# Rationalise the denominator and simplify:$\frac{2 \sqrt{3}-\sqrt{5}}{2 \sqrt{2}+3 \sqrt{3}}$

Given:

$\frac{2 \sqrt{3}-\sqrt{5}}{2 \sqrt{2}+3 \sqrt{3}}$

To do:

We have to rationalise the denominator and simplify the given expression.

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.

Therefore,

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

$=\frac{4 \sqrt{6}-6 \sqrt{3} \times \sqrt{3}-2 \sqrt{10}+3 \sqrt{15}}{(2 \sqrt{2})^{2}-(3 \sqrt{3})^{2}}$

$=\frac{4 \sqrt{6}-18-2 \sqrt{10}+3 \sqrt{15}}{8-27}$

$=\frac{-18-2 \sqrt{10}+4 \sqrt{6}+3 \sqrt{15}}{-19}$

$=\frac{18+2 \sqrt{10}-4 \sqrt{6}-3 \sqrt{15}}{19}$

Hence, $\frac{2 \sqrt{3}-\sqrt{5}}{2 \sqrt{2}+3 \sqrt{3}}=\frac{18+2 \sqrt{10}-4 \sqrt{6}-3 \sqrt{15}}{19}$.

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Updated on: 10-Oct-2022

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