# How do you simplify (sqrt(6)-2)/(11+sqrt(6))?

Sep 23, 2015

$\frac{13 \sqrt{6} - 28}{115}$

#### Explanation:

$\frac{\sqrt{6} - 2}{11 + \sqrt{6}}$
$= \frac{\sqrt{6} - 2}{11 + \sqrt{6}} \cdot \frac{11 - \sqrt{6}}{11 - \sqrt{6}}$

this is a trick to get rid of the $\sqrt{6}$ of the $\left(11 + \sqrt{6}\right)$ in the denominator. We are basically constructing what is called the "conjugate" of the denominator, $\left(11 - \sqrt{6}\right)$, which is exactly the same thing as the denominator but with the opposite sign in the middle. If the denominator is $\left(a + b\right)$, the conjugate is $\left(a - b\right)$. If the denominator is $\left(a - b\right)$, the conjugate is $\left(a + b\right)$.

Why do we want to multiply by the conjugate?
because:
$\left(a + b\right) \cdot \left(a - b\right) = {a}^{2} - {b}^{2}$
so if a or b (or both) were square-roots, the results is squared and we are thereby getting rid of the square-root.

Note that we need to multiply the top (numerator) also by the same conjugate-of-the-denominator so that
$\frac{11 - \sqrt{6}}{11 - \sqrt{6}} = 1$ (as long as it's not $\frac{0}{0}$ it is fine).
that is, we're only multiplying by 1 so we are not affecting the results at all.

Then it becomes easy:
$= \frac{\left(\sqrt{6} - 2\right) \left(11 - \sqrt{6}\right)}{{11}^{2} - 6}$
$= \frac{11 \sqrt{6} - \sqrt{6} \sqrt{6} - 22 + 2 \sqrt{6}}{121 - 6}$
$= \frac{13 \sqrt{6} - 28}{115}$

At this stage, I don't see any more ways to simplify this further, so this must be the answer.