1. Remember!
For a square root, the index of the radical is 2.

2. Example 1A: Solving Equations Containing One Radical
Solve each equation.
Check
Subtract 5.
Simplify.
Square both sides. 
Simplify.
Solve for x.

3. Example 1B: Solving Equations Containing One Radical
Solve each equation.
Check 3 7
7 5x =
Divide by 7. 7
Simplify.
Cube both sides. 
Simplify.
Solve for x.

4. Example 4A: Solving Equations with Rational Exponents
Solve each equation. 1 3
(5x + 7) = 3
Write in radical form.
Cube both sides.
5x + 7 = 27
Simplify.
5x = 20
Factor.
x = 4
Solve for x.

5. Example 2: Solving Equations Containing Two Radicals
Solve
Square both sides.
7x + 2 = 9(3x – 2)
Simplify.
7x + 2 = 27x – 18
Distribute.
20 = 20x
Solve for x.
1 = x

6. Example 3 Continued
Method 2 Use algebra to solve the equation.
Step 1 Solve for x.
Square both sides.
–3x + 33 = 25 – 10x + x2
Simplify.
0 = x2 – 7x – 8
Write in standard form.
0 = (x – 8)(x + 1)
Factor.
x – 8 = 0 or x + 1 = 0
Solve for x.
x = 8 or x = –1

7. Example 3 Continued
Method 2 Use algebra to solve the equation.
Step 2 Use substitution to check for extraneous solutions.
3 –3 x 
Because x = 8 is extraneous, the only solution is x = –1.

8. Check It Out! Example 3a Continued
Method 2 Use algebra to solve the equation.
Step 1 Solve for x.
Square both sides.
Simplify.
2x + 14 = x2 + 6x + 9
0 = x2 + 4x – 5
Write in standard form.
Factor.
0 = (x + 5)(x – 1)
x + 5 = 0 or x – 1 = 0
Solve for x.
x = –5 or x = 1

9. Check It Out! Example 3a Continued
Method 1 Use algebra to solve the equation.
Step 2 Use substitution to check for extraneous solutions. –2 x 
Because x = –5 is extraneous, the only solution is x = 1.

10. HW pg. 633
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