Question

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 :

• A. $3<a<10$
• B. $a\ge 10$
• C. $-2<a<3$
• D. $a\le -2$

Answer 1

Answer: (C)

$x^2-\left(a+1\right)x+a^2+a-8=0$
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 $a>-\frac{11}{3}$
$\therefore -\frac{11}{3}<a<3\Rightarrow -2<a<3$