Rationalise the denominator and simplify:$ \frac{1+\sqrt{2}}{3-2 \sqrt{2}} $

Given:

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

To do: 

We have to rationalise the denominator and simplify 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,

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

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

$=\frac{3+5 \sqrt{2}+4}{9-8}$

$=\frac{7+5 \sqrt{2}}{1}$

$=7+5 \sqrt{2}$

Hence, $\frac{1+\sqrt{2}}{3-2 \sqrt{2}}=7+5 \sqrt{2}$.

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Updated on: 10-Oct-2022

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