Let a,b,c be in arithmetic progression. Let the centroid of the triangle with vertices a,c,2,b     and     a,b  be   103,73 . If  α,βare the roots of the equation ax2+bx+1=0, then the value of α2+β2−αβis

Given centroid of triangle formed by the vertices (a,c),(2,b),(a,b) is 103,73a+2+a=102a+2=102a=8a=4b+c+b=72b+c=7a+c+c=72c=7−4=3c=32∣2b=a+c=4+32=112b=114 Therefore, quadratic equation is 4x2+114x+1=0 This can be written as 16x2+11x+4=0α+β=−1116,αβ=416α2+β2−αβ=(α+β)2−3αβ=121256−1216=121−12(16)256=-71256