Find the range of p? Find the range of p for which one root of the equation 
$$0= x^2 - (p+1)x  + (p^2 + p-8)$$
is greater than 2 
and of the other root is smaller than 2.
How i can achieve this? Thanks in advance.
 A: Hint:
By root, I assume you mean find $x$ such that $y = 0.$ Well, if
$$ 0 = x^2 - (p+1)x  + (p^2 + p-8) $$
then the roots $r_0, r_1 $ are given by:
$$ r_0 = \frac{p+1 + \sqrt{(p+1)^2 - 4 (p^2 +p -8)}}{2} \\
r_1 = \frac{p+1- \sqrt{(p+1)^2 - 4 (p^2 +p -8)}}{2} $$
To satisfy the conditions, set $r_0 > 2, r_1 <2 .$ So
$$ p+1 + \sqrt{33-2 p-3 p^2} > 4 \\
p+1 \color{red}{-} \sqrt{33-2 p-3 p^2} < 4$$
Try to solve for $p$ from here.
A: Now, you know that the roots of your equation are $2+a$ and $2-b$ for $a,b \gt 0$. Now, try and transform this into an equation in $a$ and $-b$. This is easy to do. 

Convince yourself that this transformation is $y=x+2$. 

Now, the only condition you have is that the product of the roots is negative.
So, you should end up with, $$\begin{align}p^2+p-8-2(p+1)+4 \lt 0\\p^2-p-6 \lt 0\\(p+2)(p-3) \lt 0\\p \in(-2,3)\end{align}$$
As André points out, there is no need to check whether the roots are real, because if they were complex, since complex roots occur in conjugate pairs we'll have their product is positive. (Thanks are due to Andre pointing this out. This makes this solution even more shorter and perfectly right.)
 $$\boxed{p \in (-2,3)}$$
A: Since the coefficient of $x^2$ is positive, the equation represents a upward opening parabola,
For one root of the equation to be greater than 2 the other root to be smaller than 2,the value of quadratic must be negative at 2.
y(2) < 0
$$2^2 - (p+1)2  + (p^2 + p-8) <0$$
$$ p^2-p-6 <0$$
$$ (p+2)(p-3) <0$$
$$\boxed{p \in (-2,3)}$$ 
But for roots to exist the discriminant of the equation $$y= x^2 - (p+1)x  + (p^2 + p-8)$$
must be positive,
$$(p+1)^2 - 4(p^2 + p-8) >0$$ 
$$-3p^2-2p+33>0$$
$$(3p+11)(3-p)>0$$
$$(3p+11)(p-3)<0$$
$$\boxed{p \in (\frac{-11}{3},3)}$$
taking union of inequalities (1) and (2) we get
$$\boxed{p \in (-2,3)}$$ 
A: A parabola $f(x)$ with a positive coefficient of $x^2$ has two roots such that one is lower than $a$ and the other is greater than $b>a$ iff $f(a)<0$ and $f(b)<0$.
This guarantees both -- the existence of two roots, and the condition above.
In your case, $a=b=2$. So, you get only one equation: $f(2)=p^2-p-6<0$, and the range for $p$: $(-2,3)$.
