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Q.
If $x^2+\frac{1}{x^2}=102$, then find the value of $x-\frac{1}{x}=?$
Polynomials
Solution:
$x^2+\frac{1}{x^2}=102$...(i)
Then
$\left(x-\frac{1}{x}\right)^2 =\left(x+\frac{1}{x}\right)^2-4 $....(ii)
$=x^2+\frac{1}{x^2}+2-4$
On putting the value of equation (i) equation (ii), we get
$ \Rightarrow\left(x-\frac{1}{x}\right)^2=102+2-4 $
$ \Rightarrow\left(x-\frac{1}{x}\right)^2=100 \Rightarrow x-\frac{1}{x}= \pm 10$