Question

Rationalize the denominator and simplify $\frac{5\sqrt{2}-\sqrt{3}}{3\sqrt{2}-\sqrt{5}}$

Answer 1

We rationalise the denominator of $\left(\frac{5\sqrt{2}-\sqrt{3}}{3\sqrt{2}-\sqrt{5}}\right)$ by multiplying its rationalising factor $(3\sqrt{2}+\sqrt{5})$ in the numerator and denominator as follows:

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

$\frac{5\sqrt{2}(3\sqrt{2}+\sqrt{5})-\sqrt{3}(3\sqrt{2}+\sqrt{5})}{(3\sqrt{2}{)}^2-(\sqrt{5}{)}^2}=\frac{(5\times 3\sqrt{2\times 2})+\left(5\sqrt{2\times 5}\right)-\left(3\sqrt{3\times 2}\right)-\left(\sqrt{3\times 5}\right)}{(9\times 2)-5}$

$=\frac{15\sqrt{4}+5\sqrt{10}-3\sqrt{6}-\sqrt{15}}{18-5}=\frac{(15\times 2)+5\sqrt{10}-3\sqrt{6}-\sqrt{15}}{13}=\frac{30+5\sqrt{10}-3\sqrt{6}-\sqrt{15}}{13}$

Hence, $\frac{5\sqrt{2}-\sqrt{3}}{3\sqrt{2}-\sqrt{5}}=\frac{30+5\sqrt{10}-3\sqrt{6}-\sqrt{15}}{13}$