# What is 12/(square root of 2 - 6)?

Sep 17, 2015

$\frac{12}{\sqrt{2} - 6} = - \frac{6 \cdot \left(\sqrt{2} + 6\right)}{17}$

#### Explanation:

I'm not quite sure on your notation here, I'm assuming you're meaning this $\frac{12}{\sqrt{2} - 6}$ and not $\frac{12}{\sqrt{2 - 6}}$.

To do this problem we just need to rationalize. The concept in rationalizing is quite simple, we know that (x-y)(x+y) = x² - y².

So to get rid of these roots on the denominator, we'll multiply it by $\sqrt{2} + 6$. Which is the same thing as the denominator but with the sign switched so we won't have any roots on the bottom to deal with.

But - and there's always a but - since this is a fraction I can't just multiply what's on the denominator. I need to multiply both the numerator and denominator by the same thing, so it goes:

$\frac{12}{\sqrt{2} - 6} = \frac{12}{\sqrt{2} - 6} \cdot \frac{\sqrt{2} + 6}{\sqrt{2} + 6}$
$\frac{12}{\sqrt{2} - 6} = 12 \cdot \frac{\sqrt{2} + 6}{{\left(\sqrt{2}\right)}^{2} - {6}^{2}}$
$\frac{12}{\sqrt{2} - 6} = \frac{12 \sqrt{2} + 12 \cdot 6}{2 - 36}$

We can put a 2 on evidence both on the numerator and on the denominator

$\frac{12}{\sqrt{2} - 6} = \frac{2 \cdot \left(6 \sqrt{2} + 6 \cdot 6\right)}{2 \cdot \left(1 - 18\right)}$
$\frac{12}{\sqrt{2} - 6} = \frac{6 \sqrt{2} + 6 \cdot 6}{- 17}$

17 is a prime number so we don't really have much more to do here. You can either put that 6 on evidence on the numerator, or evaluate ${6}^{2}$

$\frac{12}{\sqrt{2} - 6} = - \frac{6 \cdot \left(\sqrt{2} + 6\right)}{17}$ or
$\frac{12}{\sqrt{2} - 6} = - \frac{6 \sqrt{2} + 36}{17}$