# How do you rationalize the denominator and simplify (2sqrt12 - sqrt5) /( sqrt5 + 4sqrt3)?

##### 1 Answer
Aug 12, 2017

See a solution process below:

#### Explanation:

To rationalize the fraction we need to use the rule:

$\left(\textcolor{red}{x} + \textcolor{b l u e}{y}\right) \left(\textcolor{red}{x} - \textcolor{b l u e}{y}\right) = {\textcolor{red}{x}}^{2} - {\textcolor{b l u e}{y}}^{2}$

Because the denominator is the form: $\left(\textcolor{red}{\sqrt{5}} + \textcolor{b l u e}{4 \sqrt{3}}\right)$

We need to use: $\left(\textcolor{red}{\sqrt{5}} - \textcolor{b l u e}{4 \sqrt{3}}\right)$ as the numerator and denominator for the appropriate form of #1:

$\frac{2 \sqrt{12} - \sqrt{5}}{\left(\textcolor{red}{\sqrt{5}} + \textcolor{b l u e}{4 \sqrt{3}}\right)} \times \frac{\left(\textcolor{red}{\sqrt{5}} - \textcolor{b l u e}{4 \sqrt{3}}\right)}{\left(\textcolor{red}{\sqrt{5}} - \textcolor{b l u e}{4 \sqrt{3}}\right)} \implies$

$\frac{2 \sqrt{12} \sqrt{5} - \left(4 \cdot 2\right) \sqrt{12} \sqrt{3} - {\left(\sqrt{5}\right)}^{2} + 4 \sqrt{5} \sqrt{3}}{{\left(\textcolor{red}{\sqrt{5}}\right)}^{2} - {\left(\textcolor{b l u e}{4 \sqrt{3}}\right)}^{2}} \implies$

$\frac{2 \sqrt{60} - 8 \sqrt{36} - 5 + 4 \sqrt{15}}{5 - \left(16 \cdot 3\right)} \implies$

$\frac{2 \sqrt{4 \cdot 15} - \left(8 \cdot 6\right) - 5 + 4 \sqrt{15}}{5 - 48} \implies$

$\frac{2 \sqrt{4} \sqrt{15} - 48 - 5 + 4 \sqrt{15}}{- 43} \implies$

$\frac{\left(2 \cdot 2\right) \sqrt{15} - 53 + 4 \sqrt{15}}{- 43} \implies$

$\frac{4 \sqrt{15} - 53 + 4 \sqrt{15}}{- 43} \implies$

$\frac{- 53 + 4 \sqrt{15} + 4 \sqrt{15}}{- 43} \implies$

$\frac{- 53 + 8 \sqrt{15}}{- 43} \implies$

$\frac{- 53}{- 43} + \frac{8 \sqrt{15}}{- 43} \implies$

$\frac{53}{43} - \frac{8 \sqrt{15}}{43}$