Name ________________________________________ Date __________________ Class__________________
Reteach
LESSON
5-8
Solving Radical Equations and Inequalities
Solve radical equations by raising both sides of the equation to the power
of the index of the radical. For example, the index of n a is n. Therefore,
x 3
 x
2
The index of x is 2. Raise
both sides to the power of 2.
 32
x9
Solve: 3 x  2  18
Step 1
Isolate the radical.
Divide both sides of the equation by 3 and simplify.
3 x  2 18

3
3
x2 6
Step 2
Square both sides of the equation and simplify.

x2

2
 62
Remember:
x  2  36
Step 4
 a
n
n
a.
Solve.
x  38
Step 5
Check.
3 x  2  18
Always check for extraneous solutions
when solving radical equations.
3 38  2  3 36  3  6   18
Solve each equation. Check your answer.
1. 4 3 2 x  11  12
4 3 2 x  11 12

4
4
3

2 x  11  3
3

2 x  11
3
 33
2. 5  x  3  9
3. 2 x  4  10
55 x 3 95
____________________
___________________
____________________
___________________
____________________
________________________
_________________________
________________________
________________________
_________________________
________________________
________________________
_________________________
________________________
________________________
_________________________
________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
5-62
Holt McDougal Algebra 2
Name ________________________________________ Date __________________ Class__________________
LESSON
5-8
Reteach
Solving Radical Equations and Inequalities (continued)
Solving equations with rational exponents is similar to solving radical equations.
1
Solve: x   x  20  2 .
Step 1
n
Raise both sides to the reciprocal power.
1


x 2   x  20  2 


Step 2
 1
Think:  a n   a
 
2
The reciprocal of
Square both sides.
x2  x  20
Step 3
Write the quadratic equation in standard form.
x2  x  20  0
Step 4
1
is 2.
2
Set one side of the
equation equal to zero.
Factor.
(x  4) (x  5)  0
Step 5
Solve.
(x  4)  0
or
(x  5)  0
x5
x  4
Step 6
Check for extraneous solutions.
This is the only solution.
1
x   x  20  2
x5
x  4
1
4 ?  4  20  2
1
4  16  2 1
5 ?  5  20  2
1
5   25  2 
Solve each equation.
1
4.  5 x  6  4  3
4
1
5.  6 x  8  3  4
1
6. x   x  6  2
1


4
4
 5 x  6    3
____________________
____________________
________________________
_________________________
________________________
________________________
_________________________
________________________
________________________
_________________________
________________________
________________________
_________________________
________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
5-63
Holt McDougal Algebra 2
7. x =
5
2
1
⎡
⎤
6. x 2 = ⎢( x + 6) 2 ⎥
⎣
⎦
8. x = 9; x = −2 is an extraneous solution.
9. −
5
≤x≤1
4
12. x > −21
13. x > 4
14. −2 ≤ x ≤ 123
15. v =
x2 = x + 6
x2 − x − 6 = 0
10. x ≥ 5
11. 7 ≤ x ≤ 16
(x − 3)(x + 2) = 0
x=3
Challenge
3
c
2
Reteach
1. 2x + 11 = 27
1. 15.25
2. 9
3. No solution
4. 19
5. 5
6. 3
7. 5 or −
2x = 16; x = 8
4 3 2(8) + 11 = 12
8. 8 or −1
10. 6
Problem Solving
1. Directly
x −3 =4
x − 3 = 16
2. a. d =
x = 19
s2
30f
b. About 58 ft
5 + 19 − 3 = 5
c. No; possible answer: his skid marks
were only 52 ft, not 58 ft.
5 + 16 = 5 + 4
d. About 33 mi/h
=99
3.
1
9
9. 7
4 3 36 = 12 9
2.
2
3. a. About 9 ft
x +4 =5
b. By at least 15 ft
x + 4 = 25
4. B
x = 21
5. A
Reading Strategies
2 21 + 4 =
2 25 = 2 ⋅ 5
= 10 9
4. 5x + 6 = 81
1.
x = −3
2.
x+2 =6
3.
x = −2
4.
x = 18
5.
x+6 =2
6. Third power
7. Second Power
5x = 75
9. Third power
x = 15
8. Fourth power
10. x = 64
11. x = 7
3
1
⎡
⎤
5. ⎢(6 x − 8) 3 ⎥ = 43
⎣
⎦
6x − 8 = 64
6x = 72
x = 12
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A67
Holt McDougal Algebra 2