$2 \cdot 69$ For a quadratic equation with roots $\alpha, \beta$, they satisfy the relations $\frac{1+\beta}{2+\beta} = -\frac{1}{\alpha}$ and $\frac{\alpha \beta^{2}+121}{1-\alpha^{2} \beta}=1$. Then, this quadratic equation is
(A) $x^{2}+12 x-10=0$ or $x^{2}-10 x+12=0$.
(B) $x^{2}+10 x+12=0$ or $x^{2}-12 x+10=0$.
(C) $x^{2}+12 x+10=0$ or $x^{2}-10 x-12=0$.
(D) $x^{2}+10 x-12=0$ or $x^{2}-12 x-10=0$.
(Chinese Zhejiang Ningbo Junior High School Mathematics Competition, 1985)
［Solution］From the given equations, we have
$$
\left\{\begin{array}{l}
(\alpha+\beta)+\alpha \beta=-2, \\
\alpha \beta(\alpha+\beta)=-120 .
\end{array}\right.
$$

Solving these, we get $\left\{\begin{array}{l}\alpha+\beta=-12, \\ \alpha \beta=10 ;\end{array}\right.$ or $\left\{\begin{array}{l}\alpha+\beta=10, \\ \alpha \beta=-12 .\end{array}\right.$ Therefore, the required equations are
$$
x^{2}+12 x+10=0 \text { or } x^{2}-10 x-12=0 \text {. }
$$

Hence, the correct choice is $(C)$.