# What is the square root of 8 divided by square root of 5 minus square root of 2?

##### 1 Answer
Mar 23, 2018

$\frac{2 \sqrt{10} + 4}{3}$

#### Explanation:

$\frac{\sqrt{8}}{\sqrt{5} - \sqrt{2}}$

$\therefore \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}} = 1$

$\therefore = \frac{\sqrt{8}}{\sqrt{5} - \sqrt{2}} \times \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}}$

$\frac{\sqrt{8} \left(\sqrt{5} + \sqrt{2}\right)}{\left(\sqrt{5} - \sqrt{2}\right) \left(\sqrt{5} + \sqrt{2}\right)}$

$\therefore = \frac{\sqrt{8} \left(\sqrt{5} + \sqrt{2}\right)}{3}$

$\therefore = \frac{\sqrt{8} \sqrt{5} + \sqrt{8} \sqrt{2}}{3}$

$\therefore = \frac{\sqrt{8 \cdot 5} + \sqrt{8 \cdot 2}}{3}$

$\therefore = \frac{\sqrt{40} + \sqrt{16}}{3}$

$\therefore = \frac{\sqrt{2 \cdot 2 \cdot 2 \cdot 5} + \sqrt{16}}{3}$

$\therefore = \sqrt{2} \cdot \sqrt{2} = 2$

$\therefore = \frac{\sqrt{2 \cdot 2 \cdot 2 \cdot 5} + 4}{3}$

$\therefore = \frac{2 \sqrt{2 \cdot 5} + 4}{3}$

$\therefore = \frac{2 \sqrt{10} + 4}{3}$