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