Question

If $x=\frac{1}{x-5}$, find $x^2-\frac{1}{x^2}$.

• A. $\pm 5\sqrt{1}4$
• B. $\pm 5\sqrt{2}9$
• C. $\pm 5\sqrt{2}2$
• D. $\pm 5\sqrt{1}3$

Answer 1

Answer: (B)

$\pm 5\sqrt{2}9$

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