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Let f(x) = ax^(2) + bx + c, a != 0 and D...

Let `f(x) = ax^(2) + bx + c, a != 0 and Delta = b^(2) - 4ac`. If `alpha + beta, alpha^(2) + beta^(2) and alpha^(3) + beta^(3)` are in GP, then

A

`Dleta != 0`

B

`b Delta = 0`

C

`c Delta = 0`

D

`bc != 0`

Text Solution

Verified by Experts

The correct Answer is:
C
Since, `(alpha + beta), (alpha^(2) + beta^(2)), (alpha^(3) + beta^(3))` are in GP
`rArr (alpha^(2) + beta^(2))^(2) = (alpha + beta) (alpha^(3) + beta^(3))`
`rArr alpha^(4) + beta^(4) + 2 alpha^(2) beta^(2) = alpha^(4) + beta^(4) + alpha beta^(3) + beta alpha^(3)`
`rArr alpha beta (alpha^(2) + beta^(2) - 2 alpha beta) = 0`
`rArr alpha beta ( alpha - beta)^(2) = 0`
`rArr alpha beta = 0 or alpha beta`
`rArr (c)/(alpha) = 0 or Delta = 0`
`rArr cDelta = 0`
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