# $x+\frac{1}{x}=4$ find the value of the following:a) $x^{2}+\frac{1}{x^{2}}$b) $x^{4}+\frac{1}{x 4}$

Given :

The given expression is $x+\frac{1}{x}=4$.

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

$x+\frac{1}{x}=4$

Squaring on both sides,

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

$x^2 + \frac{1}{x^2} +2.x.\frac{1}{x} = 16$       $[(a+b)^2=a^2+b^2+2ab]$

$x^2 + \frac{1}{x^2} +2=16$

$x^2 + \frac{1}{x^2} =16-2$

$x^2 + \frac{1}{x^2} =14$.

Therefore, the value of $x^2 + \frac{1}{x^2}$ is 14.

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

We know that, $x^2 + \frac{1}{x^2} =14$

Squaring on both sides,

$(x^2 + \frac{1}{x^2})^2 =14^2$

$(x^2)^2 + (\frac{1}{x^2})^2 +2 .x^2.\frac{1}{x^2}=14^2$   $[(a+b)^2=a^2+b^2+2ab]$

$x^4 + \frac{1}{x^4} + 2 = 196$

$x^4 + \frac{1}{x^4}=196-2$

$x^4 + \frac{1}{x^4}=194$.

Therefore, the value of $x^4 + \frac{1}{x^4}$ is 194.

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Updated on: 10-Oct-2022

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