$2 \cdot 37$ The equation $x^{2}+1984513 x+3154891=0$ has the following root conditions:
(A) No real roots.
(B) Has integer roots.
(C) Has positive roots.
(D) The sum of the reciprocals of the roots is less than -1.
(E) None of the above conclusions are correct.
(Tianjin Junior High School Mathematics Competition; China Junior High School Mathematics League, 1984)
［Solution］From the discriminant of the equation
$$
\Delta=1984513^{2}-4 \cdot 3154891>\left(19 \cdot 10^{5}\right)^{2}-4 \cdot 33 \cdot 10^{5}>0 \text {, }
$$

we can rule out $(A)$.
If the equation has integer roots $x_{1}, x_{2}$, then by Vieta's formulas, $x_{1} \cdot x_{2}=3154891$,
which means $\left|x_{1}\right|,\left|x_{2}\right|$ are both odd, but $\left|x_{1}+x_{2}\right|=|-1984513|$, which is odd, making it impossible. Thus, (B) is incorrect.
Furthermore, since all coefficients of the equation are positive, it has no positive roots, ruling out $(C)$.
And $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=-\frac{1984513}{3154891}>-1$, ruling out (D). Therefore, the answer is $(E)$.