Question

rationalise the denominator 3√5+√3/√5-√3

Answer 1

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Answer 2

Given:

$\dfrac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

To find:

The value of given expression after rationalising the denominator.

Solution:

We have,

$\dfrac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

Multiply and divide by $(\sqrt{5}+\sqrt{3})$ to rationalise the denominator.

$=\dfrac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\times \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$=\dfrac{3(\sqrt{5})^2+3\sqrt{15}+\sqrt{15}+(\sqrt{3})^2}}{(\sqrt{5})^2-(\sqrt{3})^2}$

$=\dfrac{3(5)+4\sqrt{15}+3}{5-3}$

$=\dfrac{15+4\sqrt{15}+3}{2}$

$=\dfrac{18+4\sqrt{15}}{2}$

On further simplification, we get

$=\dfrac{18}{2}+\dfrac{4\sqrt{15}}{2}$

$=9+2\sqrt{15}$

Therefore, the required value after rationalising the denominator is $9+2\sqrt{15}$.