2.93 The negative reciprocal of the smallest root of the equation $x^{2}-3|x|-2=0$ is
(A) -1 .
(B) $-\frac{1}{2}$.
(C) $\frac{1}{2}(3+\sqrt{17})$.
(D) $\frac{1}{2}(3-\sqrt{17})$.
(E) $\frac{1}{4}(\sqrt{17}-3)$.
(Beijing, China Junior High School Mathematics Competition, 1983)
[Solution] From $x^{2}-3|x|-2=0$, we get $|x|^{2}-3|x|-2=0$. Solving this, we obtain $|x|=\frac{3+\sqrt{17}}{2}$ (where $\frac{3-\sqrt{17}}{2}<0$ is discarded).
$$
\therefore \quad x_{1}=\frac{3+\sqrt{17}}{2}, \quad x_{2}=\frac{-3-\sqrt{17}}{2} \text {. }
$$

Thus, the smallest root is $\frac{-3-\sqrt{17}}{2}$, and its negative reciprocal is:
$$
\frac{1}{-\left(\frac{-3-\sqrt{17}}{2}\right)}=\frac{2}{3+\sqrt{17}}=\frac{1}{4}(\sqrt{17}-3) \text {. }
$$

Therefore, the answer is $(E)$.