# How do you rationalize the denominator and simplify sqrt3/(2+sqrt2)?

Dec 18, 2016

You multiply by $1$, disguised as $\frac{2 - \sqrt{2}}{2 - \sqrt{2}}$

#### Explanation:

$= \frac{\sqrt{3} \times \left(2 - \sqrt{2}\right)}{\left(2 + \sqrt{2}\right) \times \left(2 - \sqrt{2}\right)}$

The denominator is now a special product of the sort:
$\left(A + B\right) \left(A - B\right) = {A}^{2} - {B}^{2}$

$= \frac{\sqrt{3} \times \left(2 - \sqrt{2}\right)}{{2}^{2} - {\sqrt{2}}^{2}} = \frac{2 \sqrt{3} - \sqrt{2} \cdot \sqrt{3}}{4 - 2}$

$= \frac{2 \sqrt{3} - \sqrt{6}}{2} = \sqrt{3} - \frac{1}{2} \sqrt{6}$