# How do you rationalize the denominator and simplify 1/(sqrt3 - sqrt5 - 2)?

Apr 22, 2016

$\frac{1}{\sqrt{3} - \sqrt{5} - 2} = - \frac{10}{47} - \frac{4}{47} \sqrt{15} + \frac{7}{47} \sqrt{3} - \frac{1}{47} \sqrt{5}$

#### Explanation:

Multiply numerator and denominator by:

$\left(\sqrt{3} + \sqrt{5} - 2\right) \left(\sqrt{3} - \sqrt{5} + 2\right) \left(\sqrt{3} + \sqrt{5} + 2\right)$

First note that:

$\left(\sqrt{3} - \sqrt{5} + 2\right) \left(\sqrt{3} + \sqrt{5} + 2\right)$

$= \left(\left(\sqrt{3} + 2\right) - \sqrt{5}\right) \left(\left(\sqrt{3} + 2\right) + \sqrt{5}\right)$

$= {\left(\sqrt{3} + 2\right)}^{2} - {\left(\sqrt{5}\right)}^{2}$

$= 2 + 4 \sqrt{3} + 4 - 5$

$= 1 + 4 \sqrt{3}$

Similarly:

$\left(\sqrt{3} - \sqrt{5} - 2\right) \left(\sqrt{3} + \sqrt{5} - 2\right)$

$= \left(\left(\sqrt{3} - 2\right) - \sqrt{5}\right) \left(\left(\sqrt{3} - 2\right) + \sqrt{5}\right)$

$= {\left(\sqrt{3} - 2\right)}^{2} - {\left(\sqrt{5}\right)}^{2}$

$= 2 - 4 \sqrt{3} + 4 - 5$

$= 1 - 4 \sqrt{3}$

So the denominator becomes:

$\left(1 - 4 \sqrt{3}\right) \left(1 + 4 \sqrt{3}\right) = {1}^{2} - {\left(4 \sqrt{3}\right)}^{2} = 1 - 48 = - 47$

Meanwhile, the numerator becomes:

$\left(\sqrt{3} + \sqrt{5} - 2\right) \left(1 + 4 \sqrt{3}\right)$

$= \sqrt{3} + \sqrt{5} - 2 + 4 \sqrt{3} \left(\sqrt{3} + \sqrt{5} - 2\right)$

$= \sqrt{3} + \sqrt{5} - 2 + 12 + 4 \sqrt{15} - 8 \sqrt{3}$

$= 10 + 4 \sqrt{15} - 7 \sqrt{3} + \sqrt{5}$

So:

$\frac{1}{\sqrt{3} - \sqrt{5} - 2} = \frac{10 + 4 \sqrt{15} - 7 \sqrt{3} + \sqrt{5}}{- 47}$

$= - \frac{10}{47} - \frac{4}{47} \sqrt{15} + \frac{7}{47} \sqrt{3} - \frac{1}{47} \sqrt{5}$