If $ x-\frac{1}{x}=5 $, find the value of
(a) $ x^{2}+\frac{1}{x^{2}} $
(b) $ x^{4}+\frac{1}{x^{4}} $

Given :

The given expression is $x-\frac{1}{x}=5$.

To do :

We have to find the values of

a) $x^{2}+\frac{1}{x^{2}}$

b) $x^{4}+\frac{1}{x 4}$.

Solution :  

a) $x^{2}+\frac{1}{x^{2}}$

 $x-\frac{1}{x}=5$

Squaring on both sides,

 $(x-\frac{1}{x})^2=(5)^2$

$x^2 + \frac{1}{x^2} -2.x.\frac{1}{x} = 25$       $[(a-b)^2=a^2+b^2-2ab]$

$x^2 + \frac{1}{x^2} -2=25$

$x^2 + \frac{1}{x^2} =25+2$

$x^2 + \frac{1}{x^2} =27$.

Therefore, the value of $x^2 + \frac{1}{x^2}$ is 27. 

b) $x^{4}+\frac{1}{x 4}$

We know that, $x^2 + \frac{1}{x^2} =27$

Squaring on both sides,

$(x^2 + \frac{1}{x^2})^2 =(27)^2$

$(x^2)^2 + (\frac{1}{x^2})^2 +2 .x^2.\frac{1}{x^2}=27^2$   $[(a+b)^2=a^2+b^2+2ab]$

$x^4 + \frac{1}{x^4} + 2 = 729$

$x^4 + \frac{1}{x^4}=729-2$

$x^4 + \frac{1}{x^4}=727$.

Therefore, the value of $x^4 + \frac{1}{x^4}$ is 727. 

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Updated on: 10-Oct-2022

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