Question

If X = √2 -1 then find the value of x^2 + 1/x^2

Answer 1

Solution:-

Given:-

$\rm \: x = \sqrt{2} - 1$

To find;

$\rm \: {x}^{2} + \frac{1}{ {x}^{2} }$

Put the value of x on equation

$\rm \: ( \sqrt{2} - 1) {}^{2} + \frac{1}{ ( \sqrt{2} - 1) {}^{2} }$

Using this identity

$\rm(a - b) {}^{2} = {a}^{2} + {b}^{2} - 2ab$

We get,

$\rm \: ( \sqrt{2} ) {}^{2} + 1 - 2 \sqrt{2} + \frac{1}{ ( \sqrt{2} ) {}^{2} + 1 - 2 \sqrt{2}}$

$\rm \: 3 - 2 \sqrt{2} + \frac{1}{ 3 - 2 \sqrt{2}}$

Taking lcm

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

$\rm \: \frac{( 3 - 2 \sqrt{2}) {}^{2} + 1 }{ 3 - 2 \sqrt{2}}$

$\rm \: \frac{(3) {}^{2} + (2 \sqrt{2} ) {}^{2} - 2 \times 3 \times 2 \sqrt{2} + 1 }{ 3 - 2 \sqrt{2}}$

$\rm \: \frac{9 + 8 - 12 \sqrt{2} + 1 }{ 3 - 2 \sqrt{2}}$

$\rm \: \frac{18 - 12 \sqrt{2} }{ 3 - 2 \sqrt{2}}$

Taking common

$\rm \: \frac{6( 3 - 2 \sqrt{2})}{ 3 - 2 \sqrt{2}}$

Answer:- 6