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Question
If a2 - 5a - 1 = 0 and a ≠ 0 ; find:
- `a - 1/a`
- `a + 1/a`
- `a^2 - 1/a^2`
Solution
(i) Consider the given equation
a2 - 5a - 1 = 0
Rewrite the given equation, we have
a2 - 1 = 5a
⇒ `[ a^2 - 1 ]/a = 5`
⇒ `[ a^2/a - 1/a ] = 5`
⇒ `a - 1/a = 5` ...(1)
(ii) We need to find `a + 1/a`:
We know the identity, (a - b)2 = a2 + b2 - 2ab
∴ `( a - 1/a )^2 = a^2 + 1/a^2 - 2`
⇒ `(5)^2 = a^2 + 1/a^2 - 2` [From(1)]
⇒ `25 = a^2 + 1/a^2 - 2`
⇒ `a^2 + 1/a^2 = 27` ...(2)
Now consider the identity (a + b)2 = a2 + b2 + 2ab
∴ `( a + 1/a )^2 = a^2 + 1/a^2 + 2`
⇒ `( a + 1/a )^2 = 27 + 2` [From (2)]
⇒ `( a + 1/a )^2 = 29`
⇒ `a + 1/a = +- sqrt29` ...(3)
(iii) We need to find `a^2 - 1/a^2`
We know the identity, a2 - b2 = (a + b)(a - b)
∴ `a^2 - 1/a^2 = ( a + 1/a )( a - 1/a )` ...(4)
From equation (3), we have,
` a + 1/a = +- sqrt29`
From equation (1), we have,
`a - 1/a = 5`;
Thus, identity (4), becomes,
`a^2 - 1/a^2 = (+- sqrt29)(5)`
⇒ `a^2 - 1/a^2 = 5(+- sqrt29 )`
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