# Thread: Solve x^.5 = 6 - x

1. ## Solve x^.5 = 6 - x

What I have done is squared both sides which gives
x=(6-x)^2 =
0 = x^2-13x+36
0=(x-4)(x-9). x=4,x=9. Of these x=4 fits the original function.

Is there a better way of doing this??

2. Originally Posted by p75213
What I have done is squared both sides which gives
x=(6-x)^2 =
0 = x^2-13x+36
0=(x-4)(x-9). x=4,x=9. Of these x=4 fits the original function.

Is there a better way of doing this??
$\sqrt{x} = 6 - x$

$x = (6 - x)^2$

$x = x^2 - 12x + 36$

$0 = x^2 - 13x + 36$

$0 = x^2 - 9x - 4x + 36$

$0 = x(x - 9) - 4(x - 9)$

$0 = (x - 9)(x - 4)$

So $x - 9 = 0$ or $x - 4 = 0$.

Therefore $x = 9$ or $x = 4$.

By the way, both values of $x$ fit the function...

3. I see your point. Thanks for that.