# 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.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

29 Views

Get certified by completing the course