Question

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

Answer 1

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

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

$\frac{\left(3\sqrt{3}\right)-(3\times 6)+\left(\sqrt{6\times 3}\right)-\left(6\sqrt{6}\right)}{3-36}=\frac{3\sqrt{3}-18+\sqrt{18}-6\sqrt{6}}{-33}=\frac{-(3\sqrt{3}+18-\sqrt{18}+6\sqrt{6})}{-33}$

$=\frac{3\sqrt{3}+18-\sqrt{2\times 9}+6\sqrt{6}}{33}=\frac{3\sqrt{3}+18-3\sqrt{2}+6\sqrt{6}}{33}=\frac{3(\sqrt{3}+6-\sqrt{2}+2\sqrt{6})}{33}=\frac{6+2\sqrt{6}-\sqrt{3}-\sqrt{2}}{11}$

Hence, $\frac{3+\sqrt{6}}{\sqrt{3}+6}=\frac{6+2\sqrt{6}-\sqrt{3}-\sqrt{2}}{11}$