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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 $119$.
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