1. 1. 64 2. (–32) Lesson 6.6, For use with pages 452-459
Evaluate the expression.
1. 64 2 3
ANSWER 16
2. (–32) 3 5
ANSWER –8

2. 3. Expand (x – 5)2 x2 – 10x + 25 4. x2 + 6x + 9 = x + 45 –9, 4
Lesson 6.6, For use with pages
Evaluate the expression.
3. Expand (x – 5)2
ANSWER
x2 – 10x + 25
Solve the equation.
4. x2 + 6x + 9 = x + 45
ANSWER
–9, 4

3. π 5. A cone-shaped cup has a height of 7 cm and a radius
Lesson 6.6, For use with pages
Solve the equation
5. A cone-shaped cup has a height of 7 cm and a radius
of 2.5 cm. Use V = r2h to find the volume of the cup. 1 – 3 π
ANSWER
about 45.8 cm3

4. Solving Radical Equations
6.6 Intermediate Algebra
Solving Radical Equations

5. Write original equation.
EXAMPLE 1
Solve a radical equation
Solve = 3. 
2x+7 
2x+7 3
= 3
Write original equation.
= 33 
2x+7 3
( )3
Cube each side to eliminate the radical.
2x+7
= 27
Simplify. 2x
= 20
Subtract 7 from each side. x
= 10
Divide each side by 2.

6. Substitute 10 for x. Simplify. Solution checks. EXAMPLE 1
Solve a radical equation
CHECK
Check x = 10 in the original equation. 
2(10)+7 3 = ?
Substitute 10 for x. 27  3 = ?
Simplify.
= 3 3
Solution checks.

7. GUIDED PRACTICE
for Example 1
Solve equation. Check your solution.
√ x – 9 = –1
3. (23 x – 3 ) = 4
ANSWER
x = 512
ANSWER
x = 11
2. ( x+25 ) = 4
x = –9
ANSWER

8. EXAMPLE 2
Solve a radical equation given a function
Wind Velocity
where p is the air pressure (in millibars) at the center of the hurricane. Estimate the air pressure at the center of a hurricane when the mean sustained wind velocity is 54.5 meters per second.
In a hurricane, the mean sustained wind velocity v (in meters per second) is given by 
v(p) = – p

9. Subtract 1013 from each side.
EXAMPLE 2
Solve a radical equation given a function
SOLUTION
= – p 
v(p)
Write given function. 
1013 – p
= 6.3
54.5
Substitute 54.5 for v(p).
8.65 
1013 – p
Divide each side by 6.3.
(8.65)2 
1013 – p 2
Square each side.
74.8
1013 –p
Simplify.
– 938.2 –p
Subtract 1013 from each side.
938.2 p
Divide each side by –1.

10. EXAMPLE 2
Solve a radical equation given a function
The air pressure at the center of the hurricane
is about 938 millibars.
ANSWER

11. GUIDED PRACTICE
for Example 2
4. What If? Use the function in Example 2 to estimate the air pressure at the center of a hurricane when the mean sustained wind velocity is 48.3 meters per second.
about 954 mille bars.
ANSWER

12. Write original equation.
EXAMPLE 3
Standardized Test Practice
SOLUTION
4x2/3 = 36
Write original equation.
x2/3 = 9
Divide each side by 4.
(x2/3)3/2 = 93/2
Raise each side to the power . 3 2
x = 27
Simplify.
ANSWER
The correct answer is D.

13. Write original equation.
EXAMPLE 4
Solve an equation with a rational exponent
Solve (x + 2)3/4 – 1 = 7.
(x + 2)3/4 – 1 = 7
Write original equation.
(x + 2)3/4 = 8
Add 1 to each side.
(x + 2)3/4
4/3
= 8 4/3
Raise each side to the power . 4 3
x + 2 = (8 1/3)4
Apply properties of exponents.
x + 2 = 24
Simplify.
x + 2 = 16
Simplify.
x = 14
Subtract 2 from each side.

14. EXAMPLE 4
Solve an equation with a rational exponent
The solution is 14. Check this in the original equation.
ANSWER

15. GUIDED PRACTICE
for Examples 3 and 4
Solve the equation. Check your solution.
x3/2 = 375
8. (x + 3)5/2 = 32
ANSWER
x = 25
ANSWER
x = 1
6. –2x3/4 = –16
9. (x – 5)5/3 = 243
ANSWER
x = 16
ANSWER
x = 32 2 3
x1/5 –
= –2 7.
(x + 2)2/3 +3 = 7
x = 243
ANSWER
ANSWER
x = –10 or 6

16. Write original equation.
EXAMPLE 5
Solve an equation with an extraneous solution 
Solve x + 1 = x + 15 
x + 1 = 7x + 15
Write original equation.
(x + 1)2 
= ( 7x + 15)2
Square each side.
x2 + 2x + 1 = 7x + 15
Expand left side and simplify
right side.
x2 – 5x – 14 = 0
Write in standard form.
(x – 7)(x + 2) = 0
Factor.
x – 7 = 0 or x + 2 = 0
Zero-product property
x = 7 or x = –2
Solve for x.

17. EXAMPLE 5
Solve an equation with an extraneous solution
CHECK
Check x = 7 in the
original equation.
Check x = –2 in the
x + 1 = 7x + 15 
x + 1 = 7x + 15 
7 + 1 ? 
= 7(7) + 15
–2 + 1 ? 
= 7(–2) + 15 8
= 64  ? –1
= 1  ?
8 = 8
= 1 / –1
The only solution is 7.
(The apparent solution –2 is extraneous.)
ANSWER

18. Write original equation.
EXAMPLE 6
Solve an equation with two radicals x 
= – x .
Solve
SOLUTION
METHOD 1
Solve using algebra. x  
= 3 – x
Write original equation. x  2 
= – x 2
Square each side.  x x
= 3 – x
Expand left side and simplify
right side. 
2 x + 2
= –2x
Isolate radical expression.

19. Zero-product property.
EXAMPLE 6
Solve an equation with two radicals 
x + 2
= –x
Divide each side by 2. 
x + 2 2
= ( –x)2
Square each side again.
x + 2
= x2
Simplify.
= x2 – x – 2
Write in standard form.
= (x – 2)(x + 1)
Factor.
x – 2 = 0 or
x + 1 = 0
Zero-product property.
x = 2 or
x = –1
Solve for x.

20. EXAMPLE 6
Solve an equation with two radicals
Check x = 2 in the
original equation.
Check x = – 1 in the
original equation. x 
= 3 – x x 
= 3 – x 
= 3 – 2 ? – 
= 3 – (–1) ? 
4 +1
= 1 ? 1  +1
= 4 ?
= –1 / 3
2 = 2
The only solution is 1.
(The apparent solution 2 is extraneous.)
ANSWER

21. EXAMPLE 6
Solve an equation with two radicals
METHOD 2
Use: a graph to solve the equation. Use a graphing calculator to graph y1 = 𝑥 and y2 = 3−𝑥 . Then find the intersection points of the two graphs by using the intersect feature. You will find that the only point of intersection is (–1, 2). Therefore, –1 is the only solution of the equation x
3 – x =

22. GUIDED PRACTICE
for Examples 5 and 6
Solve the equation. Check for extraneous solutions
11. x – 1 2 = 4 1 x
13.
2x + 5
x + 7 =
ANSWER
x = 9
ANSWER
x = 2
x + 6 – 2
x – 2
14. =
12.
10x + 9 = x + 3
ANSWER
x = 0,4
ANSWER
x = 3