Difference between revisions of "1955 AHSME Problems/Problem 39"

Latest revision as of 13:26, 19 July 2021

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

@@ Line 1: / Line 1: @@
+==Problem==
+If <math>y=x^2+px+q</math>, then if the least possible value of <math>y</math> is zero <math>q</math> is equal to:
+<math>\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</math>
 ==Solution==

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Difference between revisions of "1955 AHSME Problems/Problem 39"

Latest revision as of 13:26, 19 July 2021

Problem

Solution