Question

Rationalize the denominator and simplify $\frac{3\sqrt{5}}{\sqrt{6}-\sqrt{3}}$

Answer 1

We rationalize the denominator of $\left(\frac{3\sqrt{5}}{\sqrt{6}-\sqrt{3}}\right)$ by multiplying its rationalizing factor $(\sqrt{6}+\sqrt{3})$ in the numerator and denominator as follows:

$\frac{3\sqrt{5}}{\sqrt{6}-\sqrt{3}}\times \frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}=\frac{3\sqrt{5}(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})}$

$=\frac{3\sqrt{5\times 6}+3\sqrt{5\times 3}}{(\sqrt{6}{)}^2-(\sqrt{3}{)}^2}$

$=\frac{3\sqrt{30}+3\sqrt{15}}{6-3}$

$=\frac{3(\sqrt{30}+\sqrt{15})}{3}$

$=\sqrt{30}+\sqrt{15}$

$=\sqrt{5}(\sqrt{6}+\sqrt{3})$