1. SOLVING
RADICAL
EQUATIONS
By L.D.

2. TABLE OF CONTENTS
 Slide 3: Instructions
 Slide 4: Problem 1
 Slide 8: Example Problems
 Slide 10: Problem 2
 Slide 13: Problem 3
 Slide 17: Problem 4

3. INSTRUCTIONS
 Solve for x

4. PROBLEM 1
3 x–6=0

5. PROBLEM 1
3 x–6=0
The first step to finding x is to move the 6 to the other side of the
equation.
3 x–6=0
+6 +6
3 x=6

6. PROBLEM 1
3 x=6
The next step is to get rid of the number outside the radical x. The
way to do this is to divide out the 3.
3 x=6
÷3 ÷3
x=2

7. PROBLEM 1
x=2
The last thing we need to do is get rid of the square root symbol. To do
this we need to square the 2 and the radical x. Then we need to do the
square root.
( x)2 = (2)2
x2 = 4
x=4

8. EXAMPLE PROBLEMS
1. x – 7 = 1
2. 12 x – 3 = 0

9. EXAMPLE PROBLEMS
1. x – 7 = 1
+7 +7
x=8
( x) 2 = (8)2
x = 64
2. 12 x – 3 = 0
+3 +3
12 x = 3
÷12 ÷12
x = 1/4
( x) 2 = ( 1/4 )2
x = 1/16

10. PROBLEM 2
2 x + 6 + 9 = 21

11. PROBLEM 2
2 x + 6 + 9 = 21 First we need to do what
-9 -9 we already know.
2 x + 6 = 12
÷2 ÷2
x+6 =6

12. PROBLEM 2
( x + 6 )2 = (6)2 Now we need to square
x + 6 = 36 the WHOLE square root
and 6, then we need to use
-6 -6 the square root.
x = 30

13. PROBLEM 3
4x – 12 - x + 3 = 0

14. PROBLEM 3
4x – 12 - x + 3 = 0
First off we need to place one of the square roots on the other side
of the zero.
4x – 12 - x + 3 = 0
+ x+3 + x+3
4x – 12 = x + 3

15. PROBLEM 3
4x – 12 = x + 3
Next we need to get rid of the square root symbols.
( 4x – 12)2 = ( x + 3)2
4x – 12 = x + 3

16. PROBLEM 3
4x – 12 = x + 3
Now you need to make so like terms are on the same sides of the
equal sign.
4x – 12 = x + 3
-x + 12 -x + 12
3x = 15
Now we just need to get the x on its own.
3x = 15
÷3 ÷3
x=5

17. PROBLEM 4
4x – 3 - x = 0

18. PROBLEM 4
4x – 3 - x = 0
First we need to put them on separate sides of the equation.
4x – 3 - x = 0
+ x + x
4x – 3 = x

19. PROBLEM 4
4x – 3 = x
Now all that is left is to get rid of the square root and add the like
terms.
( 4x – 3)2 = ( x )2
4x – 3 = x
-4x -4x
-3 = -3x
÷-3 ÷-3
1=x