# How do you rationalize the denominator and simplify (2sqrt7+sqrt3) / (3sqrt7-sqrt3)?

Oct 13, 2015

$\frac{17 + 5 \sqrt{7}}{18}$

#### Explanation:

$\frac{2 \sqrt{7} + \sqrt{3}}{3 \sqrt{7} - \sqrt{3}}$

To rationalize the denominator, you need to multiply it by its conjugate. For any binomial, you get its conjugate by negating the second term.

In essence, you need to change the sign of the second term.

In your case, that would imply

${\underbrace{3 \sqrt{7} - \sqrt{3}}}_{\textcolor{b l u e}{\text{original binomial")) -> overbrace(3sqrt(7) + sqrt(3))^(color(red)("its conjugate}}}$

So, multiply the fraction by $1 = \frac{3 \sqrt{7} + \sqrt{3}}{3 \sqrt{7} + \sqrt{3}}$

$\frac{2 \sqrt{7} + \sqrt{3}}{3 \sqrt{7} - \sqrt{3}} \cdot \frac{3 \sqrt{7} + \sqrt{3}}{3 \sqrt{7} + \sqrt{3}} = \frac{\left(2 \sqrt{7} + 3\right) \left(3 \sqrt{7} + 3\right)}{\left(3 \sqrt{7} - 3\right) \left(3 \sqrt{7} + 3\right)}$

The denominator of the fraction is now in the form

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

which means that you can write it as

$\left(3 \sqrt{7} - 3\right) \left(3 \sqrt{7} + 3\right) = {\left(3 \sqrt{7}\right)}^{2} - {3}^{2}$

$= 9 \cdot 7 - 9 = 54$

For the numerator, expand the parantheses to get

$\left(2 \sqrt{7} + 3\right) \left(3 \sqrt{7} + 3\right) = 2 \sqrt{7} \cdot 3 \sqrt{7} + 6 \sqrt{7} + 9 \sqrt{7} + 9$

$= 6 \cdot 7 + 15 \sqrt{7} + 9 = 51 + 15 \sqrt{7}$

The fraction will now be

$\frac{\left(2 \sqrt{7} + 3\right) \left(3 \sqrt{7} + 3\right)}{\left(3 \sqrt{7} - 3\right) \left(3 \sqrt{7} + 3\right)} = \frac{51 + 15 \sqrt{7}}{54}$

Finally, simplify to get

$\frac{51 + 15 \sqrt{7}}{54} = \textcolor{g r e e n}{\frac{17 + 5 \sqrt{7}}{18}}$