## Elementary Algebra

$x=12$
Squaring both sides of the equation and then using the properties of equality, we obtain: \begin{array}{l}\require{cancel}\left( \sqrt{3x} \right)^2=\left( x-6 \right)^2 \\\\ 3x=(x)^2+2(x)(-6)+(-6)^2 \\\\ 3x=x^2-12x+36 \\\\ 0=x^2+(-12x-3x)+36 \\\\ x^2-15x+36=0 \\\\ (x-12)(x-3)=0 \\\\ x=\{3,12\} .\end{array} Upon checking, only $x=12$ satisfies the original equation.