2・16 Let $x_{1}$ and $x_{2}$ be the two distinct real roots of the equation $x^{2} + px + 4 = 0$. Determine which of the following conclusions is correct:
(A) $\left|x_{1}\right| > 2$ and $\left|x_{2}\right| > 2$.
(B) $\left|x_{1} + x_{2}\right| > 4$.
(C) $\left|x_{1} + x_{2}\right| < 4$.
(D) $\left|x_{1}\right| = 4$ and $\left|x_{2}\right| = 1$.
(China Beijing Junior High School Mathematics Competition, 1982)
[Solution] Since $x^{2}+p x+4=0$ has two distinct real roots $x_{1}$ and $x_{2}$, we have $x_{1} x_{2}=4, \left|x_{1}\right|\left|x_{2}\right|=4$, so (A) should be excluded.

Moreover, $\left|x_{1}\right|\left|x_{2}\right|=4$ does not necessarily imply that $\left|x_{1}\right|=4$ and $\left|x_{2}\right|=1$, so (D) should also be excluded.

From the discriminant, we know $p^{2}-16>0$, i.e., $p^{2}>16$. But $-p=x_{1}+x_{2}$, so $\left|x_{1}+x_{2}\right|^{2}>16, \left|x_{1}+x_{2}\right|>4$. Therefore, (C) should be excluded. Hence, the answer is (B).