = 2 a2 = ( 2 + 1)! = 3! = 3 ⋅ 2 ⋅ 1 = 6 a3 = (3 + 1)! = 4! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24 a4 = ( 4 + 1)! = 5! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120 a5 = (5 + 1)! = 6! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 720

a1 = a2 = a3 = a4 = a5 =

1 n2 + 1

13. an = −2n + 5 15. an = 17.

4

∑7

3n 2 n +1 2

= 7 + 7 + 7 + 7 = 28

k =1

19.

5

∑ i =1

i −2 1 1 2 1 = − +0+ + + 2 4 5 2 i +1 1 2 + 4 5 5 8 = + 20 20 13 = 20 =

3 8

a4 =

7. an =

11. an =

n! 2n 1! 1 = 2 ⋅1 2 2! 2 1 = = 2⋅2 4 2 3! 3⋅ 2 ⋅1 = =1 2⋅3 2⋅3 4! 4 ⋅ 3⋅ 2 ⋅1 = = 3 2⋅4 2⋅4 5! 5 ⋅ 4 ⋅ 3⋅ 2 ⋅1 = = 12 2⋅5 2⋅5

21.

4

∑ (5k k =1

23.

6

− 3)

1

∑ 3k k =1

25. 50, 44.5, 39, 33.5, 28, …

44.5 − 50 = −5.5 39 − 44.5 = −5.5 33.5 − 39 = −5.5 28 − 33.5 = −5.5 d = −5.5 27. a1 = 132 − 5(1) = 127 a2 = 132 − 5( 2) = 122 a3 = 132 − 5(3) = 117 a4 = 132 − 5( 4) = 112 a5 = 132 − 5(5) = 107

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

366

Chapter 11

29. a1 =

Sequences, Series, and the Binomial Theorem

(1) + 53 = 36 = 2 1 2 + 5 = 7 3( ) 3 3 1 3 + 5 = 8 3( ) 3 3 1 4 + 5 = 9 = 3 3( ) 3 3 1 5 + 5 = 10 ( ) 3 3 3

1 3

a2 = a3 = a4 = a5 =

r =

5 2

=

160 2

a3 =

155 2

−

5 2

=

150 2

a4 =

150 2

−

5 2

=

145 2

a5 =

145 2

−

5 2

=

140 2

−

5 2

=

155 2

= 75

a2 = 10(3)

= 70

a3 = 10(3)

a1

1050 = c

a2

an = −50n + 1050 − 5) =

12 2

(2 +

79) = 486

120

∑ ( 14 j + 1) = 1202 ( 54 + 31) = 60(1294 ) = 1935 j =1

1.25 = d

a5

a3

a60 = 5.25 + (60 − 1)(1.25) a60 = 79

a4

n

a5

60

∑ ai

=

n ( a 1 + an ) 2

=

60 (5.25 + 79) = 2527.5 2

i =1 50

∑ 4n n =1

45.

a4

a2

an = a1 + ( n − 1)d

43.

a3

a1

6.5 = 5.25 + d

i =1

12

=

50 (4 + 200) = 5100 2

∑ (3n + 19) n =1

= 810

( ) = 100( − 12 ) = 100( − 12 ) = 100( − 12 ) = 100( − 12 ) = 100( − 12 )

12 2

(22 + 55)

= 100

2 −1

3 −1 4 −1

5 −1

( 32 ) = 4( 32 ) = 4( 32 ) = 4( 32 ) = 4( 32 ) = 5( 32 )

= −50 = 25 = −12.5 = 6.25

n −1

1 −1

= 462 seats

( 32 ) = 4( 32 ) = 4( 32 ) = 4( 32 ) = 4( 23 )

= 4

2 −1

3 −1 4 −1

5 −1

0 1

2 3

4

= 4(1) = 4

( 32 ) = 6 = 4( 94 ) = 9 = 4( 27 = 27 8) 2 81 = 81 = 4( 16 ) 4

= 4

55. an = a1r n −1

( )

an = 1 − 23

n −1

57. an = a1r n −1 an = 24(3)

=

n −1

1 −1

an = 4

6.5 = 5.25 + ( 2 − 1)d

∑ ai

= 270

5 −1

53. an = a1r n −1

an = a1 + ( n − 1)d

41.

= 90

4 −1

an = 100 − 12

1000 = −50(1) + c

39.

= 30

51. an = a1r n −1

an = dn + c

k =1

= 10

3 −1

a5 = 10(3)

an = 4n + 6

∑ (7 k

n −1

2 −1

a4 = 10(3)

6 = c

37.

an = 10(3)

1 −1

10 = 4(1) + c

12

49. an = a1r n −1 a1 = 10(3)

33. an = dn + c

35.

5 since 2

625 20 5 50 5 125 5 2 5 = , = , = , = . 8 2 20 2 50 2 125 2

31. a1 = 80

a2 = 80 −

625 ,… 2

47. 8, 20, 50, 125,

r =

n −1

72 a2 = = 3 24 a1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Review Exercises for Chapter 11 59. an = a1r n −1

( )

an = 12 − 12

69.

n −1

280 7 = 200 5

i =1

12

i −1

∑ a1 r

i −1

⎛7⎞

∑ 200⎜⎝ 5 ⎟⎠ i =1

⎛ 4⎞ an = 27⎜ − ⎟ ⎝ 3⎠

i =1

14

i −1

∑ a1 r

⎛ 4⎞

∑ 27⎜⎝ − 3 ⎟⎠ i =1

65.

n =1

k

=

4

2 1− 3

4 = 12 1 3

=

n

5

75.

90

∑ 1000(1.125) i =1

i

⎡1.12590 − 1⎤ = 1000 ⎢ ⎥ ⎣ 1.125 − 1 ⎦ ≈ 321,222,672 visitors

36 4 = − 27 3

⎛ ⎛ 4 ⎞14 ⎞ ⎜ ⎜ − ⎟ − 1⎟ 3 ⎟ ≈ − 637.85 = 27⎜ ⎝ ⎠ ⎜ ⎛ 4⎞ ⎟ ⎜⎜ ⎜ − ⎟ − 1 ⎟⎟ 3 ⎝ ⎝ ⎠ ⎠

8! 8 ⋅ 7 ⋅ 6 ⋅ 5! = = 56 3!5! 3 ⋅ 2 ⋅ 5!

77.

8 C3

79.

15 C4

=

15! 15 ⋅ 14 ⋅ 13 ⋅ 12 = = 1365 11!4! 4 ⋅ 3⋅ 2 ⋅1

81.

40 C4

=

40! = 91,390 36!4!

83.

25 C6

=

25! = 177,100 19!6!

85.

4 C2

=

Row 4: 1 4 6 4 1 ↑ entry 2

⎛ ⎛ 3 ⎞8 ⎞ ⎜ ⎜ − ⎟ − 1⎟ 15 ⎛ 3⎞ 4 ⎟ ≈ −1.928 67. ∑ 5⎜ − ⎟ = − ⎜ ⎝ ⎠ ⎜ ⎟ 3 4 4 ⎝ ⎠ k =1 − − 1 ⎜⎜ ⎟⎟ 4 ⎝ ⎠

87.

⎛ 2⎞

1 = 8 1 8

=

7 1− 8

(b) a5 = 120,000(0.7) = $20,168.40

k

8

∞

∑ 4⎜⎝ 3 ⎟⎠

1

an = 120,000(0.7)

⎛ 212 − 1 ⎞ = 2⎜ ⎟ = 8190 ⎝ 2 −1 ⎠

12

∑ 2n

=

73. (a) an = a1r n

n −1

⎛ r n −1 ⎞ = a1 ⎜ ⎟ ⎝ r − 1⎠

i −1

i −1

k =0

⎛ ⎛ 7 ⎞12 ⎞ ⎜ ⎜ ⎟ − 1⎟ 5 ⎟ ≈ 27,846.96 = 200⎜ ⎝ ⎠ ⎜ 7 ⎟ −1 ⎟ ⎜⎜ ⎟ 5 ⎝ ⎠

63. an = a1 r n −1 ⇒ r = −

n

71.

n −1

⎛ r n −1 ⎞ = a1 ⎜ ⎟ ⎝ r − 1⎠

n

⎛7⎞

i =1

61. an = a1 r n −1 ⇒ r =

⎛7⎞ an = 200⎜ ⎟ ⎝5⎠

∞

∑ ⎜⎝ 8 ⎟⎠

367

4 C2

= 6

10 C3

Row 10: 1 10 45 120 210 252 210 120 45 10 1 ↑ entry 3 10 C3

89.

(x

= 120

− 5) = 1( x) + 4( x) ( −5) + 6( x) ( −5) + 4( x)( −5) + 1(−5) = x 4 − 20 x3 + 150 x 2 − 500 x + 625 4

4

3

2

2

3

4

91. (5 x + 2) = (1)(5 x) + (3)(5 x) ( 2) + (3)(5 x)( 2) + 1( 2) = 125 x3 + 150 x 2 + 60 x + 8 3

93.

(x

+ 1)

10

3

2

2

3

= 1x10 + 10 x9 (1) + 45 x8 (1) + 120 x 7 (1) + 210 x 6 (1) + 252 x 5 (1) + 210 x 4 (1) + 120 x3 (1) + 45 x 2 (1) 2

3

4

5

6

7

8

+ 10 x(1) + 1(1) 9

10

= x10 + 10 x 9 + 45 x8 + 120 x 7 + 210 x 6 + 252 x5 + 210 x 4 + 120 x3 + 45 x 2 + 10 x + 1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

368

Chapter 11

Sequences, Series, and the Binomial Theorem

95. (3x − 2 y ) = 1(3x) + 4(3 x) ( −2 y ) + 6(3 x) ( −2 y ) + 4(3x)(−2 y ) + (−2 y ) 4

4

3

2

2

3

4

= 81x 4 − 216 x3 y + 216 x 2 y 2 − 96 xy 3 + 16 y 4 97. (u 2 + v3 ) = (1)(u 2 ) + (5)(u 2 ) (v3 ) + (10)(u 2 ) (v3 ) + (10)(u 2 ) (v3 ) + (5)(u 2 )(v3 ) + (1)(v3 ) 5

5

4

3

2

2

3

4

5

= u10 + 5u 8v3 + 10u 6v 6 + 10u 4v9 + 5u 2v12 + v15 99.

( x + 2)10 , 7th term (r + 1)th term = nCr x n − r y r

101.

10 C5

r = 6 10 − 6 10C6 x

( 2)6

n−r

yr

n = 10, n − r = 5, r = 5, x = x, y = (−3)

r + 1 = 7, n = 10, x = x, y = 2

7th term =

n Cr x

= 252 ⋅ ( −3) = −61,236 5

= 210 x 4 (64) = 13,440 x 4

Chapter Test for Chapter 11

( ) = ( − 53 ) = ( − 53 ) = ( − 53 ) = ( − 53 ) = ( − 53 )

1. an = − 53

a1 a2 a3 a4 a5

8. an = a1 + ( n − 1)d

n −1

1−1

an = 12 + ( n − 1)4 = 12 + 4n − 4 = 4n + 8

=1

2 −1

3 −1

=

4 −1

5 −1

a1 = 4(1) + 8 = 12

= − 53

a2 = 4( 2) + 8 = 16

9 25

a3 = 4(3) + 8 = 20 a4 = 4( 4) + 8 = 24

27 = − 125

=

81 625

a5 = 4(5) + 8 = 28

9. an = a1 + ( n − 1)d

2. an = 3n 2 − n

an = 5000 + ( n − 1)( −100)

a1 = 3(1) − 1 = 2 2

an = 50000 − 100n + 100

a2 = 3( 2) − 2 = 12 − 2 = 10 2

a3 = 3(3) − 3 = 27 − 3 = 24

an = −100n + 5100

2

a4 = 3( 4) − 4 = 48 − 4 = 44 2

a5 = 3(5) − 5 = 75 − 5 = 70 2

3.

12

∑5

= 12(5) = 60

n =1

4.

8

∑ (2k − 3) = −3 + (−1) + 1 + 3 + 5 + 7 + 9 + 11 + 13

k =0

= 45 5.

5

∑ (3 − 4 n ) n =1

6.

12

n =1

7.

5 ⎡−1 2⎣

+ (−17)⎤⎦ = − 45

∑ ( 12 )

50

∑ 3n

=

n =1

50 2

(3 + 150)

= 3825

9 27 11. − 4, 3, − , , … 4 16 3 r = − since 4 27 9 − 3 3 4 3 16 3 = − , = − , = − . 4 3 4 −9 4 −4 4 12. an = a1r n −1

( 12 )

an = 4

2

∑ 3n + 1 6

=

10.

13.

8

∑ 2(2n ) n =1

n −1

⎛ 28 − 1 ⎞ = 4⎜ ⎟ = 1020 ⎝ 2 −1⎠

2k − 2

k =1

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test for Chapter 11 ⎛ 110 ⎞ n − 1⎟ 3⎜ 2 3069 ⎛1⎞ 14. ∑ 3⎜ ⎟ = ⎜ ⎟ = 1 2 2 1024 ⎜ −1⎟ n =1 ⎝ ⎠ ⎜ 2 ⎟ ⎝ ⎠

16.

10

∞

∑ 10(0.4) i =1

17.

20 C3

1 1 i ⎛1⎞ 2 = 2 =1 15. ∑ ⎜ ⎟ = 1 1 i =1 ⎝ 2 ⎠ 1− 2 2

=

i −1

=

369

10 10 100 50 = = = 1 − 0.4 0.6 6 3

20 ⋅ 19 ⋅ 18 = 1140 3 ⋅ 2 ⋅1

∞

18.

(x

− 2) = 1( x5 ) − 5 x 4 ( 2) + 10 x 3 ( 2) − 10 x 2 ( 2) + 5 x( 2) − 1( 2) 5

2

3

4

5

= x 5 − 10 x 4 + 40 x 3 − 80 x 2 + 80 x − 32

19. The coefficient of x3 y 5 in expansion of ( x + y ) is 56, since 8 C3 = 56. 8

20. 14.7 − 4.9 = 9.8 24.5 − 14.7 = 9.8

So, d = 9.8. an = a1 + ( n − 1)d an = 4.9 + ( n − 1)9.8 an = 4.9 + 9.8n − 9.8 an = 9.8n − 4.9 10

∑ 9.8n − 4.9 n =1

=

10 2

(4.9 + 93.1)

= 490 meters

1

0.048 ⎞ 0.048 ⎞ ⎛ ⎛ 21. Balance = 80⎜1 + ⎟ + … + 80⎜1 + ⎟ 12 ⎠ 12 ⎠ ⎝ ⎝

540

⎛ 1.004540 − 1 ⎞ = 80(1.004)⎜ ⎟ ⎝ 1.004 − 1 ⎠ = $153,287.87

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.