# How do you simplify (3-4sqrt3)/(4sqrt5+3sqrt2)?

Jan 24, 2017

$\frac{3 - 4 \sqrt{3}}{4 \sqrt{5} + 3 \sqrt{2}} = \frac{12 \sqrt{5} - 9 \sqrt{2} - 16 \sqrt{15} + 12 \sqrt{6}}{62}$

#### Explanation:

To simplify $\frac{3 - 4 \sqrt{3}}{4 \sqrt{5} + 3 \sqrt{2}}$, what we need to do is to multiply numerator and denominator by the conjugate of its irrational denominator.

Conjugate of a irrational number $\left(\sqrt{a} \pm \sqrt{b}\right)$ is (sqrta∓sqrtb). (If we just have $\sqrt{p}$ in denominator, then just multiplying numerator and denominator by $\sqrt{p}$ suffices.)

Hence, $\frac{3 - 4 \sqrt{3}}{4 \sqrt{5} + 3 \sqrt{2}}$

= $\frac{\left(3 - 4 \sqrt{3}\right) \left(4 \sqrt{5} - 3 \sqrt{2}\right)}{\left(4 \sqrt{5} + 3 \sqrt{2}\right) \left(4 \sqrt{5} - 3 \sqrt{2}\right)}$

= $\frac{3 \left(4 \sqrt{5} - 3 \sqrt{2}\right) - 4 \sqrt{3} \left(4 \sqrt{5} - 3 \sqrt{2}\right)}{{\left(4 \sqrt{5}\right)}^{2} - {\left(3 \sqrt{2}\right)}^{2}}$

= $\frac{12 \sqrt{5} - 9 \sqrt{2} - 16 \sqrt{15} + 12 \sqrt{6}}{80 - 18}$

= $\frac{12 \sqrt{5} - 9 \sqrt{2} - 16 \sqrt{15} + 12 \sqrt{6}}{62}$