9514 1404 393
Explanation:
Both sides of the equation are squared.
[tex]\dfrac{x-12}{4}=\sqrt{x} \qquad\text{given}\\\\ \left(\dfrac{x-12}{4}\right)^2=x \qquad\text{square both sides}\\\\ \left(\dfrac{x}{4}-3\right)^2=x \qquad\text{divide out the fraction}\\\\ \dfrac{x^2}{4^2}-\dfrac{6}{4}x+9=x \qquad\text{$(a-b)^2=a^2-2ab+b^2$}\\\\\dfrac{x^2}{16}-\dfrac{5x}{2}+9=0 \qquad\text{subtract $x$}[/tex]
I like this better without fractions, so would multiply by 16 at this point (or earlier).
x^2 -40x +144 = 0
(x -36)(x -4) = 0
x = 4 or 36 . . . . . . x = 4 is extraneous
The solution is x = 36.
