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…
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 :
$3 < a < 10$
$a \geq 10$
$-2 < a < 3$
$a \leq-2$
Solution
$
x^2-(a+1) x+a^2+a-8=0
$
Since roots are different, therefore $\mathrm{D}>0$
$
\begin{aligned}
& \Rightarrow(a+1)^2-4\left(a^2+a-8\right)>0 \\
& \Rightarrow(a-3)(3 a+1) < 0
\end{aligned}
$
There are two cases arises.
Case I. $a-3>0$ and $3 a+1 < 0$
$\Rightarrow a>3$ and $a < -\frac{11}{3}$
Hence, no solution in this case
Case II : $a-3 < 0$ and $3 a+11>0$
$\Rightarrow a < 3$ and $a>-\frac{11}{3}$
$\therefore \quad-\frac{11}{3} < a < 3 \Rightarrow-2 < a < 3$