4: Products and Quotients Involving Radicals
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398
Chapter 9 • Roots and Radicals
(d) 2x(2x ϩ 2y) ϭ 2x 2x ϩ 2x 2y
ϭ 2x2 ϩ 2xy
ϭ x ϩ 2xy
3
3
3
3
3
3
(e) 2
2( 2
4ϩ 2
10) ϭ 2
8ϩ 2
20 ϭ 2 ϩ 2
20
The distributive property plays a central role when we find the product of two binomials.
For example, (x ϩ 2)(x ϩ 3) ϭ x(x ϩ 3) ϩ 2(x ϩ 3) ϭ x2 ϩ 3x ϩ 2x ϩ 6 ϭ x2 ϩ 5x ϩ 6.
We can find the product of two binomial expressions involving radicals in a similar fashion.
Classroom Example
Multiply and simplify:
(a) ( 27 ϩ 22)( 26 ϩ 210)
(b) ( 25 Ϫ 2)(25 ϩ 8)
EXAMPLE 3
Multiply and simplify:
(a) (23 ϩ 25)(22 ϩ 26)
(b) 127 Ϫ 32127 ϩ 62
Solution
(a) (23 ϩ 25)(22 ϩ 26) ϭ 23(22 ϩ 26) ϩ 25(22 ϩ 26)
ϭ 2322 ϩ 2326 ϩ 2522 ϩ 2526
ϭ 26 ϩ 218 ϩ 210 ϩ 230
ϭ 26 ϩ 322 ϩ 210 ϩ 230
(b) (27 Ϫ 3)( 27 ϩ 6) ϭ 27( 27 ϩ 6) Ϫ 3(27 ϩ 6)
ϭ 27 27 ϩ 627 Ϫ 327 Ϫ 18
ϭ 7 ϩ 627 Ϫ 327 Ϫ 18
ϭ Ϫ11 ϩ 327
If the binomials are of the form (a ϩ b)(a Ϫ b), then we can use the multiplication pattern
(a ϩ b)(a Ϫ b) ϭ a2 Ϫ b2.
Classroom Example
Multiply and simplify:
(a) ( 212 ϩ 5)(212 Ϫ 5)
(b) (7 Ϫ 23)(7 ϩ 23)
(c) (26 ϩ 210)(26 Ϫ 210)
EXAMPLE 4
Multiply and simplify:
(a) (26 ϩ 2)(26 Ϫ 2)
(b) (3 Ϫ 25)(3 ϩ 25)
(c) (28 ϩ 25)(28 Ϫ 25)
Solution
(a) (26 ϩ 2)( 26 Ϫ 2) ϭ (26) 2 Ϫ 22 ϭ 6 Ϫ 4 ϭ 2
(b) (3 Ϫ 25)(3 ϩ 25) ϭ 32 Ϫ (25) 2 ϭ 9 Ϫ 5 ϭ 4
(c) (28 ϩ 25)( 28 Ϫ 25) ϭ ( 28) 2 Ϫ ( 25) 2 ϭ 8 Ϫ 5 ϭ 3
Note that in each part of Example 4, the final product contains no radicals. This happens
whenever we multiply expressions such as 2a ϩ 2b and 2a Ϫ 2b, where a and b are
rational numbers.
(2a ϩ 2b)(2a Ϫ 2b) ϭ ( 2a) 2 Ϫ ( 2b) 2 ϭ a Ϫ b
9.4 • Products and Quotients Involving Radicals
399
Expressions such as 28 ϩ 25 and 28 Ϫ 25 are called conjugates of each other.
Likewise, 26 ϩ 2 and 26 Ϫ 2 are conjugates, as are 3 Ϫ 25 and 3 ϩ 25. Now let’s see
how we can use conjugates to rationalize denominators.
Classroom Example
2
.
Simplify
27 ϩ 23
EXAMPLE 5
Simplify
4
25 ϩ 22
.
Solution
To simplify the expression, the denominator needs to be a rational number. Let’s multiply the
numerator and denominator by 25 Ϫ 22, which is the conjugate of the denominator.
4
25 ϩ 22
ϭ
ϭ
4
25 ϩ 22
и
25 Ϫ 22
25 Ϫ 22
25 Ϫ 22
25 Ϫ 22
is merely a form of 1
4( 25 Ϫ 22)
(25 ϩ 22)( 25 Ϫ 22)
ϭ
4(25 Ϫ 22)
5Ϫ2
ϭ
4( 25 Ϫ 22)
425 Ϫ 422
or
3
3
Either answer is acceptable
The next four examples further illustrate the process of rationalizing and simplifying
expressions that contain binomial denominators.
Classroom Example
Rationalize the denominator and
25
simplify
.
214 Ϫ 3
EXAMPLE 6
Rationalize the denominator and simplify
Solution
23
26 Ϫ 2
.
To simplify, we want to multiply numerator and denominator by the conjugate of the denominator, which is 26 ϩ 2.
23
26 Ϫ 2
ϭ
ϭ
Classroom Example
Rationalize the denominator and
20
simplify
.
225 ϩ 210
23
26 Ϫ 2
и
26 ϩ 2
26 ϩ 2
23(26 ϩ 2)
(26 Ϫ 2)(26 ϩ 2)
ϭ
218 ϩ 223
6Ϫ4
ϭ
322 ϩ 223
2
EXAMPLE 7
218 ϭ 2922 ϭ 322
Rationalize the denominator and simplify
14
223 ϩ 25
.
400
Chapter 9 • Roots and Radicals
Solution
To simplify, we want to multiply numerator and denominator by the conjugate of the denominator, which is 223 Ϫ 25.
14
223 ϩ 25
ϭ
ϭ
ϭ
14
223 ϩ 25
и
223 Ϫ 25
223 Ϫ 25
14(223 Ϫ 25)
(2 23 ϩ 25)(2 23 Ϫ 25)
14(2 23 Ϫ 25)
14(223 Ϫ 25)
ϭ
12 Ϫ 5
7
ϭ 2(223 Ϫ 25) or 423 Ϫ 225
Classroom Example
Rationalize the denominator and
2a Ϫ 4
simplify
.
2a ϩ 5
EXAMPLE 8
Rationalize the denominator and simplify
Solution
2x ϩ 2
2x Ϫ 3
.
To rationalize the denominator, we need to multiply the numerator and denominator by
1x ϩ 3, which is the conjugate of the denominator.
2x ϩ 2
2x Ϫ 3
ϭ
ϭ
Classroom Example
Rationalize the denominator and
6 ϩ 27
.
simplify
27 Ϫ 9
2x ϩ 2
2x Ϫ 3
и
2x ϩ 3
2x ϩ 3
( 2x ϩ 2)(2x ϩ 3)
( 2x Ϫ 3)(2x ϩ 3)
ϭ
x ϩ 22x ϩ 32x ϩ 6
xϪ9
ϭ
x ϩ 52x ϩ 6
xϪ9
EXAMPLE 9
Rationalize the denominator and simplify
Solution
3 ϩ 22
22 Ϫ 6
.
To change the denominator to a rational number, we can multiply the numerator and denominator by 12 ϩ 6, which is the conjugate of the denominator.
3 ϩ 22
22 Ϫ 6
ϭ
ϭ
3 ϩ 22
22 Ϫ 6
и
22 ϩ 6
22 ϩ 6
(3 ϩ 22)( 22 ϩ 6)
( 22 Ϫ 6)( 22 ϩ 6)
ϭ
322 ϩ 18 ϩ 2 ϩ 622
2 Ϫ 36
ϭ
922 ϩ 20
Ϫ34
ϭϪ
922 ϩ 20
34
a
a
ϭϪ
Ϫb
b
9.4 • Products and Quotients Involving Radicals
401
Concept Quiz 9.4
For Problems 1–10, answer true or false.
n
n
n
1. The property 2x 2y ϭ 2xy can be used to express the product of two radicals as
one radical.
2. The product of two radicals always results in an expression that has a radical even
after simplifying.
3. The conjugate of 5 ϩ 23 is -5 Ϫ 23.
4. The product of (2 Ϫ 27) and (2 ϩ 27) is a rational number.
5. To rationalize the denominator for the expression
28 ϩ 212
6.
22
22
7.
28 ϩ 212
225
4 Ϫ 25
, we would multiply by
25
25
.
ϭ 2 ϩ 26.
ϭ
1
2 ϩ 26
8. The product of (5 ϩ 23) and ( -5 Ϫ 23) is Ϫ28.
9. The product of (223 ϩ 25) and (223 Ϫ 25) is 7.
3
3
3
10. ( 28)( 23) ϭ 223
Problem Set 9.4
For Problems 1– 20, multiply and simplify where possible.
(Objective 1)
1. 27 25
2. 23 210
3. 26 28
4. 26 212
5. 25 210
6. 22 212
3
3
7. 29 26
3
3
8. 29 29
9. 28 212
30. 322(325 Ϫ 227)
31. ( 22 ϩ 6) ( 22 ϩ 9)
32. ( 23 ϩ 4) ( 23 ϩ 7)
33. ( 26 Ϫ 5) ( 26 ϩ 3)
34. ( 27 Ϫ 6) ( 27 ϩ 1)
10. 212 220
11. (3 23) (527)
3
29. 423( 22 Ϫ 225)
35. ( 23 ϩ 26)( 26 ϩ 28)
3
12. (5210)(723)
36. ( 22 ϩ 26)( 28 ϩ 25)
13. (Ϫ26)(524)
14. (523)(Ϫ6210)
37. (5 ϩ 210) (5 Ϫ 210)
15. (326) (426)
16. (227)(5 27)
38. ( 211 Ϫ 2) ( 211 ϩ 2)
17. (522) (4212)
18. (322)(2 227)
39. (322 Ϫ 23)(322 ϩ 23)
19. (4 23) (2215)
20. (727)(2 214)
40. (526 Ϫ 22)(526 ϩ 22)
3
3
For Problems 21– 42, find the products by applying the
distributive property. Express your answers in simplest radical
form. (Objective 1)
41. (523 ϩ 226)(523 Ϫ 226)
42. (425 ϩ 527)(425 Ϫ 527)
21. 22( 23 ϩ 25)
22. 23( 25 ϩ 27)
23. 26( 22 Ϫ 5)
24. 28( 23 Ϫ 4)
For Problems 43– 54, find each product and express your
answers in simplest radical form. All variables represent
nonnegative real numbers. (Objective 1)
25. 22( 24 ϩ 210)
3
3
3
26. 23( 29 Ϫ 28)
43. 2xy 2x
44. 2x 2 y 2y
27. 212( 26 Ϫ 28)
28. 215( 23 Ϫ 25)
3
3
45. 225x 2 25x
3
3
46. 29x 23x
3
3
3
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402
Chapter 9 • Roots and Radicals
47. (42a)(32ab)
59.
48. (52a)(62a)
49. 22x( 23x Ϫ 26y)
61.
50. 26x( 22x Ϫ 24y)
51. ( 2x ϩ 5)( 2x Ϫ 3)
63.
52. (3 ϩ 2x) (7 Ϫ 2x)
53. ( 2x ϩ 7)( 2x Ϫ 7)
65.
54. ( 2x ϩ 9)( 2x Ϫ 9)
For Problems 55–70, rationalize the denominators and simplify. All variables represent positive real numbers.
(Objective 2)
55.
57.
67.
69.
3
56.
22 ϩ 4
8
58.
26 Ϫ 2
5
2
25 ϩ 23
10
2 Ϫ 323
4
2x Ϫ 2
2x
2x ϩ 3
2a ϩ 2
2a Ϫ 5
2 ϩ 23
3 Ϫ 22
60.
62.
64.
66.
68.
70.
3
26 ϩ 25
5
322 Ϫ 4
7
2x ϩ 4
2y
2y Ϫ 6
2a Ϫ 3
2a ϩ 1
3 Ϫ 25
4 ϩ 28
23 ϩ 7
10
3 Ϫ 27
Thoughts Into Words
71. Explain how the distributive property has been used in
this chapter.
72. How would you help someone rationalize the denom24
inator and simplify the expression
?
212 ϩ 28
Further Investigations
that you obtained when you rationalized the denominators.
73. Return to Problems 55–62 and use a calculator to
evaluate each expression. Then evaluate the result
Answers to the Concept Quiz
2. False
3. False
4. True
1. True
9.5
5. False
6. True
7. False
8. False
9. True
10. True
Solving Radical Equations
OBJECTIVES
1
Find the solution sets for radical equations
2
Check the solutions for radical equations
3
Solve formulas involving radicals
Equations that contain radicals with variables in the radicand are called radical equations.
Here are some examples of radical equations that involve one variable.
2x ϭ 3
22x ϩ 1 ϭ 5
23x ϩ 4 ϭ - 4
25s Ϫ 2 ϭ 22s ϩ 19
22y Ϫ 4 ϭ y Ϫ 2
2x ϩ 6 ϭ x
9.5 • Solving Radical Equations
403
In order to solve such equations we need the following property of equality.
Property 9.4
For real numbers a and b, if a ϭ b, then
a2 ϭ b2
Property 9.4 states that we can square both sides of an equation. However, squaring both
sides of an equation sometimes produces results that do not satisfy the original equation. Let’s
consider two examples to illustrate the point.
Classroom Example
Solve 2m ϭ 5.
EXAMPLE 1
Solve 2x ϭ 3.
Solution
2x ϭ 3
( 2x) 2 ϭ 32
Square both sides
xϭ9
Since 29 ϭ 3, the solution set is 596.
Classroom Example
Solve 2m ϭ Ϫ4.
EXAMPLE 2
Solve 2x ϭ - 3.
Solution
2x ϭ Ϫ3
( 2x) 2 ϭ (Ϫ3) 2
xϭ9
Square both sides
Because 29 ϶ Ϫ3, 9 is not a solution, and the solution set is л.
In general, squaring both sides of an equation produces an equation that has all of the
solutions of the original equation, but it may also have some extra solutions that do not satisfy
the original equation. Such extra solutions are called extraneous solutions or extraneous
roots. Therefore, when using the “squaring” property (Property 9.4), you must check each
potential solution in the original equation.
We now consider some examples to demonstrate different situations that arise when solving radical equations.
Classroom Example
Solve 27x Ϫ 5 ϭ 3.
EXAMPLE 3
Solve 22x ϩ 1 ϭ 5.
Solution
22x ϩ 1 ϭ 5
( 22x ϩ 1) 2 ϭ 52
2x ϩ 1 ϭ 25
2x ϭ 24
x ϭ 12
Square both sides
404
Chapter 9 • Roots and Radicals
✔ Check
22x ϩ 1 ϭ 5
22(12) ϩ 1 5
224 ϩ 1 5
225 5
5ϭ5
The solution set is 5126.
Classroom Example
Solve 26x ϩ 4 ϭ - 8.
EXAMPLE 4
Solve 23x ϩ 4 ϭ -4.
Solution
23x ϩ 4 ϭ Ϫ4
( 23x ϩ 4) 2 ϭ (Ϫ4) 2
3x ϩ 4 ϭ 16
3x ϭ 12
xϭ4
Square both sides
✔ Check
23x ϩ 4 ϭ Ϫ4
23(4) ϩ 4 Ϫ4
216 Ϫ4
4 ϶ Ϫ4
Since 4 does not check (4 is an extraneous root), the equation 23x ϩ 4 ϭ Ϫ4 has no real
number solutions. The solution set is л.
Classroom Example
Solve 327x ϩ 1 ϭ 4.
EXAMPLE 5
Solve 322y ϩ 1 ϭ 5.
Solution
322y ϩ 1 ϭ 5
5
22y ϩ 1 ϭ
3
5 2
( 22y ϩ 1) 2 ϭ a b
3
25
2y ϩ 1 ϭ
9
25
2y ϭ
Ϫ1
9
25
9
2y ϭ
Ϫ
9
9
16
2y ϭ
9
1
1 16
(2y) ϭ a b
2
2 9
8
yϭ
9
Divided both sides by 3
Multiplied both sides by
1
2
9.5 • Solving Radical Equations
✔ Check
322y ϩ 1 ϭ 5
3
B
3
8
2a b ϩ 1 5
9
16
9
ϩ 5
B9
9
25
5
B9
5
3a b 5
3
5ϭ5
3
8
The solution set is e f .
9
Classroom Example
Solve 22x ϩ 11 ϭ 27x Ϫ 9
EXAMPLE 6
Solve 25s Ϫ 2 ϭ 22s ϩ 19.
Solution
25s Ϫ 2 ϭ 22s ϩ 19
( 25s Ϫ 2) 2 ϭ ( 22s ϩ 19) 2
5s Ϫ 2 ϭ 2s ϩ 19
Square both sides
3s ϭ 21
sϭ7
✔ Check
25s Ϫ 2 ϭ 22s ϩ 19
25(7) Ϫ 2 22(7) ϩ 19
233 ϭ 233
The solution set is 576.
Classroom Example
Solve 23x Ϫ 14 ϭ x Ϫ 4.
EXAMPLE 7
Solve 22y Ϫ 4 ϭ y Ϫ 2.
Solution
22y Ϫ 4 ϭ y Ϫ 2
(22y Ϫ 4) 2 ϭ (y Ϫ 2) 2
Square both sides
2y Ϫ 4 ϭ y Ϫ 4y ϩ 4
2
0 ϭ y2 Ϫ 6y ϩ 8
0 ϭ (y Ϫ 4) (y Ϫ 2)
yϪ4ϭ0
or
yϪ2ϭ0
yϭ4
or
yϭ2
Factor the right side
Remember the property: ab ϭ 0 if and only
if a ϭ 0 or b ϭ 0
405
406
Chapter 9 • Roots and Radicals
✔ Check
if y ؍4
if y ؍2
22y Ϫ 4 ϭ y Ϫ 2 22y Ϫ 4 ϭ y Ϫ 2
22(4) Ϫ 4 4 Ϫ 2 22(2) Ϫ 4 2 Ϫ 2
24 2
20 0
2 ϭ 2
The solution set is 52, 46.
Classroom Example
Solve 22m ϩ 4 ϭ m.
EXAMPLE 8
0ϭ0
Solve 2x ϩ 6 ϭ x.
Solution
2x ϩ 6 ϭ x
2x ϭ x Ϫ 6
( 2x) 2 ϭ (x Ϫ 6) 2
x ϭ x2 Ϫ 12x ϩ 36
0 ϭ x2 Ϫ 13x ϩ 36
0 ϭ (x Ϫ 4)(x Ϫ 9)
xϪ4ϭ0
or
xϪ9ϭ0
xϭ4
or
xϭ9
We added Ϫ6 to both sides so that the term with
the radical is alone on one side of the equation
Apply ab ϭ 0 if and only if a ϭ 0 or b ϭ 0
✔ Check
if x ؍4
if x ؍9
2x ϩ 6 ϭ x 2x ϩ 6 ϭ x
24 ϩ 6 4 29 ϩ 6 9
2 ϩ 6 4 3 ϩ 6 9
8 ϶ 4
9ϭ9
The solution set is 596.
Note in Example 8 that we changed the form of the original equation, 2x ϩ 6 ϭ x, to
2x ϭ x Ϫ 6 before we squared both sides. Squaring both sides of 2x ϩ 6 ϭ x produces
x ϩ 122x ϩ 36 ϭ x 2, a more complex equation that still contains a radical. So again, it pays
to think ahead a few steps before carrying out the details of the problem.
Another Look at Applications
In Section 9.1 we used the formula S ϭ 230Df to approximate how fast a car was traveling,
based on the length of skid marks. (Remember that S represents the speed of the car in miles
per hour, D the length of skid marks measured in feet, and f a coefficient of friction.) This
same formula can be used to estimate the lengths of skid marks that are produced by cars traveling at different rates on various types of road surfaces. To use the formula for this purpose,
we change the form of the equation by solving for D.
230Df ϭ S
30Df ϭ S 2
S2
Dϭ
30f
The result of squaring both sides of the original equation
D, S, and f are positive numbers, so this final equation
and the original one are equivalent
9.5 • Solving Radical Equations
Classroom Example
Suppose that for a parttcular road
surface the coefficient of friction is
0.45. How far will a car traveling at
75 miles per hour skid when the
brakes are applied?
407
EXAMPLE 9
Suppose that for a particular road surface the coefficient of friction is 0.35. How far will a car
traveling at 60 miles per hour skid when the brakes are applied?
Solution
We substitute 0.35 for f and 60 for S in the formula D ϭ
Dϭ
602
ϭ 343
30(0.35)
S2
.
30f
to the nearest whole number
The car will skid approximately 343 feet.
Remark: Pause for a moment and think about the result in Example 9. The coefficient of fric-
tion of 0.35 applies to a wet concrete road surface. Note that a car traveling at 60 miles per
hour will skid farther than the length of a football field.
Concept Quiz 9.5
For Problems 1–10, answer true or false.
1. To solve a radical equation, we can raise each side of the equation to a positive integer
power.
2. Solving the equation that results from squaring each side of an original equation may
not give all the solutions of the original equation.
3
3. The equation 2x Ϫ 1 ϭ - 2 has a solution.
4. Potential solutions that do not satisfy the original equation are called extraneous
solutions.
5. The equation 2x ϩ 1 ϭ - 2 has no solutions.
6. The solution set for 2x ϩ 2 ϭ x is {1, 4}.
7. The solution set for 2x ϩ 1 ϩ 2x Ϫ 2 ϭ -3 is the null set.
3
8. The solution set for 2x ϩ 2 ϭ - 2 is the null set.
9. The solution set for 2x ϭ x is {0}.
3
10. The solution set for 2x ϭ x is {Ϫ1, 0, 1}.
Problem Set 9.5
For Problems 1– 40, solve each equation. Be sure to check
all potential solutions in the original equation. (Objective 1)
1. 2x ϭ 7
2. 2x ϭ 12
3. 22x ϭ 6
4. 23x ϭ 9
5. 23x ϭ -6
6. 22x ϭ -8
7. 24x ϭ 3
8. 25x ϭ 4
9. 32x ϭ 2
10. 42x ϭ 3
11. 22n Ϫ 3 ϭ 5
12. 23n ϩ 1 ϭ 7
13. 25y ϩ 2 ϭ - 1
14. 24n Ϫ 3 Ϫ 4 ϭ 0
15. 26x Ϫ 5 Ϫ 3 ϭ 0
16. 25x ϩ 3 ϭ -4
17. 52x ϭ 30
18. 62x ϭ 42
19. 23a Ϫ 2 ϭ 22a ϩ 4
20. 24a ϩ 3 ϭ 25a Ϫ 4
21. 27x Ϫ 3 ϭ 24x ϩ 3
408
Chapter 9 • Roots and Radicals
Solve Problems 41– 43 by using formulas that are radical
equations. (Objective 3)
22. 28x Ϫ 6 ϭ 24x ϩ 11
23. 22y ϩ 1 ϭ 5
24. 32y Ϫ 2 ϭ 4
25. 2x ϩ 3 ϭ x ϩ 3
26. 2x ϩ 7 ϭ x ϩ 7
27. 2-2x ϩ 28 ϭ x Ϫ 2
28. 2-2x ϭ x ϩ 4
29. 23n Ϫ 4 ϭ 2n
30. 25n Ϫ 1 ϭ 22n
31. 23x ϭ x Ϫ 6
32. 22x ϭ x Ϫ 3
33. 42x ϩ 5 ϭ x
34. 2-x Ϫ 6 ϭ x
35. 2x2 ϩ 27 ϭ x ϩ 3
41. Go to the formula given in the solution to Example 9;
use a coefficient of friction of 0.95. How far will a car
skid at 40 miles per hour? At 55 miles per hour? At 65
miles per hour? Express the answers to the nearest foot.
L
for L. (Remember that
B 32
in this formula, which was introduced in Section 9.1,
T represents the period of a pendulum expressed in
seconds, and L represents the length of the pendulum in
feet.)
42. Solve the formula T ϭ 2p
36. 2x2 Ϫ 35 ϭ x Ϫ 5
37. 2x2 ϩ 2x ϩ 3 ϭ x ϩ 2
43. In Problem 42 you should have obtained the equation
8T 2
L ϭ 2 . What is the length of a pendulum that has
p
a period of 2 seconds? Of 2.5 seconds? Of 3 seconds?
Use 3.14 as an approximation for p, and express the
answers to the nearest tenth of a foot.
38. 2x2 ϩ x ϩ 4 ϭ x ϩ 3
39. 28x Ϫ 2 ϭ x
40. 22x Ϫ 4 ϭ x Ϫ 6
Thoughts Into Words
44. Explain in your own words why possible solutions for
radical equations must be checked.
as follows:
(3 ϩ 22x)2 ϭ x 2
45. Your friend attempts to solve the equation
9 ϩ 122x ϩ 4x ϭ x 2
3 ϩ 22x ϭ x
At this step, she stops and doesn’t know how to proceed.
What help can you give her?
Further Investigations
To solve an equation with cube roots, we raise each side of
the equation to the third power. Consider the following
example:
For Problems 46 – 49, solve each of the equations.
46. 2x ϩ 1 ϭ 5 Ϫ 2x Ϫ 4
47. 2x Ϫ 2 ϭ 2x ϩ 7 Ϫ 1
3
2
xϪ4ϭ5
48. 22n Ϫ 1 ϩ 2n Ϫ 5 ϭ 3
49. 22n ϩ 1 Ϫ 2n Ϫ 3 ϭ 2
50. Suppose that we solve the equation x 2 ϭ a2 for x as
follows:
3
(2
x Ϫ 4) 3 ϭ 53
x Ϫ 4 ϭ 125
x ϭ 129
For Problems 51–55, solve each equation.
x2 Ϫ a2 ϭ 0
1x ϩ a21x Ϫ a2 ϭ 0
x ϩ a ϭ 0
or x Ϫ a ϭ 0
x ϭ -a or
xϭa
3
52. 2x Ϫ 12 ϭ -2
3
3
54. 24x Ϫ 5 ϭ 3
51. 2x ϩ 7 ϭ 4
3
53. 22x Ϫ 4 ϭ 6
3
55. 24x Ϫ 8 ϭ 10
What is the significance of this result?
Answers to the Concept Quiz
1. True
2. False
3. True
4. True
5. True
6. False
7. True
8. False
9. False
10. True