chapter 1 equations, inequalities, and mathematical modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17....

70
CHAPTER 1 Equations, Inequalities, and Mathematical Modeling Section 1.1 Graphs of Equations . . . . . . . . . . . . . . . . . . . . 39 Section 1.2 Linear Equations in One Variable . . . . . . . . . . . . . 43 Section 1.3 Modeling with Linear Equations . . . . . . . . . . . . . . 51 Section 1.4 Quadratic Equations and Applications . . . . . . . . . . . 58 Section 1.5 Complex Numbers . . . . . . . . . . . . . . . . . . . . . 67 Section 1.6 Other Types of Equations . . . . . . . . . . . . . . . . . 71 Section 1.7 Linear Inequalities in One Variable . . . . . . . . . . . . 82 Section 1.8 Other Types of Inequalities . . . . . . . . . . . . . . . . 87 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

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Page 1: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

C H A P T E R 1Equations, Inequalities, and Mathematical Modeling

Section 1.1 Graphs of Equations . . . . . . . . . . . . . . . . . . . . 39

Section 1.2 Linear Equations in One Variable . . . . . . . . . . . . . 43

Section 1.3 Modeling with Linear Equations . . . . . . . . . . . . . . 51

Section 1.4 Quadratic Equations and Applications . . . . . . . . . . . 58

Section 1.5 Complex Numbers . . . . . . . . . . . . . . . . . . . . . 67

Section 1.6 Other Types of Equations . . . . . . . . . . . . . . . . . 71

Section 1.7 Linear Inequalities in One Variable . . . . . . . . . . . . 82

Section 1.8 Other Types of Inequalities . . . . . . . . . . . . . . . . 87

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

Page 2: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

39

C H A P T E R 1Equations, Inequalities, and Mathematical Modeling

Section 1.1 Graphs of Equations

■ You should be able to use the point-plotting method of graphing.

■ You should be able to find x- and y-intercepts.

(a) To find the x-intercepts, let and solve for x.

(b) To find the y-intercepts, let and solve for y.

■ You should be able to test for symmetry.

(a) To test for x-axis symmetry, replace y with

(b) To test for y-axis symmetry, replace x with

(c) To test for origin symmetry, replace x with and y with

■ You should know the standard equation of a circle with center and radius r:

�x � h�2 � �y � k�2 � r 2

�h, k�

�y.�x

�x.

�y.

x � 0

y � 0

1.

(a)

Yes, the point is on the graph.

(b)

Yes, the point is on the graph.

3 � �9

�5, 3�: 3 �? �5 � 4

2 � 2

�0, 2�: 2 �? �0 � 4

y � �x � 4 3.

(a)

No, the point is not on the graph.

(b)

Yes, the point is on the graph.

0 � 4 � 4

�6, 0�: 0 �? 4 � �6 � 2�

5 � 4 � 1

�1, 5�: 5 �? 4 � �1 � 2�

y � 4 � �x � 2�

Vocabulary Check

1. solution or solution point 2. graph 3. intercepts

4. axis 5. circle; 6. point-plotting�h, k�; ry-

5.

x541−1−2−3

4

7

2

1

−1

2

3

5

yy � �2x � 5

x 0 1 2

y 7 5 3 1 0

�52, 0��2, 1��1, 3��0, 5���1, 7��x, y�

52�1

Page 3: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

40 Chapter 1 Equations, Inequalities, and Mathematical Modeling

7.

1 2

3

4

5

4 5x

−2 −1−1

−2

yy � x2 � 3x

x 0 1 2 3

y 4 0 0

�3, 0��2, �2��1, �2��0, 0���1, 4��x, y�

�2�2

�1

9.

x-intercepts:

y-intercept:

�0, 16�

y � 16 � 4�0�2 � 16

��2, 0�, �2, 0�

x � ±2

x2 � 4

4x2 � 16

0 � 16 � 4x2

y � 16 � 4x2 11.

x-intercepts:

or

y-intercept:

�0, 0�

y � 0

y � 2�0�3 � 4�0�2

�0, 0�, �2, 0�

x � 2 x � 0

0 � 2x2�x � 2�

0 � 2x3 � 4x2

y � 2x3 � 4x2

13. y-axis symmetry

–4 –3 –1 1 3 4

1

2

−2

3

4

x

y

15. Origin symmetry

–4 –3 –2 1 2 3 4

–4

–3

–2

1

2

3

4

x

y

17.

y-axis symmetry

No x-axis symmetry

No origin symmetry ��x�2 � ��y� � 0 ⇒ x2 � y � 0 ⇒

x2 � ��y� � 0 ⇒ x2 � y � 0 ⇒

��x�2 � y � 0 ⇒ x2 � y � 0 ⇒

x2 � y � 0 19.

No y-axis symmetry

No x-axis symmetry

Origin symmetry�y � ��x�3 ⇒ �y � �x3 ⇒ y � x3 ⇒

�y � x3 ⇒ y � �x3 ⇒

y � ��x�3 ⇒ y � �x3 ⇒

y � x3

21.

No y-axis symmetry

No x-axis symmetry

Origin symmetry�y ��x

��x�2 � 1 ⇒ �y �

�xx2 � 1

⇒ y �x

x2 � 1 ⇒

�y �x

x2 � 1 ⇒ y �

�xx2 � 1

y ��x

��x�2 � 1 ⇒ y �

�xx2 � 1

y �x

x2 � 1

23.

No y-axis symmetry

x-axis symmetry

No origin symmetry��x���y�2 � 10 � 0 ⇒ �xy2 � 10 � 0 ⇒

x��y�2 � 10 � 0 ⇒ xy2 � 10 � 0 ⇒

��x�y2 � 10 � 0 ⇒ �xy2 � 10 � 0 ⇒

xy2 � 10 � 0

Page 4: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.1 Graphs of Equations 41

25.

x-intercept:

y-intercept:

No axis or origin symmetry

�0, 1�

�13, 0�

( (x

(0, 1) 13

y

−1−2−3−4 1 2 3 4−1

−2

−3

1

4

5

, 0

y � �3x � 1 27.

Intercepts:

No axis or origin symmetry

�0, 0�, �2, 0�

−1

2

3

4

−1 1 2 3 4x

−2

−2

(0, 0) (2, 0)

yy � x2 � 2x

x 0 1 2 3

y 3 0 0 3�1

�1

29.

Intercepts:

No axis or origin symmetry

−4 −3 −2−1

1

2

4

5

6

7

1 2 3 4x

(0, 3)

−3, 03( (

y�0, 3�, � 3��3, 0�y � x3 � 3

x 0 1 2

y 2 3 4 11�5

�1�2

31.

Domain:

Intercept:

No axis or originsymmetry

�3, 0��3, ��

1 2 3 4 5 6–1

1

2

3

4

5

x(3, 0)

yy � �x � 3

x 3 4 7 12

y 0 1 2 3

33.

Intercepts:

No axis or origin symmetry

�0, 6�, �6, 0�

2−2

2

−2

4

6

8

10

12

4 6 8 10 12x

(6, 0)

(0, 6)

yy � �x � 6�

x 0 2 4 6 8 10

y 8 6 4 2 0 2 4

�2

35.

Intercepts:

x-axis symmetry

–2 1 2 3 4

–3

–2

2

3

x

(0, −1)

(−1, 0) (0, 1)

y

�0, �1�, �0, 1�, ��1, 0�

x � y2 � 1

x 0 3

y 0 ±2±1

�1

37.

Intercepts: �6, 0�, �0, 3�

−10

−10

10

10

y � 3 �12x

39.

Intercepts: �3, 0�, �1, 0�, �0, 3�

−10 10

−10

10

y � x2 � 4x � 3 41.

Intercept: �0, 0�

−10

−10

10

10

y �2x

x � 143.

Intercept: �0, 0�

−10

−10

10

10

y � 3�x

Page 5: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

42 Chapter 1 Equations, Inequalities, and Mathematical Modeling

45.

Intercepts: �0, 0�, ��6, 0�

−10

−10

10

10

y � x�x � 6 47.

Intercepts: ��3, 0�, �0, 3�

−10

−10

10

10

y � �x � 3� 49. Center: radius: 4

Standard form:

x2 � y2 � 16

�x � 0�2 � �y � 0�2 � 42

�0, 0�;

53. Center: solution point:

Standard form: �x � 1�2 � �y � 2�2 � 5

�0 � 1�2 � �0 � 2�2 � r 2 ⇒ 5 � r2

�x � ��1��2 � �y � 2�2 � r2

�0, 0���1, 2�;51. Center: radius: 4

Standard form:

�x � 2�2 � �y � 1�2 � 16

�x � 2�2 � �y � ��1��2 � 42

�2, �1�;

55. Endpoints of a diameter:

Center:

Standard form: �x � 3�2 � �y � 4�2 � 25

�0 � 3�2 � �0 � 4�2 � r 2 ⇒ 25 � r 2

�x � 3�2 � �y � 4�2 � r2

�0 � 62

, 0 � 8

2 � � �3, 4�

�0, 0�, �6, 8� 57.

Center: radius: 5�0, 0�,

1−1−2

−2−3−4

−6

−3−4

1234

6

2 3 4 6x

y

(0, 0)

x2 � y2 � 25

59.

Center: radius: 3

−3 −2 1

1

2 4 5x

−1

−2

−3

−4

−5

−6

−7

y

(1, −3)

�1, �3�,

�x � 1�2 � �y � 3�2 � 9 61.

Center: radius:

–1 1 2 3

1

3

x

y

( )12

, 12

32�1

2, 12�,�x �

12�2

� �y �12�2

�94 63.

250,000

200,000

150,000

100,000

50,000

1 2 3 4 5 6 7 8t

Year

Dep

reci

ated

val

ue

y

y � 225,000 � 20,000t, 0 ≤ t ≤ 8

65. (a)

(c)

0 1800

8000

y

x

(b)

(d) When yards, the area is a maximum of square yards.

(e) A regulation NFL playing field is 120 yards long and yards wide. The actual area is 6400 square yards.531

3

751119x � y � 862

3

A � xy � x�5203 � x�

y �5203 � x

2y �1040

3 � 2x

2x � 2y �1040

3

Page 6: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.2 Linear Equations in One Variable 43

71. The viewing window is incorrect. Change the viewingwindow. Examples will vary. For example,will not appear in the standard window setting.

y � x2 � 20

73.

Terms: 9x5, 4x3, �7

9x5 � 4x3 � 7 75. �18x � �2x � 3�2x � �2x � 2�2x

77. 6�t2 � t2�6 � �t�1�3 � 3��t�

Section 1.2 Linear Equations in One Variable

■ You should know how to solve linear equations.

■ An identity is an equation whose solution consists of every real number in its domain.

■ To solve an equation you can:

(a) Add or subtract the same quantity from both sides.

(b) Multiply or divide both sides by the same nonzero quantity.

(c) Remove all symbols of grouping and all fractions.

(d) Combine like terms.

(e) Interchange the two sides.

■ Check the answer!

■ A “solution” that does not satisfy the original equation is called an extraneous solution.

■ Be able to find intercepts algebraically.

a � 0ax � b � 0,

Vocabulary Check

1. equation 2. solve 3. identities; conditional

4. 5. extraneousax � b � 0

67.

(a) and (b)

Year (20 ↔ 1920)

Lif

e ex

pect

ancy

20 40 60 80 100

100

80

60

40

20

t

y

y � �0.0025t 2 � 0.574t � 44.25, 20 ≤ t ≤ 100

(c) For the year 1948, let years.

(d) For the year 2005, let years.

For the year 2010, let years.

(e) No. The graph reaches a maximum of yearswhen or during the year 2014. After this time,the model has life expectancy decreasing, which is notrealistic.

t � 114.8,y � 77.2

t � 110: y � 77.1

t � 105: y � 77.0

t � 48: y � 66.0

69. False. A graph is symmetric with respect to the axis if,whenever is on the graph, is also on thegraph.

�x, �y��x, y�x-

Page 7: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

44 Chapter 1 Equations, Inequalities, and Mathematical Modeling

1.

(a)

is not a solution.

(c)

is a solution.x � 4

17 � 17

5�4� � 3 �?

3�4� � 5

x � 0

�3 � 5

5�0� � 3 �?

3�0� � 5

5x � 3 � 3x � 5

(b)

is not a solution.

(d)

is not a solution.x � 10

47 � 35

5�10� � 3 �?

3�10� � 5

x � �5

�28 � �10

5��5� � 3 �?

3��5� � 5

3.

(a)

is a solution.

(c)

is not a solution.x � 4

51 � 30

48 � 8 � 5 �?

32 � 2

3�4�2 � 2�4� � 5 �?

2�4�2 � 2

x � �3

16 � 16

27 � 6 � 5 �?

18 � 2

3��3�2 � 2��3� � 5 �?

2��3�2 � 2

3x2 � 2x � 5 � 2x2 � 2 (b)

is a solution.

(d)

is not a solution.x � �5

60 � 48

75 � 10 � 5 �?

50 � 2

3��5�2 � 2��5� � 5 �?

2��5�2 � 2

x � 1

0 � 0

3 � 2 � 5 �?

2 � 2

3�1�2 � 2�1� � 5 �?

2�1�2 � 2

5.

(a)

is a solution.x � �12

3 � 3

�5 � 8 �?

3

52��1�2� �

4��1�2� �

?3

52x

�4x

� 3

(c) is undefined.

is not a solution.x � 0

52�0� �

40

(b)

is not a solution.

(d)

is not a solution.x �14

�6 � 3

10 � 16 �?

3

52�1�4� �

41�4

�?

3

x � 4

�38

� 3

58

� 1 �?

3

52�4� �

44

�?

3

7.

(a)

is not a solution.x � 3

�7 � 4

�3�3� � 2 �?

4

�3x � 2 � 4

(b)

is not a solution.x � 2

�4 � 4

�3�2� � 2 �?

4 (c)

is not a solution.x � 9

�25 � 4

�3�9� � 2 �?

4 (d)

is not a solution.x � �6

��20 � 4

�3��6� � 2 �?

4

Page 8: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.2 Linear Equations in One Variable 45

11. is an identity by the DistributiveProperty. It is true for all real values of x.2�x � 1� � 2x � 2 13. is conditional. There are

real values of x for which the equation is not true.�6�x � 3� � 5 � �2x � 10

15.

This is an identity by simplification. It is true for all realvalues of x.

4�x � 1� � 2x � 4x � 4 � 2x � 2x � 4 � 2�x � 2� 17.

Thus, is an identity bysimplification. It is true for all real values of x.

x2 � 8x � 5 � �x � 4�2 � 11

�x � 4�2 � 11 � x2 � 8x � 16 � 11 � x2 � 8x � 5

19. is conditional. There are real values

of x for which the equation is not true.

3 �1

x � 1�

4xx � 1

21. Original equation

Subtract 32 from both sides.

Simplify.

Divide both sides by 4.

Simplify. x �514

4x4

�514

4x � 51

4x � 32 � 32 � 83 � 32

4x � 32 � 83

9.

(a)

is a solution.

(c)

is a solution.x �72

0 � 0

1472 �

772 �

702 �

?0

6�72�2

� 11�72� � 35 �

?0

x � �53

0 � 0

503 �

553 �

1053 �

?0

6��53�2

� 11��53� � 35 �

?0

6x2 � 11x � 35 � 0

(b)

is not a solution.

(d)

is not a solution.x �53

�1103 � 0

503 �

553 �

1053 �

?0

6�53�2

� 11�53� � 35 �

?0

x � �27

�153749 � 0

2449 �

15449 �

171549 �

?0

6��27�2

� 11��27� � 35 �

?0

23.

x � 4

x � 11 � 11 � 15 � 11

x � 11 � 15 25.

x � �9

�2x�2

�18�2

�2x � 18

7 � 7 � 2x � 25 � 7

7 � 2x � 25 27.

x � 5

5x5

�255

5x � 25

5x � 5 � 5 � 20 � 5

5x � 5 � 20

8x � 3x � 5 � 3x � 3x � 20

8x � 5 � 3x � 20

29.

x � 9

�x � �9

�x � 3 � 3 � �6 � 3

�x � 3 � �6

2x � 3x � 3 � 3x � 3x � 6

2x � 3 � 3x � 6

2x � 10 � 7 � 3x � 6

2�x � 5� � 7 � 3�x � 2� 31.

No solution

�9 � 8

�5x � 5x � 9 � 8 � 5x � 5x

�5x � 9 � 8 � 5x

x � 6x � 9 � 8 � 5x

x � 3�2x � 3� � 8 � 5x 33.

x � �4

5x � 2 � 4x � 2

4�5x4 � � 4�1

2� � 4�x� � 4�12�

4�5x4

�12� � 4�x �

12�

5x4

�12

� x �12

Page 9: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

46 Chapter 1 Equations, Inequalities, and Mathematical Modeling

37.

x � 9

�0.50x � �4.5

�0.50x � 7.5 � 3

0.25x � 7.5 � 0.75x � 3

0.25x � 0.75�10 � x� � 3

39. or

The second way is easier since you are not working withfractions until the end of the solution.

x �73 x �

73

3x � 7 x �43 � 1

3x � 3 � 4 x � 1 �43

3�x � 1� � 4 3�x � 1� � 4 41. or

The first way is easier. The fraction is eliminated in thefirst step.

x � 13

x � 3�133 �

13x �133 x � 13

13x � 5 �23 x � 2 � 15

13x �23 � 53�1

3��x � 2� � 3�5�

13�x � 2� � 5 13�x � 2� � 5

43.

Contradiction; no solution

8 � �4

x � 8 � x � 4

x � 8 � 2x � 4 � x

x � 8 � 2�x � 2� � x 45.

x � 10

�31x � �310

400 � 16x � 15x � 18 � 72

4�100 � 4x� � 3�5x � 6� � 72

12�100 � 4x3 � � 12�5x � 6

4 � � 12�6�

100 � 4x

3�

5x � 64

� 6

35.

z � �65

5z � �6

6z � 30 � z � 24 � 0

6�z � 5� � �z � 24� � 0

4�32��z � 5� � 4�1

4��z � 24� � 4�0�

432�z � 5� �

14�z � 24� � 4�0�

32�z � 5� �14�z � 24� � 0

47.

x � 4

5x � 20

15x � 12 � 10x � 8

3�5x � 4� � 2�5x � 4�

5x � 45x � 4

�23

49.

x � 3

6x � 18

10x � 13 � 4x � 5

10x � 13

x�

4x � 5x

10 �13x

� 4 �5x

51.

z � 0

3z � 6 � 2z � 4 � 2

3�z � 2� � �2 �2

z � 2��z � 2�

3 � 2 �2

z � 253.

Contradiction; no solution

3 � 0

1 � 2 � 0

x � 4x � 4

� 2 � 0

x

x � 4�

4x � 4

� 2 � 0

Page 10: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.2 Linear Equations in One Variable 47

57.

Multiply both sides by

x � 5

2x � 10

1�x � 3� � 1�x � 3� � 10

�x � 3��x � 3�. 1

x � 3�

1x � 3

�10

�x � 3��x � 3�

1

x � 3�

1x � 3

�10

x2 � 9

59.

Multiply both sides by

A check reveals that is an extraneous solution since it makes the denominator zero, so there is no solution.x � 3

x � 3

3x � 9

3 � 4x � 12 � x

3 � 4�x � 3� � x

x�x � 3�.3x�x � 3� �

4x

�1

x � 3

3

x2 � 3x�

4x

�1

x � 3

55. Multiply both sides by

A check reveals that is an extraneous solution—it makes the denominator zero. There is no real solution.x � 4

4 � x

12 � 3x

2 � 3x � 10

2 � x � 2 � 2x � 8

2 � 1�x � 2� � 2�x � 4�

�x � 4��x � 2�. 2

�x � 4��x � 2� �1

x � 4�

2x � 2

61.

x � 0

�2x � 0

4x � 9 � 6x � 9

x2 � 4x � 4 � 5 � x2 � 6x � 9

�x � 2�2 � 5 � �x � 3�2 63.

The equation is an identity; every real number is a solution.

4 � 4

x2 � 4x � 4 � x2 � 4x � 4

�x � 2�2 � x2 � 4�x � 1�

65.

The x-intercept is at 3. The solution to and the x-intercept of are the same.They are both The x-intercept is �3, 0�.x � 3.

y � 2�x � 1� � 40 � 2�x � 1� � 4

x � 3

3 � x

6 � 2x

0 � 2x � 6

0 � 2x � 2 � 4

−6 12

−8

4

0 � 2�x � 1� � 4y � 2�x � 1� � 4 67.

The x-intercept is at 10. The solution toand the x-intercept of are the same. They are both The x-intercept is �10, 0�.x � 10.

y � 20 � �3x � 10�0 � 20 � �3x � 10�

x � 10

3x � 30

0 � 30 � 3x

0 � 20 � 3x � 10

−20 40

35

−5

0 � 20 � �3x � 10�y � 20 � �3x � 10�

Page 11: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

48 Chapter 1 Equations, Inequalities, and Mathematical Modeling

73.

x-intercept:

y-intercept:

The x-intercept is and the y-intercept is �0, �3�.��12, 0�

y � �3�2�0� � 1� ⇒ y � �3

0 � �3�2x � 1� ⇒ 0 � 2x � 1 ⇒ x � �12

y � �3�2x � 1� 75.

x-intercept:

y-intercept:

The x-intercept is and the y-intercept is �0, 103 �.�5, 0�

2�0� � 3y � 10 ⇒ 3y � 10 ⇒ y �103

2x � 3�0� � 10 ⇒ 2x � 10 ⇒ x � 5

2x � 3y � 10

77. Multiply both sides by 5.

x-intercept:

y-intercept:

The x-intercept is and the y-intercept is �0, 83�.��20, 0�

2�0� � 40 � 15y � 0 ⇒ 40 � 15y � 0 ⇒ y �4015

�83

2x � 40 � 15�0� � 0 ⇒ 2x � 40 � 0 ⇒ x � �20

2x5

� 8 � 3y � 0 ⇒ 2x � 40 � 15y � 0

79.

x-intercept:

y-intercept:

The x-intercept is and the y-intercept is �0, �0.3�.�1.6, 0�

4y � 0.75�0� � 1.2 � 0 ⇒ 4y � 1.2 � 0 ⇒ y ��1.2

4� �0.3

4�0� � 0.75x � 1.2 � 0 ⇒ � 0.75x � 1.2 � 0 ⇒ x �1.2

0.75� 1.6

4y � 0.75x � 1.2 � 0

71.

x-intercept:

y-intercept:

The x-intercept is and the y-intercept is �0, 12�.�125 , 0�

y � 12 � 5�0� ⇒ y � 12

0 � 12 � 5x ⇒ 5x � 12 ⇒ x �125

y � 12 � 5x69.

The x-intercept is at The solution to and the x-intercept of are the same.They are both The x-intercept is �7

5, 0�.x �75.

y � �38 � 5�9 � x�0 � �38 � 5�9 � x�7

5.

x �75

5x � 7

0 � 7 � 5x

0 � �38 � 45 � 5x

−6 10

−2

10

0 � �38 � 5�9 � x�y � �38 � 5�9 � x�

81.

x �1

3 � a, a � 3

x�3 � a� � 1

3x � ax � 1

4x � 4 � ax � x � 5

4�x � 1� � ax � x � 5 83.

x �5

4 � a, a � �4

x�4 � a� � 5

4x � ax � 5

6x � ax � 2x � 5

85.

Multiply both sides by 2.

x �18

36 � a, a � �36

x�36 � a� � 18

36x � ax � 18

18x �12

ax � 9

19x �12

ax � x � 9 87.

x ��17

�2a � 10�

�1710 � 2a

, a � 5

x��2a � 10� � �17

�2ax � 10x � �17

�2ax � 10x � 18 � 1

�2ax � 6x � 18 � �4x � 1

�2ax � 6�x � 3� � �4x � 1

Page 12: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.2 Linear Equations in One Variable 49

91. Multiply both sides by

x ��5.405��7.398�

2� 19.993

2x � �5.405��7.398�

2x � �4.405��7.398� � 7.398

2x � �4.405��7.398� � 7.398

7.398x. 2

7.398�

4.405x

�1x

93.

h � 10 feet

h �471 � 50�

10��

471 � 50(3.14)10(3.14)

� 10

471 � 50� � 10�h

471 � 50� � 10�h

471 � 2��25� � 2��5h�

95. (a) Female:

inches

(c)

The lengths of the male and female femurs areapproximately equal when the lengths are 100 inches.

x � 61.2

26.440.432

� x

26.44 � 0.432x

For y � 16: 16 � 0.432x � 10.44

y � 0.432x � 10.44 (b) Male:

Yes, it is likely that both bones came from the sameperson because the estimated height of a male with a 19-inch thigh bone is 69.4 inches.

(d)

inches

It is unlikely that a female would be over 8 feet tall,so if a femur of this length was found, it most likelybelonged to a very tall male.

x � 100.59

1.71 � 0.017x

0.432x � 10.44 � 0.449x � 12.15

69.4 � x

31.15 � 0.449x

For y � 19: 19 � 0.449x � 12.15

y � 0.449x � 12.15

Height x Female Femur Male FemurLength Length

60 15.48 14.79

70 19.80 19.28

80 24.12 23.77

90 28.44 28.26

100 32.76 32.75

110 37.08 37.24

89.

� 138.889

x �62.50.45

�0.45x � �62.5

0.275x � 362.5 � 0.725x � 300

0.275x � 0.725�500 � x� � 300

97.

(a)

The intercept is �0, 36.8�.y-

y

t20 4 6 8 10 12

32

40

44

48

52

56

60

Year (0 ↔ 1990)

Con

sum

ptio

n of

gas

(in

billi

ons

of g

allo

ns)

y � 1.64t � 36.8, �1 ≤ t ≤ 12 (b) Let

intercept:

(c)

This corresponds with the year 2017. Explanations will vary.

t �28.21.64

� 17.2

28.2 � 1.64t

65 � 1.64t � 36.8

�0, 36.8�y-

y � 1.64�0� � 36.8 � 36.8t � 0:

Page 13: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

50 Chapter 1 Equations, Inequalities, and Mathematical Modeling

99.

m � 23,437.5 miles

75000.32

� m

7500 � 0.32m

10,000 � 0.32m � 2500 101. False.

This is a quadratic equation. The equation cannot be written in the form ax � b � 0.

x�3 � x� � 10 ⇒ 3x � x2 � 10

103. Equivalent equations are derived from the substitution principle and simplification techniques.They have the same solution(s).

and are equivalent equations.2x � 52x � 3 � 8

105. (a)

(b) Since the sign changes from negative at 1 to positive at 2, the root is somewhere between 1 and 2.

(c)

(d) Since the sign changes from negative at to positive at the root is somewhere between and

To improve accuracy, evaluate the expression at subintervals within this interval and determine where the sign changes.

1.8 < x < 1.9

1.9.1.81.9,1.8

1 < x < 2

x 0 1 2 3 4

73.80.6�2.6�5.8�93.2x � 5.8

�1

x 2

0.60.28�0.04�0.36�0.68�13.2x � 5.8

1.91.81.71.61.5

107.x2 � 5x � 362x2 � 17x � 9

��x � 9��x � 4��2x � 1��x � 9� �

x � 42x � 1

, x � �9

109.

Intercepts: �0, �5�, �53, 0�

x−3 −2 −1 1 2 3 4 5

−1

1

2

−2

−3

−4

−5

y

y � 3x � 5 111.

Intercepts: �0, 0�, ��5, 0�

x1

1

4

5

6

7

2−7 −6 −4 −3 −2 −1−1

−2

y

y � �x2 � 5x � �x�x � 5�

Page 14: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.3 Modeling with Linear Equations 51

Section 1.3 Modeling with Linear Equations

■ You should be able to set up mathematical models to solve problems.

■ You should be able to translate key words and phrases.

(a) Equality: (b) Addition:

Equals, equal to, is, are, was, will be, represents Sum, plus, greater, increased by, more than,exceeds, total of

(c) Subtraction: (d) Multiplication:

Difference, minus, less than, decreased by, Product, multiplied by, twice, times, percent ofsubtracted from, reduced by, the remainder

(e) Division: (f) Consecutive:

Quotient, divided by, ratio, per Next, subsequent

■ You should know the following formulas:

(a) Perimeter: (b) Area:

1. Square: 1. Square:

2. Rectangle: 2. Rectangle:

3. Circle: 3. Circle:

4. Triangle: 4. Triangle:

(c) Volume

1. Cube:

2. Rectangular solid:

3. Cylinder:

4. Sphere:

(d) Simple Interest: (e) Compound Interest:

(f) Distance: (g) Temperature:

■ You should be able to solve word problems. Study the examples in the text carefully.

F �9

5 C � 32d � rt

A � P�1 �r

n�nt

I � Prt

V � �4

3��r 3

V � �r 2 h

V � lwh

V � s3

A � �1

2�bhP � a � b � c

A � �r2C � 2�r

A � lwP � 2l � 2w

A � s2P � 4s

1.

The sum of a number and 4 A number increased by 4

x � 4 3.

The ratio of a number and 5The quotient of a number and 5A number divided by 5

u

55.

The difference of a number and 4 is divided by 5. A number decreased by 4 is divided by 5.

y � 4

5

Vocabulary Check

1. mathematical modeling 2. verbal model; 3. 4.algebraic equation

5. 6. 7. 8. I � PrtA � P�1 �r

12�12t

V � �r2hV � s3

P � 2l � 2wA � �r2

Page 15: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

52 Chapter 1 Equations, Inequalities, and Mathematical Modeling

7.

The product of and the sum of a number and 2.Negative 3 is multiplied by a number increased by 2.

�3

�3�b � 2� 9.

The difference of a number and 5 is multiplied by 12times the number. 12 is multiplied by a number and thatproduct is multiplied by the number decreased by 5.

12x�x � 5�

11. Verbal Model: (Sum) (first number) (second number)

Labels: Sum first number second number

Expression: S � n � �n � 1� � 2n � 1

� n � 1� n,� S,

��

13. Verbal Model: Product (first odd integer)(second odd integer)

Labels: Product first odd integer second odd integer

Expression: P � �2n � 1��2n � 1� � 4n2 � 1

� 2n � 1 � 2 � 2n � 1� 2n � 1,� P,

15. Verbal Model: (Distance) (rate) (time)

Labels: Distance rate time

Expression: d � 50t

� t� 50 mph,� d,

��

17. Verbal Model: (Amount of acid) (amount of solution)

Labels: Amount of acid (in gallons) amount of solution (in gallons)

Expression: A � 0.20x

� x� A,

� 20% �

19. Verbal Model:

Labels: Perimeter widthlength (width)

Expression: P � 2x � 2�2x� � 6x

� 2x� 2� x,� P,

Perimeter � 2�width� � 2�length�

21. Verbal Model: (Total cost) (unit cost)(number of units) (fixed cost)

Labels: Total cost fixed cost unit cost number of units

Expression: C � 25x � 1200

� x� $25,� $1200,� C,

��

23. Verbal Model: Thirty percent of the list price L

Expression: 0.30L

25. Verbal Model: percent of 500 that is represented by thenumber N

Equation: , p is in decimal formN � p�500�

27.

A � 4x � 8x � 12x

Area � Area of top rectangle � Area of bottom rectangle

4

42x

x

x

8

Page 16: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.3 Modeling with Linear Equations 53

31. Verbal Model: Difference (one number) (another number)

Labels: Difference one number another number

Equation:

Answer: The two numbers are 37 and 185.

5x � 185

x � 37

148 � 4x

148 � 5x � x

� x� 5x,� 148,

��

33. Verbal Model: Product (smaller number) (larger number)

Labels: Smaller number larger number

Equation:

Answer: Smaller number larger number � n � 1 � �4� n � �5,

n � �5

n2 � n � n2 � 5

n�n � 1� � n2 � 5

� n � 1� n,

� �smaller number)2 � 5��

29. Verbal Model: Sum (first number) (second number)

Labels: Sum first number second number

Equations:

Answer: First number second number � n � 1 � 263� n � 262,

n � 262

524 � 2n

525 � 2n � 1

525 � n � �n � 1�

� n � 1� n,� 525,

��

35.

� 13.5

� 0.30�45�

� �30%��45�

x � percent � number 37.

p � 0.27 � 27%

4321600 � p

432 � p�1600�

432 � percent � 1600 39.

x � 2400

12

0.005� x

12 � 0.005x

12 �12

% � number

41. Verbal Model: Loan payments

Labels: Loan payments

Annual income

Equation:

The family’s annual income is $22,316.98.

I � 22,316.98

13,077.75 � 0.586I

� I

� 13,077.75

� 58.6% � Annual income

Page 17: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

54 Chapter 1 Equations, Inequalities, and Mathematical Modeling

45. Verbal Model:

Labels: 2002 price for a pound of tomatoes: $1.66

1990 price for a pound of tomatoes: $0.86

Percentage increase:

Equation:

Answer: Percentage increase � 93%

0.930 � p

0.80 � 0.86p

1.66 � 0.86p � 0.86

p

� �1990 price for a pound of tomatoes�

� �percentage increase��1990 price for a pound of tomatoes� �2002 price for a pound of tomatoes�

43. Verbal Model:

Labels: 2002 price for a gallon of unleaded gasoline: $1.36

1990 price for a gallon of unleaded gasoline: $1.16

Percentage increase:

Equation:

Answer: Percentage increase � 17.2%

0.172 � p

0.20 � 1.16p

1.36 � 1.16p � 1.16

p

� �1990 price for a gallon of unleaded gasoline�

� �percentage increase��1990 price for a gallon of unleaded gasoline� �2002 price for a gallon of unleaded gasoline�

47. Verbal Model: (Sale price) (list price) (discount)

Labels: Sale price list price discount

Equation:

Answer: The list price of the pool is $1450.

1450 � L

1210.75 � 0.835L

1210.75 � L � 0.165L

� 0.165L� L,� $1210.75,

��

49. (a)

w

l

(b)

� 5w

� 2�1.5w� � 2w

P � 2l � 2w

l � 1.5w (c)

Width:

Length:

Dimensions: 7.5 meters � 5 meters

l � 1.5w � 7.5 meters

w � 5 meters

5 � w

25 � 5w

51. Verbal Model: Average

Labels: Average

Equation:

Answer: You must score 97 (or better) on test #4 to earn an A for the course.

x � 97

360 � 263 � x

90 �87 � 92 � 84 � x

4

test #4 � xtest #3 � 84,test #2 � 92,test #1 � 87,� 90,

��test #1� � �test #2� � �test #3� � �test #4�

4

Page 18: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.3 Modeling with Linear Equations 55

55.

Multiply both sides by

The salesman averaged 58 miles per hour for 3 hours and 52 miles per hour for 2 hours and 45 minutes.

t2 �317 � 174

52� 2.75 hours

t1 �17458

� 3 hours

x � 174 miles

�6x � �1044

52x � 18,386 � 58x � 17,342

52x � 58�317 � x� � 5.75�58��52�

58�52�. x

58�

317 � x52

� 5.75

d1

r1�

d2

r2� T

t1 � t2 � T

�time on first part� � �time on second part) � �Total time�

57. (a) Time for the first family:

Time for the other family: hours

(b)

(c) d � r t � 42�16042

�16050 � � 25.6 miles

t �dr

�100

42 � 50�

10092

� 1.08 hours

t2 �dr2

�16050

� 3.2

t1 �dr1

�16042

� 3.81 hours

53.

Total time �total distance

rate�

300 kilometers100 kilometers�hour

� 3 hours

Rate �distance

time�

50 kilometers12 hour

� 100 kilometers�hour

59. Verbal Model:

Labels: Let wind speed, then the rate to thecity the rate from the city the distance to thecity kilometers, the distance traveled so far in the returntrip kilometers.

Equation:

Wind speed: kilometers per hour6623

6623

� x

180,000 � 2700x

900,000 � 1500x � 720,000 � 1200x

1500�600 � x� � 1200�600 � x�

1500600 � x

�1200

600 � x

� 1500 � 300 � 1200

� 1500� 600 � x,

� 600 � x,x �

time �distance

rate61. Verbal Model:

Equation:

The radio wave travels from Mission Control to the moonin 1.28 seconds.

t � 1.28 seconds

t �3.84 � 108 meters

3.0 � 108 meters per second

time �distance

rate

Page 19: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

56 Chapter 1 Equations, Inequalities, and Mathematical Modeling

63. Verbal Model:

Label: Let h height of the building in feet.

Equation:

The Chrysler building is 1044 feet tall.

h � 1044 feet

13

h � 348

h

87�

41�3

h feet

87 feet�

4 feet1�3 foot

�height of building��length of building's shadow�

��height of stake�

�length of stake's shadow�

65. Verbal Model:

Label: Let length of person’s shadow.

Equation:

x � 4.36 feet

44x � 192

50x � 192 � 6x

50x � 6�32 � x�

50

32 � x�

6x

x �

50 ft

6 ft

32 ft x

�height of silo��length of silo’s shadow�

��height of person�

�height of person’s shadow�

67. Verbal Model: interest in (total interest)

Labels: Let x amount in the 3% fund. Then amount in the fund.

Equation:

25,000 � x � $16,666.67 at 412%

x � $8333.33 at 3%

�125 � �0.015x

1000 � 0.03x � 1125 � 0.045x

1000 � 0.03x � 0.045�25,000 � x�41

2%25,000 � x ��

�412% fund��Interest in 3% fund� � �

69. Verbal Model: (profit on minivans) (profit on SUVs) (total profit)

Labels: Let x amount invested in minivans. Then, amount invested in SUVs.

Equation:

$450,000 is invested in minivans and is invested in SUVs.600,000 � x � $150,000

x � 450,000

�0.04x � �18,000

0.24x � 168,000 � 0.28x � 150,000

0.24x � 0.28�600,000 � x� � 0.25�600,000�

600,000 � x ��

��

71. Verbal Model:

Label: Let x amount of 100% concentrate

Equation:

Approximately 32.1 gallons of the 100% concentrate will be needed.

x � 32.1 gallons

41.25 � 0.60x � 22

41.25 � 22 � 0.40x � 1.00x

0.75�55� � 0.40�55 � x� � 1.00x

�Final concentration��Amount� � �Solution 1 concentration��Amount� � �Solution 2 concentration��Amount�

Page 20: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.3 Modeling with Linear Equations 57

73. Verbal Model: (price per pound of peanuts)(number of pounds of peanuts) (price per pound of walnuts)(number ofpounds of walnuts) (price per pound of nut mixture)(number of pounds of nut mixture)

Labels: Let x number of pounds of $2.49 peanuts. Then number of pounds of $3.89 walnuts.

Equation:

Use 50 pounds of peanuts and 50 pounds of walnuts.

100 � x � 50 pounds of $3.89 walnuts

x � 50 pounds of $2.49 peanuts

x ��70

�1.40

�1.40x � �70

2.49x � 389 � 3.89x � 319

2.49x � 3.89�100 � x� � 3.19�100�

100 � x ��

��

75. Verbal Model:

Labels:

Equation:

At most the company can manufacture 8064 units.

x �75,000

9.3� 8064.52 units

$85,000 � $10,000 � $9.30x

Total cost � $85,000, variable costs � $9.30x, fixed costs � $10,000

Total cost � �Fixed cost� � �Variable cost per unit� � �Number of units�

77.

2Ab

� h

2A � bh

A �12

bh 79.

S

1 � R� C

S � C�1 � R�

S � C � RC 81.

3V

4�a2 � b

�3�4�V

�a2 � b

34

V � �a2b

V �43

�a2b 83.

2�h � v0t�

t2� a

2�h � v0t� � at2

h � v0t �1

2at2

h � v0t �1

2at2

85.

CC2

C2 � C� C1

C2 � C

CC2�

1C1

1C

�1

C2�

1C1

1C

�1

C1�

1C2

C �1

1C1

�1

C2

89.

x � 6 feet from 50-pound child.

125x � 750

50x � 750 � 75x

50x � 75�10 � x�

W1x � W2�L � x�87.

L � a � d

d� n

L � a � d � nd

L � a � nd � d

L � a � �n � 1�d

Page 21: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

58 Chapter 1 Equations, Inequalities, and Mathematical Modeling

Section 1.4 Quadratic Equations and Applications

■ You should be able to solve a quadratic equation by factoring, if possible.

■ You should be able to solve a quadratic equation of the form by extracting square roots.

■ You should be able to solve a quadratic equation by completing the square.

■ You should know and be able to use the Quadratic Formula: For

■ You should be able to determine the types of solutions of a quadratic equation by checking the discriminant

(a) If there are two distinct real solutions. The graph has two x-intercepts.

(b) If there is one repeated real solution. The graph has one x-intercept.

(c) If there is no real solution. The graph has no x-intercepts.

■ You should be able to use your calculator to solve quadratic equations.

■ You should be able to solve applications involving quadratic equations. Study the examples in the text carefully.

b2 � 4ac < 0,

b2 � 4ac � 0,

b2 � 4ac > 0,

b2 � 4ac.

x ��b ± �b2 � 4ac

2a.

ax2 � bx � c � 0, a � 0,

u2 � d

93.

When

59�64.4 � 32� � 18�C.

F � 64.4�,

C �59�F � 32� 95.

When

95�50� � 32 � 122�F.

C � 50�,

F �95C � 32

97. False, it should be written as

z3 � 8z2 � 9

.

99.

(a) If then a and b have the same sign andis negative.

(b) If then a and b have opposite signs andis positive.x � �b�a

ab < 0,

x � �b�aab > 0,

ax � b � 0 ⇒ x � �ba

101. �5x4��25x2��1 �5x4

25x2�

x2

5, x � 0

91.

r � 3�4.47�

� 1.12 inches

17.88

4�� r3

17.88 � 4�r3

5.96 �43

�r3

V �43

�r3

103. 3

x � 5�

2

5 � x�

3

x � 5�

��1��2�x � 5

�3 � 2

x � 5�

1

x � 5

105.10

7�3�

107�3

��3�3

�10�3

21 107.

� �11 � �6

� ���6 � �11 �

�5�6 � 5�11

�5

5

�6 � �11�

5

�6 � �11��6 � �11

�6 � �11

Page 22: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.4 Quadratic Equations and Applications 59

1.

General form: 2x2 � 8x � 3 � 0

2x2 � 3 � 8x 3.

General form: x2 � 6x � 6 � 0

x2 � 6x � 9 � 3

�x � 3�2 � 3 5.

General form: 3x2 � 90x � 10 � 0

3x2 � 10 � 90x

15�3x2 � 10� � 18x

7.

x � 0 or x � �12

3x � 0 or 2x � 1 � 0

3x�2x � 1� � 0

6x2 � 3x � 0 9.

x � 4 or x � �2

x � 4 � 0 or x � 2 � 0

�x � 4��x � 2� � 0

x2 � 2x � 8 � 0 11.

x � �5

x � 5 � 0

�x � 5�2 � 0

x2 � 10x � 25 � 0

13.

x � 3 or x � �12

3 � x � 0 or 1 � 2x � 0

�3 � x��1 � 2x� � 0

3 � 5x � 2x2 � 0 15.

x � �6 or x � 2

x � 6 � 0 or x � 2 � 0

�x � 6��x � 2� � 0

x2 � 4x � 12 � 0

x2 � 4x � 12 17.

x � �203 or x � �4

3x � 20 � 0 or x � 4 � 0

�3x � 20��x � 4� � 0

3x2 � 32x � 80 � 0

4�34x2 � 8x � 20� � 4�0�

34x2 � 8x � 20 � 0

Vocabulary Check

1. quadratic equation 2. factoring, extracting square roots, completing the square,and the Quadratic Formula

3. discriminant 4. position equation; velocity of theobject; initial height of the object

5. Pythagorean Theorem

�16t 2 � v0t � s0;

19.

x � �a

x � a � 0

�x � a�2 � 0

x2 � 2ax � a2 � 0 21.

x � ±7

x2 � 49 23.

x � ±�11

x2 � 11

25.

x � ±3�3

x2 � 27

3x2 � 81 27.

x � 16 or x � 8

x � 12 ± 4

x � 12 � ±4

�x � 12�2 � 16 29.

x � �2 ± �14

x � 2 � ±�14

�x � 2�2 � 14

31.

x �1 ± 3�2

2

2x � 1 ± 3�2

2x � 1 � ±�18

�2x � 1�2 � 18 33.

The only solution to the equation is x � 2.

x � 2

�7 � 3 or 2x � 4

x � 7 � x � 3 or x � 7 � �x � 3

x � 7 � ± �x � 3�

�x � 7�2 � �x � 3�2

Page 23: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

60 Chapter 1 Equations, Inequalities, and Mathematical Modeling

39.

x � 1 ±�63

x � 1 ± �23

x � 1 � ±�23

�x � 1�2 �23

x2 � 2x � 12 � �13

� 12

x2 � 2x � �13

9x2 � 18x � �3 41.

x � 2 ± 2�3

x � 2 � ±�12

�x � 2�2 � 12

x2 � 4x � 22 � 8 � 22

x2 � 4x � 8

x2 � 4x � 8 � 0

�x2 � 4x � 8 � 0

8 � 4x � x2 � 0

45.

�1

�x � 1�2 � 4

1

x2 � 2x � 5�

1x2 � 2x � 12 � 12 � 5

43.

x ��5 ± �89

4

x � �54

±�89

4

x �54

� ±�89

4

�x �54�

2

�8916

x2 �52

x � �54�

2

� 4 � �54�

2

x2 �52

x � 4

2x2 � 5x � 8

2x2 � 5x � 8 � 0

35.

x � 4 or x � �8

x � �2 ± 6

x � 2 � ±6

�x � 2�2 � 36

x2 � 4x � 22 � 32 � 22

x2 � 4x � 32

x2 � 4x � 32 � 0 37.

x � �6 ± �11

x � 6 � ±�11

�x � 6�2 � 11

x2 � 12x � 62 � �25 � 62

x2 � 12x � �25

x2 � 12x � 25 � 0

47.

�4

�x � 2�2 � 7

4

x2 � 4x � 3�

4x2 � 4x � 4 � 4 � 3

49.

�1

���x � 3�2 � 9�

1�9 � �x � 3�2

�1

��1��x � 3�2 � 9

1

�6x � x2�

1��1�x2 � 6x � 32 � 32�

51. (a)

(b) The x-intercepts are

(c)

(d) The x-intercepts of the graph are solutions to theequation 0 � �x � 3�2 � 4.

x � �1 or x � �5

�3 ± 2 � x

±�4 � x � 3

4 � �x � 3�2

0 � �x � 3�2 � 4

��1, 0� and ��5, 0�.

−10 5

−6

4

y � �x � 3�2 � 4

Page 24: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.4 Quadratic Equations and Applications 61

53. (a)

(b) The x-intercepts are �1, 0� and �3, 0�.

−3

−4

6

2

y � 1 � �x � 2�2 (c)

(d) The x-intercepts of the graph are solutions to the equation 0 � 1 � �x � 2�2.

x � 3 or x � 1

x � 2 ± 1

x � 2 � ±1

�x � 2�2 � 1

0 � 1 � �x � 2�2

55. (a)

(b) The x-intercepts are and

(d) The x-intercepts of the graph are solutions to theequation 0 � �4x2 � 4x � 3.

�32, 0�.��1

2, 0�

−5 7

−3

5

y � �4x2 � 4x � 3 (c)

x �32 or x � �

12

x �12 ± 1

x �12 � ±�1

�x �12�2

� 1

x2 � x � �12�2 �

34 � �1

2�2

x2 � x �34

4�x2 � x� � 3

4x2 � 4x � 3

0 � �4x2 � 4x � 3

57. (a)

(b) The x-intercepts are and �1, 0�.��4, 0�

−8 4

−7

1

y � x2 � 3x � 4 (c)

(d) The x-intercepts of the graph are solutions to the equation 0 � x2 � 3x � 4.

x � �4 or x � 1

x � 4 � 0 or x � 1 � 0

0 � �x � 4��x � 1�

0 � x2 � 3x � 4

59.

No real solution

b2 � 4ac � ��5�2 � 4�2��5� � �15 < 0

2x2 � 5x � 5 � 0 61.

Two real solutions

b2 � 4ac � ��1�2 � 4�2���1� � 9 > 0

2x2 � x � 1 � 0

63.

No real solution

b2 � 4ac � ��5�2 � 4�13��25� � �

253 < 0

13x2 � 5x � 25 � 0 65.

Two real solutions

b2 � 4ac � �1.2�2 � 4�0.2���8� � 7.84 > 0

0.2x2 � 1.2x � 8 � 0

67.

��1 ± 3

4�

12

, �1

��1 ± �12 � 4�2���1�

2�2�

x ��b ± �b2 � 4ac

2a

2x2 � x � 1 � 0 69.

��8 ± 16

32�

14

, �34

��8 ± �82 � 4�16���3�

2�16�

x ��b ± �b2 � 4ac

2a

16x2 � 8x � 3 � 0 71.

��2 ± 2�3

�2� 1 ± �3

��2 ± �22 � 4��1��2�

2��1�

x ��b ± �b2 � 4ac

2a

�x2 � 2x � 2 � 0

2 � 2x � x2 � 0

Page 25: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

62 Chapter 1 Equations, Inequalities, and Mathematical Modeling

77.

��12 ± 6�7

�18�

23

±�73

��12 ± �122 � 4��9��3�

2��9�

x ��b ± �b2 � 4ac

2a

�9x2 � 12x � 3 � 0

12x � 9x2 � �3

79.

� �43

��24 ± 0

18

��24 ± �242 � 4�9��16�

2�9�

x ��b ± �b2 � 4ac

2a

9x2 � 24x � 16 � 0 81.

��4 ± 8�2

8� �

12

± �2

��4 ± �42 � 4�4���7�

2�4�

x ��b ± �b2 � 4ac

2a

4x2 � 4x � 7 � 0

4x2 � 4x � 7 83.

��28 ± 0

�98�

27

��28 ± �282 � 4��49���4�

2��49�

x ��b ± �b2 � 4ac

2a

�49x2 � 28x � 4 � 0

28x � 49x2 � 4

85.

��8 ± 2�6

�4� 2 ±

�62

��8 ± �82 � 4��2���5�

2��2�

t ��b ± �b2 � 4ac

2a

�2t2 � 8t � 5 � 0

8t � 5 � 2t2 87.

�12 ± 2�11

2� 6 ± �11

����12� ± ���12�2 � 4�1��25�

2�1�

y ��b ± �b2 � 4ac

2a

y2 � 12y � 25 � 0

�y � 5�2 � 2y

73.

��14 ± 2�5

2� �7 ± �5

��14 ± �142 � 4�1��44)

2�1�

x ��b ± �b2 � 4ac

2a

x2 � 14x � 44 � 0 75.

��8 ± 4�5

2� �4 ± 2�5

��8 ± �82 � 4�1���4�

2�1�

x ��b ± �b2 � 4ac

2a

x2 � 8x � 4 � 0

89.

� �38

±�265

8

��3 ± �265

8

��3 ± �32 � 4�4���16�

2�4�

x ��b ± �b2 � 4ac

2a

4x2 � 3x � 16 � 0

4x2 � 3x � 16

12

x2 �38

x � 2

91.

x � 0.976, �0.643

x �1.7 ± ���1.7�2 � 4�5.1���3.2�

2�5.1�

5.1x2 � 1.7x � 3.2 � 0 93.

x � �14.071, 1.355

x ����0.852� ± ���0.852�2 � 4��0.067��1.277�

2��0.067�

�0.067x2 � 0.852x � 1.277 � 0

95.

x � 1.687, �0.488

x �506 ± ���506�2 � 4�422���347�

2�422�

422x2 � 506x � 347 � 0 97.

x � �2.200, �0.290

x ��31.55 ± ��31.55�2 � 4�12.67��8.09�

2�12.67�

12.67x2 � 31.55x � 8.09 � 0

Page 26: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.4 Quadratic Equations and Applications 63

99. Complete the square.

x � 1 ± �2

x � 1 � ± �2

�x � 1�2 � 2

x2 � 2x � 12 � 1 � 12

x2 � 2x � 1

x2 � 2x � 1 � 0 101.

x � 6 or x � �12

x � 3 � 9 or x � 3 � �9

x � 3 � ±9

�x � 3�2 � 81 Extract square roots.

103. Complete the square.

x �12 ± �3

x �12 � ±�12

4

�x �12�2

�124

x2 � x � �12�2

�114 � �1

2�2

x2 � x �114

x2 � x �114 � 0 105. Extract square roots.

x � �12

2x � �1

For x � ��x � 1�:

0 � 1 No solution

For x � ��x � 1�:

x � ± �x � 1�

x2 � �x � 1�2

�x � 1�2 � x2

107. Quadratic Formula

�34

±�97

4

�3 ± �97

4

x ����3� ± ���3�2 � 4�2���11�

2�2�

0 � 2x2 � 3x � 11

3x � 4 � 2x2 � 7 109. (a)

(b)

(c)

Since w must be greater than zero, we have feet and the length is w � 14 � 48 feet.

w � 34

w � �48 or w � 34

�w � 48��w � 34� � 0

w2 � 14w � 1632 � 0

w�w � 14� � 1632

w

w + 14

111.

Since x must be positive, we have inches. The dimensions of the box are 6 inches � 6 inches � 2 inches.x � 6

x � �14 or x � 6

0 � �x � 14��x � 6�

0 � x2 � 8x � 84

84 � x2 � 4x�2�

S � x2 � 4xh

113.

Thus,

—CONTINUED—

a � 1, b � �150, and c � 2500.

4�x2 � 150x � 2500� � 0

4x2 � 600x � 10,000 � 0

20,000 � 600x � 4x2 � 10,000100 ft

200 ft

100 2− x

200 2− x

x

x

�200 � 2x��100 � 2x� �1

2�100��200�

Page 27: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

64 Chapter 1 Equations, Inequalities, and Mathematical Modeling

113. —CONTINUED—

not possible since the lot is only 100 feet wide.

The person must go around the lot 19.098 feet

24 inches�

19.098 feet

2 feet� 9.5 times.

x �150 � 111.8034

2� 19.098 feet

x �150 � 111.8034

2� 130.902 feet,

x �150 ± ���150�2 � 4�1��2500�

2�1��

150 ± 111.8034

2

115.

(a)

(b) Model:

Labels: Rate 500 miles per hour

Distance

Equation:

The bomb will travel approximately 6.2 miles horizontally.

d � 500�0.0124� � 6.2 miles

� d

Time �20�53600

� 0.0124 hour

�Rate� � �time� � �distance�

� 44.72 seconds

� 20�5 seconds

t � �2000

t2 � 2000

�16t2 � 32,000 � 0

s � �16t2 � 32,000 117.

(a)

(b) When

When

When

During the interval the baseball’s speeddecreased due to gravity.

(c) The ball hits the ground when

By the Quadratic Formula, or Assuming that the ball is not caught and drops to theground, it will be in the air for approximately 9.209seconds.

t � 9.209.t � �0.042

�16t 2 � 146 23t � 6 1

4 � 0

s � 0.

3 ≤ t ≤ 5,

s�5� � 339.58 feett � 5:

s�4� � 336.92 feett � 4:

s�3� � 302.25 feett � 3:

s � �16t 2 � 14623t � 61

4

s0 � 614 feet

v0 � 100 mph ��100��5280�

3600� 1462

3 ft�sec

s � �16t 2 � v0t � s0

119.

(a)

The average admission price reached or surpassed $5.00 in 1999.

(b)

By the Quadratic Formula, or 42.80. Since we choose which corresponds to 1999.

(c) For 2008, let Answers will vary.P�18� � $6.87.t � 18:

t � 8.68 � 97 ≤ t ≤ 13,t � 8.68

�0.0081t 2 � 0.417t � 3.01 � 0

�0.0081t 2 � 0.417t � 1.99 � 5.00

P � �0.0081t 2 � 0.417t � 1.99, 7 ≤ t ≤ 13

7 8 9 10 11 12 13

$4.51 $4.81 $5.09 $5.35 $5.60 $5.83 $6.04P

t

Page 28: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.4 Quadratic Equations and Applications 65

121. Pythagorean Theorem

Each leg in the right triangle is approximately 3.54 centimeters.

�5�2

2� 3.54 centimeters

�5

�2

x ��25

2

x2 �25

2x

x

5

2x2 � 25

x2 � x2 � 52

123.

Using the positive value for we have one plane moving northbound at miles per hour and one plane moving eastbound at miles per hour.r � 550

r � 50 � 600r,

r ��900 ± �9002 � 4�18���5,931,100�

2�18��

�900 ± 60�118,847

36

18r2 � 900r � 5,931,100 � 0

9�r � 50�2 � 9r2 � 24402

dN2 � dE

2 � 24402

dE � �3 hours��r mph�

2440 miles

3r

3(r + 50)

dN � �3 hours��r � 50 mph�

125.

x � 50,000 units

0 � �x � 50,000�2

0 � x2 � 100,000x � 2,500,000,000

0 � 0.0002x2 � 20x � 500,000

x�20 � 0.0002x� � 500,000 127.

Using the positive value for x, we have

x ��20 � �7150

0.25� 258 units.

x ��20 ± �202 � 4�0.125���13,500�

2�0.125�

0.125x2 � 20x � 13,500 � 0

0.125x2 � 20x � 500 � 14,000

129.

Choosing the positive value for x, we have

x ��0.04 � �7.0416

0.004� 653 units.

��0.04 ± �7.0416

0.004

x ��0.04 ± ��0.04�2 � 4�0.002���880�

2�0.002�

0.002x2 � 0.04x � 880 � 0

800 � 0.04x � 0.002x2 � 1680

Page 29: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

66 Chapter 1 Equations, Inequalities, and Mathematical Modeling

131.

(a)

The total money in circulation reached or surpassed $600 billion in 2001.

(b)

By the Quadratic Formula, or Since we choose which corresponds to 2001.

(c) For 2008, let billionAnswers will vary.

M�18� � $991.98t � 18:

t � 11.15 ≤ t ≤ 13,�13.1.t � 11.1

1.835t 2 � 3.58t � 267.0 � 0

1.835t 2 � 3.58t � 333.0 � 600

M � 1.835t 2 � 3.58t � 333.0, 5 ≤ t ≤ 13

5 6 7 8 9 10 11 12 13

(in billions) $396.78 $420.54 $447.98 $479.08 $513.86 $552.30 $594.42 $640.20 $689.66M

t

133.

The other two distances are 354.5 miles and 433.5 miles.

x �1576 ± �15762 � 4�2��307,344�

2�2� � 354.5 or 433.5

0 � 2x2 � 1576x � 307,344

313,600 � x2 � 620,944 � 1576x � x2

5602 � x2 � (788 � x�2

Distance from New Orleans to Austin � 1348 � 560 � x � 788 � x

Distance from Oklahoma City to Austin � x

Distance from Oklahoma City to New Orleans � 560

�Distance from OklahomaCity to New Orleans �

2

� �Distance from OklahomaCity to Austin �

2

� � Distance from New Orleans to Austin�

2

135. False

so, the quadratic equation has no real solutions.

b2 � 4ac � ��1�2 � 4��3���10� < 0,

137. The student should have subtracted 15x from both sidesso that the equation is equal to zero. By factoring out anx, there are two solutions, and x � 6.x � 0

139.

(a) Let

or

or x � �5 x � �103

x � 4 � �1 x � 4 �23

u � �1 u �23

u � 1 � 0 3u � 2 � 0

�3u � 2��u � 1� � 0

3u2 � u � 2 � 0

u � x � 4

3�x � 4�2 � �x � 4� � 2 � 0

(b)

or

(c) The method of part (a) reduces the number of algebraicsteps.

x � �5x � �103

x � 5 � 03x � 10 � 0

�3x � 10��x � 5� � 0

3x2 � 25x � 50 � 0

3x2 � 24x � 48 � x � 4 � 2 � 0

3�x2 � 8x � 16� � �x � 4� � 2 � 0

141.

One possible equation is:

Any non-zero multiple of this equation would also havethese solutions.

x2 � 3x � 18 � 0

�x � 3��x � 6� � 0

�x � ��3���x � 6� � 0

�3 and 6 143. 8 and 14

One possible equation is:

Any non-zero multiple of this equation would also havethese solutions.

x2 � 22x � 112 � 0

�x � 8��x � 14� � 0

Page 30: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.5 Complex Numbers 67

149. by the Additive Inverse Property.7x 4 � ��7x 4� � 0

159. Answers will vary.

151.

� x2 � 3x � 18

�x � 3��x � 6� � x2 � 6x � 3x � 18

153.

� x3 � 3x2 � 2x � 8

� x3 � x2 � 2x � 4x2 � 4x � 8

�x � 4��x2 � x � 2� � x�x2 � x � 2� � 4�x2 � x � 2� 155.

� x2�x � 3��x2 � 3x � 9�

� x2�x3 � 33�

x5 � 27x2 � x2�x3 � 27�

157.

� �x � 5��x2 � 2�

x3 � 5x2 � 2x � 10 � x2�x � 5� � 2�x � 5�

Section 1.5 Complex Numbers

■ Standard form: .

If then is a real number.

If and then is a pure imaginary number.

■ Equality of Complex Numbers: if and only if and

■ Operations on complex numbers

(a) Addition:

(b) Subtraction:

(c) Multiplication:

(d) Division:

■ The complex conjugate of

■ The additive inverse of

■ ��a � �a i for a > 0.

a � bi is �a � bi.

�a � bi��a � bi� � a2 � b2

a � bi is a � bi:

a � bi

c � di�

a � bi

c � di�

c � di

c � di�

ac � bd

c2 � d 2�

bc � ad

c2 � d 2i

�a � bi��c � di� � �ac � bd� � �ad � bc�i

�a � bi� � �c � di� � �a � c� � �b � d�i

�a � bi� � �c � di� � �a � c� � �b � d�i

b � da � ca � bi � c � di

a � bib � 0,a � 0

a � bib � 0,

a � bi

145. and

One possible equation is:

Any non-zero multiple of this equation would also havethese solutions.

x2 � 2x � 1 � 0

x2 � 2x � 1 � 2 � 0

�x � 1�2 � ��2 �2� 0

��x � 1� � �2� ��x � 1� � �2� � 0

�x � �1 � �2�� �x � �1 � �2 �� � 0

1 � �21 � �2 147. by the Associative Property ofMultiplication.�10x�y � 10�xy�

Page 31: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

68 Chapter 1 Equations, Inequalities, and Mathematical Modeling

1.

b � 6

a � �10

a � bi � �10 � 6i 3.

b � 3 � 8 ⇒ b � 5

a � 1 � 5 ⇒ a � 6

�a � 1� � �b � 3�i � 5 � 8i 5. 4 � ��9 � 4 � 3i

7.

� 2 � 3�3 i

2 � ��27 � 2 � �27 i 9. ��75 � �75 i � 5�3 i 11. 8 � 8 � 0i � 8

13.

� �1 � 6i

�6i � i2 � �6i � 1 15.

� 0.3i

��0.09 � �0.09 i 17. �5 � i� � �6 � 2i� � 11� i

19.

� 4

�8 � i� � �4 � i� � 8 � i � 4 � i

21.

� 3 � 3�2 i

��2 � ��8 � � �5 � ��50 � � �2 � 2�2 i � 5 � 5�2 i

23.

� �14 � 20i

13i � �14 � 7i� � 13i � 14 � 7i 25.

� 16 �

76i

� �96 �

156 i �

106 �

226 i

��32 �

52i� � �5

3 �113 i� � �

32 �

52i �

53 �

113 i

27.

� 3 � i � 2 � 5 � i

�1 � i��3 � 2i� � 3 � 2i � 3i � 2i2 29.

� 12 � 30i

6i�5 � 2i� � 30i � 12i2 � 30i � 12

31.

� 14 � 10 � 24

��14 � �10 i���14 � �10 i� � 14 � 10i2

Vocabulary Check

1. (a) iii (b) i (c) ii 2.

3. principal square 4. complex conjugates

��1; �1

33.

� �9 � 40i

� 16 � 40i � 25

�4 � 5i�2 � 16 � 40i � 25i 2

35.

� �10

� 4 � 12i � 9 � 4 � 12i � 9

�2 � 3i�2 � �2 � 3i�2 � 4 � 12i � 9i2 � 4 � 12i � 9i2 37. The complex conjugate of is

�6 � 3i��6 � 3i� � 36 � �3i�2 � 36 � 9 � 45

6 � 3i.6 � 3i

39. The complex conjugate of is

� 1 � 5 � 6

��1 � �5i���1 � �5i� � ��1�2 � ��5i�2

�1 � �5i.�1 � �5i 41. The complex conjugate of is

�2�5i���2�5i� � �20i2 � 20

�2�5i.��20 � 2�5i

43. The complex conjugate of is

��8���8� � 8

�8.�8 45.5i

�5i

��i�i

��5i

1� �5i

Page 32: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.5 Complex Numbers 69

47.

�2�4 � 5i�16 � 25

�8 � 10i

41�

841

�1041

i

2

4 � 5i�

24 � 5i

�4 � 5i4 � 5i

49.

�9 � 6i � i 2

9 � 1�

8 � 6i10

�45

�35

i

3 � i3 � i

�3 � i3 � i

�3 � i3 � i

51.

��6i � 5i 2

1� �5 � 6i

6 � 5i

i�

6 � 5ii

��i�i

53.

� �1201681

�27

1681i

��27i � 120i2

81 � 1600�

�120 � 27i1681

3i

�4 � 5i�2 �3i

16 � 40i � 25i2�

3i�9 � 40i

��9 � 40i�9 � 40i

55.

� �1

2�

5

2i

��1 � 5i

2

�2 � 2i � 3 � 3i

1 � 1

2

1 � i�

3

1 � i�

2�1 � i� � 3�1 � i��1 � i��1 � i�

57.

�62 � 297i

949�

62

949�

297

949i

��100 � 297i � 162

949

��100 � 72i � 225i � 162i2

625 � 324

��4 � 9i

25 � 18i�

25 � 18i

25 � 18i

�4i2 � 9i

9 � 18i � 16

�3i � 8i2 � 6i � 4i2

9 � 24i � 6i � 16i2

i

3 � 2i�

2i

3 � 8i�

i�3 � 8i� � 2i�3 � 2i��3 � 2i��3 � 8i�

59.

� �2�3

��6 � ��2 � ��6i���2i� � �12i 2 � �2�3 ���1� 61. ���10 �2� ��10 i �2

� 10i2 � �10

63.

� �21 � 5�2 � � �7�5 � 3�10 �i � �21 � �50 � � �7�5 � 3�10 �i � 21 � 3�10 i � 7�5 i � �50 i2

�3 � ��5 ��7 � ��10 � � �3 � �5 i��7 � �10 i�

65.

� 1 ± i

�2 ± 2i

2

�2 ± ��4

2

x ����2� ± ���2�2 � 4�1��2�

2�1�

x2 � 2x � 2 � 0; a � 1, b � �2, c � 2 67.

��16 ± 4i

8� �2 ±

1

2i

��16 ± ��16

8

x ��16 ± ��16�2 � 4�4��17�

2�4�

4x2 � 16x � 17 � 0; a � 4, b � 16, c � 17

Page 33: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

70 Chapter 1 Equations, Inequalities, and Mathematical Modeling

69.

x � �12

8� �

3

2 or x � �

20

8� �

5

2

��16 ± �16

8�

�16 ± 4

8

x ��16 ± ��16�2 � 4�4��15�

2�4�

4x2 � 16x � 15 � 0; a � 4, b � 16, c � 15 71. Multiply both sides by 2.

�12 ± 6�2i

6� 2 ± �2i

�12 ± ��72

6

x ����12� ± ���12�2 � 4�3��18�

2�3�

3x2 � 12x � 18 � 0

32

x2 � 6x � 9 � 0

73. Multiply both sides by 5.

�5 ± 5�15

7�

57

±5�15

7

�10 ± �1500

14�

10 ± 10�1514

x ����10� ± ���10�2 � 4�7���50�

2�7�

7x2 � 10x � 50 � 0

1.4x2 � 2x � 10 � 0 75.

� �1 � 6i

� 6i � 1

� �6��1�i � ��1�

�6i 3 � i 2 � �6i2i � i2

77.

� �5i

� �5��1���1�i

�5i5 � �5i2i2i 79.

� �375�3 i

� 125�3�3 � ��1�i

� 53��3 �3i 3

���75 �3 � �5�3 i�3 81.1

i3�

1

�i�

1

�i�

i

i�

i

�i2�

i

1� i

83. (a)

(b)

�11,240 � 4630i

877�

11,240877

�4630877

i z � �340 � 230i29 � 6i ��29 � 6i

29 � 6i�

�29 � 6i

340 � 230i�

20 � 10i � 9 � 16i�9 � 16i��20 � 10i��

19 � 16i

�1

20 � 10i 1z

�1z1

�1z2

z1 � 9 � 16i, z2 � 20 � 10i

85. (a)

(b)

(c)

(d) ��2i�4 � ��2�4i 4 � 16�1� � 16

�2i�4 � 24i 4 � 16�1� � 16

��2�4 � 16

24 � 16 87. False, if then

That is, if the complex number is real, the number equalsits conjugate.

a � bi � a � bi � a.b � 0

89. False

� 1 � ��1� � 1 � i � i � 1

� �1�11 � �1�37��1� � �1�18��1� � �1�27�i� � �1�15�i�

i44 � i150 � i 74 � i109 � i61 � �i4�11 � �i4�37�i2� � �i4�18�i2� � �i4�27�i� � �i4�15�i�

Page 34: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.6 Other Types of Equations 71

Section 1.6 Other Types of Equations

■ You should be able to solve certain types of nonlinear or nonquadratic equations by rewriting them in a form in whichyou can factor, extract square roots, complete the square, or use the Quadratic Formula.

■ For equations involving radicals or rational exponents, raise both sides to the same power.

■ For equations that are of the quadratic type, use either factoring, the Quadratic Formula,or completing the square.

■ For equations with fractions, multiply both sides by the least common denominator to clear the fractions.

■ For equations involving absolute value, remember that the expression inside the absolute value can be positive or negative.

■ Always check for extraneous solutions.

au2 � bu � c � 0, a � 0,

Vocabulary Check

1. polynomial 2. extraneous 3. quadratic type

91.

The complex conjugate of this product is

The product of the complex conjugates is:

Thus, the complex conjugate of the product of two complex numbers is the product of their complex conjugates.

� �a1a2 � b1b2� � �a1b2 � a2b1�i

�a1 � b1i��a2 � b2i� � a1a2 � a1b2i � a2b1i � b1b2i2

�a1a2 � b1b2� � �a1b2 � a2b1�i.

� �a1a2 � b1b2� � �a1b2 � a2b1�i

�a1 � b1i��a2 � b2i� � a1a2 � a1b2i � a2b1i � b1b2i2

93. �4 � 3x� � �8 � 6x � x2� � �x2 � 3x � 12 95.

� 3x2 �232 x � 2

�3x �12��x � 4� � 3x2 � 12x �

12x � 2

97.

x � �31

�x � 31

�x � 12 � 19 99.

x �272

2x � 27

2x � 27 � 0

20x � 24 � 18x � 3 � 0

4�5x � 6� � 3�6x � 1� � 0 101.

a �1

2�3V

�b�

�3V�b

2�b

� 3V

4�b� a

3V

4�b� a2

3V � 4�a2b

V �4

3�a2b

103. Let liters withdrawn and replaced.

x � 1 liter

0.50x � 0.50

2.50 � 0.50x � 1.00x � 3.00

0.50�5 � x� � 1.00x � 0.60�5�

x � #

Page 35: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

72 Chapter 1 Equations, Inequalities, and Mathematical Modeling

5.

x2 � 6x � 36 � 0 ⇒ x � 3 ± 3�3i

x � 6 � 0 ⇒ x � �6

�x � 6��x2 � 6x � 36� � 0

x3 � 63 � 0

x3 � 216 � 0

7.

x � 3 � 0 ⇒ x � �3

5x � 0 ⇒ x � 0

5x�x � 3�2 � 0

5x�x2 � 6x � 9� � 0

5x3 � 30x2 � 45x � 0 9.

x � 1 � 0 ⇒ x � 1

x � 1 � 0 ⇒ x � �1

x � 3 � 0 ⇒ x � 3

�x � 3��x � 1��x � 1� � 0

�x � 3��x2 � 1� � 0

x2�x � 3� � �x � 3� � 0

x3 � 3x2 � x � 3 � 0

11.

x2 � x � 1 � 0 ⇒ x �1

2±�3

2i

x � 1 � 0 ⇒ x � �1

x � 1 � 0 ⇒ x � 1

�x � 1��x � 1��x2 � x � 1� � 0

�x � 1��x3 � 1� � 0

x3�x � 1� � �x � 1� � 0

x4 � x3 � x � 1 � 0

1.

2x2 � 9 � 0 ⇒ x � ±3�2

2

2x2 � 0 ⇒ x � 0

2x2�2x2 � 9� � 0

4x4 � 18x2 � 0 3.

2x � 3 � 0 ⇒ x � 3

2x � 3 � 0 ⇒ x � �3

x2 � 9 � 0 ⇒ x � ±3i

�x2 � 9��x � 3��x � 3� � 0

x4 � 81 � 0

13.

x � 1 � 0 ⇒ x � 1

x � 1 � 0 ⇒ x � �1

x � �3 � 0 ⇒ x � �3

x � �3 � 0 ⇒ x � ��3

�x � �3 ��x � �3 ��x � 1��x � 1� � 0

�x2 � 3��x2 � 1� � 0

x4 � 4x2 � 3 � 0

15.

1x � 4 � 0 ⇒ x � 4

1x � 4 � 0 ⇒ x � �4

2x � 1 � 0 ⇒ x �12

2x � 1 � 0 ⇒ x � �12

�2x � 1��2x � 1��x � 4��x � 4� � 0

�4x2 � 1��x2 � 16� � 0

4x4 � 65x2 � 16 � 0

Page 36: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.6 Other Types of Equations 73

17.

x2 � x � 1 � 0 ⇒ x � �1

2 ±

�3

2i

x � 1 � 0 ⇒ x � 1

x2 � 2x � 4 � 0 ⇒ x � 1 ± �3i

x � 2 � 0 ⇒ x � �2

�x � 2��x2 � 2x � 4��x � 1��x2 � x � 1� � 0

�x3 � 8��x3 � 1� � 0

x6 � 7x3 � 8 � 0

19.

1 � 5x � 0 ⇒ x � �1

5

1 � 3x � 0 ⇒ x � �1

3

�1 � 3x��1 � 5x� � 0

1 � 8x � 15x2 � 0

1

x2�

8

x� 15 � 0 21.

��x � �5 is not a solution.� �x �

12 ⇒ x �

14

�2�x � 1���x � 5� � 0

2��x �2� 9�x � 5 � 0

2x � 9�x � 5 � 0

2x � 9�x � 5

23.

x1�3 � 1 � 0 ⇒ x1�3 � 1 ⇒ x � �1�3 � 1

2x1�3 � 5 � 0 ⇒ x1�3 � �52 ⇒ x � �� 5

2�3� �

1258

�2x1�3 � 5��x1�3 � 1� � 0

2�x1�3�2 � 3x1�3 � 5 � 0

2x2�3 � 3x1�3 � 5 � 0

3x1�3 � 2x2�3 � 5

25.

(a)

(b) x-intercepts: ��1, 0�, �0, 0�, �3, 0�

−9 9

−7

5

y � x3 � 2x2 � 3x

(c)

(d) The intercepts of the graph are the same as the solutions to the equation.x-

x � 3 � 0 ⇒ x � 3

x � 1 � 0 ⇒ x � �1

x � 0

0 � x�x � 1��x � 3�

0 � x3 � 2x2 � 3x

27.

(a)

(b) x-intercepts: �±1, 0�, �±3, 0�

−5 5

20

−20

y � x4 � 10x2 � 9

(c)

(d) The x-intercepts of the graph are the same as the solutions to the equation.

x � 3 � 0 ⇒ x � 3

x � 3 � 0 ⇒ x � �3

x � 1 � 0 ⇒ x � 1

x � 1 � 0 ⇒ x � �1

0 � �x � 1��x � 1��x � 3��x � 3�

0 � �x2 � 1��x2 � 9�

0 � x4 � 10x2 � 9

Page 37: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

74 Chapter 1 Equations, Inequalities, and Mathematical Modeling

29.

x � 50

2x � 100

�2x � 10

�2x � 10 � 0 31.

x � 26

x � 10 � 16

�x � 10 � 4

�x � 10 � 4 � 0 33.

x � �16

2x � �32

2x � 5 � �27

33�2x � 5 � �3

3�2x � 5 � 3 � 0

35.

x � 2 � 0 ⇒ x � 2

x � 5 � 0 ⇒ x � �5

�x � 5��x � 2� � 0

x2 � 3x � 10 � 0

16 � 8x � x2 � 26 � 11x

4 � x � �26 � 11x

��26 � 11x � 4 � x 37.

x � 0

�2x � 0

x � 1 � 3x � 1

�x � 1 � �3x � 1

39.

9 � x

4 � x � 5

2 � �x � 5

4 � 2�x � 5

x � 1 � 2�x � 5 � x � 5

��x �2� �1 � �x � 5 � 2

�x � 1 � �x � 5

�x � �x � 5 � 1 41.

1014 � x

101 � 4x

81 � 4x � 20

81 � 4�x � 5�

9 � 2�x � 5

�90 � �20�x � 5

x � 5 � 100 � 20�x � 5 � x � 5

��x � 5 �2� �10 � �x � 5 �2

�x � 5 � 10 � �x � 5

�x � 5 � �x � 5 � 10

43.

x � 14 � 0 ⇒ x � 14

x � 2 � 0 ⇒ x � 2, extraneous

�x � 2��x � 14� � 0

x2 � 16x � 28 � 0

x2 � 8x � 16 � 8x � 12

x2 � 8x � 16 � 4�2x � 3�

�x � 4�2 � �2�2x � 3�2

x � 4 � 2�2x � 3

�x � 4 � �2�2x � 3

x � 2 � 2x � 2 � 2�2x � 3

x � 2 � 2x � 3 � 2�2x � 3 � 1

��x � 2�2� ��2x � 3 � 1�2

�x � 2 � �2x � 3 � 1

�x � 2 � �2x � 3 � �1 45.

x � 5 � 4 � 9

x � 5 � 3�64

�x � 5�3 � 82

�x � 5�3�2 � 8

Page 38: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.6 Other Types of Equations 75

47.

x � �3 ± 16�2

x � 3 � ±�512

x � 3 � ±�83

�x � 3�2 � 83

�x � 3�2�3 � 8 49.

x � ±�14

x2 � 14

x2 � 5 � 9

x2 � 5 � 3�272

�x2 � 5�3 � 272

�x2 � 5�3�2 � 27

51.

5x � 2 � 0 ⇒ x �25, extraneous

�x � 1�1�2 � 0 ⇒ x � 1 � 0 ⇒ x � 1

�x � 1�1�2�5x � 2� � 0

�x � 1�1�2�3x � 2�x � 1�� � 0

3x�x � 1�1�2 � 2�x � 1�3�2 � 0

53.

(a)

(b) x-intercepts:

(c)

(d) The x-intercepts of the graph are the same as the solutions to the equation.

x � 6 � 0 ⇒ x � 6

x � 5 � 0 ⇒ x � 5

�x � 5��x � 6� � 0

x2 � 11x � 30 � 0

x2 � 11x � 30

x � �11x � 30

0 � �11x � 30 � x

�5, 0�, �6, 0�

−1 11

−6

2

y � �11x � 30 � x 55.

(a)

(b) x-intercepts:

(c)

(d) The x-intercepts of the graph are the same as thesolutions to the equation.

x � 4 � 0 ⇒ x � 4

x � 0

x�x � 4� � 0

x2 � 4x � 0

x2 � 16x � 64 � 20x � 64

x2 � 16x � 64 � 4�5x � 16�

�x � 8�2 � �2�5x � 16�2

x � 8 � 2�5x � 16

2x � 16 � 4�5x � 16

7x � 36 � 5x � 20 � 4�5x � 16

7x � 36 � 4 � 4�5x � 16 � 5x � 16

��7x � 36 �2� �2 � �5x � 16 �2

�7x � 36 � 2 � �5x � 16

��7x � 36 � ��5x � 16 � 2

0 � �7x � 36 � �5x � 16 � 2

�0, 0�, �4, 0�−0.5

0.5

5−3

y � �7x � 36 � �5x � 16 � 2

Page 39: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

76 Chapter 1 Equations, Inequalities, and Mathematical Modeling

57.

x � 2 � 0 ⇒ x � 2

2x � 3 � 0 ⇒ x � �3

2

�2x � 3��x � 2� � 0

2x2 � x � 6 � 0

2x2 � 6 � x

�2x��x� � �2x��3

x� � �2x��1

2�

x �3

x�

1

259.

x ��3 ± ��3�2 � 4�3���1�

2�3��

�3 ± �21

6

a � 3, b � 3, c � �1

0 � 3x2 � 3x � 1

1 � 3x2 � 3x

x � 1 � x � 3x�x � 1�

x�x � 1�1

x� x�x � 1�

1

x � 1� x�x � 1��3�

1

x�

1

x � 1� 3

61.

x � 4 � 0 ⇒ x � 4

x � 5 � 0 ⇒ x � �5

0 � �x � 5��x � 4�

0 � x2 � x � 20

20 � x � x2

20 � x

x� x 63.

�2 ± �124

6�

2 ± 2�31

6�

1 ± �31

3

x ����2� ± ���2�2 � 4�3���10�

2�3�

a � 3, b � �2, c � �10

3x2 � 2x � 10 � 0

x � x � 2 � 3x2 � 12

�x � 2��x � 2�x

x2 � 4� �x � 2��x � 2�

1

x � 2� 3�x � 2��x � 2�

x

x2 � 4�

1

x � 2� 3

65.

��2x � 1� � 5 ⇒ x � �2

2x � 1 � 5 ⇒ x � 3

2x � 1 � 5

67.

First equation:

Only are solutions to the original equation. and are extraneous.x � 1x � ��3 x � �3 and x � �3

x � ±�3

x2 � 3 � 0

x � x2 � x � 3

x � x2 � x � 3

Second equation:

x � 3 � 0 ⇒ x � �3

x � 1 � 0 ⇒ x � 1

�x � 1��x � 3� � 0

x2 � 2x � 3 � 0

�x � x2 � x � 3

Page 40: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.6 Other Types of Equations 77

69.

First equation:

Only are solutions to the original equation. and are extraneous.x ��1 � �17

2x � �2x � 3 and x �

�1 � �17

2

x � 2 � 0 ⇒ x � �2

x � 3 � 0 ⇒ x � 3

�x � 3��x � 2� � 0

x2 � x � 6 � 0

x � 1 � x2 � 5

x � 1 � x2 � 5

Second equation:

x ��1 ± �17

2

x2 � x � 4 � 0

�x � 1 � x2 � 5

��x � 1� � x2 � 5

71.

(a)

(b) x-intercept:

(c)

(d) The x-intercept of the graph is the same as the solution to the equation.

x � 1 � 0 ⇒ x � �1

0 � �x � 1�2

0 � x2 � 2x � 1

0 � �x2 � 2x � 1

0 � x � 1 � 4x � x2 � x

0 � �x � 1� � 4x � x�x � 1�

0 �1

x�

4

x � 1� 1

��1, 0�

4−4

−24

24

y �1

x�

4

x � 1� 1 73.

(a)

(b) x-intercepts:

(c)

OR

OR

(d) The x-intercepts of the graph are the same asthe solutions to the equation.

x � �3

�x � 3

�x � 1 � 2 x � 1

��x � 1� � 2 x � 1 � 2

2 � x � 1 0 � x � 1 � 2

�1, 0�, ��3, 0�

−10 8

−4

8

y � x � 1 � 2

75.

Using the positive value for we have x � ±�1.5 � �29.13

6.4 ±1.038.x2,

x2 �1.5 ± �1.52 � 4�3.2���2.1�

2�3.2�

3.2x4 � 1.5x2 � 2.1 � 0

77.

Use the Quadratic Formula with and

Considering only the positive value for we have:

x 16.756.

�x 4.0934

�x ,

�x �6 ± �36 � 4�1.8���5.6�

2�1.8�

6 ± 8.7361

3.6

c � �5.6.b � �6,a � 1.8,

1.8��x �2� 6�x � 5.6 � 0

1.8x � 6�x � 5.6 � 0 Given equation 79. and 5

One possible equation is:

Any non-zero multiple of this equation would also havethese solutions.

x2 � 3x � 10 � 0

�x � 2��x � 5� � 0

�x � ��2���x � 5� � 0

�2

Page 41: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

78 Chapter 1 Equations, Inequalities, and Mathematical Modeling

81. and

One possible equation is:

Any non-zero multiple of this equation would also havethese solutions.

21x2 � 31x � 42 � 0

�3x � 7��7x � 6� � 0

x �67 ⇒ 7x � 6 ⇒ 7x � 6 is a factor.

x � �73 ⇒ 3x � �7 ⇒ 3x � 7 is a factor.

67�

73 83. and 4

One possible equation is:

Any non-zero multiple of this equation would also havethese solutions.

x3 � 4x2 � 3x � 12 � 0

�x2 � 3��x � 4� � 0

�x � �3��x � �3��x � 4� � 0

�x � �3��x � ���3���x � 4� � 0

�3, ��3,

85. and

One possible equation is:

Any non-zero multiple of this equation would also have these solutions.

x4 � 1 � 0

�x2 � 1��x2 � 1� � 0

�x � 1��x � 1��x � i��x � i� � 0

�x � ��1���x � 1��x � i��x � ��i�� � 0

�i�1, 1, i,

87. Let the number of students in the original group. Then, the original cost per student.

When six more students join the group, the cost per student becomes

Model:

Multiply both sides by to clear fractions.

Using the positive value for x we conclude that the original number was students.x � 34

x �90 ± ���90�2 � 4��15��20,400�

2��15��

90 ± 1110

�30

�15x2 � 90x � 20,400 � 0

2x �3400 � 15x��x � 6� � 3400x

�1700

x� 7.5��x � 6� � 1700

�Cost per student� � �Number of students� � Total cost

1700x

� 7.50.

1700x

�x �

89. Model:

Labels: Let average speed of the plane. Then wehave a travel time of If the averagespeed is increased by 40 mph, then

Now, we equate these two equations and solve for x.

t �145

x � 40�

1

5.

t �12

60�

145

x � 40

t � 145�x.x �

Time �Distance

RateEquation:

Using the positive value for x found by the Quadratic Formula, we have mph and

mph. The airspeed required to obtain the decrease in travel time is 191.5 miles per hour.x � 40 � 191.5

x 151.5

0 � x2 � 40x � 29,000

725x � 29,000 � 725x � x2 � 40x

145�5��x � 40� � 145�5�x � x�x � 40�

145

x�

145

x � 40�

1

5

Page 42: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.6 Other Types of Equations 79

91.

r 0.04 � 4%

��1.220996�1�60 � 1��12� � r

�1.220996�1�60 � 1 �r

12

1.220996 � �1 �r

12�60

3052.49 � 2500�1 �r

12��12��5�

A � P�1 �r

n�nt

93.

(a)

The number of medical doctors reached 816,000 lateduring the year 1999.

(b)

The model predicts the number of medical doctorswill reach 900,000 during the year 2005.

The actual number of medical doctors in 2005 isabout 800,000.

t 15.27

t � �436.03111.6 �

2

111.6�t � 436.03

463.97 � 111.6�t � 900

t 9.95

t � �352.03111.6 �

2

111.6�t � 352.03

463.97 � 111.6�t � 816

M � 463.97 � 111.6�t, 4 ≤ t ≤ 12

95.

(a)

(b) when pounds per square inch.x 15T � 212�

T � 75.82 � 2.11x � 43.51�x, 5 ≤ x ≤ 40

(c)

By the Quadratic Formula, we have

Since x is restricted to let pounds per square inch.

(d)

05 40

300

(14.806119, 212)

x � 14.8065 ≤ x ≤ 40,

�x 3.84787 ⇒ x 14.806

�x 16.77298 ⇒ x 281.333

0 � �2.11x � 43.51�x � 136.18

212 � 75.82 � 2.11x � 43.51�xAbsolute TemperaturePressure,

5 162.56

10 192.31

15 212.68

20 228.20

25 240.62

30 250.83

35 259.38

40 266.60

Tx

97.

Rounding x to the nearest whole unit yields units.x 500

� 10.01x � 500.25

� 10.01x � 5.0025

0.01x � 1 � 6.0025

�0.01x � 1 � 2.45

1x � 37.55 � 40 � �0.01x � 1

Page 43: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

80 Chapter 1 Equations, Inequalities, and Mathematical Modeling

99. Model:

Labels:

Equation:

The distance between bases is approximately 90 feet.

x � ±�8128.125 ±90

x2 �16,256.25

2

2x2 � 16,256.25

x2 � x2 � �127.5�2

distance from home to 2nd � 127.5distance from 1st to 2nd � x,Distance from home to 1st � x,

�Distance fromhome to 1st �

2

� �distance from1st to 2nd �

2

� �distance fromhome to 2nd �

2

101.

(a)

When

(b)

when h is between 11 and 12 inches.S � 350

S � 350, h 11.4.

0150

450

S � 8��64 � h2

(c)

(d) Solving graphically or numerically yields anapproximate solution. An exact solution is obtained algebraically.

h 11.4

h2 129.935

64 � h2 193.935

122,500 � 64�2�64 � h2�

�350�2 � �8��64 � h2 �2

350 � 8��64 � h2

8 9 10 11 12 13

284.3 302.6 321.9 341.8 362.5 383.6S

h

103. Model:

Labels: Work done rate of first person time worked by first person

rate of second person time worked by second person

Equation:

(Choose the positive value for r.)

It would take approximately 23 hours and 26 hours individually.

r 23

r ����21� ± ���21�2 � 4�1���36�

2�1��

21 ± �585

2

0 � r2 � 21r � 36

12r � 36 � 12r � r2 � 3r

r�r � 3�12

r� r�r � 3�

12

r � 3� r�r � 3�

12

r�

12

r � 3� 1

� 12�1

r � 3,

� 12,�1

r,� 1,

� Portion doneby first person� � � portion done

by second person� � �workdone�

105.

v2�s

R� g

v2�s � gR

v2 �gR�s

v ��gR�s

107. False—See Example 7 on page 137.

Page 44: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.6 Other Types of Equations 81

109. The distance between is 13.

Both and are a distance of 13from �1, 2�.

�6,�10���4, �10�

x � 6 � 0 ⇒ x � 6

x � 4 � 0 ⇒ x � �4

�x � 4��x � 6� � 0

x2 � 2x � 24 � 0

x2 � 2x � 1 � 144 � 169

�x � 1�2 � ��12�2 � 132

��x � 1�2 � ��10 � 2�2 � 13

�1, 2� and �x, �10� 111. The distance between is 17.

Both and are a distance of 17from �0, 0�.

�8, �15��8, 15�

� ±15

y2 � ±�225

y2 � 225

64 � y2 � 289

�8�2 � �y�2 � 172

��8 � 0�2 � �y � 0�2 � 17

�0, 0� and �8, y�

113.

OR

Thus, or From the original equationwe know that

Some possibilities are:

or

or

or

or

or a � 14a � 4b � 14,

a � 13a � 5b � 13,

a � 12a � 6b � 12,

a � 11a � 7b � 11,

a � 10a � 8b � 10,

a � 9b � 9,

b ≥ 9.a � b.a � 18 � b

a � 18 � b

�a � b � 18

9 � a � b � 9

9 � a � b � 9

9 � 9 � a � b

a � b

�a � �b

9 � a � �b � 9

9 � a � ��b � 9�

115.

This formula gives the relationship between a and b. From the original equation we know that and

Choose a b value, where and then solve for a, keeping in mind that

Some possibilities are:

b � 25, a � �5

b � 24, a � 14

b � 23, a � 11

b � 22, a � 16

b � 21, a � 19

b � 20, a � 20

a ≤ 20.b ≥ 20b ≥ 20.

a ≤ 20

a � �b2 � 40b � 380

�a � b2 � 40b � 380

20 � a � b2 � 40b � 400

�20 � a � b � 20

20 � �20 � a � b

117.83x

�32x

�166x

�96x

�256x 119.

��3z2 � 2z � 4

z�z � 2�

�2z � 3z2 � 6z � 2z � 4

z�z � 2�

�2z � 3z�z � 2� � 2�z � 2�

z�z � 2�

2

z � 2� �3 �

2z� �

2z � 2

� 3 �2z

121.

x � 11

x � 11 � 0

�x � 11�2 � 0

x2 � 22x � 121 � 0

Page 45: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

82 Chapter 1 Equations, Inequalities, and Mathematical Modeling

Section 1.7 Linear Inequalities in One Variable

■ You should know the properties of inequalities.

(a) Transitive: and implies

(b) Addition: and implies

(c) Adding or Subtracting a Constant: if

(d) Multiplying or Dividing a Constant: For

1. If then and

2. If then and

■ You should be able to solve absolute value inequalities.

(a) if and only if

(b) if and only if x < �a or x > a.�x� > a

�a < x < a.�x� < a

a

c>

b

c.ac > bcc < 0,

a

c<

b

c.ac < bcc > 0,

a < b,

a < b.a ± c < b ± c

a � c < b � d.c < da < b

a < c.b < ca < b

1. Interval:

(a) Inequality:

(b) The interval is bounded.

�1 ≤ x ≤ 5

��1, 5� 3. Interval:

(a) Inequality:

(b) The interval is unbounded.

x > 11

�11, �� 5. Interval:

(a) Inequality:

(b) The interval is unbounded.

x < �2

���, �2�

7.

Matches (b).

x < 3 9.

Matches (d).

�3 < x ≤ 4

Vocabulary Check

1. solution set 2. graph 3. negative

4. equivalent 5. double 6. union

11.

Matches (e).

�x� < 3 ⇒ �3 < x < 3

13.

(a)

Yes, is asolution.

x � 3

3 > 0

5�3� � 12 >?

0

x � 3

5x � 12 > 0

(b)

No, is not asolution.

x � �3

�27 >� 0

5��3� � 12 >?

0

x � �3 (c)

Yes, is a solution.x �52

12 > 0

5�52� � 12 >

?0

x �52 (d)

No, is not a solution.

x �32

�92 >� 0

5�32� � 12 >

?0

x �32

15.

(a)

Yes, is asolution.

x � 4

0 <12

< 2

0 <? 4 � 2

4<?

2

x � 4

0 <x � 2

4< 2

(b)

No, is not asolution.

x � 10

0 < 2 <� 2

0 <? 10 � 2

4<?

2

x � 10 (c)

No, is not a solution.

x � 0

0 <� �12

< 2

0 <? 0 � 2

4<?

2

x � 0 (d)

Yes, is a solution.x �72

0 <38

< 2

0 <? �7�2� � 2

4<?

2

x �7

2

Page 46: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.7 Linear Inequalities in One Variable 83

17.

(a)

Yes, is asolution.

x � 13

3 ≥ 3

�13 � 10� ≥?

3

x � 13

�x � 10� ≥ 3

(b)

Yes, is a solution.

x � �1

11 ≥ 3

��1 � 10� ≥?

3

x � �1 (c)

Yes, is a solution.

x � 14

4 ≥ 3

�14 � 10� ≥?

3

x � 14 (d)

No, is not a solution.

x � 9

1 ≥� 3

�9 � 10� ≥?

3

x � 9

19.

1 2 3 4 5

x

x < 3

14�4x� < 14�12�

4x < 12 21.

2 310−1

x

−2

32

x < 32

�12��2x� < ��1

2���3�

�2x > �3 23.

10 11 12 13 14

x

x ≥ 12

x � 5 ≥ 7

25.

0 1 2 3 4

x

x > 2

�2x < �4

2x � 7 < 3 � 4x 27.

−2 −1 10 2

x

27

x ≥ 27

7x ≥ 2

2x � 1 ≥ 1 � 5x 29.

43 5 6 7

x

x < 5

4 � 2x < 9 � 3x

4 � 2x < 3�3 � x�

31.

432 5 6

x

x ≥ 4

�14x ≤ �1

34x � 6 ≤ x � 7 33.

210 3 4

x

x ≥ 2

4x �12 ≥ 3x �

52

12�8x � 1� ≥ 3x �52 35.

−2−3−4−5−6

x

x ≥ �4

3.6x ≥ �14.4

3.6x � 11 ≥ �3.4

37.

30 1 2−1

x

�1 < x < 3

�2 < 2x < 6

�1 < 2x � 3 < 9 39.

−2−4−6

0 2 64 8

x29

215

1�92 < x < 15

2

1�9 < 2x < 15

�12 < 2x � 3 < 12

1�4 <2x � 3

3< 4 41.

0 1−1

x

−43

41

� 34

< x < �14

�14

> x > �34

34

> x � 1 >14

43.

121110 13 14

x

13.510.5

10.5 ≤ x ≤ 13.5

4.2 ≤ 0.4x ≤ 5.4

3.2 ≤ 0.4x � 1 ≤ 4.4 45.

−2−4−6 0 2 4 6

x

�6 < x < 6

�x� < 6 47.

−2−3 −1 0 1 2 3

x

x < �2 x > 2

x2

< �1 or x2

> 1

�x2� > 1

Page 47: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

84 Chapter 1 Equations, Inequalities, and Mathematical Modeling

49.

No solution. The absolute value of a number cannot beless than a negative number.

�x � 5� < � 1 51.

201510 25 30

x

14 26

14 ≤ x ≤ 26

�6 ≤ x � 20 ≤ 6

�x � 20� ≤ 6

53.

or

−2 −1 10 2 3 4

x

32

x ≤ �32 x ≥ 3

�4x ≥ 6 �4x ≤ �12

3 � 4x ≥ 9 3 � 4x ≤ �9

�3 � 4x� ≥ 9 55.

or

0−5−10−15 5 10 15

x

11

x ≥ 11 x ≤ �5

x � 3 ≥ 8 x � 3 ≤ �8

x � 3

2≥ 4

x � 3

2≤ �4

�x � 32 � ≥ 4

57.

x

3 654

4 < x < 5

5 > x > 4

�10 < �2x < �8

�1 < 9 � 2x < 1

�9 � 2x� < 1

�9 � 2x� � 2 < �1 59.

or

−12

− −

−16 −4−8

x

112

292

x ≥ �112 x ≤ �

292

x � 10 ≥ 92 x � 10 ≤ �

92

�x � 10� ≥ 92

2�x � 10� ≥ 9

61.

−10 10

−10

10

6x > 2

6x > 12 63.

−10 10

−10

10

5 � 2x ≤ 2

5�2x ≥ �4

5 � 2x ≥ 1 65.

10

−10

−10 24

�6 ≤ x ≤ 22

�14 ≤ x � 8 ≤ 14

�x � 8� ≤ 14

67.

or

−15 1

−10

10

x ≥ �12 x ≤ �

272

x � 7 ≥ 132 x � 7 ≤ �

132

�x � 7� ≥ 132

2�x � 7� ≥ 13 69.

(a)

(b)

2 � 3x ≤ 32

� 32x ≤ 3

2x � 3 ≤ 0

2 � 3y ≤ 0

2 � 3x ≥ 2

� 32x ≥ 4

2x � 3 ≥ 1

2 � 3y ≥ 1−4 8

−5

3y � 2x � 3

Page 48: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.7 Linear Inequalities in One Variable 85

71.

(a)

(b)

x ≤ 4

�12x ≥ �2

�12x � 2 ≥ 0

y ≥ 0

4 ≥ x ≥ �2

�2 ≤ �12x ≤ 1

0 ≤ �12x � 2 ≤ 3

0 ≤ y ≤ 3

−6 6

−2

6y � �12x � 2 73.

(a)

(b)

or

or x ≥ 7 x ≤ �1

x � 3 ≥ 4 x � 3 ≤ �4

�x � 3� ≥ 4

y ≥ 4

1 ≤ x ≤ 5

�2 ≤ x � 3 ≤ 2

�x � 3� ≤ 2

y ≤ 2

−5 10

−2

8y � �x � 3�

75.

�5, ��

x ≥ 5

x � 5 ≥ 0 77.

��3, ��

x ≥ �3

x � 3 ≥ 0 79.

���, 72� x ≤ 7

2

�2x ≥ �7

7 � 2x ≥ 0 81.

All real numbers within8 units of 10.

�x � 10� < 8

83. The midpoint of the interval is 0. The interval represents all realnumbers x no more than 3 unitsfrom 0.

�x� ≤ 3

�x � 0� ≤ 3

��3, 3� 85. The graph shows all real numbersat least 3 units from 7.

�x � 7� ≥ 3

87. All real numbers within 10 units of 12

�x � 12� < 10

89. All real numbers more than 4 units from

�x � 3� > 4

�x � ��3�� > 4

�3 91. Let

Type A account charges:

Type B account charges:

If you write more than six checks a month, then thecharges for the type A account are less than the chargesfor the type B account.

6 < x

1.50 < 0.25x

6.00 � 0.25x < 4.50 � 0.50x

4.50 � 0.50x

6.00 � 0.25x

x � the number of checks written in a month.

93.

r > 3.125%

r > 0.03125

2r > 0.0625

1 � 2r > 1.0625

1000�1 � r�2�� > 1062.50 95.

120.95x ≥ 36 units

120.95x > 35.7995

120.95x > 750

115.95x > 95x � 750

115.95R > C

Page 49: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

86 Chapter 1 Equations, Inequalities, and Mathematical Modeling

97. Let

Revenue:

Cost:

Profit:

In whole dozens, 134 ≤ x ≤ 234.

13313 ≤ x ≤ 2331

3

200 ≤ 1.50x ≤ 350

50 ≤ 1.50x � 150 ≤ 200

50 ≤ P ≤ 200

� 1.50x � 150

� 2.95x � �150 � 1.45x�

P � R � C

C � 150 � 1.45x

R � 2.95x

x � daily sales level (in dozens) of doughnuts. 99. (a)

(b) From the graph we see that when Algebraically we have:

IQ scores are not a good predictor of GPAs. Otherfactors include study habits, class attendance, andattitude.

x ≥ 129

8.638 ≤ 0.067x

3 ≤ 0.067x � 5.638

x ≥ 129.y ≥ 3

075 150

5

y � 0.067x � 5.638

101.

(a)

Rounding to the nearest year, The average salary was at least $32,000 but not morethan $42,000 between 1991 and 2000.

(b)

According to the model, the average salary willexceed $48,000 in 2006.

t > 16

1.05t > 17

1.05t � 31 > 48

1 ≤ t ≤ 10.

0.95 ≤ t ≤ 10.48

1 ≤ 1.05t ≤ 11

32 ≤ 1.05t � 31 ≤ 42

S � 1.05t � 31.0, 0 ≤ t ≤ 12 103.

Since we have

106.864 ≤ area ≤ 109.464.

�10.3375�2 ≤ area ≤ �10.4625�2

A � s2,

10.3375 ≤ s ≤ 10.4625

�0.0625 ≤ s � 10.4 ≤ 0.0625

� 116 ≤ s � 10.4 ≤ 1

16

�s � 10.4� ≤ 116

105.

You might have been undercharged or overcharged by $0.19.

110�$1.89� � $0.19

14.9 ≤ x ≤ 15.1 gallons

� 110 ≤ x � 15 ≤ 1

10

�x � 15� ≤ 110 107.

Two-thirds of the workers could perform the task in thetime interval between 13.7 minutes and 17.5 minutes.

13.7 < t < 17.5

�1.9 < t � 15.6 < 1.91615141312 17 18 19

t

17.513.7 �1 <t � 15.6

1.9 < 1

�t � 15.6

1.9 � < 1

109.

The minimum relative humidity is20 and the maximum is 80.

20 ≤ h ≤ 80

�30 ≤ h � 50 ≤ 30

�h � 50� ≤ 30 111. False. If c is negative, thenac ≥ bc.

113.

or

Matches (b).

x ≥ a � 2

x � a ≥ 2

x ≤ a � 2

x � a ≤ �2

�x � a� ≥ 2

Page 50: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.8 Other Types of Inequalities 87

117. and

Midpoint: �3 � ��5�2

, 6 � ��8�

2 � � ��1, �1�

d � ���5 � 3�2 � ��8 � 6�2 � ���8�2 � ��14�2 � �260 � 2�65

��5, �8��3, 6�

119.

x � 10

6x � 60

6x � 24 � 36

�12 � 6x � 12 � 36

�6�2 � x� � 12 � 36 121.

or

x � �12 x �

17

2x � �1 7x � 1

2x � 1 � 0 7x � 1 � 0

�7x � 1��2x � 1� � 0

14x2 � 5x � 1 � 0

123. ��3, 10� 125. Answers will vary.

Section 1.8 Other Types of Inequalities

■ You should be able to solve inequalities.

(a) Find the critical number.

1. Values that make the expression zero

2. Values that make the expression undefined

(b) Test one value in each test interval on the real number line resulting from the critical numbers.

(c) Determine the solution intervals.

Vocabulary Check

1. critical; test intervals 2. zeroes; undefined values 3. P � R � C

115. and

Midpoint: ��4 � 12

, 2 � 12

2 � � ��32

, 7�d � ��1 � ��4��2 � �12 � 2�2 � �52 � 102 � �125 � 5�5

�1, 12���4, 2�

1.

(a)

No, is not asolution.

x � 3

6 <� 0

�3�2 � 3 <?

0

x � 3

x2 � 3 < 0

(b)

Yes, is a solution.x � 0

�3 < 0

�0�2 � 3 <?

0

x � 0 (c)

Yes, is a solution.x �32

�34 < 0

�32�2

� 3 <?

0

x �32 (d)

No, is not asolution.

x � �5

22 <� 0

��5�2 � 3 <?

0

x � �5

Page 51: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

88 Chapter 1 Equations, Inequalities, and Mathematical Modeling

3.

(a)

Yes, is asolution.

x � 5

7 ≥ 3

5 � 25 � 4

≥?

3

x � 5

x � 2x � 4

≥ 3

(b)

is undefined.

No, is not asolution.

x � 4

60

4 � 24 � 4

≥?

3

x � 4 (c)

No, is not asolution.

x � �92

517

≥� 3

92 � 2

�92 � 4

≥?

3

x � �9

2(d)

Yes, is a solution.x �92

13 ≥ 3

92 � 292 � 4

≥?

3

x �9

2

5.

Critical numbers: x � �32

, x � 2

x � 2 � 0 ⇒ x � 2

2x � 3 � 0 ⇒ x � �32

2x2 � x � 6 � �2x � 3��x � 2� 7.

Critical numbers: x �72

, x � 5

⇒ x � 5 x � 5 � 0

⇒ x �72

2x � 7 � 0

�2x � 7x � 5

2 �3

x � 5�

2�x � 5� � 3x � 5

9.

Critical numbers:

Test intervals:

Test: Is

Interval x-Value Value of Conclusion

Positive

Negative

Positive

Solution set:

−1−2−3 0 1 2 3

x

��3, 3�

16 � 9 � 7x � 4�3, ��

0 � 9 � �9x � 0��3, 3�

16 � 9 � 7x � �4���, �3�

x2 � 9

�x � 3��x � 3� ≤ 0?

���, �3�, ��3, 3�, �3, ��

x � ±3

�x � 3��x � 3� ≤ 0

x2 � 9 ≤ 0

x2 ≤ 9 11.

Critical numbers:

Test intervals:

Test: Is

Interval x-Value Value of Conclusion

Positive

Negative

Positive

Solution set:

−8 −4−6

−7

0 2

3

4

x

−2

��7, 3�

�12��2� � 24x � 5�3, ��

�7���3� � �21x � 0��7, 3�

��3���13� � 39x � �10���, �7�

�x � 7��x � 3�

�x � 7��x � 3� < 0?

���, �7�, ��7, 3�, �3, ��

x � �7, x � 3

�x � 7��x � 3� < 0

x2 � 4x � 21 < 0

x2 � 4x � 4 < 25

�x � 2�2 < 25

Page 52: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.8 Other Types of Inequalities 89

13.

Critical numbers:

Test intervals:

Test: Is

Interval x-Value Value of Conclusion

Positive

Negative

Positive

Solution set:

x

210−1−2−3−4−5−6

���, �5� � �1, ��

�7��1� � 7x � 2�1, ��

�5���1� � �5x � 0��5, 1�

��1���7� � 7x � �6���, �5�

�x � 5��x � 1�

�x � 5��x � 1� ≥ 0?

���, �5�, ��5, 1�, �1, ��

x � �5, x � 1

�x � 5��x � 1� ≥ 0

x2 � 4x � 5 ≥ 0

x2 � 4x � 4 ≥ 9 15.

Critical numbers:

Test intervals:

Test: Is

Interval x-Value Value of Conclusion

Positive

Negative

Positive

Solution set:

0−1−2−3 1 2

x

��3, 2�

�6��1� � 6x � 3�2, ��

�3���2� � �6x � 0��3, 2�

��1���6� � 6x � �4���, �3�

�x � 3��x � 2�

�x � 3��x � 2� < 0?

���, �3�, ��3, 2�, �2, ��

x � �3, x � 2

�x � 3��x � 2� < 0

x2 � x � 6 < 0

x2 � x < 6

17.

Critical numbers:

Test intervals:

Test: Is

Interval x-Value Value of Conclusion

Positive

Negative

Positive

Solution set:

10−1−2−3

x

��3, 1�

�5��1� � 5x � 2�1, ��

�3���1� � �3x � 0��3, 1�

��1���5� � 5x � �4���, �3�

�x � 3��x � 1�

�x � 3��x � 1� < 0?

���, �3�, ��3, 1�, �1, ��

x � �3, x � 1

�x � 3��x � 1� < 0

x2 � 2x � 3 < 0

19.

Complete the square.

Critical numbers:

Test intervals:

Test: Is

Interval x-Value Value of Conclusion

Positive

Negative

Positive

Solution set: ��� < �4 � �21� � ��4 � �21, �� 20

x

−8 −6 −4 −2−10

−4 + 21−4 − 214 � 16 � 5 � 15x � 2��4 � �21, ��0 � 0 � 5 � �5x � 0��4 � �21, �4 � �21 �100 � 80 � 5 � 15x � �10���, �4 � �21 �

x2 � 8x � 5

x2 � 8x � 5 ≥ 0?

��4 � �21, ����4 � �21, �4 � �21�,���, �4 � �21�,x � �4 ± �21

x � �4 ± �21

x � 4 � ±�21

�x � 4�2 � 21

x2 � 8x � 16 � 5 � 16

x2 � 8x � 5 � 0

x2 � 8x � 5 ≥ 0

Page 53: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

90 Chapter 1 Equations, Inequalities, and Mathematical Modeling

21.

Critical numbers:

Test intervals:

Test: Is ?

Interval x-Value Value of Conclusion

Negative

Positive

Negative

Positive

Solution set: ��1, 1� � �3, ��

�5��3��1� � 15x � 4�3, ��

�3��1���1� � �3x � 2�1, 3�0 1 2 43

x

−1

�1���1���3� � 3x � 0��1, 1�

��1���3���5� � �15x � �2���, �1�

�x � 1��x � 1��x � 3�

�x � 1��x � 1��x � 3� > 0

���, �1�, ��1, 1�, �1, 3�, �3, ��

x � ±1, x � 3

�x � 1��x � 1��x � 3� > 0

�x2 � 1��x � 3� > 0

x2�x � 3� � 1�x � 3� > 0

x3 � 3x2 � x � 3 > 0

23.

Critical numbers:

Test intervals:

Test: Is

Interval x-Value Value of Conclusion

Negative

Positive

Negative

Positive

Solution set: ��3, 2� � �3, ��

�2��7��1� � 14x � 4�3, ��

�0.5��5.5���0.5� � �1.375x � 2.5�2, 3�

x

−4 −3 −2 −1 0 1 2 3 4 5

��2��3���3� � 18x � 0��3, 2�

��6���1���7� � �42x � �4���, �3�

�x � 2��x � 3��x � 3�

�x � 2��x � 3��x � 3� ≥ 0?

���, �3�, ��3, 2�, �2, 3�, �3, ��

x � 2, x � ±3

�x � 2��x � 3��x � 3� ≥ 0

�x � 2��x2 � 9� ≥ 0

x2�x � 2� � 9�x � 2� ≥ 0

x3 � 2x2 � 9x � 18 ≥ 0

x3 � 2x2 � 9x � 2 ≥ �20

25.

Critical number:

Test intervals:

Test: Is

Interval x-Value Value of Conclusion

Positive

Positive

Solution set:

210−1

x

−2

12

x �12

�1�2 � 1x � 1�12, ��

��1�2 � 1x � 0���, 12��2x � 1�2

�2x � 1�2 ≤ 0?

���, 12�, �12, ��

x �12

�2x � 1�2 ≤ 0

4x2 � 4x � 1 ≤ 0 27.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality,we see that the solution set is: ���, 0� � �0, 32�

2x2�2x � 3� < 0?

���, 0�, �0, 32�, �32, ��

x � 0, x �32

2x2�2x � 3� < 0

4x3 � 6x2 < 0

Page 54: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.8 Other Types of Inequalities 91

29.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality,we see that the solution set is: ��2, 0� � �2, ��

x�x � 2��x � 2� ≥ 0?

���, �2�, ��2, 0�, �0, 2�, �2, ��

x � 0, x � ±2

x�x � 2��x � 2� ≥ 0

x3 � 4x ≥ 0 31.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality, we see that the solution set is: ��2, ��

�x � 1�2�x � 3�3 ≥ 0?

���, �2�, ��2, 1�, �1, �)

x � 1, x � �2

�x � 1�2�x � 2�3 ≥ 0

33.

(a)

(b) y ≥ 3 when 0 ≤ x ≤ 2.

y ≤ 0 when x ≤ �1 or x ≥ 3.

−5 7

−2

6

y � �x2 � 2x � 3 35.

(a)

(b) y ≤ 6 when x ≤ 4.

y ≥ 0 when �2 ≤ x ≤ 0, 2 ≤ x < �.

−12 12

−8

8

y �18x3 �

12x

37.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality,we see that the solution set is:

−1−2 210

x

���, �1� � �0, 1�

1 � x2

x> 0?

���, �1�, ��1, 0�, �0, 1�, �1, ��

x � 0, x � ±1

1 � x2

x> 0

1

x� x > 0 39.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality,we see that the solution set is:

−2 −1 10 2 3 4 5

x

���, �1� � �4, ��

4 � x

x � 1< 0?

���, �1�, ��1, 4�, �4, ��

x � �1, x � 4

4 � x

x � 1< 0

x � 6 � 2�x � 1�x � 1

< 0

x � 6

x � 1� 2 < 0

41.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality, we see that the solution set is: �5, 15�

15 � x

x � 5> 0?

���, 5�, �5, 15�, �15, ��

x � 5, x � 153 6 9

5

12 15 18

x 15 � x

x � 5> 0

3x � 5 � 4�x � 5�x � 5

> 0

3x � 5

x � 5� 4 > 0

3x � 5

x � 5> 4

Page 55: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

92 Chapter 1 Equations, Inequalities, and Mathematical Modeling

43.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality, we see that the solution set is: ��5, �32� � ��1, ��

7�x � 1��x � 5��2x � 3�

> 0?

��3

2, �1�, ��1, ��

���, �5�, ��5, �3

2�,

2

−1−2−3−4−5

0

3

x

x � �1, x � �5, x � �3

2

7x � 7

�x � 5��2x � 3�> 0

4�2x � 3� � �x � 5��x � 5��2x � 3�

> 0

4

x � 5�

1

2x � 3> 0

4

x � 5>

1

2x � 3

45.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality, we see that the solution set is: ��34, 3� � �6, ��

30 � 5x

�x � 3��4x � 3�≤ 0?

−2−4

−33

4 6 820

4x

���, �3

4�, ��3

4, 3�, �3, 6�, �6, ��

x � 3, x � �3

4, x � 6

30 � 5x

�x � 3��4x � 3�≤ 0

4x � 3 � 9�x � 3��x � 3��4x � 3�

≤ 0

1

x � 3�

9

4x � 3≤ 0

1

x � 3≤

9

4x � 3

47.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality, we see that the solution set is:

−1−2−3 0 1 2 3

x

��3, �2� � �0, 3�

x�x � 2��x � 3��x � 3�

≤ 0?

���, �3�, ��3, �2�, ��2, 0�, �0, 3�, �3, ��

x � 0, x � �2, x � ±3

x�x � 2��x � 3��x � 3�

≤ 0

x2 � 2x

x2 � 9≤ 0

Page 56: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.8 Other Types of Inequalities 93

51.

(a)

(b) y ≥ 6 when 2 < x ≤ 4.

y ≤ 0 when 0 ≤ x < 2.

−6 12

8

−4

y �3x

x � 253.

(a) when or

This can also be expressed as

(b) for all real numbers

This can also be expressed as �� < x < �.

x.y ≤ 2

�x� ≥ 2.

x ≥ 2.x ≤ �2y ≥ 1

−6 6

−2

6

y �2x2

x2 � 4

49.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality, we see that the solution set is:

x

−1

43210

23

���, �1� � ��23, 1� � �3, ��

��3x � 2��x � 3��x � 1��x � 1� < 0?

���, �1�, ��1, �23�, ��2

3, 1�, �1, 3�, �3, ��

x � �23

, x � 3, x � ±1

��3x � 2��x � 3�

�x � 1��x � 1� < 0

�3x2 � 7x � 6�x � 1��x � 1� < 0

5x � 5 � 2x2 � 2x � x2 � 1

�x � 1��x � 1� < 0

5�x � 1� � 2x�x � 1� � �x � 1��x � 1�

�x � 1��x � 1� < 0

5

x � 1�

2xx � 1

� 1 < 0

5

x � 1�

2xx � 1

< 1

55.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality,we see that the domain set is: ��2, 2�

4 � x2 ≥ 0?

���, �2�, ��2, 2�, �2, ��

x � ±2

�2 � x��2 � x� ≥ 0

4 � x2 ≥ 0 57.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality,we see that the domain set is: ���, 3� � �4, ��

�x � 3��x � 4� ≥ 0?

���, 3�, �3, 4�, �4, ��

x � 3, x � 4

�x � 3��x � 4� ≥ 0

x2 � 7x � 12 ≥ 0

Page 57: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

94 Chapter 1 Equations, Inequalities, and Mathematical Modeling

63.

The zeros are

Critical numbers: ,

Test intervals:

By testing an x-value in each test interval in the inequality,we see that the solution set is: ��0.13, 25.13�

�25.13, ��

���, �0.13�, ��0.13, 25.13�,

x 25.13 x �0.13

x ��12.5 ± ��12.5�2 � 4��0.5��1.6�

2��0.5�.

�0.5x2 � 12.5x � 1.6 > 0 65.

Critical numbers:

Test intervals:

By testing an x-value in each test interval in the inequality,we see that the solution set is: �2.26, 2.39�

���, 2.26�, �2.26, 2.39�, �2.39, ��

x 2.39, x 2.26

�7.82x � 18.68

2.3x � 5.2> 0

1 � 3.4�2.3x � 5.2�

2.3x � 5.2> 0

1

2.3x � 5.2� 3.4 > 0

1

2.3x � 5.2> 3.4

67.

(a)

It will be back on the ground in 10 seconds.

(b)

4 < t < 6 seconds

�t � 4��t � 6� < 0

t2 � 10t � 24 < 0

�16�t2 � 10t � 24� > 0

�16t2 � 160t � 384 > 0

�16t2 � 160t > 384

t � 0, t � 10

�16t�t � 10� � 0

�16t2 � 160t � 0

s � �16t2 � v0t � s0 � �16t2 � 160t 69.

By the Quadratic Formula we have:

Critical numbers:

Test: Is

Solution set:

13.8 meters ≤ L ≤ 36.2 meters

25 � 5�5 ≤ L ≤ 25 � 5�5

�L2 � 50L � 500 ≥ 0?

L � 25 ± 5�5

�L2 � 50L � 500 ≥ 0

L�50 � L� ≥ 500

LW ≥ 500

2L � 2W � 100 ⇒ W � 50 � L

59.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequali-ty, we see that the domain set is: ��5, 0� � �7, ��

x�x � 5��x � 7� ≥ 0?

���, �5�, ��5, 0�, �0, 7�, �7, ��

x � 0, x � �5, x � 7

x

�x � 5��x � 7� ≥ 0

x

x2 � 2x � 35≥ 0 61.

Critical numbers:

Test intervals:

By testing an x-value in each test interval in the inequality,we see that the solution set is: ��3.51, 3.51�

���, �3.51�, ��3.51, 3.51�, �3.51, ��

x ±3.51

0.4�x2 � 12.35� < 0

0.4x2 � 4.94 < 0

0.4x2 � 5.26 < 10.2

Page 58: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Section 1.8 Other Types of Inequalities 95

71. and

Critical numbers: (These were obtained by using the Quadratic Formula.)

Test intervals:

By testing values in each test interval in the inequality, we see that the solution set is orThe price per unit is

For For Therefore, for 40,000 ≤ x ≤ 50,000, $50.00 ≤ p ≤ $55.00.x � 50,000, p � $50.x � 40,000, p � $55.

p �Rx

� 75 � 0.0005x.

40,000 ≤ x ≤ 50,000.�40,000, 50,000�x-

�50,000, ���0, 40,000�, �40,000, 50,000�,

x � 40,000, x � 50,000

�0.0005x2 � 45x � 1,000,000 ≥ 0

�0.0005x2 � 45x � 250,000 ≥ 750,000

P ≥ 750,000

� �0.0005x2 � 45x � 250,000

� �75x � 0.0005x2� � �30x � 250,000�

P � R � C

C � 30x � 250,000R � x�75 � 0.0005x�

73.

(a)

(b) will be greater than 75% whenwhich corresponds to 2011.

(c) when t 30.41.C � 75

t 31,C

0 230

80

C � 0.0031t3 � 0.216t 2 � 5.54t � 19.1, 0 ≤ t ≤ 23

(d)

will be between 85% and 100% when is between 37and 42. These values correspond to the years 2017 to2022.

(e) when or

(f ) The model is a third-degree polynomial and ast →�, C →�.

37 ≤ t ≤ 42.36.82 ≤ t ≤ 41.8985 ≤ C ≤ 100

tC

24 70.5

26 71.6

28 72.9

30 74.6

32 76.8

34 79.6

Ct

36 83.2

37 85.4

38 87.8

39 90.5

40 93.5

41 96.8

42 100.4

43 104.4

Ct

Page 59: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

96 Chapter 1 Equations, Inequalities, and Mathematical Modeling

75.

Since we have

Since the only critical number is Theinequality is satisfied when R1 ≥ 2 ohms.

R1 � 2.R1 > 0,

R1 � 2

2 � R1

≥ 0.

2R1

2 � R1

� 1 ≥ 0

2R1

2 � R1

≥ 1

R ≥ 1,

2R1

2 � R1

� R

2R1 � R�2 � R1�

2R1 � 2R � RR1

1

R�

1

R1

�1

277. True

The test intervals are and�4, ��.

���, �3�, ��3, 1�, �1, 4�,

x3 � 2x2 � 11x � 12 � �x � 3��x � 1��x � 4�

79.

To have at least one real solution, Thisoccurs when This can be written as���, �4� � �4, ��.

b ≤ �4 or b ≥ 4.b2 � 16 ≥ 0.

x2 � bx � 4 � 0 81.

To have at least one real solution,

Critical numbers:

Test intervals:

Test: Is

Solution set: ���, �2�30� � �2�30, ��b2 � 120 ≥ 0?

���, �2�30 �, ��2�30, 2�30 �, �2�30, ��

b � ±�120 � ±2�30

�b � �120 ��b � �120 � ≥ 0

b2 � 120 ≥ 0

b2 � 4�3��10� ≥ 0.

3x2 � bx � 10 � 0

83. (a) If a > 0 and c ≤ 0, then b can be any real number. Ifand then for to be greater than

or equal to zero, b is restricted to or

(b) The center of the interval for b in Exercises 79–82 is 0.

b > 2�ac.b < �2�ac

b2 � 4acc > 0,a > 085. 4x2 � 20x � 25 � �2x � 5�2

87.

� �x � 2��x � 2��x � 3�

x2�x � 3) � 4�x � 3� � �x2 � 4��x � 3� 89.

� 2x2 � x

� �2x � 1��x�

Area � �length��width�

Page 60: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Review Exercises for Chapter 1

Review Exercises for Chapter 1 97

1.

–3 –2 –1 1 2 3

–5

–4

–3

–2

–1

1

x

y

y � 3x � 5

x 0 1 2

y 1�2�5�8�11

�1�2

5.

Line with x-intercept and y-intercept �0, 3�

��32, 0�

y � 2x � 3

–5 –4 –3 –1 1 2 3

–2

1

3

4

5

6

x

y y � 2x � 3 � 0

3.

–3 –2 –1 1 2 54

–3

–2

4

5

x

y

y � x2 � 3x

x 0 1 2 3 4

y 4 0 0 4�2�2

�1

7.

Domain: ���, 5�

–2 –1 1 2 3 4 5 6

–2

1

3

4

5

6

x

yy � �5 � x

x 5 4 1

y 0 1 2 3

�4

x 0 1

y 1 3 5�1

�1�2

9.

is a parabola. y � �2x2

–3 –2 –1 1 2 3

–5

–4

–3

–2

1

x

y y � 2x2 � 0

x 0

y 0 �8�2

±2±1

11.

x-intercepts:

y-intercept:

The intercepts are and The intercept is �0, 5�.y-�5, 0�.�1, 0�x-

y � �0 � 3�2 � 4 ⇒ y � 9 � 4 ⇒ y � 5

⇒ x � 5 or x � 1

0 � �x � 3�2 � 4 ⇒ �x � 3�2 � 4 ⇒ x � 3 � ±2 ⇒ x � 3 ± 2

y � �x � 3�2 � 4

13.

Intercepts:

No y-axis symmetry

No x-axis symmetry

No origin symmetry�y � �4��x� � 1 ⇒ y � �4x � 1 ⇒

�y � �4x � 1 ⇒ y � 4x � 1 ⇒

y � �4��x� � 1 ⇒ y � 4x � 1 ⇒

�14, 0�, �0, 1�

x1−1−2−3−4

4

2−1

3

1

4

−2

−3

−4

yy � �4x � 1

Page 61: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

98 Chapter 1 Equations, Inequalities, and Mathematical Modeling

15.

Intercepts:

y-axis symmetry

No x-axis symmetry

No origin symmetry

x1

6

−13 42−1−3−4

3

−2

1

2

4

y

�y � 5 � ��x�2 ⇒ y � �5 � x2 ⇒

�y � 5 � x2 ⇒ y � �5 � x2 ⇒

y � 5 � ��x�2 ⇒ y � 5 � x2 ⇒

�±�5, 0�, �0, 5�

y � 5 � x2 17.

Intercepts:

No y-axis symmetry

No x-axis symmetry

No origin symmetry

x1 2 4

7

−13−2−3−4

1

2

4

5

6

y

�y � ��x�3 � 3 ⇒ y � x3 � 3 ⇒

�y � x3 � 3 ⇒ y � �x3 � 3 ⇒

y � ��x�3 � 3 ⇒ y � �x3 � 3 ⇒

�� 3�3, 0�, �0, 3�

y � x3 � 3

19.

Domain:

Intercepts:

No y-axis symmetry

No x-axis symmetry

No origin symmetry�y � ��x � 5 ⇒ y � ���x � 5 ⇒

�y � �x � 5 ⇒ y � ��x � 5 ⇒

y � ��x � 5 ⇒

��5, 0�, �0, �5���5, ��

x1

7

−6−1

−−5

1

3

4

5

6

−1−24 2−3

yy � �x � 5

21.

Center:

Radius: 3

�0, 0�

–4 –2 –1 1 2 4

–4

–2

–1

1

2

4

x(0, 0)

yx2 � y2 � 9 23.

Center:

Radius: 4

��2, 0�

�x � ��2��2 � �y � 0�2 � 42

–8 –4 –2 4

–6

–2

2

6

x(−2, 0)

y�x � 2�2 � y2 � 16

25.

Center:

Radius: 6

x2

8

8−2

−8

2

4

−4

−2−4−6 4

( (12, −1

y

�12, �1�

�x �12�

2� �y � ��1��2 � 62

�x �12�

2

� �y � 1�2 � 36 27. Endpoints of a diameter: and

Center:

Radius:

Standard form:

�x � 2�2 � �y � 3�2 � 13

�x � 2�2 � �y � ��3��2 � ��13�2

r � ��2 � 0�2 � ��3 � 0�2 � �4 � 9 � �13

�0 � 42

, 0 � ��6�

2 � � �2, �3�

�4, �6��0, 0�

Page 62: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Review Exercises for Chapter 1 99

29.

(a)

(c) When pounds.F �504 � 12.5x � 10,

0 ≤ x ≤ 20F �54x,

(b)

x

F

4 8 12 2416 20

5

10

15

20

25

Length (in inches)

Forc

e (i

n po

unds

)

30x 0 4 8 12 16 20

F 0 5 10 15 20 25

31.

All real numbers are solutions.

0 � 0 Identity

2 � 4x � x2 � 2 � 4x � x2

6 � �x2 � 4x � 4� � 2 � 4x � x2

6 � �x � 2�2 � 2 � 4x � x2 33.

All real numbers are solutions.

0 � 0 Identity

�x3 � x2 � 7x � 3 � �x3 � x2 � 7x � 3

�x3 � 7x � x2 � 3 � �x3 � x2 � 7x � 7 � 4

�x3 � x�7 � x� � 3 � x��x2 � x� � 7�x � 1� � 4

35.

x � 20

3x � 2x � 10 � 10

3x � 2�x � 5� � 10 37.

x � �1

2

10x � �5

4x � 9 � �6x � 4

4x � 12 � 3 � 8 � 6x � 4

4�x � 3� � 3 � 2�4 � 3x� � 4 39.

x � �5

�4x � 20

x � 15 � 5x � 5

5�x5

� 3� � �x � 1�5

x5

� 3 �2x2

� 1

41.

x � 9

8x � 72

18x � 72 � 10x

18�x � 4� � 10x

18x

�10

x � 443.

x-intercept:

y-intercept:

The x-intercept is and the y-intercept is �0, �1�.�13, 0�

y � 3�0� � 1 ⇒ y � �1

0 � 3x � 1 ⇒ x �13

y � 3x � 1

45.

x-intercept:

y-intercept:

The x-intercept is and the y-intercept is �0, �8�.�4, 0�

y � 2�0 � 4� ⇒ y � �8

0 � 2�x � 4� ⇒ x � 4

y � 2�x � 4� 47.

x-intercept:

y-intercept:

The x-intercept is and the y-intercept is �0, 23�.�43, 0�

y � �12

�0� �23

⇒ y �23

0 � �12

x �23

⇒ x �2�31�2

�43

y � �12

x �23

49.

x-intercept:

y-intercept:

The x-intercept is and the y-intercept is �0, � 519�.�2, 0�

3.8y � 0.5�0� � 1 � 0 ⇒ y ��13.8

� �5

19

3.8�0� � 0.5x � 1 � 0 ⇒ x �1

0.5� 2

3.8y � 0.5x � 1 � 0 51.

The height is 10 inches.

10 � h

188.40 � 18.84h

244.92 � 56.52 � 18.84h

244.92 � 2�3.14��3�2 � 2�3.14��3�h

Page 63: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

100 Chapter 1 Equations, Inequalities, and Mathematical Modeling

53. Verbal Model:

Labels: Let September’s profit. Then

Equation:

Answer: September profit: $325,000, October profit: $364,000

x � 0.12x � 364,000

x � 325,000

2.12x � 689,000

x � �x � 0.12x� � 689,000

x � 0.12x � October’s profit.x �

September’s profit � October’s profit � 689,000

55. Let height of the streetlight.

By similar triangles we have:

The streetlight is 24 feet tall.

x � 24

5x � 120

x

20�

65

x

15 ft5 ft

6 ft

x � 57. Let the number of original investors.

Each person’s share is

If three more people invest, each person’s share is

Since this is $2500 less than the original cost, we have:

There are currently nine investors.

x � �12, extraneous or x � 9

�2500�x � 12��x � 9� � 0

�2500�x2 � 3x � 108� � 0

�2500x2 � 7500x � 270,000 � 0

90,000x � 270,000 � 2500x2 � 7500x � 90,000x

90,000�x � 3� � 2500x�x � 3� � 90,000x

90,000

x� 2500 �

90,000x � 3

90,000x � 3

.

90,000x

.

x �

59. Let the number of liters of pure antifreeze.

x �2

0.70�

20

7� 2

6

7 liters

0.70x � 2

3 � 0.30x � 1.00x � 5

0.30�10 � x� � 1.00x � 0.50�10�

30% of �10 � x� � 100% of x � 50% of 10

x � 61.

3V

�r 2� h

3V � �r 2 h

V �1

3�r 2h

63.

t �23 hour or 40 minutes

15t � 10

55t � 40t � 10

rate time distance

1st car 40 mph t 40t

2nd car 55 mph t 55t

65.

x � 3 � 0 ⇒ x � 3

2x � 5 � 0 ⇒ x � �52

0 � �2x � 5��x � 3�

0 � 2x2 � x � 15

15 � x � 2x2 � 0

Page 64: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Review Exercises for Chapter 1 101

67.

±�2 � x

2 � x2

6 � 3x2 69.

x � �4 ± 3�2

x � 4 � ±�18

�x � 4�2 � 18 71.

x � 6 ± �6

x � 6 � ±�6

�x � 6�2 � 6

x2 � 12x � 36 � �30 � 36

x2 � 12x � �30

x2 � 12x � 30 � 0

73.

��5 ± �241

4

x ��5 ± �52 � 4�2���27�

2�2�

2x2 � 5x � 27 � 0

�2x2 � 5x � 27 � 0 75.

(a) when feet and feet.

(b)

(c) The bending moment is greatest when feet.x � 10

00

22

55,000

x � 20x � 0500x�20 � x� � 0

M � 500x�20 � x�

77. 6 � ��4 � 6 � 2i 79. i2 � 3i � �1 � 3i

81. �7 � 5i� � ��4 � 2i� � �7 � 4� � �5i � 2i� � 3 � 7i 83. 5i�13 � 8i� � 65i � 40i2 � 40 � 65i

85.

� �4 � 46i

�10 � 8i��2 � 3i� � 20 � 30i � 16i � 24i2 87.

�2317

�1017

i

�23 � 10i

17

�24 � 10i � i2

16 � 1

6 � i4 � i

�6 � i4 � i

�4 � i4 � i

89.

�2113

�1

13i

� � 813

� 1� � �1213

i � i�

�8

13�

1213

i � 1 � i

�8 � 12i4 � 9

�2 � 2i1 � 1

4

2 � 3i�

21 � i

�4

2 � 3i�

2 � 3i2 � 3i

�2

1 � i�

1 � i1 � i

91.

� ±�1

3 i � ±

�3

3i

x � ±��1

3

x2 � �1

3

3x2 � �1

3x2 � 1 � 0

93.

x � 1 ± 3i

x � 1 � ±��9

�x � 1�2 � �9

x2 � 2x � 1 � �10 � 1

x2 � 2x � 10 � 0 95.

x � 0 or 5 � x �12

5

x3 � 0 or 5x � 12 � 0

x3�5x � 12� � 0

5x4 � 12x3 � 0

Page 65: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

102 Chapter 1 Equations, Inequalities, and Mathematical Modeling

97.

or

x � ±�3 x � ±�2

x2 � 3 x2 � 2

x2 � 3 � 0 x2 � 2 � 0

�x2 � 2��x2 � 3� � 0

x4 � 5x2 � 6 � 0 99.

x � 5

x � 4 � 9

��x � 4 �2� �3�2

�x � 4 � 3

101.

No solution

x � 3, extraneous or x � 11, extraneous

�x � 3��x � 11� � 0

x2 � 14x � 33 � 0

x2 � 2x � 1 � 16�x � 2�

�x � 1�2 � ��4�x � 2 �2

x � 1 � �4�x � 2

2x � 3 � 4 � 4�x � 2 � x � 2

��2x � 3 �2� �2 � �x � 2 �2

�2x � 3 � �x � 2 � 2 103.

x � 126 or x � �124

x � 1 ± 125

x � 1 � ±�253

�x � 1�2 � 253

�x � 1�2�3 � 25

�x � 1�2�3 � 25 � 0

105.

or

x � �4

5x2 � 20x � 1 � 0 �x � 4�1�2 � 0

�x � 4�1�2�5x2 � 20x � 1� � 0

�x � 4�1�2�1 � 5x�x � 4�� � 0

�x � 4�1�2 � 5x�x � 4�3�2 � 0

x � �2 ±�95

5

x ��20 ± 2�95

10

x ��20 ± �400 � 20

10

x ��20 ± �400 � 20

10

107.

±�10 � x

10 � x2

5x � 10 � x2 � 2x � 3x

5�x � 2� � 1�x��x � 2� � 3x

5x

� 1 �3

x � 2109.

x �43

9x � 12

15x � 5x � 10 � x � 2

3�5x� � 5�x � 2� � x � 2

3

x � 2�

1x

�15x

111.

x � �51 or 5 � x � 15

x � 5 � �10 or x � 5 � 10

x � 5 � 10

Page 66: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Review Exercises for Chapter 1 103

113.

or

The only solutions to the original equation are or �x � �3 and x � �1 are extraneous.�x � 1.

x � 3

x � �3 or x � 1 x � 3 or x � �1

�x � 3��x � 1� � 0 �x � 3��x � 1� � 0

x2 � 2x � 3 � 0 x2 � 2x � 3 � 0

x2 � 3 � �2x x2 � 3 � 2x

x2 � 3 � 2x 115.

143,203 units

x � 143,202.5

0.001x � 143.2025

0.001x � 2 � 145.2025

�0.001x � 2 � 12.05

�12.05 � ��0.001x � 2

29.95 � 42 � �0.001x � 2

117. Interval:

Inequality:

The interval is bounded.

�7 < x ≤ 2

��7, 2� 119. Interval:

Inequality:

The interval is unbounded.

x ≤ �10

���, �10� 121.

���, 12�

x ≤ 12

2x ≤ 24

9x � 8 ≤ 7x � 16

123.

�3215, ��

x ≥ 3215

�152 x ≤ �16

20 � 8x ≤ 4 �12x

4�5 � 2x� ≤ 12�8 � x� 125.

��23, 17�

�23 < x ≤ 17

�2 < 3x ≤ 51

�19 < 3x � 17 ≤ 34 127.

��4, 4�

�4 ≤ x ≤ 4

x ≤ 4

129.

or

or

���, �1� � �7, ��

x > 7 x < �1

x � 3 > 4 x � 3 < �4

x � 3 > 4 131. If the side is 19.3 cm, then with the possible error of 0.5 cm we have:

353.44 cm2 ≤ area ≤ 392.04 cm2

18.8 ≤ side ≤ 19.8

133.

Critical numbers:

Test intervals:

Test: Is ?

By testing an x-value in each test interval in the inequality, we see that the solution set is: ��3, 9�

�x � 3��x � 9� < 0

���, �3�, ��3, 9�, �9, ��

x � �3, x � 9

�x � 3��x � 9� < 0

x2 � 6x � 27 < 0 135.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality, we see that the solution set is: ��4

3, 12�

�3x � 4��2x � 1� < 0?

���, �43, �, ��4

3, 12�, �12, ��

x � �43, x �

12

�3x � 4��2x � 1� < 0

6x2 � 5x � 4 < 0

6x2 � 5x < 4

137.

Critical numbers:

Test intervals:

Test: Is

By testing an value in each test interval in the inequality,we see that the solution set is: ��4, 0� � �4, ��

x-

x�x � 4��x � 4� ≥ 0?

�4, ���0, 4�,��4, 0�,���, �4�,

x � 0, x � ±4

x�x � 4��x � 4� ≥ 0

x3 � 16x ≥ 0 139.

Critical numbers:

Test intervals:

By testing an value in each test interval in the inequality,we see that the solution set is: ���, �5� � ��2, ��

x-

��2, ����5, �2�,���, �5�,

x � �2, x � �5

�x � 2x � 5

< 0

x � 8 � 2�x � 5�

x � 5< 0

x � 8x � 5

� 2 < 0

Page 67: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

104 Chapter 1 Equations, Inequalities, and Mathematical Modeling

143.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality,we see that the solution set is: ��4, �3� � �0, ��

�x � 4��x � 3�x

≥ 0?

���, �4�, ��4, �3�, ��3, 0�, �0, ��

x � �4, x � �3, x � 0

�x � 4��x � 3�x

≥ 0

x2 � 7x � 12

x≥ 0

145.

r > 4.9%

r > 0.0488

1 � r > 1.0488

�1 � r�2 > 1.1

5000�1 � r�2 > 5500 147. False

���18���2� � �36 � 6

��18��2 � ��18i���2i� � �36i2 � �6

149. Rational equations, equations involving radicals, and absolute value equations, may have “solutions” that are extraneous. So checking solutions, in the original equations, is crucial to eliminate these extraneous values.

Problem Solving for Chapter 1

1.

Time (in seconds)

Dis

tanc

e (i

n fe

et)

y

x

141.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality,we see that the solution set is: ��5, �1� � �1, ��

��x � 5��x � 1��x � 1� ≤ 0?

���, �5�, ��5, �1�, ��1, 1�, �1, ��

x � �5, x � ±1

��x � 5�

�x � 1��x � 1�≤ 0

2x � 2 � 3x � 3

�x � 1��x � 1)≤ 0

2�x � 1� � 3�x � 1�

�x � 1��x � 1�≤ 0

2

x � 1≤

3

x � 1

3. (a)

thus:

—CONTINUED—

A � �a�20 � a�

a � b � 20 ⇒ b � 20 � a,

A � �ab (b)a 4 7 10 13 16

A 64 91 100 91 64�����

Page 68: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Problem Solving for Chapter 1 105

3. —CONTINUED—

(c)

or a � 7.877a � 12.123

� 10 ±10����� � 3�

�20� ± 20���� � 3�

2�

�20� ± �400�2 � 1200�

2�

a �20� ± ���20��2 � 4��300�

2�

�a2 � 20�a � 300 � 0

300 � 20�a � �a2

300 � �a�20 � a� (d)

(e) The a-intercepts occur at and Both yield anarea of 0. When and you have a verticalline of length 40. Likewise when and youhave a horizontal line of length 40. They represent theminimum and maximum values of a.

(f) The maximum value of A is This occurswhen and the ellipse is actually a circle.a � b � 10

100� � 314.159.

a � 20, b � 0a � 0, b � 20

a � 20.a � 0

00 20

350

A � na�20 � a�

5.

(a)

(b)

seconds

(c) The speed at which the water drains decreases as theamount of the water in the bathtub decreases.

t �225�12.5

��3� 146.2

0 � ��12.5 ���3225

t�2

t �225

��3�5 � �12.5 � � 60.6 seconds

�12.5 � 5 ���3225

t

12.5 � �5 ���3225

t�2

h � �5 �8��31800

t�2

� �5 ���3225

t�2

l � 60�, w � 30�, h0 � 25�, d � 2�

h � ��h0 �2�d 2�3

lw t�2

7. (a) 5, 12, and 13; 8, 15, and 17

7, 24, and 25

(b) which is divisible by 3, 4, and 5

which is divisible by 3, 4, and 5

which is also divisible by 3, 4,and 5

(c) Conjecture: If where a, b, and c arepositive integers, then is divisible by 60.abc

a2 � b2 � c2

7 � 24 � 25 � 4200

8 � 15 � 17 � 2040

5 � 12 � 13 � 780

9.

x1 � x2 �ca

x1 � x2 � �ba

x2 �ba

x �ca

� 0

ax2 � bx � c � 0

Page 69: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

106 Chapter 1 Equations, Inequalities, and Mathematical Modeling

13. (a)

The terms are:

The sequence is bounded so is in the Mandelbrot Set.

(b)

The terms are:

The sequence is unbounded so is not in the Mandelbrot Set.

(c)

The terms are:

The sequence is bounded so is in the Mandelbrot Set.c � �2

�2, 2, 2, 2, 2, . . .

c � �2

c � 1 � i

1 � i, 1 � 3i, �7 � 7i, 1 � 97i, �9407 � 193i, . . .

c � 1 � i

c � i

i, �1 � i, �i, �1 � i, �i, �1 � i, �i,�1 � i, �i, . . .

c � i

15.

From the graph we see that on the intervals ���, �2� � ��1, 2� � �2, ��.x4 � x3 � 6x2 � 4x � 8 > 0

−5

−5 5

10

� �x � 2�2�x � 1��x � 2�

y � x4 � x3 � 6x2 � 4x � 8

11. (a)

�1 � i

2�

12

�12

i

�1

1 � i�

11 � i

�1 � i1 � i

zm �1z

(b)

�3 � i

10�

310

�1

10i

�1

3 � i�

13 � i

�3 � i3 � i

zm �1z

(c)

��2 � 8i

68� �

134

�2

17i

�1

�2 � 8i�

�2 � 8i�2 � 8i

�1

�2 � 8i

zm �1z

Page 70: CHAPTER 1 Equations, Inequalities, and Mathematical Modeling · 4 x 1 2x 4x 4 2x 2x 4 2 x 2 17. Thus, is an identity by simplification. It is true for all real values of x. x2 8x

Practice Test for Chapter 1 107

Practice Test for Chapter 1

1. Graph 3x � 5y � 15. 2. Graph y � �9 � x.

3. Solve 5x � 4 � 7x � 8. 4. Solve x

3� 5 �

x

5� 1.

5. Solve 3x � 1

6x � 7�

2

5. 6. Solve �x � 3�2 � 4 � �x � 1�2.

7. Solve A �12�a � b�h for a. 8. 301 is what percent of 4300?

9. Cindy has in quarters and nickels. How many of each coin does she have if there are 53 coins in all?$6.05

10. Ed has $15,000 invested in two funds paying and 11% simple interest, respectively. How much is invested in each if the yearly interest is $1582.50?

912%

11. Solve by factoring.28 � 5x � 3x2 � 0

12. Solve by taking the square root of both sides.�x � 2�2 � 24

13. Solve by completing the square.x2 � 4x � 9 � 0

14. Solve by the Quadratic Formula.x2 � 5x � 1 � 0

15. Solve by the Quadratic Formula.3x2 � 2x � 4 � 0

16. The perimeter of a rectangle is 1100 feet. Find the dimensions so that the enclosed area will be 60,000 square feet.

17. Find two consecutive even positive integers whose product is 624.

18. Solve by factoring.x3 � 10x2 � 24x � 0 19. Solve 3�6 � x � 4.

20. Solve �x2 � 8�2�5 � 4. 21. Solve x4 � x2 � 12 � 0.

22. Solve 4 � 3x > 16. 23. Solve �x � 3

2 � < 5.

24. Solve x � 1

x � 3< 2. 25. Solve �3x � 4� ≥ 9.