432
(8-20)
Chapter 8
Powers and Roots
Simplify each expression. See Example 8.
5
3
79. 80. 3
2
6 2
36 32
52
53
4
3
2
81. 82. 5
3
2 5
15 3
2 10
2
3
2 3
2 3
83. 84. 5 3
3 1
6 3 2 3
13 73
2
22
5 4
7 5
85. 86. 27 1
32 5
310 122 5 45
19 117
13
27
Simplify.
87. 5
a 20a
35a
88. a6 a12
6a22
89. 7
5 6
90. 24 150
52
2
2
92. (6 5 )(6
5 )
70 305
20
93. 5
3
55
3
1.866
5 2
98. 3 7
1.465
Solve each problem.
99. Find the exact area and perimeter of the given rectangle.
12 ft2, 102 ft
––
√ 8 ft
—–
√18 ft
FIGURE FOR EXERCISE 99
100. Find the exact area and perimeter of the given triangle.
9, 53
39
—–
√12
—–
√27
FIGURE FOR EXERCISE 100
101. Find the exact volume in cubic meters of the given rectangular box.
6 m3
1
5
94. 8 3
2 3
96. 6
In this
8.4
section
––
√6 m
––
√2 m
3
22
––
√3 m
Use a calculator to find the approximate value of each
expression to three decimal places.
2 3
95. 2
1.974
—–
√39
36
91. (5 35 )
4 6
97. 5 3
0.045
FIGURE FOR EXERCISE 101
102. Find the exact surface area of the box in Exercise 101.
26
43 62 m2
SOLVING EQUATIONS WITH
RADICALS AND EXPONENTS
Equations involving radicals and exponents occur in many applications. In this section you will learn to solve equations of this type, and you will see how these equations occur in applications.
●
The Square Root Property
●
Obtaining Equivalent
Equations
●
Squaring Each Side of an
Equation
The Square Root Property
●
Solving for the Indicated
Variable
An equation of the form x 2 k can have two solutions, one solution, or no solutions, depending on the value of k. For example,
●
Applications
x2 4
8.4
Solving Equations with Radicals and Exponents
(8-21)
433
has two solutions because (2)2 4 and (2)2 4. So x 2 4 is equivalent to the
compound equation
x2
x 2,
or
which is also written as x 2 and is read as “x equals positive or negative 2.” The
only solution to x 2 0 is 0. The equation
x 2 4
has no real solution because the square of every real number is nonnegative.
These examples illustrate the square root property.
Square Root Property (Solving x2 k)
For k 0 the equation x 2 k is equivalent to the compound equation
x k or
x k.
(Also written x k.)
For k 0 the equation x 2 k is equivalent to x 0.
For k 0 the equation x 2 k has no real solution.
CAUTION
The expression 9 has a value of 3 only, but the equation
x 2 9 has two solutions: 3 and 3.
E X A M P L E
helpful
1
hint
We do not say “take the square
root of each side.” We are not
doing the same thing to each
side of x 2 12 when we write
x 12
. This is the second
time that we have seen a rule
for obtaining an equivalent
equation without “doing the
same thing to each side.”
(What was the first?)
Using the square root property
Solve each equation.
a) x 2 12
c) x 2 9
b) 2(x 1)2 18 0
d) (x 16)2 0
Solution
a) x 2 12
x 12 Square root property
x 23
Check: (23)2 4 3 12, and (23
)2 4 3 12.
b) 2(x 1)2 18 0
Add 18 to each side.
2(x 1)2 18
(x 1)2 9
Divide each side by 2.
x 1 9 Square root property
x13
or
x 1 3
x2
or
x 4
Check 2 and 4 in the original equation. Both 4 and 2 are solutions to the
equation.
c) The equation x 2 9 has no real solution because no real number has a square
that is negative.
d) (x 16)2 0
x 16 0 Square root property
x 16
■
Check: (16 16)2 0. The equation has only one solution, 16.
434
(8-22)
Chapter 8
Powers and Roots
Obtaining Equivalent Equations
When solving equations, we use a sequence of equivalent equations with each one
simpler than the last. To get an equivalent equation, we can
1.
2.
3.
4.
add the same number to each side,
subtract the same number from each side,
multiply each side by the same nonzero number, or
divide each side by the same nonzero number.
However, “doing the same thing to each side” is not the only way to obtain an
equivalent equation. In Chapter 5 we used the zero factor property to obtain equivalent equations. For example, by the zero factor property the equation
(x 3)(x 2) 0
is equivalent to the compound equation
x30
study
tip
Make sure that you know
what your instructor expects
from you. You can determine what your instructor
feels is important by looking
at the examples that your
instructor works in class and
the homework assigned.
When in doubt, ask your
instructor what you will be
responsible for and write
down the answer.
or
x 2 0.
In this section you just learned how to obtain equivalent equations by the square
root property. This property tells us how to write an equation that is equivalent to
the equation x2 k. Note that the square root property does not tell us to “take the
square root of each side.” To become proficient at solving equations, we must understand these methods. One of our main goals in algebra is to keep expanding our
skills for solving equations.
Squaring Each Side of an Equation
Some equations involving radicals can be solved by squaring each side:
x 5
(x)2 52
x 25
Square each side.
All three of these equations are equivalent. Because 2
5 5 is correct, 25 satisfies the original equation.
However, squaring each side does not necessarily produce an equivalent equation. For example, consider the equation
x 3.
Squaring each side, we get
x 2 9.
Both 3 and 3 satisfy x 2 9, but only 3 satisfies the original equation x 3. So
x 2 9 is not equivalent to x 3. The extra solution to x 2 9 is called an extraneous solution.
These two examples illustrate the squaring property of equality.
Squaring Property of Equality
When we square each side of an equation, the solutions to the new equation
include all of the solutions to the original equation. However, the new equation might have extraneous solutions.
8.4
Solving Equations with Radicals and Exponents
(8-23)
435
This property means that we may square each side of an equation, but we must
check all of our answers to eliminate extraneous solutions.
E X A M P L E
2
Using the squaring property of equality
Solve each equation.
6
1 3
b) x 2x
3
a) x2
Solution
a) x2
6
1 3
2
6
1)2 32
(x
x 2 16 9
x 2 25
x 5
Check each solution:
c) x2
x
4 2
x
3
Square each side.
Square root property
2
6
1 2
5
6
1 9 3
5
2
5
)
6
1 25
6
1 9
3
(
study
tip
Most instructors believe that
what they do in class is important. If you miss class, you miss
what is important to your instructor.You miss what is most
likely to appear on the test.
So both 5 and 5 are solutions to the equation.
b)
x 2x
3
2
x (2x
)
3 2 Square each side.
x 2 2x 3
Solve by factoring.
x 2 2x 3 0
(x 3)(x 1) 0
Factor.
x30
or
x10
Zero factor property
x3
or
x 1
Check in the original equation:
Check x 3:
Check x 1:
3 2
3
3
1 2(
1)
3
3 9
1 1
Correct
Incorrect
Because 1 does not satisfy the original equation, 1 is an extraneous solution. The only solution is 3.
x
4 2
x
3
c)
x2
x 2 4x 2 3x Square each side.
x2 x 2 0
(x 2)(x 1) 0
Zero factor property
x20
or
x10
x2
or
x 1
Check each solution in the original equation:
Check x 2:
2
42 2
32
2
4
4
Check x 1:
(
1
)2
(
41) 2
(
31)
5 5
Because 4 is not a real number, 2 is an extraneous solution. The only
■
solution to the original equation is 1.
436
(8-24)
Chapter 8
Powers and Roots
In the next example, one of the sides of the equation is a binomial. When we
square each side, we must be sure to square the binomial properly.
E X A M P L E
3
Squaring each side of an equation
2
x
3.
Solve the equation x 2 Solution
x 2 2
x
3
2
(x 2) (
2
x
3)2
2
x 4x 4 2 3x
x 2 7x 6 0
(x 6)(x 1) 0
x60
or
x10
x 6 or
x 1
Square each side.
Square the binomial on the
left side.
Factor.
Check these solutions in the original equation:
Check x 6:
2
(
36)
6 2 4 1
6 Incorrect
Check x 1:
1 2 2
(
31)
1 1
Correct
The solution 6 does not check. The only solution to the original equation
■
is 1.
Solving for the Indicated Variable
In the next example we use the square root property to solve a formula for an indicated variable.
E X A M P L E
4
Solving for a variable
Solve the formula A r 2 for r.
Solution
A r 2
A
r 2
A
r
Divide each side by .
Square root property
The formula solved for r is
A
r .
If r is the radius of a circle with area A, then r is positive and
r
.
A
■
Applications
Equations involving exponents occur in many applications. If the exact answer to a
problem is an irrational number in radical notation, it is usually helpful to find a
decimal approximation for the answer.
8.4
E X A M P L E
10 ft
5
x ft
x ft
FIGURE 8.1
WARM-UPS
(8-25)
437
Finding the side of a square with a given diagonal
If the diagonal of a square window is 10 feet long, then what are the exact and approximate lengths of a side? Round the approximate answer to two decimal places.
Solution
First make a sketch as in Fig. 8.1. Let x be the length of a side. The Pythagorean
theorem tells us that the sum of the squares of the sides is equal to the diagonal
squared:
x 2 x 2 102
2x 2 100
x 2 50
x 50
52
Because the length of a side must be positive, we disregard the negative solution.
feet. Use a calculator to get 52 7.07. The
The exact length of a side is 52
symbol means “is approximately equal to.” The approximate length of a side is
■
7.07 feet.
True or false? Explain your answer.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
8. 4
Solving Equations with Radicals and Exponents
The equation x 2 9 is equivalent to the equation x 3. False
The equation x 2 16 has no real solution. True
The equation a2 0 has no solution. False
and 5
are solutions to x 2 5 0. False
Both 5
2
The equation x 9 has no real solution. True
4 2x,
9 first take the square root of each side.
To solve x
False
All extraneous solutions give us a denominator of zero. False
Squaring both sides of x 1 will produce an extraneous solution.
True
The equation x 2 3 0 is equivalent to x 3. True
x2
x
8 has no solution. True
The equation 2 6
EXERCISES
Reading and Writing After reading this section, write out the
answers to these questions. Use complete sentences.
1. What is the square root property?
The square root property says that x 2 k for k 0 is
equivalent to x k.
2. When do we take the square root of each side of an
equation?
We do not take the square root of each side of an equation.
3. What new techniques were introduced in this section for
solving equations?
The square root property and squaring each side are two
new techniques used for solving equations.
4. Is there any way to obtain an equivalent equation other than
doing the same thing to each side?
The square root property and the zero factor property produce equivalent equations without doing the same thing to
each side.
5. Which property for solving equations can give extraneous
roots?
Squaring each side can produce extraneous roots.
6. Which property of equality does not always give you an
equivalent equation?
Squaring each side does not always give equivalent
equations.
438
(8-26)
Chapter 8
Powers and Roots
Solve each equation. See Example 1.
7. x 16
4, 4
8. x 49
7, 7
2
2
9. x 2 40 0
210, 210
10. x 2 24 0
26, 26
11. 3x 2
6 6
, 3
3
12. 2x 3
6 6
, 2
2
13. 9x 4
No solution
14. 25x 1 0
No solution
15. (x 1)2 4
1, 3
16. (x 3)2 9
6, 0
17. 2(x 5)2 1 7
5 3, 5 3
18. 3(x 6)2 4 11
6 5, 6 5
19. (x 19)2 0
19
20. 5x 2 5 5
0
2
2
2
2
Solve each equation. See Example 2.
21. x9 9
90
22. x
3 4
13
23. 2
x3 4
No solution
24. 3
x5 9
No solution
25. 4 x2
9
5, 5
26. 1 x2
1
2, 2
27. x 18
x
3
3
28. x 6x
7
2
9
29. x x
0, 1
30. x 2x
0, 2
1 2x5
31. x
6
32. 1
x
3 x
5
1
33. 32x1 3 5
13
18
34. 4x
5 3 9
4
Solve each equation. See Example 3.
35. x 3 2x
6
36. x 1 3x5
3, 5
2, 3
3
1 x 1
37. x
3
38. x 1 22
x
2
3
39. 10x4
4 x 2
6, 8
40. 8x7 x 1
2, 4
45. b2 4ac 0 for b
c
b 2a
1
46. s gt 2 v for t
2
2s 2v
t g
47. v 2pt for t
v2
t 2p
48. y 2
x for x
y2
x 2
Solve each equation.
49. 3x 2 6 0
2, 2
50. 5x 2 3 0
No solution
x3 3x
1
51. 2
No solution
53. (2x 1)2 8
1 22
1 22
, 2
2
55. 2
x9 0
9
2
57. x 1 2x
0
1
3
59. 3(x 1)2 27 0
4, 2
61. (2x 5)2 0
5
2
52. 2x4 x9
No solution
54. (3x 2)2 18
2 32
2 32
, 3
3
56. 5
x
3 0
5
3
58. x 3 2x
8
1
9
60. 2(x 3)2 50 0
2, 8
62. (3x 1)2 0
1
3
Use a calculator to find approximate solutions to each
equation. Round your answers to three decimal places.
63. x2 3.25
1.803, 1.803
65. x
2 1.73
0.993
67. 1.3(x 2.4)2 5.4
0.362, 4.438
64. (x 1)2 20.3
5.506, 3.506
66. 2
.3
x.4
1 3.3
5.343
68. 2.4x 2 9.55
1.995, 1.995
Find the exact answer to each problem. If the answer is irrational, then find an approximation to three decimal places. See
Example 5.
69. Side of a square. Find the length of the side of a square
whose area is 18 square feet. 32
or 4.243 ft
70. Side of a field. Find the length of the side of a square wheat
field whose area is 75 square miles. 53
or 8.660 mi
Solve each formula for the indicated variable. See Example 4.
41. V r2h for r
h
V
r
43. a b c for b
2
2
4
42. V r 2h for r
3
3V
r 4h
2
b c2
a2
44. y ax 2 c for x
yc
x a
FIGURE FOR EXERCISE 70