Question

The sum of all possible integral values of $p$, for which the equation $x^2-(p+1)x+p^2+p-8=0$ has one root greater than $2$ and the other root less than $2$, will be

Answer 1

Let $f\left(x\right)=x^2-(p+1)x+p^2+p-8$.
Since it is an upward shaped parabola,
$f\left(2\right)=4-(p+1)2+p^2+p-8<0$
i.e., $(p-3)(p+2)<0$,
i.e., $-2<p<3$.
Their intersection will be:
$-2<p<3$
So, the integral values of $p$ are $-1,0,1,2$.
Hence, their sum is $2$.
(We can check that the real roots exist for the given value of $p$).