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??
$\displaystyle \sqrt{x} = 6 - x$
$\displaystyle x = (6 - x)^2$
$\displaystyle x = x^2 - 12x + 36$
$\displaystyle 0 = x^2 - 13x + 36$
$\displaystyle 0 = x^2 - 9x - 4x + 36$
$\displaystyle 0 = x(x - 9) - 4(x - 9)$
$\displaystyle 0 = (x - 9)(x - 4)$
So $\displaystyle x - 9 = 0$ or $\displaystyle x - 4 = 0$.
Therefore $\displaystyle x = 9$ or $\displaystyle x = 4$.
By the way, both values of $\displaystyle x$ fit the function...