$2 \cdot 32$ Given the equation $a x^{2}+b x+c=0$ has two positive roots, then among the following conclusions:
(1) $a, b, c>0$.
(2) $a, b, c<0$.
(3) $a>0, b, c<0$.
(4) $a<0, b, c>0$.

the number of conclusions that are definitely incorrect is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
[Solution] Let $x_{1}, x_{2}$ be the roots of the equation. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{b}{a}, \\
x_{1} x_{2}=\frac{c}{a} .
\end{array}\right.
$$

Given $x_{1}>0, x_{2}>0$, it follows that $a, b$ have opposite signs; $a, c$ have the same sign.
Thus, conclusions (1), (2) contradict the fact that $a, b$ have opposite signs, and conclusions (3), (4) contradict the fact that $a, c$ have the same sign. Therefore, none of the four conclusions are true.
Hence, the correct choice is $(D)$.