# How do you rationalize the denominator and simplify 3/(2-sqrt6)?

Jun 16, 2016

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

#### Explanation:

To rationalize denominator and simplify $\frac{3}{2 - \sqrt{6}}$,

one should multiply numerator and denominator of the fraction by conjugate irrational number of denominator $2 - \sqrt{6}$, which is $2 + \sqrt{6}$.

Hence $\frac{3}{2 - \sqrt{6}} = \frac{3}{2 - \sqrt{6}} \times \frac{2 + \sqrt{6}}{2 + \sqrt{6}}$

= $\frac{3 \left(2 + \sqrt{6}\right)}{\left(2 - \sqrt{6}\right) \left(2 + \sqrt{6}\right)}$

= $\frac{6 + 3 \sqrt{6}}{{2}^{2} - {\left(\sqrt{6}\right)}^{2}}$

= $\frac{6 + 3 \sqrt{6}}{4 - 6}$

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

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