# How do you evaluate 5√1/2 - 2√1/8?

Nov 9, 2015

First, $\sqrt{\frac{1}{2}} = \frac{\sqrt{1}}{\sqrt{2}} = \frac{1}{\sqrt{2}}$ and also $\sqrt{\frac{1}{8}} = \frac{1}{\sqrt{8}}$:

$5 \cdot \sqrt{\frac{1}{2}} - 2 \cdot \sqrt{\frac{1}{8}}$
$= \frac{5}{\sqrt{2}} - \frac{2}{\sqrt{8}}$
$= \frac{5}{\sqrt{2}} - \frac{2}{\sqrt{4 \cdot 2}}$
$= \frac{5}{\sqrt{2}} - \frac{2}{\sqrt{4} \cdot \sqrt{2}}$
$= \frac{5}{\sqrt{2}} - \frac{2}{2 \cdot \sqrt{2}}$
$= \frac{5}{\sqrt{2}} - \frac{\cancel{2}}{\cancel{2} \cdot \sqrt{2}}$
$= \frac{5}{\sqrt{2}} - \frac{1}{\sqrt{2}}$
$= \frac{5 - 1}{\sqrt{2}}$
$= \frac{4}{\sqrt{2}}$

This might already be the final solution. However, it is nicer not to have any radicals in the denominator, so you could transform it further:

$\frac{4}{\sqrt{2}} = \frac{4 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{4 \cdot \sqrt{2}}{2} = 2 \cdot \sqrt{2}$