SOLUTION: Solve by taking square root of BOTH sides of the equation. 1) x^2=16 2) x^2-36=0 3) (x-4)^2=25 4) 3(x+4)^2=12

Algebra ->  Square-cubic-other-roots -> SOLUTION: Solve by taking square root of BOTH sides of the equation. 1) x^2=16 2) x^2-36=0 3) (x-4)^2=25 4) 3(x+4)^2=12      Log On


   

Question 320117: Solve by taking square root of BOTH sides of the equation.
1) x^2=16
2) x^2-36=0
3) (x-4)^2=25
4) 3(x+4)^2=12
Found 2 solutions by Fombitz, nyc_function:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
1) x%5E2=16+
x=0+%2B-+4
.
.
.
2)+x%5E2-36=0
x%5E2=36
x=0+%2B-+6
.
.
.
3) %28x-4%29%5E2=25++
x-4=0+%2B-+5
x=4+%2B-+5
x=-1 and +x=9+
.
.
.
4) 3%28x%2B4%29%5E2=12
%28x%2B4%29%5E2=4
x%2B4=+0+%2B-+2
x=-4+%2B-+2
+x=-6 and x=-2

Answer by nyc_function(2741) About Me  (Show Source):
You can put this solution on YOUR website!
Post one question at a time or else no one will reply.
(1) x^2 = 16
Let sqrt = square root for short
sqrt{x^2} = sqrt{16}
x = 4
Done!
================================
(2) x^2 - 36 = 0
Add 36 to both sides.
x^2 - 36 + 36 = 0 + 36
x^2 = 36
Take square root of both sides.
sqrt{x^2} = sqrt{36}
x = 6
Done!
Now post questions 3 and 4 individually.