If $x\ =\ 2\ +\ 3\sqrt{2}$

Find $x\ + \frac{4}{x}$.

Given: $x\ =\ 2\ +\ 3\sqrt{2}$

To find: Here we have to find the value of $x\ + \frac{4}{x}$.

Solution:

$x\ +\ \frac{4}{x}$

Putting value of $x\ =\ 2\ +\ 3\sqrt{2}$.

$=\ \left( 2\ +\ 3\sqrt{2}\right) \ +\ \frac{4}{2\ +\ 3\sqrt{2}}$

$=\ \frac{\left( 2\ +\ 3\sqrt{2}\right)^{2} \ +\ 4}{2\ +\ 3\sqrt{2}}$

$=\ \frac{( 2)^{2} \ +\ \left( 3\sqrt{2}\right)^{2} \ +\ 2( 2)\left( 3\sqrt{2}\right) \ +\ 4}{2\ +\ 3\sqrt{2}}$

$=\ \frac{4\ +\ 18\ +\ 12\sqrt{2} \ +\ 4}{2\ +\ 3\sqrt{2}}$

$=\ \frac{26\ +\ 12\sqrt{2}}{2\ +\ 3\sqrt{2}}$

$=\ \frac{26\ +\ 12\sqrt{2}}{2\ +\ 3\sqrt{2}} \ \times \ \frac{2\ -\ 3\sqrt{2}}{2\ -\ 3\sqrt{2}}$

$=\ \frac{\left( 26\ +\ 12\sqrt{2}\right)\left( 2\ -\ 3\sqrt{2}\right)}{( 2)^{2} \ -\ \left( 3\sqrt{2}\right)^{2}}$

$=\ \frac{2( 26) \ +\ 2\left( 12\sqrt{2}\right) \ -\ 3\sqrt{2}( 26) \ -\ 3\sqrt{2}\left( 12\sqrt{2}\right)}{4\ -\ 18}$

$=\ \frac{52\ +\ 24\sqrt{2} \ -\ 78\sqrt{2} \ -\ 72}{-\ 14}$

$=\ \frac{-\ 20\ -\ 54\sqrt{2}}{-\ 14}$

$=\ \mathbf{\frac{10\ +\ 27\sqrt{2}}{7}}$

So, value of $x\ + \frac{4}{x}$ is $\frac{10\ +\ 27\sqrt{2}}{7}$.

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Updated on: 10-Oct-2022

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