# How do you simplify 6/(2+sqrt12)?

Jun 11, 2017

See a solution process below:

#### Explanation:

Use this rule of radicals to rewrite the expression:

$\sqrt{\textcolor{red}{a}} \cdot \sqrt{\textcolor{b l u e}{b}} = \sqrt{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}}$

$\frac{6}{2 + \sqrt{12}} \implies \frac{6}{2 + \sqrt{\textcolor{red}{4} \cdot \textcolor{b l u e}{3}}} \implies \frac{6}{2 + \left(\sqrt{\textcolor{red}{4}} \cdot \sqrt{\textcolor{b l u e}{3}}\right)} \implies \frac{6}{2 + 2 \sqrt{3}}$

Now, we can factor the numerator and denominator and cancel common terms to complete the simplification:

$\frac{6}{2 + 2 \sqrt{3}} \implies \frac{2 \times 3}{\left(2 \times 1\right) + \left(2 \times \sqrt{3}\right)} \implies$

$\frac{2 \times 3}{2 \left(1 + \sqrt{3}\right)} \implies \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \times 3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \left(1 + \sqrt{3}\right)} \implies$

$\frac{3}{1 + \sqrt{3}}$

If necessary and required we can modify this result by rationalizing the denominator or, in other words, removing all the radicals from the denominator:

$\frac{3}{1 + \sqrt{3}} \implies \frac{1 - \sqrt{3}}{1 - \sqrt{3}} \times \frac{3}{1 + \sqrt{3}} \implies$

$\frac{\left(3 \times 1\right) - \left(3 \times \sqrt{3}\right)}{{1}^{2} - \sqrt{3} + \sqrt{3} - \left({\sqrt{3}}^{2}\right)} \implies$

$\frac{3 - 3 \sqrt{3}}{1 - 3} \implies$

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

Or

$- \frac{3}{2} \left(1 - \sqrt{3}\right)$

Or

$\frac{3}{2} \left(\sqrt{3} - 1\right)$