Let n be the number of real values of $p$ for which the roots of 
$x^2-px+p=0$
are equal. Then n equals:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \text{a finite number greater than 2}\qquad \textbf{(E)}\ \infty$
If the roots of the quadratic $Ax^2 + Bx + C = 0$ are equal, then $B^2 - 4AC = 0$.  Plugging in $A=1, B=-p, C = p$ into the equation gives $p^2 - 4p = 0$.  This leads to $p = 0, 4$, so there are two values of $p$ that work, giving answer $\boxed{\textbf{(C)}}$.