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

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) (Show Source): You can put this solution on YOUR website!
1)

.
.
.
2)

.
.
.
3)

and
.
.
.
4)

and
Answer by nyc_function(2741) (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.

RELATED QUESTIONS
Solve by taking the square roots of both sides. (x + 4)^2 - 5 = 2 (answered by josgarithmetic)
solve and check the quadratic equation below by taking the square root of both sides.... (answered by stanbon)
Which of the following points is on the graph of the equation y = x2 ? A. (-2, 4) (answered by rfer)
Solve by completing the square: 4x^2 + 2x - 3 = 0 4x^2 + 2x = 3 x^2 + 2/4x = 3/4... (answered by vertciel)
Solve by completing the square: 4x^2 + 2x - 3 = 0 4x^2 + 2x = 3 x^2 + 2/4x = 3/4... (answered by stanbon)
Solve by completing the square: 4x^2 + 2x - 3 = 0 4x^2 + 2x = 3 x^2 + 2/4x = 3/4... (answered by bucky)
Please help me solve by taking roots of of both sides: y^2=16... (answered by Earlsdon,solver91311)
Solve the equation 4(x+6)^2=160 by taking the square... (answered by rfer)
Solve by completing the square. x^2 + (2/3)x - 1/3 = 0 Let me see. x^2 +... (answered by timofer,MathTherapy)