The roots of $Ax^2 + Bx + C = 0$ are $r$ and $s$. For the roots of
$x^2+px +q =0$
to be $r^2$ and $s^2$, $p$ must equal:
$\textbf{(A)}\ \frac{B^2 - 4AC}{A^2}\qquad  \textbf{(B)}\ \frac{B^2 - 2AC}{A^2}\qquad  \textbf{(C)}\ \frac{2AC - B^2}{A^2}\qquad \\ \textbf{(D)}\ B^2 - 2C\qquad  \textbf{(E)}\ 2C - B^2$
By Vieta's, $r + s = -\frac{B}{A}$, $rs = \frac{C}{A}$, and $r^2 + s^2 = -p$. Note that $(r+s)^2 = r^2 + s^2 + 2rs$.
Therefore, $(r + s)^2 - 2rs = r^2 + s^2$, or $-\left(\frac{B}{A}\right)^2 - \frac{2C}{A} = -p$.
Simplifying, $\frac{B^2 - 2CA}{A^2} = -p$. 
Finally, multiply both sides by $-1$ to get $p = \frac{2CA - B^2}{A^2}$, making the answer $\fbox{C}$.