Question

Rationalize the denominator and simplify $\frac{4\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

Answer 1

We rationalise the denominator of $\left(\frac{4\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right)$ by multiplying its rationalising factor $(\sqrt{3}-\sqrt{2})$ in the numerator and denominator as follows:

$\frac{4\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{(4\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\frac{4\sqrt{3}(\sqrt{3}-\sqrt{2})+\sqrt{2}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}=\frac{\left(4\sqrt{3\times 3}\right)-\left(4\sqrt{3\times 2}\right)+\left(\sqrt{2\times 3}\right)-\left(\sqrt{2\times 2}\right)}{3-2}=\frac{4\sqrt{9}-4\sqrt{6}+\sqrt{6}-\sqrt{4}}{1}$

$=(4\times 3)-3\sqrt{6}-2=12-2-3\sqrt{6}=10-3\sqrt{6}$

Hence, $\frac{4\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=10-3\sqrt{6}$