Question

If $x^2+\frac{1}{x^2}=102$, then find the value of $x-\frac{1}{x}=?$

• A. 10
• B. -10
• C. 4
• D. None of these

Answer 1

Answer: (A), (B)

$x^2+\frac{1}{x^2}=102$...(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
$\Rightarrow {\left(x-\frac{1}{x}\right)}^2=102+2-4$
$\Rightarrow {\left(x-\frac{1}{x}\right)}^2=100\Rightarrow x-\frac{1}{x}=\pm 10$