# How do you rationalize the denominator and simplify 1/(sqrt2-sqrt8)?

Mar 6, 2018

See a solution process below:

#### Explanation:

Multiply the fractions by this form of $1$ to rationalize the denominator by removing the radicals from the denominator:

$\frac{\textcolor{red}{\sqrt{2}} + \textcolor{red}{\sqrt{8}}}{\textcolor{red}{\sqrt{2}} + \textcolor{red}{\sqrt{8}}}$

This gives:

$\frac{\textcolor{red}{\sqrt{2}} + \textcolor{red}{\sqrt{8}}}{\textcolor{red}{\sqrt{2}} + \textcolor{red}{\sqrt{8}}} \times \frac{1}{\sqrt{2} - \sqrt{8}} \implies$

$\frac{\sqrt{2} + \sqrt{8}}{\left(\textcolor{red}{\sqrt{2}} + \textcolor{red}{\sqrt{8}}\right) \times \left(\sqrt{2} - \sqrt{8}\right)} \implies$

$\frac{\sqrt{2} + \sqrt{8}}{\left(\textcolor{red}{\sqrt{2}} \sqrt{2} - \textcolor{red}{\sqrt{2}} \sqrt{8} + \textcolor{red}{\sqrt{8}} \sqrt{2} - \textcolor{red}{\sqrt{8}} \sqrt{8}\right)} \implies$

$\frac{\sqrt{2} + \sqrt{8}}{{\sqrt{2}}^{2} - \textcolor{red}{\sqrt{2}} \sqrt{8} + \sqrt{2} \textcolor{red}{\sqrt{8}} - {\sqrt{8}}^{2}} \implies$

$\frac{\sqrt{2} + \sqrt{8}}{2 - 0 - 8} \implies$

$\frac{\sqrt{2} + \sqrt{8}}{- 6} \implies$

$- \frac{\sqrt{2} + \sqrt{8}}{6}$