Answer
$48\sqrt{2x}$
Work Step by Step
Factor each radicand so that one factor is a perfect square:
$5\sqrt{16\cdot2x}+4\sqrt{49\cdot2x}$
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:
$5\sqrt{16\cdot2x}+4\sqrt{49\cdot2x}$
$=5\sqrt{16}\cdot \sqrt{2x}+4\sqrt{49}\cdot \sqrt{2x}$
Recall that $4^2=16$ and $7^2=49$:
Thus, the expression above simplifies to:
$5\cdot4\sqrt{2x}+4\cdot7\sqrt{2x}$
$=20\sqrt{2x}+28\sqrt{2x}$
$=(20+28)\sqrt{2x}$
$=48\sqrt{2x}$
