$2 \cdot 62$ The roots of the equation $x^{2}+b x+c=0$ are both real numbers greater than 1. Let $s=b+c+1$, then $s$
（A）may be less than zero.
（B）may be equal to zero.
(C) should be greater than zero.
(D) should be less than zero.
(E) is between -1 and 1.
（10th American High School Mathematics Examination, 1959）
[Solution] Let the two roots of the equation be $1+\alpha$ and $1+\beta$, where $\alpha>0$ and $\beta>0$. By Vieta's formulas,
$$
\left\{\begin{array}{l}
(1+\alpha)+(1+\beta)=-b, \\
(1+\alpha)(1+\beta)=c .
\end{array}\right.
$$

Then,
$$
\begin{aligned}
s & =b+c+1 \\
& =-(\alpha+\beta+2)+(1+\alpha)(1+\beta)+1 \\
& =\alpha \beta>0 .
\end{aligned}
$$

Therefore, the answer is $(C)$.