# How do you rationalize the denominator and simplify 2/(sqrt6-sqrt5)?

Apr 10, 2017

$\frac{2}{\sqrt{6} - \sqrt{5}} = 2 \sqrt{6} + 2 \sqrt{5}$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Using this with $a = \sqrt{6}$ and $b = \sqrt{5}$, we can multiply both numerator and denominator by $\sqrt{6} + \sqrt{5}$ to eliminate the square roots from the denominator...

$\frac{2}{\sqrt{6} - \sqrt{5}} = \frac{2 \left(\sqrt{6} + \sqrt{5}\right)}{\left(\sqrt{6} - \sqrt{5}\right) \left(\sqrt{6} + \sqrt{5}\right)}$

$\textcolor{w h i t e}{\frac{2}{\sqrt{6} - \sqrt{5}}} = \frac{2 \sqrt{6} + 2 \sqrt{5}}{{\left(\sqrt{6}\right)}^{2} - {\left(\sqrt{5}\right)}^{2}}$

$\textcolor{w h i t e}{\frac{2}{\sqrt{6} - \sqrt{5}}} = \frac{2 \sqrt{6} + 2 \sqrt{5}}{6 - 5}$

$\textcolor{w h i t e}{\frac{2}{\sqrt{6} - \sqrt{5}}} = \frac{2 \sqrt{6} + 2 \sqrt{5}}{1}$

$\textcolor{w h i t e}{\frac{2}{\sqrt{6} - \sqrt{5}}} = 2 \sqrt{6} + 2 \sqrt{5}$