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

Jan 15, 2017

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

#### Explanation:

Simplify $\frac{2}{\sqrt{3} + \sqrt{2}}$

Rationalize the denominator using the difference of squares, ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, where $a = \sqrt{3}$ and $b = \sqrt{2}$.

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

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

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

Apply the rule ${\sqrt{a}}^{2} = a$

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

$\frac{2 \left(\sqrt{3} - \sqrt{2}\right)}{1} =$

$2 \left(\sqrt{3} - \sqrt{2}\right)$