Question

x=1-√2 find (x-1/x)^4​

Answer 1

Given :

$x = 1 - \sqrt{2}$

To find :

${(x - \dfrac{1}{x}) }^{4}$

Solution :

We know that :-

$x = 1 - \sqrt{2}$

Now :-

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

Now rationalise the denominator :-

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

we know that :- (a-b) (a+b) = a²-b²

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

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

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

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

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

Now we will find :- x - 1/x

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

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

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

$x - \dfrac{1}{x} = 2$

Now ,

${(x - \dfrac{1}{x} )}^{4} = {2}^{4}$

${(x - \dfrac{1}{x} )}^{4} = 2 \times 2 \times 2 \times 2$

${(x - \dfrac{1}{x} )}^{4} = 16$

Answer : 16