Question

If $x=\frac{1}{x-5}$ and $x\ne 5$, find the value of $x^2-\frac{1}{x^2}$.

• A. $\pm 2\sqrt{29}$
• B. $\pm 5\sqrt{29}$
• C. $\pm 9\sqrt{29}$
• D. $\pm 7\sqrt{29}$

Answer 1

The correct option is B $\pm 5\sqrt{29}$

$x=\frac{1}{x-5}$
$=>x-5=\frac{1}{x}$
$=>x-\frac{1}{x}=5$
Now,
Squaring both sides
$(x-\frac{1}{x}{)}^2=(5{)}^2$
$=>x^2+\frac{1}{x^2}-2\left(x\right)\left(\frac{1}{x}\right)=25$
$=>x^2+\frac{1}{x^2}=25+2$
$=>x^2+\frac{1}{x^2}=27$
Now,
$(x+\frac{1}{x}{)}^2=x^2+\frac{1}{x^2}+2(x\left)\right(\frac{1}{x})$
$=27+2$
$=>x+\frac{1}{x}=\sqrt{29}$
Again,
$x^2-\frac{1}{x^2}=(x+\frac{1}{x})(x-\frac{1}{x})$
$=\left(\sqrt{29}\right)\left(5\right)$
$=\pm 5\sqrt{29}$