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1 Solving Quadratic Equations: Factoring and Special Forms

# 1 Solving Quadratic Equations: Factoring and Special Forms

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Section 10.1

2

᭤ Solve quadratic equations by the

Square Root Property.

Solving Quadratic Equations: Factoring and Special Forms

619

The Square Root Property

Consider the following equation, where d > 0 and u is an algebraic expression.

u2 ϭ d

Original equation

u2 Ϫ d ϭ 0

Write in general form.

͑u ϩ Ίd ͒͑u Ϫ Ίd ͒ ϭ 0

Technology: Tip

To check graphically the solutions

of an equation written in general

form, graph the left side of the

equation and locate its x-intercepts.

For instance, in Example 2(b), write

the equation as

͑x Ϫ 2͒2 Ϫ 10 ϭ 0

and then use a graphing calculator

to graph

Factor.

u ϩ Ίd ϭ 0

u ϭ Ϫ Ίd

Set 1st factor equal to 0.

u Ϫ Ίd ϭ 0

u ϭ Ίd

Set 2nd factor equal to 0.

Because the solutions differ only in sign, they can be written together using a

“plus or minus sign”: u ϭ ± Ίd. This form of the solution is read as “u is equal

to plus or minus the square root of d.” Now you can use the Square Root

Property to solve an equation of the form u2 ϭ d without going through the steps

of factoring.

Square Root Property

The equation u2 ϭ d, where d > 0, has exactly two solutions:

u ϭ Ίd and u ϭ Ϫ Ίd.

These solutions can also be written as u ϭ ± Ίd. This solution process is

also called extracting square roots.

y ϭ ͑x Ϫ 2͒2 Ϫ 10

as shown below. You can use the

zoom and trace features or the

zero or root feature to approximate

the x-intercepts of the graph to be

x Ϸ 5.16 and x Ϸ Ϫ1.16.

5

Square Root Property

EXAMPLE 2

a. 3x2 ϭ 15

Original equation

x2 ϭ 5

Divide each side by 3.

x ϭ ± Ί5

−5

8

Square Root Property

The solutions are x ϭ Ί5 and x ϭ Ϫ Ί5. Check these in the original equation.

b. ͑x Ϫ 2͒2 ϭ 10

−10

Original equation

x Ϫ 2 ϭ ± Ί10

Square Root Property

x ϭ 2 ± Ί10

The solutions are x ϭ 2 ϩ Ί10 Ϸ 5.16 and x ϭ 2 Ϫ Ί10 Ϸ Ϫ1.16.

c. ͑3x Ϫ 6͒2 Ϫ 8 ϭ 0

Original equation

͑3x Ϫ 6͒2 ϭ 8

3x Ϫ 6 ϭ ± 22Ί2

3x ϭ 6 ± 2Ί2

xϭ2 ±

2Ί2

3

Square Root Property and

rewrite Ί8 as 2Ί2.

Divide each side by 3.

The solutions are x ϭ 2 ϩ 2Ί2͞3 Ϸ 2.94 and x ϭ 2 Ϫ 2Ί2͞3 Ϸ 1.06.

CHECKPOINT Now try Exercise 23.

620

Chapter 10

3 ᭤ Solve quadratic equations with

complex solutions by the Square Root

Property.

Technology: Discovery

below algebraically. Then use

a graphing calculator to check the

solutions. Which equations have

real solutions and which have

complex solutions? Which graphs

have x-intercepts and which have

no x-intercepts? Compare the

type(s) of solution(s) of each

x-intercept(s) of the graph of

the equation.

a. y ϭ 2x 2 ϩ 3x Ϫ 5

b. y ϭ 2x 2 ϩ 4x ϩ 2

Prior to Section 9.6, the only solutions you could find were real numbers. But

now that you have studied complex numbers, it makes sense to look for other

types of solutions. For instance, although the quadratic equation x2 ϩ 1 ϭ 0 has

no solutions that are real numbers, it does have two solutions that are complex

numbers: i and Ϫi. To check this, substitute i and Ϫi for x.

͑i ͒2 ϩ 1 ϭ Ϫ1 ϩ 1 ϭ 0

Solution checks.

͑Ϫi ͒2 ϩ 1 ϭ Ϫ1 ϩ 1 ϭ 0

Solution checks.

One way to find complex solutions of a quadratic equation is to extend the Square

Root Property to cover the case in which d is a negative number.

Square Root Property (Complex Square Root)

The equation u2 ϭ d, where d < 0, has exactly two solutions:

ԽԽ

u ϭ Ί d i and

ԽԽ

u ϭ Ϫ Ί d i.

d. y ϭ ͑x ϩ 7͒ ϩ 2

2

ԽԽ

These solutions can also be written as u ϭ ± Ί d i.

Square Root Property

EXAMPLE 3

a. x2 ϩ 8 ϭ 0

Original equation

x2 ϭ Ϫ8

c. y ϭ x 2 ϩ 4

Subtract 8 from each side.

x ϭ ± Ί8i ϭ ± 2Ί2i

Square Root Property

The solutions are x ϭ 2Ί2i and x ϭ Ϫ2Ί2i. Check these in the original

equation.

b. ͑x Ϫ 4͒2 ϭ Ϫ3

Original equation

x Ϫ 4 ϭ ± Ί3i

Square Root Property

x ϭ 4 ± Ί3i

The solutions are x ϭ 4 ϩΊ3i and x ϭ 4 Ϫ Ί3i. Check these in the original

equation.

c. 2͑3x Ϫ 5͒2 ϩ 32 ϭ 0

2͑3x Ϫ 5͒2 ϭ Ϫ32

͑3x Ϫ 5͒ ϭ Ϫ16

Original equation

Subtract 32 from each side.

2

Divide each side by 2.

3x Ϫ 5 ϭ ± 4i

Square Root Property

3x ϭ 5 ± 4i

5 4

± i

3 3

Divide each side by 3.

The solutions are x ϭ 5͞3 ϩ 4͞3i and x ϭ 5͞3 Ϫ 4͞3i. Check these in the

original equation.

CHECKPOINT Now try Exercise 45.

Section 10.1

4

᭤ Use substitution to solve equations

Solving Quadratic Equations: Factoring and Special Forms

621

Both the factoring method and the Square Root Property can be applied to

nonquadratic equations that are of quadratic form. An equation is said to be of

quadratic form if it has the form

au2 ϩ bu ϩ c ϭ 0

where u is an algebraic expression. Here are some examples.

Equation

x4 ϩ 5x2 ϩ 4 ϭ 0

͑x2͒2 ϩ 5͑x2͒ ϩ 4 ϭ 0

x Ϫ 5Ίx ϩ 6 ϭ 0

2x2͞3 ϩ 5x1͞3 Ϫ 3 ϭ 0

͑Ίx ͒2 Ϫ 5͑Ίx ͒ ϩ 6 ϭ 0

2

2͑x1͞3 ͒ ϩ 5͑x1͞3 ͒ Ϫ 3 ϭ 0

18 ϩ 2x2 ϩ ͑x2 ϩ 9͒2 ϭ 8

͑x2 ϩ 9͒2 ϩ 2͑x2 ϩ 9͒ Ϫ 8 ϭ 0

To solve an equation of quadratic form, it helps to make a substitution and rewrite

the equation in terms of u, as demonstrated in Examples 4 and 5.

Technology: Tip

EXAMPLE 4

You may find it helpful to graph the

equation with a graphing calculator

before you begin. The graph will

indicate the number of real

solutions an equation has. For

instance, shown below is the graph

of the equation in Example 4. You

can see from the graph that there

are four x-intercepts and so there

are four real solutions.

4

−6

Solve x 4 Ϫ 13x2 ϩ 36 ϭ 0.

Solution

Begin by writing the original equation in quadratic form, as follows.

x4 Ϫ 13x2 ϩ 36 ϭ 0

Write original equation.

͑x2͒2 Ϫ 13͑x2͒ ϩ 36 ϭ 0

Next, let u ϭ x2 and substitute u into the equation written in quadratic form.

Then, factor and solve the equation.

u2 Ϫ 13u ϩ 36 ϭ 0

Substitute u for x2.

͑u Ϫ 4͒͑u Ϫ 9͒ ϭ 0

Factor.

6

−4

Solving an Equation of Quadratic Form

uϪ4ϭ0

uϭ4

Set 1st factor equal to 0.

uϪ9ϭ0

uϭ9

Set 2nd factor equal to 0.

At this point you have found the “u-solutions.” To find the “x-solutions,” replace

u with x2 and solve for x.

uϭ4

x2 ϭ 4

x ϭ ±2

uϭ9

x2 ϭ 9

x ϭ ±3

The solutions are x ϭ 2, x ϭ Ϫ2, x ϭ 3, and x ϭ Ϫ3. Check these in the

original equation.

CHECKPOINT Now try Exercise 101.

Be sure you see in Example 4 that the u-solutions of 4 and 9 represent only

a temporary step. They are not solutions of the original equation and cannot be

substituted into the original equation.

622

Chapter 10

Study Tip

When solving equations involving

square roots, be sure to check for

extraneous solutions.

EXAMPLE 5

a. x Ϫ 5Ίx ϩ 6 ϭ 0

Original equation

This equation is of quadratic form with u ϭ Ίx.

͑Ίx͒2 Ϫ 5͑Ίx͒ ϩ 6 ϭ 0

u2

Ϫ 5u ϩ 6 ϭ 0

Substitute u for Ίx.

͑u Ϫ 2͒͑u Ϫ 3͒ ϭ 0

Factor.

uϪ2ϭ0

uϭ2

Set 1st factor equal to 0.

uϪ3ϭ0

uϭ3

Set 2nd factor equal to 0.

Now, using the u-solutions of 2 and 3, you obtain the x-solutions as follows.

2͞3

b. x

uϭ2

Ίx ϭ 2

xϭ4

uϭ3

Ίx ϭ 3

xϭ9

Ϫ

x1͞3

Ϫ6ϭ0

Original equation

This equation is of quadratic form with u ϭ

x1͞3.

͑x1͞3͒2 Ϫ ͑x1͞3͒ Ϫ 6 ϭ 0

u2 Ϫ u Ϫ 6 ϭ 0

Substitute u for x1͞3.

͑u ϩ 2͒͑u Ϫ 3͒ ϭ 0

Factor.

uϩ2ϭ0

u ϭ Ϫ2

Set 1st factor equal to 0.

uϪ3ϭ0

uϭ3

Set 2nd factor equal to 0.

Now, using the u-solutions of Ϫ2 and 3, you obtain the x-solutions as follows.

u ϭ Ϫ2

x1͞3 ϭ Ϫ2

uϭ3

x1͞3

ϭ3

x ϭ Ϫ8

x ϭ 27

CHECKPOINT Now try Exercise 107.

Surface Area of a Softball

EXAMPLE 6

The surface area of a sphere of radius r is given by S ϭ 4␲r 2. The surface area of

a softball is 144͞␲ square inches. Find the diameter d of the softball.

Solution

144

ϭ 4␲r2

36

ϭ r2

␲2

Substitute 144͞␲ for S.

±

Ί␲36 ϭ r

2

Divide each side by 4␲ and

use Square Root Property.

Choose the positive root to obtain r ϭ 6͞␲. The diameter of the softball is

d ϭ 2r ϭ 2

΂␲6 ΃ ϭ 12␲ Ϸ 3.82 inches.

CHECKPOINT Now try Exercise 131.

Section 10.1

Solving Quadratic Equations: Factoring and Special Forms

623

Concept Check

1. Explain the Zero-Factor Property and how it can be

used to solve a quadratic equation.

3. Does the equation 4x 2 ϩ 9 ϭ 0 have two real

solutions or two complex solutions? Explain your

reasoning.

2. Determine whether the following statement is true

The only solution of the equation x 2 ϭ 25 is x ϭ 5.

4. Is the equation x 6 Ϫ 6x 3 ϩ 9 ϭ 0 of quadratic

Go to pages 676–677 to

10.1 EXERCISES

Developing Skills

In Exercises 1–20, solve the equation by factoring. See

Example 1.

1. x2 Ϫ 15x ϩ 54 ϭ 0

2. x2 ϩ 15x ϩ 44 ϭ 0

27.

29. 4x2 Ϫ 25 ϭ 0

31.

3. x 2 Ϫ x Ϫ 30 ϭ 0

4. x 2 Ϫ 2x Ϫ 48 ϭ 0

5. x2 ϩ 4x ϭ 45

6. x2 Ϫ 7x ϭ 18

7.

8.

9.

10.

x2 Ϫ 16x ϩ 64 ϭ 0

x2 ϩ 60x ϩ 900 ϭ 0

9x2 Ϫ 10x Ϫ 16 ϭ 0

8x2 Ϫ 10x ϩ 3 ϭ 0

11.

4x2

Ϫ 12x ϭ 0

12.

25y2

Ϫ 75y ϭ 0

13. u͑u Ϫ 9͒ Ϫ 12͑u Ϫ 9͒ ϭ 0

14. 16x͑x Ϫ 8͒ Ϫ 12͑x Ϫ 8͒ ϭ 0

w2

ϭ 49

4

4u2

Ϫ 225 ϭ 0

28.

x2

ϭ 24

6

30. 16y2 Ϫ 121 ϭ 0

32. 16x2 Ϫ 1 ϭ 0

͑x ϩ 4͒2 ϭ 64

͑m Ϫ 12͒2 ϭ 400

͑x Ϫ 3͒2 ϭ 0.25

͑x ϩ 2͒2 ϭ 0.81

37. ͑x Ϫ 2͒2 ϭ 7

38. ͑y ϩ 4͒2 ϭ 27

39. ͑2x ϩ 1͒2 ϭ 50

40. ͑3x Ϫ 5͒2 ϭ 48

33.

34.

35.

36.

41. ͑9m Ϫ 2͒2 Ϫ 108 ϭ 0 42. ͑5x ϩ 11͒2 Ϫ 300 ϭ 0

15. 2x͑x Ϫ 5͒ ϩ 9͑x Ϫ 5͒ ϭ 0

16. 3͑4 Ϫ x͒ Ϫ 2x͑4 Ϫ x͒ ϭ 0

17. ͑ y Ϫ 4͒͑ y Ϫ 3͒ ϭ 6

18. ͑5 ϩ u͒͑2 ϩ u͒ ϭ 4

19. 2x͑3x ϩ 2͒ ϭ 5 Ϫ 6x2

20. ͑2z ϩ 1͒͑2z Ϫ 1͒ ϭ Ϫ4z2 Ϫ 5z ϩ 2

In Exercises 43– 64, solve the equation by using the

Square Root Property. See Example 3.

43. z2 ϭ Ϫ36

44. x2 ϭ Ϫ16

45. x2 ϩ 4 ϭ 0

46. p2 ϩ 9 ϭ 0

47. 9u2 ϩ 17 ϭ 0

48. 25x2 ϩ 4 ϭ 0

In Exercises 21– 42, solve the equation by using the

Square Root Property. See Example 2.

21. x2 ϭ 49

22. p2 ϭ 169

23. 6x2 ϭ 54

24. 5t2 ϭ 5

25. 25x ϭ 16

26. 9z2 ϭ 121

2

49.

50.

51.

52.

͑t Ϫ 3͒2 ϭ Ϫ25

͑x ϩ 5͒2 ϭ Ϫ81

͑3z ϩ 4͒2 ϩ 144 ϭ 0

͑2y Ϫ 3͒2 ϩ 25 ϭ 0

624

Chapter 10

53. ͑4m ϩ 1͒2 ϭ Ϫ80

55. 36͑t ϩ 3͒2 ϭ Ϫ100

54. ͑6y Ϫ 5͒2 ϭ Ϫ8

56. 4͑x Ϫ 4͒2 ϭ Ϫ169

57. ͑x Ϫ 1͒2 ϭ Ϫ27

58. ͑2x ϩ 3͒2 ϭ Ϫ54

59. ͑x ϩ 1͒2 ϩ 0.04 ϭ 0

60. ͑y Ϫ 5͒2 ϩ 6.25 ϭ 0

61. ͑c Ϫ 23 ͒ ϩ 19 ϭ 0

62. ͑u ϩ 58 ͒ ϩ 49

16 ϭ 0

63. ͑x ϩ 73 ͒ ϭ Ϫ 38

9

64. ͑y Ϫ 58 ͒ ϭ Ϫ 54

2

2

2

2

In Exercises 65– 80, find all real and complex solutions

65. 2x2 Ϫ 5x ϭ 0

66. 4t2 ϩ 20t ϭ 0

67. 2x2 ϩ 5x Ϫ 12 ϭ 0

68. 3x2 ϩ 8x Ϫ 16 ϭ 0

69. x2 Ϫ 900 ϭ 0

70. z2 Ϫ 256 ϭ 0

71. x2 ϩ 900 ϭ 0

72. z2 ϩ 256 ϭ 0

73. 23x2 ϭ 6

74. 13x2 ϭ 4

84.

85.

86.

87.

88.

89.

90.

y ϭ x2 ϩ 3x Ϫ 40

y ϭ 4 Ϫ ͑x Ϫ 3͒2

y ϭ 4͑x ϩ 1͒2 Ϫ 9

y ϭ 2x2 Ϫ x Ϫ 6

y ϭ 4x2 Ϫ x Ϫ 14

y ϭ 3x2 Ϫ 13x Ϫ 10

y ϭ 5x2 ϩ 9x Ϫ 18

In Exercises 91–96,use a graphing calculator to graph

the function and observe that the graph has

no x-intercepts. Set y ϭ 0 and solve the resulting

equation. Of what type are the solutions of the equation?

91. y ϭ x2 ϩ 7

92. y ϭ x2 ϩ 5

93.

94.

95.

96.

y ϭ ͑x

y ϭ ͑x

y ϭ ͑x

y ϭ ͑x

Ϫ 4͒2 ϩ 2

ϩ 2͒2 ϩ 3

ϩ 3͒2 ϩ 5

Ϫ 2͒2 ϩ 3

In Exercises 97–100, solve for y in terms of x. Let f and

g be functions representing, respectively, the positive

square root and the negative square root.Use a graphing

calculator to graph f and g in the same viewing window.

97. x2 ϩ y2 ϭ 4

99. x2 ϩ 4y2 ϭ 4

75. ͑p Ϫ 2͒2 Ϫ 108 ϭ 0

76. ͑ y ϩ 12͒2 Ϫ 400 ϭ 0

77. ͑p Ϫ 2͒2 ϩ 108 ϭ 0

78. ͑ y ϩ 12͒2 ϩ 400 ϭ 0

79. ͑x ϩ 2͒2 ϩ 18 ϭ 0

80. ͑x ϩ 2͒2 Ϫ 18 ϭ 0

In Exercises 81–90, use a graphing calculator to

graph the function. Use the graph to approximate

any x-intercepts. Set y ϭ 0 and solve the resulting

equation. Compare the result with the x-intercepts of

the graph.

81. y ϭ x2 Ϫ 9

82. y ϭ 5x Ϫ x2

83. y ϭ x2 Ϫ 2x Ϫ 15

98. x2 Ϫ y2 ϭ 4

100. x Ϫ y2 ϭ 0

In Exercises 101–130, solve the equation of quadratic

form. (Find all real and complex solutions.) See

Examples 4 and 5.

101. x 4 Ϫ 5x2 ϩ 4 ϭ 0

102. x 4 Ϫ 10x2 ϩ 25 ϭ 0

103. x 4 Ϫ 5x2 ϩ 6 ϭ 0

104. x 4 Ϫ 10x2 ϩ 21 ϭ 0

105. ͑x2 Ϫ 4͒2 ϩ 2͑x2 Ϫ 4͒ Ϫ 3 ϭ 0

106. ͑x2 Ϫ 1͒2 ϩ ͑x2 Ϫ 1͒ Ϫ 6 ϭ 0

107. x Ϫ 3Ίx Ϫ 4 ϭ 0

Section 10.1

Solving Quadratic Equations: Factoring and Special Forms

625

121. x1͞2 Ϫ 3x1͞4 ϩ 2 ϭ 0

122. x1͞2 Ϫ 5x1͞4 ϩ 6 ϭ 0

3

1

123. 2 Ϫ ϩ 2 ϭ 0

x

x

1

1

124. 2 Ϫ Ϫ 6 ϭ 0

x

x

108. x Ϫ Ίx Ϫ 6 ϭ 0

109. x Ϫ 7Ίx ϩ 10 ϭ 0

110. x Ϫ 11Ίx ϩ 24 ϭ 0

111. x2͞3 Ϫ x1͞3 Ϫ 6 ϭ 0

112. x2͞3 ϩ 3x1͞3 Ϫ 10 ϭ 0

113. 2x2͞3 Ϫ 7x1͞3 ϩ 5 ϭ 0

125. 4xϪ2 Ϫ xϪ1 Ϫ 5 ϭ 0

114. 5x2͞3 Ϫ 13x1͞3 ϩ 6 ϭ 0

115. x2͞5 Ϫ 3x1͞5 ϩ 2 ϭ 0

116. x2͞5 ϩ 5x1͞5 ϩ 6 ϭ 0

126. 2xϪ2 Ϫ xϪ1 Ϫ 1 ϭ 0

127. ͑x 2 Ϫ 3x͒2 Ϫ 2͑x 2 Ϫ 3x͒ Ϫ 8 ϭ 0

128. ͑x 2 Ϫ 6x͒2 Ϫ 2͑x 2 Ϫ 6x͒ Ϫ 35 ϭ 0

117. 2x2͞5 Ϫ 7x1͞5 ϩ 3 ϭ 0

΂xx ϪϪ 18΃ ϩ 8΂xx ϪϪ 18΃ ϩ 1 ϭ 0

xϩ2

xϩ2

130. 9΂

Ϫ 6΂

ϩ1ϭ0

x ϩ 3΃

x ϩ 3΃

2

118. 2x2͞5 ϩ 3x1͞5 ϩ 1 ϭ 0

129. 16

119. x1͞3 Ϫ x1͞6 Ϫ 6 ϭ 0

120. x1͞3 ϩ 2x1͞6 Ϫ 3 ϭ 0

2

Solving Problems

131. Unisphere The Unisphere is the world’s largest

man-made globe. It was built as the symbol of the

1964–1965 New York World’s Fair. A sphere with

the same diameter as the Unisphere globe would

have a surface area of 45,239 square feet. What is

the diameter of the Unisphere? (Source: The

World’s Fair and Exposition Information and

Reference Guide)

Designing the Unisphere was an

engineering challenge that at one

point involved simultaneously solving

670 equations.

132.

Geometry The surface area S of a basketball

is 900͞␲ square inches. Find the radius r of the

Free-Falling Object In Exercises 133–136, find the

time required for an object to reach the ground when it

is dropped from a height of s0 feet.The height h (in feet)

is given by h ϭ Ϫ16t2 ϩ s0, where t measures the time

(in seconds) after the object is released.

133. s0 ϭ 256

134. s0 ϭ 48

135. s0 ϭ 128

136. s0 ϭ 500

137. Free-Falling Object The height h (in feet) of

an object thrown vertically upward from the

top of a tower 144 feet tall is given by

h ϭ 144 ϩ 128t Ϫ 16t2, where t measures the time

in seconds from the time when the object is

released. How long does it take for the object to

reach the ground?

138. Profit The monthly profit P (in dollars) a company

makes depends on the amount x (in dollars) the

company spends on advertising according to the

model

1

P ϭ 800 ϩ 120x Ϫ x2.

2

Find the amount spent on advertising that will yield

a monthly profit of \$8000.

Chapter 10

Compound Interest The amount A in an account after

2 years when a principal of P dollars is invested at annual

interest rate r compounded annually is given by

A ϭ P͑1 ϩ r͒2. In Exercises 139 and 140, find r.

139. P ϭ \$1500, A ϭ \$1685.40

140. P ϭ \$5000, A ϭ \$5724.50

National Health Expenditures In Exercises 141 and

142, the national expenditures for health care in the

United States from 1997 through 2006 are given by

y ϭ 4.95t2 ϩ 876, 7 Յ t Յ 16.

y

Expenditures

(in billions of dollars)

626

2200

2000

1800

1600

1400

1200

1000

t

7

8

9

10 11 12 13 14 15 16

Year (7 ↔ 1997)

Figure for 141 and 142

In this model, y represents the expenditures (in billions

of dollars) and t represents the year, with t ϭ 7

corresponding to 1997 (see figure). (Source: U.S.

Centers for Medicare & Medicaid Services)

141. Algebraically determine the year when expenditures

were approximately \$1500 billion. Graphically

confirm the result.

142. Algebraically determine the year when expenditures

were approximately \$1850 billion. Graphically

confirm the result.

Explaining Concepts

143.

For a quadratic equation ax 2 ϩ bx ϩ c ϭ 0,

where a, b, and c are real numbers with a 0,

explain why b and c can equal 0, but a cannot.

145.

Describe the steps you would use to solve a

quadratic equation when using the Square Root

Property.

146.

Describe a procedure for solving an equation of

144. Is it possible for a quadratic equation of the form

x2 ϭ m to have one real solution and one complex

Cumulative Review

In Exercises 147–150, solve the inequality and sketch

the solution on the real number line.

147.

148.

149.

150.

3x Ϫ 8 > 4

4 Ϫ 5x Ն 12

2x Ϫ 6 Յ 9 Ϫ x

x Ϫ 4 < 6 or x ϩ 3 > 8

In Exercises 151 and 152, solve the system of linear

equations.

151. x ϩ y Ϫ z ϭ 4

2x ϩ y ϩ 2z ϭ 10

x Ϫ 3y Ϫ 4z ϭ Ϫ7

152. 2x Ϫ y ϩ z ϭ Ϫ6

x ϩ 5y Ϫ z ϭ 7

Ϫx Ϫ 2y Ϫ 3z ϭ 8

In Exercises 153–158, combine the radical expressions,

if possible, and simplify.

153. 5Ί3 Ϫ 2Ί3

154. 8Ί27 ϩ 4Ί27

3 y Ϫ 9Ί

3 x

155. 16 Ί

156. 12Ίx Ϫ 1 ϩ 6Ίx Ϫ 1

157. Ί16m 4 n 3 ϩ mΊm2 n

4 32x 2 ϩ x Ί

4 2x 6 y 4 Ϫ yΊ

4 162x10

158. x 2 yΊ

Section 10.2

Completing the Square

627

10.2 Completing the Square

What You Should Learn

1 ᭤ Rewrite quadratic expressions in completed square form.

2 ᭤ Solve quadratic equations by completing the square.

Constructing Perfect Square Trinomials

Why You Should Learn It

You can use techniques such as

completing the square to solve quadratic

equations that model real-life situations.

For instance, in Exercise 90 on page 633,

you will find the dimensions of an

outdoor enclosure of a kennel by

completing the square.

1

completed square form.

͑x Ϫ 2͒2 ϭ 10.

Completed square form

You know from Example 2(b) in the preceding section that this equation has two

solutions: x ϭ 2 ϩ Ί10 and x ϭ 2 Ϫ Ί10. Suppose you were given the

equation in its general form

x2 Ϫ 4x Ϫ 6 ϭ 0.

General form

How could you solve this form of the quadratic equation? You could try factoring,

but after attempting to do so you would find that the left side of the equation is not

factorable using integer coefficients.

In this section, you will study a technique for rewriting an equation in a

completed square form. This technique is called completing the square. Note that

prior to completing the square, the coefficient of the second-degree term must be 1.

Completing the Square

To complete the square for the expression x2 ϩ bx, add ͑b͞2͒2, which is

the square of half the coefficient of x. Consequently,

x2 ϩ bx ϩ

΂b2΃ ϭ ΂x ϩ 2b΃ .

2

2

͑half͒2

EXAMPLE 1

Constructing a Perfect Square Trinomial

What term should be added to x2 Ϫ 8x so that it becomes a perfect square

trinomial? To find this term, notice that the coefficient of the x-term is Ϫ8. Take

half of this coefficient and square the result to get ͑Ϫ4͒2 ϭ 16. Add this term to

the expression to make it a perfect square trinomial.

x2 Ϫ 8x ϩ ͑Ϫ4͒2 ϭ x2 Ϫ 8x ϩ 16

Add ͑Ϫ4͒2 ϭ 16 to the expression.

You can then rewrite the expression as the square of a binomial, ͑x Ϫ 4͒2.

CHECKPOINT Now try Exercise 3.

628

Chapter 10

2

completing the square.

Solving Equations by Completing the Square

Completing the square can be used to solve quadratic equations. When using this

procedure, remember to preserve the equality by adding the same constant to

each side of the equation.

Study Tip

Completing the Square: Leading Coefficient Is 1

EXAMPLE 2

In Example 2, completing the square

is used for the sake of illustration.

This particular equation would be

easier to solve by factoring. Try

reworking the problem by factoring

to see that you obtain the same

two solutions.

Solve x2 ϩ 12x ϭ 0 by completing the square.

Solution

x2 ϩ 12x ϭ 0

Write original equation.

x2 ϩ 12x ϩ 62 ϭ 36

Add 62 ϭ 36 to each side.

͑122 ͒2

͑x ϩ 6͒2 ϭ 36

Completed square form

x ϩ 6 ϭ ± Ί36

Square Root Property

x ϭ Ϫ6 ± 6

Subtract 6 from each side.

x ϭ Ϫ6 ϩ 6 or x ϭ Ϫ6 Ϫ 6

Separate solutions.

xϭ0

Simplify.

x ϭ Ϫ12

The solutions are x ϭ 0 and x ϭ Ϫ12. Check these in the original equation.

Technology: Tip

You can use a graphing calculator

to check the solution to Example 3.

Graph

y ϭ x 2 Ϫ 6x ϩ 7

as shown below. Then use the

zero or root feature of the graphing

calculator to approximate the

x-intercepts to be x Ϸ 4.41 and

x Ϸ 1.59, which are the same

solutions obtained in Example 3.

CHECKPOINT Now try Exercise 17.

Completing the Square: Leading Coefficient Is 1

EXAMPLE 3

Solve x2 Ϫ 6x ϩ 7 ϭ 0 by completing the square.

Solution

x2 Ϫ 6x ϩ 7 ϭ 0

x2 Ϫ 6x ϭ Ϫ7

x Ϫ 6x ϩ ͑Ϫ3͒ ϭ Ϫ7 ϩ 9

2

2

Write original equation.

Subtract 7 from each side.

Add ͑Ϫ3͒2 ϭ 9 to each side.

3

͑Ϫ 62 ͒2

−2

7

͑x Ϫ 3͒2 ϭ 2

x Ϫ 3 ϭ ± Ί2

−3

Completed square form

Square Root Property

x ϭ 3 ± Ί2

x ϭ 3 ϩ Ί2 or x ϭ 3 Ϫ Ί2

Separate solutions.

The solutions are x ϭ 3 ϩ Ί2 Ϸ 4.41 and x ϭ 3 Ϫ Ί2 Ϸ 1.59. Check these in

the original equation.

CHECKPOINT Now try Exercise 33.

Section 10.2

Completing the Square

629

If the leading coefficient of a quadratic equation is not 1, you must divide

each side of the equation by this coefficient before completing the square.

EXAMPLE 4

Completing the Square: Leading Coefficient Is Not 1

2x2 Ϫ x Ϫ 2 ϭ 0

Original equation

2x2 Ϫ x ϭ 2

1

x2 Ϫ x ϭ 1

2

Divide each side by 2.

΂ ΃

2

΂x Ϫ 41΃

2

1

1

x2 Ϫ x ϩ Ϫ

2

4

ϭ1ϩ

ϭ

1

16

17

16

The solutions are x ϭ

2

Completed square form

Ί17

1

ϭ±

4

4

1

1

Add ͑Ϫ 4 ͒ ϭ 16 to each side.

1 Ί17

±

4

4

Square Root Property

1

1 Ί17

1 Ί17

ϩ

Ϸ 1.28 and x ϭ Ϫ

Ϸ Ϫ0.78.

4

4

4

4

CHECKPOINT Now try Exercise 61.

EXAMPLE 5

Completing the Square: Leading Coefficient Is Not 1

3x 2 Ϫ 6x ϩ 1 ϭ 0

Original equation

3x 2 Ϫ 6x ϭ Ϫ1

x 2 Ϫ 2x ϭ Ϫ

Subtract 1 from each side.

1

3

Divide each side by 3.

1

x2 Ϫ 2x ϩ ͑Ϫ1͒2 ϭ Ϫ ϩ 1

3

͑x Ϫ 1͒2 ϭ

2

3

xϪ1ϭ±

xϪ1ϭ±

Add ͑Ϫ1͒2 ϭ 1 to each side.

Completed square form

Ί23

Ί6

Rationalize the denominator.

3

xϭ1 ±

Square Root Property

Ί6

3

The solutions are x ϭ 1 ϩ Ί6͞3 Ϸ 1.82 and x ϭ 1 Ϫ Ί6͞3 Ϸ 0.18.

CHECKPOINT Now try Exercise 59. ### Tài liệu bạn tìm kiếm đã sẵn sàng tải về

1 Solving Quadratic Equations: Factoring and Special Forms

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