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Question

The equation x2+pxq=0 has one root as a square root of the other.
If p3+q2=q(1+kp), find the value of k.


A

4

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B

-1

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C

-2

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D

-3

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Solution

The correct option is D

-3


Let the roots of the equation be α,α2.
α+α2=p -(i)
α3=q-(ii)
Cubing (i) gives
(α+α2)3=[α(α+1)]3
p3=α3(α3+3α(α+1)+1)
=α3(α3+3(α+α2)+1)
=q(q+3(p)+1)
p3=q2+3pqq
p3+q2=q3pq
=q(13p)
p3+q3=q(1+kp) (given)
k=3


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