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Question
If a + `1/a`= 6 and a ≠ 0 find :
(i) `a - 1/a (ii) a^2 - 1/a^2`
Solution
We know that,
( a + b )2 = a2 + 2ab + b2
and
( a - b )2 = a2 - 2ab + b2
Thus,
`( a + 1/a )^2 = a^2 + 1/a^2 + 2 xx a xx 1/a`
= `a^2 + 1/a^2 + 2` .....(1)
Given that `a + 1/a` = 6; Substitute in equation (1), we have
`(6)^2 = a^2 + 1/a^2 + 2`
⇒ `a^2 + 1/a^2 = 36 - 2`
⇒ `a^2 + 1/a^2 = 34` ....(2)
Similarly, consider
`( a - 1/a )^2 = a^2 + 1/a^2 - 2 xx a xx 1/a`
= `a^2 + 1/a^2 - 2`
= 34 - 2 [ from (2) ]
⇒ `( a - 1/a )^2` = 32
⇒ `( a - 1/a ) = +- sqrt32`
⇒ `( a - 1/a ) = +- 4sqrt2` ....(3)
(ii) We need to find `a^2 - 1/a^2`
We know that, `a^2 - 1/a^2 = ( a - 1/a )( a + 1/a )`
`a - 1/a = +- 4sqrt2 ; a + 1/a = 6`
Thus,
`a^2 - 1/a^2 = (+- 4sqrt2 )(6)`
⇒ `a^2 - 1/a^2 = (+- 24sqrt2 )`
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