Question

If x= 1+√2, find the value of x square+1/x square​

Answer 1

Given -

$x = 1 + \sqrt{2}$

To find -

$The \: value \: of \: \: {x}^{2} + \frac{1}{ {x}^{2} }$

Solution -

$x = 1 + \sqrt{2}$

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

By rationalising the denominator we will get -

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

$\frac{1}{x } = \frac{1 - \sqrt{2} }{ {(1)}^{2} - {( \sqrt{2}) }^{2} }$

$\frac{1}{x} = \frac{1 - \sqrt{2} }{1 - 2}$

$\frac{1}{x} = \frac{1 - \sqrt{2} }{ - 1}$

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

$\frac{1}{x} = \sqrt{2} - 1$

Now,

$= {x}^{2} + \frac{1}{ {x}^{2} }$

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

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

$= 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2}$

$= 3 + 3$

$= 6$

Therefore the value of x²+1/x² is 6