9.101. For what values of $p$ are both roots of the quadratic trinomial $x^{2}+2(p+1) x+9 p-5$ negative?
## Solution.

According to Vieta's theorem, the roots of the quadratic trinomial $A x^{2}+B x+C$ exist and are both negative if and only if

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ D = B ^ { 2 } - 4 A C \geq 0 , } \\
{ \frac { B } { A } > 0 , } \\
{ \frac { C } { A } > 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
4(p+1)^{2}-4(9 p-5) \geq 0, \\
9 p-5>0, \\
p+1>0
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ p ^ { 2 } - 7 p + 6 \geq 0 , } \\
{ p > \frac { 5 } { 9 } , } \\
{ p > - 1 , }
\end{array} \quad \text { from which we have } \left[\begin{array}{l}
\frac{5}{9}<p \leq 1, \\
p \geq 6
\end{array}\right.\right.
\end{aligned}
$$

Answer: $p \in\left(\frac{5}{9} ; 1\right] \cup[6 ; \infty)$.