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Express each one of the following with rational denominator:$ \frac{\sqrt{3}+1}{2 \sqrt{2}-\sqrt{3}} $
Given:
\( \frac{\sqrt{3}+1}{2 \sqrt{2}-\sqrt{3}} \)
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,
$\frac{\sqrt{3}+1}{2 \sqrt{2}-\sqrt{3}}=\frac{(\sqrt{3}+1)(2 \sqrt{2}+\sqrt{3})}{(2 \sqrt{2}-\sqrt{3})(2 \sqrt{2}+\sqrt{3})}$
$=\frac{\sqrt{3} \times 2 \sqrt{2}+\sqrt{3} \times \sqrt{3}+2 \sqrt{2}+\sqrt{3}}{(2 \sqrt{2})^{2}-(\sqrt{3})^{2}}$ [Since $(a+b)(a-b)=a^2-b^2$]
$=\frac{2 \sqrt{6}+3+2 \sqrt{2}+\sqrt{3}}{8-3}$
$=\frac{2 \sqrt{6}+2 \sqrt{2}+\sqrt{3}+3}{5}$
Hence, $\frac{\sqrt{3}+1}{2 \sqrt{2}-\sqrt{3}}=\frac{2 \sqrt{6}+2 \sqrt{2}+\sqrt{3}+3}{5}$.
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