## Algebra 2 Common Core

$7\sqrt{2}$
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: $\sqrt{9\cdot2}+\sqrt{16\cdot2}$ $=\sqrt{9}\cdot\sqrt{2}+\sqrt{16}\cdot \sqrt{2}$ $=3\sqrt{2}+4\sqrt{2}$ $=7\sqrt{2}$