Express each one of the following with rational denominator:$ \frac{16}{\sqrt{41}-5} $

Given:

\( \frac{16}{\sqrt{41}-5} \)

To do: 

We have to express the given fraction with rational denominator.

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,

$\frac{16}{\sqrt{41}-5}=\frac{16(\sqrt{41}+5)}{(\sqrt{41}-5)(\sqrt{41}+5)}$

$=\frac{16(\sqrt{41}+5)}{(\sqrt{41})^{2}-(5)^{2}}$          [Since $(a+b)(a-b)=a^{2}-b^{2}$]

$=\frac{16(\sqrt{41}+5)}{41-25}$

$=\frac{16(\sqrt{41}+5)}{16}$

$=\sqrt{41}+5$

Hence, $\frac{16}{\sqrt{41}-5}=\sqrt{41}+5$. 

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Updated on: 10-Oct-2022

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