Question

Rationalise the denominator of $\frac{3}{\sqrt{3}-\sqrt{2}+\sqrt{5}}$

• A. $\frac{2\sqrt{3}-\sqrt{30}+3\sqrt{2}}{4}$
• B. $\frac{2\sqrt{3}-\sqrt{30}+3\sqrt{2}}{2}$
• C. $\frac{2\sqrt{3}+\sqrt{30}-3\sqrt{2}}{2}$
• D. $\frac{2\sqrt{3}+\sqrt{30}-3\sqrt{2}}{4}$

Answer 1

The correct option is D $\frac{2\sqrt{3}+\sqrt{30}-3\sqrt{2}}{4}$

We will rationalize the denominator of the given expression $\frac{3}{(\sqrt{3}-\sqrt{2})+\sqrt{5}}$and solve as follows:

$\frac{3}{(\sqrt{3}-\sqrt{2})+\sqrt{5}}\times \frac{(\sqrt{3}-\sqrt{2})-\sqrt{5}}{(\sqrt{3}-\sqrt{2})-\sqrt{5}}=\frac{3[(\sqrt{3}-\sqrt{2})-\sqrt{5}]}{[(\sqrt{3}-\sqrt{2})+\sqrt{5}][(\sqrt{3}-\sqrt{2})-\sqrt{5}]}=\frac{3[(\sqrt{3}-\sqrt{2})-\sqrt{5}]}{{(\sqrt{3}-\sqrt{2})}^2-{\left(\sqrt{5}\right)}^2}(\because (a-b\left)\right(a+b)=a^2-b^2)=\frac{3[(\sqrt{3}-\sqrt{2})-\sqrt{5}]}{[{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2-2\left(\sqrt{3}\right)\left(\sqrt{2}\right)]-5}(\because (a-b{)}^2=a^2+b^2-2ab)$

$=\frac{3[(\sqrt{3}-\sqrt{2})-\sqrt{5}]}{(3+2-2\sqrt{6})-5}=\frac{3[(\sqrt{3}-\sqrt{2})-\sqrt{5}]}{5-2\sqrt{6}-5}=\frac{3[(\sqrt{3}-\sqrt{2})-\sqrt{5}]}{-2\sqrt{6}}=-\frac{3(\sqrt{3}-\sqrt{2}-\sqrt{5})}{2\sqrt{6}}\times \frac{2\sqrt{6}}{2\sqrt{6}}\left(Rationlizedenominatoragain\right)$

$=-\frac{6\sqrt{6}(\sqrt{3}-\sqrt{2}-\sqrt{5})}{{\left(2\sqrt{6}\right)}^2}=-\frac{6\sqrt{6}(\sqrt{3}-\sqrt{2}-\sqrt{5})}{24}=-\frac{\sqrt{6}(\sqrt{3}-\sqrt{2}-\sqrt{5})}{4}=\frac{-(\sqrt{18}-\sqrt{12}-\sqrt{30})}{4}$

$=\frac{\sqrt{12}+\sqrt{30}-\sqrt{18}}{4}=\frac{\sqrt{3\times 2^2}+\sqrt{30}-\sqrt{2\times 3^2}}{4}=\frac{2\sqrt{3}+\sqrt{30}-3\sqrt{2}}{4}$

Hence, $\frac{3}{\sqrt{3}-\sqrt{2}+\sqrt{5}}=\frac{2\sqrt{3}+\sqrt{30}-3\sqrt{2}}{4}$