# If $x - \frac{1}{x} = 3$, find the values of $x^2 + \frac{1}{x^2}$ and $x^4 + \frac{1}{x^4}$.

Given:

$x - \frac{1}{x} = 3$

To do:

We have to find the values of $x^2 + \frac{1}{x^2}$ and $x^4 + \frac{1}{x^4}$.

Solution:

The given expression is $x - \frac{1}{x} = 3$. Here, we have to find the values of $x^2 + \frac{1}{x^2}$ and $x^4 + \frac{1}{x^4}$. So, by squaring the given expression and using the identities $(a+b)^2=a^2+2ab+b^2$...................(i) and $(a-b)^2=a^2-2ab+b^2$.............(ii), we can find the required values.

Let us consider,

$x - \frac{1}{x} = 3$

Squaring on both sides, we get,

$(x - \frac{1}{x})^2 = 3^2$                 [Using (ii)]

$x^2-2\times x \times \frac{1}{x}+\frac{1}{x^2}=9$

$x^2-2+\frac{1}{x^2}=9$

$x^2+\frac{1}{x^2}=9+2$                     (Transposing $-2$ to RHS)

$x^2+\frac{1}{x^2}=11$

Now,

$x^2+\frac{1}{x^2}=11$

Squaring on both sides, we get,

$(x^2+\frac{1}{x^2})^2 = (11)^2$                 [Using (i)]

$x^4+2\times x^2 \times \frac{1}{x^2}+\frac{1}{x^4}=121$

$x^4+2+\frac{1}{x^4}=121$

$x^4+\frac{1}{x^4}=121-2$                     (Transposing $2$ to RHS)

$x^4+\frac{1}{x^4}=119$

Hence, the value of $x^2+\frac{1}{x^2}$ is $11$ and the value of $x^4+\frac{1}{x^4}$ is $119$.

Updated on: 01-Apr-2023

62 Views 