# How do you evaluate square root of 9x^4 + square root of 16x?

$\sqrt{9 {x}^{4}} + \sqrt{16 x}$
$3 {x}^{2} + 4 \sqrt{x}$
$\sqrt{9 {x}^{4}}$ is a perfect square which is $\sqrt{\left(3 {x}^{2}\right) \left(3 {x}^{2}\right)}$ and the $\sqrt{16 x}$ is equal to $\sqrt{\left({4}^{2}\right) \left(x\right)}$. Therefore the answer is $3 {x}^{2}$ + $4 \sqrt{x}$.