Question 334 of 455MediumMCQ
The values of 'a' for which one root of x2−(a+1)x+a2+a−8=0 exceeds 2 and the other is less than 2, are given by: Solutions · 1
Official solution
Need f(2)<0: 4−2(a+1)+a2+a−8<0⟹a2−a−6<0⟹(a−3)(a+2)<0⟹a∈(−2,3). Answer the question to unlock