Question

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

Answer 1

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

$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=\frac{(\sqrt{3}+\sqrt{2}{)}^2}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}=\frac{(\sqrt{3}{)}^2+(\sqrt{2}{)}^2+(2\times \sqrt{3}\times \sqrt{2})}{3-2}=\frac{3+2+2\sqrt{6}}{1}=5+2\sqrt{6}$

Hence, $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=5+2\sqrt{6}$