Answer
$13\sqrt{2}$
Work Step by Step
Factor each radicand so that one factor is a perfect square:
$\sqrt{36\cdot2}+\sqrt{16\cdot2}+\sqrt{9\cdot2}$
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:
$\sqrt{36\cdot 2}+\sqrt{16\cdot 2}+\sqrt{9\cdot2}$
$=\sqrt{36}\cdot \sqrt{2}+\sqrt{16}\cdot \sqrt{2}+\sqrt{9}\cdot \sqrt{2}$
Recall that $6^2=36$, $4^2=16$, and $3^2=9$.
Thus, the expression above simplifies to:
$6\sqrt{2}+4\sqrt{2}+3\sqrt{2}$
$=(6+4+3)\sqrt{2}$
$=13\sqrt{2}$
