$4 \cdot 26$ Given that the two roots of the equation $x^{2}+p x+1=0$ are the lengths of two sides of a triangle, and the third side is 2. Then the range of the real number $p$ is
(A) $p<-2$.
(B) $p \leqslant-2$ or $p \geqslant 2$.
(C) $-2 \sqrt{2}<p<2 \sqrt{2}$.
(D) $-2 \sqrt{2}<p<-2$.
(China Sichuan Province High School Mathematics Competition, 1991)
[Solution] Let the two roots of the equation be $r, s$. According to the problem, they are two sides of a triangle, so
$$
r+s>2, \quad r+2>s, \quad s+2>r .
$$

The last two inequalities can be reduced to $|r-s|<2$, \\
{ p ^ { 2 } < 8 . }
\end{array} \text { which gives } \quad \left\{\begin{array}{l}
p<-2, \\
-2 \sqrt{2}<p<2 \sqrt{2} .
\end{array}\right.\right.
$$

In summary, we have $\quad-2 \sqrt{2}<p<-2$.
Therefore, the correct choice is ( $D$ ).