Rationalize the following :
$\frac{1}{\sqrt{3}+\sqrt{4}}$

Given :

The given term is $\frac{1}{\sqrt{3}+\sqrt{4}}$

To do :

We have to rationalize the given term.

Solution :

$\frac{1}{\sqrt{3} +\sqrt{4}}$

$=\frac{1}{\sqrt{3} +\sqrt{4}} =\frac{1}{\sqrt{3} +2}$

Rationalizing factor of the denominator $\sqrt{3} +2$ is $\sqrt{3} -2$.

Therefore,

$= \frac{1}{\sqrt{3} +\sqrt{4}} =\frac{1}{\sqrt{3} +2}$

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

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

$=\frac{\sqrt{3} -2}{3-4}$

$=\frac{\sqrt{3} -2}{-1}$

$=2-\sqrt{3}$ .

The Rationalized term of $\frac{1}{\sqrt{3}+\sqrt{4}}$ is  $2-\sqrt{3}$

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Updated on: 10-Oct-2022

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