If $ x+\frac{1}{x}=11 $, find the value of
(a) $ x^{2}+\frac{1}{x^{2}} $
(b) $ x^{4}+\frac{1}{x^{4}} $

Given :

The given expression is $x+\frac{1}{x}=11$.

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

 $x+\frac{1}{x}=11$

Squaring on both sides,

 $(x+\frac{1}{x})^2=(11)^2$

$x^2 + \frac{1}{x^2} +2.x.\frac{1}{x} = 121$       $[(a+b)^2=a^2+b^2+2ab]$

$x^2 + \frac{1}{x^2} +2=121$

$x^2 + \frac{1}{x^2} =121-2$

$x^2 + \frac{1}{x^2} =119$.

Therefore, the value of $x^2 + \frac{1}{x^2}$ is 119. 

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

We know that, $x^2 + \frac{1}{x^2} =119$

Squaring on both sides,

$(x^2 + \frac{1}{x^2})^2 =(119)^2$

$(x^2)^2 + (\frac{1}{x^2})^2 +2 .x^2.\frac{1}{x^2}=14161$   $[(a+b)^2=a^2+b^2+2ab]$

$x^4 + \frac{1}{x^4} + 2 = 14161$

$x^4 + \frac{1}{x^4}=14161-2$

$x^4 + \frac{1}{x^4}=14159$.

Therefore, the value of $x^4 + \frac{1}{x^4}$ is 14159. 

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Updated on: 10-Oct-2022

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