$2 \cdot 47$ If the two roots of the equation $x^{2}+p x-q=0$ are $x_{1}, x_{2}$, and $x_{1}>1$, and $p+q+3>0$, then $x_{2}$
(A) is less than 1.
(B) equals 1.
(C) is greater than 1.
(D) is uncertain.
(China Sichuan Province Junior High School Mathematics Competition, 1990)
[Solution] By Vieta's formulas, we have $p=-(x_{1}+x_{2})$, and $q=-x_{1} x_{2}$. Also, $p+q+3>0$, then

or
$$
\begin{array}{c}
-(x_{1}+x_{2})-x_{1} x_{2}+3>0, \\
4-(x_{1}+1)(x_{2}+1)>0, \\
(x_{1}+1)(x_{2}+1)<1$, then $x_{1}+1>2$, thus $x_{2}+1<2$, i.e., $x_{2}<1$. Therefore, the answer is $(A)$.