$2 \cdot 53$ Let the roots of $A x^{2}+B x+C=0$ be $r$ and $s$, and the roots of $x^{2}+p x+q=0$ be $r^{2}$ and $s^{2}$, then $p$ must equal
(A) $\frac{B^{2}-4 A C}{A^{2}}$.
(B) $\frac{B^{2}-2 A C}{A^{2}}$.
(C) $\frac{2 A C-B^{2}}{A^{2}}$.
(D) $B^{2}-2 C$.
(E) $2 C-B^{2}$.
(9th American High School Mathematics Examination, 1958)
［Solution］From the given conditions and Vieta's formulas, we have
$$
\begin{aligned}
r^{2}+s^{2} & =-p, \quad r+s=-\frac{B}{A}, \quad r s=\frac{C}{A}, \\
\therefore \quad p & =-\left(r^{2}+s^{2}\right)=-(r+s)^{2}+2 r s \\
& =-\frac{B^{2}}{A^{2}}+\frac{2 C}{A}=\frac{2 A C-B^{2}}{A^{2}} .
\end{aligned}
$$

Therefore, the correct choice is $(C)$.