Answer
$7\sqrt{2}$
Work Step by Step
Factor each radicand so that one factor is a perfect square:
$\sqrt{9\cdot2}+\sqrt{16\cdot2}$
Recall the property (pg. 367):
$\sqrt[n]{a}\cdot\sqrt[n]{b}=\sqrt[n]{ab}$ (if $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers)
Applying this property, we get:
$\sqrt{9\cdot2}+\sqrt{16\cdot2}$
$=\sqrt{9}\cdot\sqrt{2}+\sqrt{16}\cdot \sqrt{2}$
$=3\sqrt{2}+4\sqrt{2}$
$=7\sqrt{2}$
