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Question

Let α and β be the roots of equation px2+qx+r=0,0. If p,q,r are in A.P. and 1α+1β=4, then the value of |αβ|is :

A
349
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B
2139
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C
619
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D
2179
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Solution

The correct option is B 2139
Let p,q,r are in AP
2q=p+r....(i)
Given 1α+1β=4=α+βαβ=4
We have α+β=qp and αβ=rp
qprp = 4 q=4r
From (i), we have
2(4r)=p+r p=9r
q=4r
Now |αβ|=(α+β)24αβ
= (qp)24rp=q24pr|p|
= 16r2+36r2|9r|=2139

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