)+β=0 (l2y2+m2+2lmy)−α(ly+m)(y+l)+β(y+l)2=0 (l2−αl+β)y2+(2lm−αl2−αm+2βl)‌y+(m2−αml+βl2)=0 Above equation and 2x2+αx+β=0 has same roots so,
l2−αl+β
1
=
2lm−αl2−αm+2βl
α
=
m2−αml+βl2
β
βl2−βαl+β2=m2−αml+βl2 β2−βm+β(m−αl)−m(m−αl)=0 Further simplify the above, β(β−m)+(m−αl)(β−m)=0 β=m,αl−m