# How do you solve sqrt(x+4) = sqrt(3x) and find any extraneous solutions?

Jun 12, 2016

$\sqrt{x + 4} = \sqrt{3 x}$

${\left(\sqrt{x + 4}\right)}^{2} = {\left(\sqrt{3 x}\right)}^{2}$

$x + 4 = 3 x$

$4 = 3 x - x$

$4 = 2 x$

$2 = x$

Checking back in the original equation, we have that $\sqrt{6} = \sqrt{6}$, so the solution works.

Our solution set is $\left\{x = 2\right\}$.

Hopefully this helps!