Question

rationalise the dinominator (\sqrt(2))/(\sqrt(2)+\sqrt(3)-\sqrt(5))

Answer 1

Answer:

(3 + √15 + √6) / 6

For solution , please refer to the attachment .

Answer 2

Concept:

We need to first recall the concept of Rationalisation to solve this question.

• The expressions which contain square roots in denominators, have to be rationalise first to make the denominator free from square root and make the calculation easy.

• To rationalise it we multiply both numerator and denominator by an irrational number which is called rationalisation factor.

Given:

The expression : $\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$

To find :

We have to rationalise the denominator of given expression. .

Solution:

We multiply the given expression with the rationalise factor to make it rationalise.

$\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$  $=\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}} \times \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$

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

                 $=\frac{(2+\sqrt{6}+\sqrt{10})}{(2+3+2\sqrt{6} -5)}$

                 $=\frac{(2+\sqrt{6}+\sqrt{10})}{2\sqrt{6} }$

                 $=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{5})}{2\sqrt{3} }$

                 $=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{5})}{2\sqrt{3} }\times \frac{\sqrt{3}}{\sqrt{3}}$

                 $=\frac{(\sqrt{6}+3+\sqrt{15})}{6}$.

Hence the rationalise form is   $\frac{(\sqrt{6}+3+\sqrt{15})}{6}$ .