Values of $p$ such that the roots of $x^2+px+p$ are greater than $1$

Question

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?

Answer 1

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 $p>-0.5$.

Answer 2

Hint: $$p^2-4p\geq 0$$ means that $$p(p-4)\geq 0$$ and this means $$p\geq 4$$ or $$p\le 0$$

Answer 3

Write $$ p={-x^2\over x+1}$$ Problem ask for which $p$ the line $y=p$ cuts the graph of $f(x)= {-x^2\over x+1}$ only for $x>1$. If you draw a graph for $f$ you'l see that is never a case. So there is no such $p$.