Question

If x = ${\displaystyle \frac{1}{x-5}\text{and x ≠ 5. Find}:x^2-\frac{1}{x^2}}$

Answer 1

Given x = ${\displaystyle \frac{1}{x-5}}$;
By cross multiplication,
⇒ x (x - 5) = 1
⇒ x² - 5x = 1
⇒ x² - 1 = 5x                                     ....(1)

Dividing both sides by x,
${\displaystyle {(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2}$
⇒ ${\displaystyle {\left(5\right)}^2=x^2+\frac{1}{x^2}-2}$
⇒ ${\displaystyle x^2+\frac{1}{x^2}=25+2=27}$           ....(2)

Let us consider the expansion of ${\displaystyle {(x+\frac{1}{x})}^2}$ :
${\displaystyle {(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2}$

⇒ ${\displaystyle {(x+\frac{1}{x})}^2=27+2}$                   [from(2)]

⇒ ${\displaystyle {(x+\frac{1}{x})}^2=29}$

⇒ ${\displaystyle (x+\frac{1}{x})=\pm \sqrt{29}}$               ....(3)
We know that,

${\displaystyle x^2-\frac{1}{x^2}=(x+\frac{1}{x})(x-\frac{1}{x})}$= ${\displaystyle (\pm \sqrt{29})\left(5\right)}$  [From equation (1) and (3)] 

${\displaystyle x^2-\frac{1}{x^2}=\pm 5\sqrt{29}}$