Answer
$\{3\}$
Work Step by Step
Square both sides of the equation.
$\left(\sqrt{12-x}\right)^2=(x)^2$
$12-x=x^2$
Add $x-12$ to both sides.
$12-x+x-12=x^2+x-12$
Simplify.
$0=x^2+x-12$
Rewrite $x$ as $4x-3x$.
$0=x^2+4x-3x-12$
Group the terms.
$0=(x^2+4x)+(-3x-12)$
Factor out $x$ in the first group and $-3$ in the second group.
$0=x(x+4)-3(x+4)$
Factor out $(x+4)$.
$0=(x+4)(x-3)$
Use Zero-Product Property, which states that if $ab=0$, then either $a=0$, $b=0$, or both are $0$.
$x+4=0\quad $ or $\quad x-3=0$
Solve for $x$ in each equation:
$x=-4\quad $ or $\quad x=3$
Check:
For $x=-4$:
$\sqrt{12-(-4)}=-4$
$\sqrt{12+4}=-4$
$\sqrt{16}=-4$
$\sqrt{4^2}=-4$
$4=-4$ False. This means $-4$ is an extraneous solution.
For $x=3$:
$\sqrt{12-(3)}=3$
$\sqrt{12-3}=3$
$\sqrt{9}=3$
$\sqrt{3^2}=3$
$3=3$ True.
Hence, the solution set is $\{3\}$.
