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Question

If a root of the equation ax2+bx+c=0 be reciprocal of the equation then ax2+bx+c=0, then


A

(ccaa)2=(bacb)(abbc).

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B

(cc+aa)2=(bacb)(abbc).

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C

(ccaa)2=(bacb)(abbc).

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D

(bbaa)2=(cabc)(abbc).

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Solution

The correct option is C

(ccaa)2=(bacb)(abbc).


Let α be a root of first equation, then 1α be a root of second equation.

therefore aα2+bα+c=0 and a1α2+b1α+c=0 or cα2+bα+d=0

Hence α2babc=αccaa=1abbc

(ccaa)2=(bacb)(abbc).


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