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
Add 2 to each side.
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
Add 8 to each side.
3x Ϫ 6 ϭ ± 22Ί2
3x ϭ 6 ± 2Ί2
xϭ2 ±
2Ί2
3
Square Root Property and
rewrite Ί8 as 2Ί2.
Add 6 to each side.
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
Quadratic Equations, Functions, and Inequalities
3 ᭤ Solve quadratic equations with
complex solutions by the Square Root
Property.
Technology: Discovery
Solve each quadratic equation
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
quadratic equation with the
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
Quadratic Equations with Complex Solutions
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
Add 4 to each side.
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
xϭ
5 4
± i
3 3
Add 5 to each side.
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
of quadratic form.
Solving Quadratic Equations: Factoring and Special Forms
621
Equations of Quadratic Form
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
Written in Quadratic Form
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
Write in quadratic form.
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
Quadratic Equations, Functions, and Inequalities
Study Tip
When solving equations involving
square roots, be sure to check for
extraneous solutions.
Solving Equations of Quadratic Form
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
Write in quadratic form.
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
Write in quadratic form.
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 4r 2. The surface area of
a softball is 144͞ square inches. Find the diameter d of the softball.
Solution
144
ϭ 4r2
36
ϭ r2
2
Substitute 144͞ for S.
±
Ί36 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
or false. Justify your answer.
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
form? Explain your reasoning.
Go to pages 676–677 to
record your assignments.
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
Quadratic Equations, Functions, and Inequalities
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
of the quadratic equation.
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 8xx 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
© Rudy Sulgan/CORBIS
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
basketball.
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
Quadratic Equations, Functions, and Inequalities
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
quadratic form. Give an example.
144. Is it possible for a quadratic equation of the form
x2 ϭ m to have one real solution and one complex
solution? Explain your reasoning.
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
© Sean Cayton/The Image Works
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
᭤ Rewrite quadratic expressions in
completed square form.
Consider the quadratic equation
͑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
Quadratic Equations, Functions, and Inequalities
2
᭤ Solve quadratic equations by
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
Add 3 to each side.
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
Add 2 to each side.
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
xϪ
ϭ1ϩ
ϭ
1
16
17
16
The solutions are x ϭ
2
Completed square form
Ί17
1
ϭ±
4
4
xϭ
1
1
Add ͑Ϫ 4 ͒ ϭ 16 to each side.
1 Ί17
±
4
4
Square Root Property
1
Add 4 to each side.
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
Add 1 to each side.
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.