Express each one of the following with rational denominator:$ \frac{b^{2}}{\sqrt{a^{2}+b^{2}}+a} $

Given:

\( \frac{b^{2}}{\sqrt{a^{2}+b^{2}}+a} \)

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 ${\sqrt{a}-\sqrt{b}}$.

This implies,

Rationalising factor of the fraction with denominator $\sqrt{a^{2}+b^{2}}+a$ is $\sqrt{a^{2}+b^{2}}-a$.

Therefore,

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

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

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

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

$=\sqrt{a^{2}+b^{2}}-a$

Hence, $\frac{b^{2}}{\sqrt{a^{2}+b^{2}}+a}=\sqrt{a^{2}+b^{2}}-a$.

Tutorialspoint

Updated on: 10-Oct-2022

30 Views

Kickstart Your Career

Get certified by completing the course

Get Started