1964 AHSME Problems/Problem 7

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Problem

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:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \text{a finite number greater than 2}\qquad \textbf{(E)}\ \infty$

Solution

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 $\boxed{\textbf{(C)}}$.