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Question

Let α and β be the roots of equation px2+qx+r=0,p0. If p,q and r in AP and 1α+1β=4, then the value of |αβ| is


A
619
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B
2179
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C
349
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D
2139
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Solution

The correct option is D 2139

PLAN: If ax2+bx+c=0 has roots α and β, then α+β=ba and αβ=ca. Find the values of α+β and αβ and then put in (αβ)2=(α+β)24αβ to get required value.
Given, α and β are roots of px2+qx+t=0,p0.
α+β=qp,αβ=rp...(i)
Since, p,q and r are in AP.
2q=p+r....(ii)
Also, 1α+1β=4α+βαβ=4
α+β=4αβqp=4rp [from Eq.(i)]
q=4r
On putting the value of q in Eq. (ii), we get
2(4r)=p+rp=9r
Now, α+β=qp=4rp=4r9r=49
and αβ=rp=r9r=19
(αβ)2=(α+β)24αβ=1681+49=16+3681(αβ)2=5281|αβ|=2913


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