For t > –1, let αt and βt be the roots of the equation ((t+2)1/7–1)x2+((t+2)1/6–1)x+((t+2)1/21–1)=0. If limt→–1+αt=a and limt→–1+βt=b, then 72(a+b)2 is equal to __________. [2025]
(98)
a+b=limt→–1+(αt+βt)
=limt→–1+–[(t+2)16–1](t+2)17–1 [Sum of roots]
Let t + 2 = y, we get
a+b=limy→1+–y1/6–1y1/7–1=-76
So, 72(a+b)2=72(4936)=98.