Find $p$ and $q$ such that $x^2+px+q
Find $p$ and $q$ such that $$x^2+px+q<x$$ iff $$x \in (1,5)$$
I tried the following:
$$x^2+px+q = (x+\frac{p}{2})^2+q-\frac{p^2}{4}$$
where the global minimum is $$q-\frac{p^2}{4}$$ if $$x = \frac{-p}{2}$$
However this doesn't seem to help with the problem at hand...

First rearrange as follows: $$x^2+px+q<x\iff x^2+(p-1)x+q<0$$ Recall from the properties of parabolas that $$ax^2+bx+c<0$$ between the (real) roots of the equation if $a>0$. So since in our case $a=1>0$, the parabola $$x^2+(p-1)x+q<0$$ between its zeros. So since we are given that $x\in(1,5)$ is the solution set, we surmise from above that $x^2+(p-1)x+q=(x-1)(x-5)=x^2-6x+5\implies p=-5$ and $q=5$
