If $ x=\frac{1}{3+2 \sqrt{2}}, $ then the value of $ x-\frac{1}{x} $ is


Given: $x\ =\ \frac{1}{3\ +\ 2\sqrt{2}}$


To find: Here we have to find the value of $x\ -\ \frac{1}{x}$.



Solution:

Given that,

$x\ =\ \frac{1}{3\ +\ 2\sqrt{2}}$

$x\ =\ \frac{1}{3\ +\ 2\sqrt{2}} \ \times \ \frac{3\ -\ 2\sqrt{2}}{3\ -\ 2\sqrt{2}}$

$x\ =\ \frac{3\ -\ 2\sqrt{2}}{(3)^{2} \ -\ \left( 2\sqrt{2}\right)^{2}}$

$x\ =\ \frac{3\ -\ 2\sqrt{2}}{9\ -\ 8}$

$\mathbf{x\ =\ 3\ -\ 2\sqrt{2}}$

Now, put this value of $x$ in $x\ -\ \frac{1}{x}$:

$x\ -\ \frac{1}{x}$

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

$=\ \frac{\left( 3\ -\ 2\sqrt{2}\right)^{2} \ -\ 1}{\left( 3\ -\ 2\sqrt{2}\right)}$

$=\ \frac{\left( 9\ +\ 8\ -\ 12\sqrt{2}\right) \ -\ 1}{\left( 3\ -\ 2\sqrt{2}\right)}$

$=\ \frac{17\ -\ 12\sqrt{2} \ -\ 1}{\left( 3\ -\ 2\sqrt{2}\right)}$

$=\ \frac{16\ -\ 12\sqrt{2}}{3\ -\ 2\sqrt{2}}$

$=\ \frac{16\ -\ 12\sqrt{2}}{3\ -\ 2\sqrt{2}} \ \times \ \frac{3\ +\ 2\sqrt{2}}{3\ +\ 2\sqrt{2}}$

$=\ \frac{\left( 16\ -\ 12\sqrt{2}\right)\left( 3\ +\ 2\sqrt{2}\right)}{( 3)^{2} \ -\ \left( 2\sqrt{2}\right)^{2}}$

$=\ \frac{16\left( 3\ +\ 2\sqrt{2}\right) \ -\ 12\sqrt{2}\left( 3\ +\ 2\sqrt{2}\right)}{9\ -\ 8}$

$=\ 48\ +\ 32\sqrt{2} \ -\ 36\sqrt{2} \ -\ 48$

$=\ \mathbf{-\ 4\sqrt{2}}$

So, the value of ($x\ -\ \frac{1}{x}$)  is  $-\ 4\sqrt{2}$.

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Updated on: 10-Oct-2022

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