Question

If x = `1/( x - 5 ) "and x ≠ 5. Find" : x^2 - 1/x^2`

Answer 1

Given x = `1/( x - 5 )`;
By cross multiplication,
⇒ x (x - 5) = 1
⇒ x² - 5x = 1
⇒ x² - 1 = 5x                                     ....(1)

Dividing both sides by x,
`( x - 1/x )^2 = x^2 + 1/x^2 - 2`
⇒ `(5)^2 = x^2 + 1/x^2 - 2`
⇒ `x^2 + 1/x^2 = 25 + 2 = 27`           ....(2)

Let us consider the expansion of `( x + 1/x )^2` :
`( x + 1/x )^2 = x^2 + 1/x^2 + 2`

⇒ `( x + 1/x )^2 = 27 + 2`                   [from(2)]

⇒ `( x + 1/x )^2 = 29`

⇒ `( x + 1/x ) = +- sqrt29`               ....(3)
We know that,

`x^2 - 1/x^2 = ( x + 1/x )( x - 1/x )`= `( +- sqrt29 )(5)`  [From equation (1) and (3)] 

`x^2 - 1/x^2 = +- 5sqrt29`