Question

If $x^2+\frac{1}{x^2}=27$, find the value of $x-\frac{1}{x}$.

• A. $+5$
• B. $-5$
• C. $+6$
• D. $-6$

Answer 1

Answer: (A), (B)

The correct options are
A $+5$
B $-5$

$(x-\frac{1}{x}{)}^2=x^2-2\times x\times \frac{1}{x}+\frac{1}{x^2}$
$=x^2-2+\frac{1}{x^2}$
$=x^2+\frac{1}{x^2}-2$
$=27-2$ [$\because x^2+\frac{1}{x^2}=27\left(given\right)]$
$=25$
$\Rightarrow {(x-\frac{1}{x})}^2=(\pm 5{)}^2\Rightarrow x-\frac{1}{x}=\pm 5$