［Solution］ $\because m 、 n$ have opposite signs and $m0 \text {. }
$$

Thus, $m n=\frac{c}{a}0$
For the equation $c x^{2}+(m-n) a x-a=0$, the discriminant is
$$
\begin{aligned}
\Delta & =a^{2}(m-n)^{2}+4 a c \\
& =a^{2}\left[(m-n)^{2}+4 \cdot \frac{c}{a}\right] \\
& =a^{2}(m+n)^{2} \geqslant 0 .
\end{aligned}
$$

Therefore, it must have two real roots. Let these two real roots be $x_{1} 、 x_{2}$. Then, from $x_{1}+x_{2}=\frac{a}{c}(n-m)0$, we know that $x_{1} 、 x_{2}$ are both negative. Hence, the answer is $(A)$.
$2 \cdot 14$ Given the quadratic equation $a x^{2}+b x+c=0(a c \neq 0)$ has two real roots of opposite signs $m$ and $n$, and $m<|m|$. Then the root situation of the quadratic equation $c x^{2}+(m-n) a x-a=0$ is
(A) It has two negative roots.
(B) It has two positive roots.
(C) The roots are of opposite signs.
(D) It has no real roots.
(Shaanxi Province, China Junior High School Mathematics Competition, 1997)