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

Given:

\( \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \)

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,

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

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

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

$=\frac{5-2 \sqrt{6}}{1}$

$=5-2 \sqrt{6}$

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

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Updated on: 10-Oct-2022

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