JEE Main 2013 (Online) 9th AprilQuestion 1
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 :
Quadratic Equation · Location of Roots
Solution
Concept. For an upward parabola, $2$ lies between the roots exactly when $f(2)<0$. Formulas used. (1) $a>0$: $k$ lies between the roots $\iff f(k)<0$Step 1 - Compute $f(2)$. $$f(2)=4-2(a+1)+a^2+a-8=a^2-a-6.$$ Step 2 - Use (1). $a^2-a-6=(a-3)(a+2)<0$, so $-2<a<3$. Answer: (C)