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Question

Let p and q be real numbers such that p0, p3q and p3q. If α and β are nonzero complex numbers satisfying α+β=p and α3+β3=q, then a quadratic equation having αβ and βα as its roots is

A
(p3q)x2(p32q)x+(p3q)=0
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B
(p3q)x2(p3+2q)x+(p3q)=0
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C
(p3+q)x2(5p3+2q)x+(p3+q)=0
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D
(p3+q)x2(5p32q)x+(p3+q)=0
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Solution

The correct option is B (p3q)x2(p3+2q)x+(p3q)=0
Let the new equation be x2+Bx+C=0
B=(αβ+βα)
C=αβ.βα
x2(αβ+βα)x+αββα=0
x2(α2+β2)αβx+1=0
x2((α+β)22αβαβ)x+1=0

Now, we have α3+β3=q
(α+β)33αβ(α+β)=q
p3+3pαβ=q
αβ=q+p33p
Substituting the value of αβ in the equation,

x2p22(p3q3p)p3q3px+1=0
(p3q)x2(3p32p3+2q)x+(p3q)=0
(p3q)x2(p3+2q)x+(p3q)=0

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