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Q.
If $x^2+\frac{1}{x^2}=102$, then find the value of $\frac{x^2-5 x-1}{x}$
Pair of Linear Equations in Two Variables
Solution:
$x^2+\frac{1}{x^2}=102$
Subtract 2 on both sides:
$x^2+\frac{1}{x^2}-2=102-2$
$ \Rightarrow\left(x-\frac{1}{x}\right)^2=100 \Rightarrow x-\frac{1}{x}= \pm 10 $
$ \Rightarrow \frac{x^2+5 x-1}{x}=\frac{\frac{x^2}{x}-\frac{5}{x}+\frac{1}{x}}{\frac{x}{x}}$
$\{$ Divide the numerator and denominator by $x \}$
$\Rightarrow \frac{\left(x-\frac{1}{x}\right)+5}{1}$
When $x-\frac{1}{x}=10$
Then $\frac{10+5}{1}=15$
When $x-\frac{1}{x}=-10 \Rightarrow \frac{-10+5}{1}=-5$
Possible values of $x=15,-5$