Let α and β be the roots of equation px2+ qx+ r = 0 p≠0 If p ,q and r, n AP and 1α+1β=4 then the  value of |α−β| is

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

  1. A

    619

  2. B

    2179

  3. C

    349

  4. D

    2139

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    Solution:

    Given αand β are roots of px2 + qx + r =0, p0
     α+β=qp and αβ=rp…………………….(i)
    Since, p, q and r are in AP
     2q=p+r……………………………………….(ii)
     Also,     1α+1β=4    α+βαβ=4                       [given]
    α+β=4αβqp=4rpq=4r........................[from Eq.(i)]
    On putting the value of qin 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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