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Question

If α and β are the roots of x2p(x+1)q=0, then


A

(α+1)(β+1)=1q

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B

(α+1)(β+1)=1+q

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C

(α+1)2(α+1)2+q1+(β+1)2(β+1)2+q1=q

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D

α2+2α+1α2+2α+q+β2+2β+1β2+2β+q=1

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Solution

The correct options are
A

(α+1)(β+1)=1q


D

α2+2α+1α2+2α+q+β2+2β+1β2+2β+q=1


x2p(x+1)q=0x2pxpq=0α+β=p,αβ=pqα+β+αβ=qα+β+αβ+1=1q(1+α)(1+β)=1q(1)α2+2α+1α2+2α+q+β2+2β+1β2+2β+q
On putting the value of q - 1 from equation (1)
(α+1)2(α+1)2(1+α)(1+β)+(β+1)2(β+1)2(1+α)(1+β)α+1αββ+1αβ(αβ)(αβ)=1


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