2.52 For the equation $x^{2}+b x+c=0$ with roots $r$ and $s$, and the equation $x^{2}+p x+q=0$ with roots $r^{2}$ and $s^{2}$, then $p$ equals
(A) $4 c-b^{2}$.
(B) $b^{2}-4 c$.
(C) $b^{2}-2 c$.
(D) $2 c-b^{2}$.
(China Junior High School Mathematics Correspondence Competition, 1989)
［Solution］By the given and Vieta's formulas, we have
$$
\left\{\begin{array}{l}
r + s = -b, \\
r s = c;
\end{array} \quad \left\{\begin{array}{l}
r^{2}+s^{2}=-p, \\
r^{2} s^{2}=q .
\end{array}\right.\right.
$$

Thus, $p=-\left(r^{2}+s^{2}\right)=-\left[(r+s)^{2}-2 r s\right]=-b^{2}+2 c$, hence the answer is $(D)$.