9.102. For what values of $n$ are both roots of the equation $(n-2) x^{2}-$ $-2 n x+n+3=0$ positive?
## Solution.

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

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ D = B ^ { 2 } - 4 A C \geq 0 , } \\
{ \frac { B } { A } < 0, } \\
{ \frac { C } { A } > 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
n^{2}-(n-2)(n+3) \geq 0, \\
\frac{n}{n-2}>0, \\
\frac{n+3}{n-2}>0
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ n^2 - (n^2 + n - 6) \geq 0, } \\
{ \frac{n}{n-2}>0, } \\
{ \frac{n+3}{n-2}>0 }
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 6 - n \geq 0, } \\
{ n(n-2)>0, } \\
{ (n+3)(n-2)>0 }
\end{array}\right.\right. \\
& 2(n \in(-\infty ;-3) \cup(2 ; 6) .
\end{aligned}
$$

Answer: $n \in(-\infty ;-3) \cup(2 ; 6]$.