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If $x + \frac{1}{x} =20$, find the value of $x^2 + \frac{1}{x^2}$.
Given:
$x + \frac{1}{x} =20$
To do:
We have to find the value of $x^2 + \frac{1}{x^2}$.
Solution:
The given expression is $x + \frac{1}{x} =20$. Here, we have to find the value of $x^2 + \frac{1}{x^2}$. So, by squaring the given expression and using the identity $(a+b)^2=a^2+2ab+b^2$, we can find the value of $x^2 + \frac{1}{x^2}$.
$(a+b)^2=a^2+2ab+b^2$...................(i)
Now,
$x + \frac{1}{x} =20$
Squaring on both sides, we get,
$(x+\frac{1}{x})^2=(20)^2$
$x^2+2\times x \times \frac{1}{x}+(\frac{1}{x})^2=400$ [Using (i)]
$x^2+2+\frac{1}{x^2}=400$
$x^2+\frac{1}{x^2}=400-2$ (Transposing $2$ to RHS)
$x^2+\frac{1}{x^2}=398$
Hence, the value of $x^2+\frac{1}{x^2}$ is $398$.
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