Rationalise the denominators of each of the following:$ \frac{\sqrt{3}+1}{\sqrt{2}} $

Given:

\( \frac{\sqrt{3}+1}{\sqrt{2}} \)

To do: 

We have to rationalise the denominator of 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,

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

$=\frac{\sqrt{3\times2}+1\times\sqrt{2}}{(\sqrt{2})^2}$

$=\frac{\sqrt{6}+\sqrt{2}}{2}$

Hence, $\frac{\sqrt{3}+1}{\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{2}$.