1955 AHSME Problems/Problem 39

Problem

If $y=x^2+px+q$ , then if the least possible value of $y$ is zero $q$ is equal to:

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac{p^2}{4}\qquad\textbf{(C)}\ \frac{p}{2}\qquad\textbf{(D)}\ -\frac{p}{2}\qquad\textbf{(E)}\ \frac{p^2}{4}-q$

Solution

The least possible value of $y$ is given at the $y$ coordinate of the vertex. The $x$ - coordinate is given by $\frac{-p}{(2)(1)} = \frac{-p}{2}$ Plugging this into the quadratic, we get $y = \frac{p^2}{4} - \frac{p^2}{2} + q$ $0 = \frac{p^2}{4} - \frac{2p^2}{4} + q$ $0 = \frac{-p^2}{4} + q$ $q = \frac{p^2}{4} = \boxed{B}$

~JustinLee2017

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1955 AHSME Problems/Problem 39

Problem

Solution