is integer = λ1 If b=aλ1 then, c=a(2λ1−1) {because a, b, c are in A.P.} ∴P(x)=ax2+aλ1x+a(2λ1−1) =a[x2+λ1x+(2λ1−1)] D=λ12−4(2λ1−1) is perfect square for integral roots ⇒D=λ12−8λ1+4 is perfect square Let D=(λ1−4)2−12=k2 {where k ∈ I} ⇒(λ1−4−k)(λ1−4+k)=12 This gives λ1−4−k=2 & λ1−4+k=6 ⇒λ1=8 & k=1 ∴α+β+αβ=8−1=7