JEE Main2013MathematicsQuadratic EquationActual
The values of ' a ' for which one root of the equation x^2-(a+1) x+a^2+a-8=0 exceeds 2 and the other is lesser than 2 , are given by :
Options
- A3 < a < 10
- Ba 10
- C-2 < a < 3
- Da -2
Correct answer
C. -2 < a < 3
Step-by-step solution
x^2-(a+1) x+a^2+a-8=0 Since roots are different, therefore D >0 aligned & (a+1)^2-4 (a^2+a-8 )>0 & (a-3)(3 a+1) 0 and 3 a+1 3 and a 0 a - 11 3 - 11 3 < a < 3 -2 < a < 3