# Thread: [SOLVED] need help now

1. ## [SOLVED] need help now

How would you solve(for "x"):
the square root of 4x-12 = x

I know that you could square both sides to get rid of the square root on the left side, but I don't know how to finish the problem. (Hint: The small 2's means squared. x2 means x squared.)

square root of(4x-12)2 = (x)2

4x-12 = x2
4x = x2-12
4x-x2 = -12

What do you do after the last step?

2. Originally Posted by bubba1
How would you solve(for "x"):
the square root of 4x-12 = x

I know that you could square both sides to get rid of the square root on the left side, but I don't know how to finish the problem. (Hint: The small 2's means squared. x2 means x squared.)

square root of(4x-12)2 = (x)2

4x-12 = x2
4x = x2-12
4x-x2 = -12

What do you do after the last step?
You have,
$\displaystyle \sqrt{4x-12}=x$
Square both sides,
$\displaystyle 4x-12=x^2$
Thus,
$\displaystyle x^2-4x+12=0$
Thus, but this problem has no real solutions (only complex). Thus, there is no solution.