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प्रश्न
If a - `1/a`= 8 and a ≠ 0 find :
(i) `a + 1/a (ii) a^2 - 1/a^2`
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उत्तर
We know that,
( a + b )2 = a2 + 2ab + b2
Given that `a - 1/a` = 8 ; Substitute in equation (1), we have
`(8)^2 = a^2 + 1/a^2 - 2`
⇒ `a^2 + 1/a^2 = 64 + 2`
⇒ `a^2 + 1/a^2 = 66`
⇒ `(a + 1/a)^2 = a^2 + 1/a^2 + 2`
⇒ `(a + 1/a)^2 = 66 + 2`
⇒ `(a + 1/a)^2 = 68`
i) `a + 1/a = sqrt68 `
⇒ `sqrt(17xx4 )= _-^+2sqrt17`
ii) `a^2 - 1/a^2 = (a+1/a) (a - 1/a)`
⇒ `a^2 - 1/a^2 = _-^+2sqrt17 xx 8`
⇒ `a^2 - 1/a^2 = _-^+16sqrt17`
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