WHEN SOLVING EQUATIONS WE PERFORM THE INVERSE OPERATION.

x + 5 = 10 5x = 10 x = 5 - 5 - 5 5 5 ( x ) 2 = ( 5 )2

x = 5 x = 2 x = 25

(5) + 5 = 10 5(2) = 10 25 = 5 10 = 10 10 = 10 5 = 5

Recall:

Solve: a1/2 = 8

a1 = a

8

Check[(-2x+6)1/5]5 = [(-8+10x) 1/5]5

-2x + 6 = -8 + 10x -12x = -14 x = 7/6

Raise both sides to the 9th power.

Raise both sides to the 2/3 power.

-4x – 6 = -2 + 3x-7x = 4x = -4/7

Check: [-4(-4/7)-6]1/9 =[(-2+3(-4/7)] 1/9

[(3x-8)3/2]2/3= 82/3

3x-8 = 43x = 12x = 4

Check: [3(4)-8]3/2 =8(4)3/2 =8 8 = 8

After you add the radical to the Right Side,

Square Both Sides

3x+2 = 5x -10 -2x = -12 x = 6

Check: (3*6 + 2) =(5*6-10)(20) =(20)

Square Both Sides

x2 = -3x + 40

Solve the quadratic equation! x2 + 3x – 40 = 0(x+8)(x-5) = 0x = -8 and 5Check: (Use the Original Equation)-8 = (-3*-8+40)-8 = 44

5= (-3*5+40)5= 25

The only solution is x = 5. x=-8 is an extraneous solution!

Example 9 Example 10Example 9 Example 10