# If $x + \frac{1}{x} = 9$ find the value of $x^4 + \frac{1}{x^4}$.

Given:

$x + \frac{1}{x} = 9$

To do:

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

Solution:

The given expression is $x + \frac{1}{x} = 9$. Here, we have to find the value of $x^4 + \frac{1}{x^4}$. So, by squaring the given expression and using the identity $(a+b)^2=a^2+2ab+b^2$, we can find the required value.

$(a+b)^2=a^2+2ab+b^2$...................(i)

Let us consider,

$x + \frac{1}{x} = 9$

Squaring on both sides, we get,

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

$x^2+2\times x \times \frac{1}{x}+\frac{1}{x^2}=81$            [Using (I)]

$x^2+2+\frac{1}{x^2}=81$

$x^2+\frac{1}{x^2}=81-2$                       (Transposing $2$ to RHS)

$x^2+\frac{1}{x^2}=79$

Now,

$x^2+\frac{1}{x^2}=79$

Squaring on both sides, we get,

$(x^2+\frac{1}{x^2})^2 = (79)^2$

$x^4+2\times x^2 \times \frac{1}{x^2}+\frac{1}{x^4}=6241$            [Using (I)]

$x^4+2+\frac{1}{x^4}=6241$

$x^4+\frac{1}{x^4}=6241-2$                       (Transposing $2$ to RHS)

$x^4+\frac{1}{x^4}=6239$

Hence, the value of $x^4+\frac{1}{x^4}$ is $6239$.