Find the value of $p$ for which $px^2+4x+p$ is greater than zero for all real value of $x$.
I know that if the value of $p$ is $(-\infty,-2)$ and $(2,+\infty)$ the quadratic equation is equal to zero for all real value of $x$ then for which value of p does $px^2+4x+p$ is greater than zero for all real value of $x$?
For $px^2+4x+p $ to be always positive, $p >0 \tag 1$ and, it must have no real roots (Floating above the $x$- axis, without touching it) $$\implies \Delta=4^2-4 \cdot p \cdot p <0 $$$$ \implies p^2 >4\implies p \in (-\infty,-2) \cup (2, \infty) \tag 2 $$
Therefore, taking intersection of $(1)$ and $(2)$, we've $$\color{red}{p \in (2, \infty)}$$
For $p\leq 0$, we have that for $x=0$, then $px^2+4x+p=p\leq 0$ which is not positive. On the other hand, for $p>0$, and for $x\in\mathbb{R}$, $$px^2+4x+p=\left(\sqrt{p}x+\frac{2}{\sqrt{p}}\right)^2+p-\frac{4}{p}\geq p-\frac{4}{p}.$$ Can you take it from here?
P.S. "I know that if the value of $p$ is $(-\infty,-2)$ and $(2,+\infty)$ the quadratic equation is equal to zero for NO real value of $x$".
I know that if the value of p is .. the quadratic equation is equal to zero for all real value of xSorry, but that makes no sense, and it's hard to guess what you really meant to say. – dxiv Feb 07 '18 at 07:00