Question

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

Answer 1

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

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

Hence, $\frac{2}{\sqrt{3}+\sqrt{2}}=2(\sqrt{3}-\sqrt{2})$