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Q.
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 :
JEE MainJEE Main 2013Complex Numbers and Quadratic Equations
Solution:
$x^{2}-\left(a + 1\right)x + a^{2} + a-8 = 0$
Since roots are different, therefore $D > 0$
$\Rightarrow \left(a+ 1\right)^{2}-4\left(a^{2} + a-8\right) > 0$
$\Rightarrow \left(a-3\right)\left(3a+l\right)<0$
There are two cases arises.
Case I. $a - 3 > 0$ and $3a +1 < 0$
$\Rightarrow a > 3$ and $a <-\frac{11}{3}$
Hence, no solution in this case
Case II : $a - 3 < 0$ and $3a +11 > 0$
$\Rightarrow a < 3$ and $a >-\frac{11}{3}$
$\therefore -\frac{11}{3} < a < 3 \Rightarrow -2 < a < 3$