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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 $6239$.
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