Question
If $x^2+\frac{1}{x^2}=18$; find : $x-\frac{1}{x}$
✓
Answer
$x^2+\frac{1}{x^2}=18 $
Using$\left(x-\frac{1}{x}\right)^2$
$ =x^2+\frac{1}{x^2}-2 $
$ \Rightarrow\left(x-\frac{1}{x}\right)^2$
$=18-2 $
$ =16$
$ \Rightarrow x-\frac{1}{x}$
$ =4 .$
Using$\left(x-\frac{1}{x}\right)^2$
$ =x^2+\frac{1}{x^2}-2 $
$ \Rightarrow\left(x-\frac{1}{x}\right)^2$
$=18-2 $
$ =16$
$ \Rightarrow x-\frac{1}{x}$
$ =4 .$
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