If $r$ and $s$ are the roots of $x^2-px+q=0$, then $r^2+s^2$ equals: 
$\textbf{(A)}\ p^2+2q\qquad\textbf{(B)}\ p^2-2q\qquad\textbf{(C)}\ p^2+q^2\qquad\textbf{(D)}\ p^2-q^2\qquad\textbf{(E)}\ p^2$
We can write $r^2+s^2$ in terms of the sum of the roots $(r+s)$ and the products of the roots $(rs):$
\[r^2 + s^2 = (r+s)^2 - 2rs = p^2 - 2q\]
The answer is $\boxed{\textbf{(B)}}.$