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 less than $2$, are given by

• A. $a>3$
• B. $9<a<10$
• C. $-2<a<3$
• D. none of these

Answer 1

The correct option is C $-2<a<3$

Let

$x^2-\left(a+1\right)x+a^2+a-8=0=f\left(x\right)$

If $D>0$

$\Rightarrow {\left(a+1\right)}^2-4.1\left(a^2+a-8\right)>0$

$\Rightarrow a^2+2a+1-4a^2-4a+32>0$

$\Rightarrow -3a^2-2a+33>0$

$\Rightarrow 3a^2+2a-33<0$

$\Rightarrow \left(3a+11\right)\left(a-3\right)<0$

$\Rightarrow a\in \left(\frac{-11}{3},3\right)$

Now,

$A.f\left(x\right)<0$

$\Rightarrow 1\left[4-\left(a+1\right).2+a^2+a-8\right]<0$

$\Rightarrow a^2-a-6<0$

$\Rightarrow \left(a+2\right)\left(a-3\right)<0$

$\Rightarrow a\in \left(-2,3\right)$

$So,commonregion\Rightarrow -2<a<3$

Hence, the option $\left(C\right)$ is the correct answer.