Values of $p$ such that the roots of $x^2+px+p$ are greater than $1$ I found this question in a book, the answer is given as $(-\infty,-5)$. 
I know that for both roots of a quadratic polynomial to be greater than 1, $D \ge 0$, which implies that $p^2-4p \ge 0$, which means $p \ge 0$ and $p \ge 4$, or $p < 0$ and $p < 4$.  
Also $f(1) >0$, which implies $2p>0$ or $p>0$. 
Also $(-p/2)>1$ or $p<-2$. 
This is not only contradictory but also doesn't match with the answer. What should I do? And if the answer is wrong what is the right answer?

Let $x=a+1$, then the equation is equivalent to
$(a+1)^2+p(a+1)+p=0$
$\Leftrightarrow a^2+2a+1+ap+p+p=0$
$\Leftrightarrow a^2+(p+2)a+2p+1=0$
We need to find $p$ so that $x>1$, or $a+1>1$, or $a>0$, which means we need to find $a$ so that the equation $a^2+(p+2)a+2p+1=0$ has two roots and both of them are positive roots.
This happens if and only if $\begin{cases}\Delta \ge 0 \\ a_1+a_2>0 \\a_1a_2>0\end{cases}\Leftrightarrow \begin{cases}(p+2)^2-4(2p+1) \ge 0 \\ -p-2>0 \\2p+1>0\end{cases}\Leftrightarrow\begin{cases}p^2-4p\ge0\\p<-2\\p>-0.5\end{cases}$
There are no real numbers $p$ satisfy the condition because $p<-2$ and $p>-0.5$.
