3a+2+a=310
a=4
and 3c+b+b=37
c+2b=7
also 2b=a+c
2b−a+2b=7
b=411
now 4x2+411x+1=0

α2+β2−αβ=(α+β)2−3αβ
=(16−11)2−3(41)
=256121−43=256−71
JEE Main 2021 — Mathematics Algebra
Let a,b,c be in arithmetic progression. Let the centroid of the triangle with vertices (a,c),(2,b) and (a,b) be (310,37). If α,β are the roots of the equation ax2+bx+1=0, then the value of α2+β2−αβ is:
Held on 24 Feb 2021 · Verified 6 Jul 2026.
−25671
25669
25671
−25669
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