$2 \cdot 46$ If $p, q$ are odd numbers, and the equation $x^{2}+2 p x+2 q=0$ has two real roots, then the two roots are
(A) irrational numbers.
(B) odd numbers.
(C) even numbers.
(D) rational numbers, but not integers.
(China Shanghai Junior High School Mathematics Competition, 1983; China Partial Provinces and Cities Junior High School Mathematics Correspondence Competition, 1985)
[Solution] By the quadratic formula, we have $x=-p \pm \sqrt{p^{2}-2 q}$. If $x$ is a rational number, then $\sqrt{p^{2}-2 q}=r$ is a rational number.
$$
\therefore \quad p^{2}-2 q=r^{2} \text {, }
$$

which means $p^{2}-r^{2}=2 q$, or $(p+r)(p-r)=2 q$.
Since $r$ is a rational number, and $r^{2}=p^{2}-2 q$, then $r$ is an integer and odd $(\because p, q$ are odd).

Thus, $p+r, p-r$ are both even, so $(p+r)(p-r)$ is a multiple of 4. But when $q$ is odd, $2 q$ is not a multiple of 4.
$\therefore \quad$ The assumption that the roots of the equation $x$ are rational is incorrect.
Hence, the answer is $(A)$.