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

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==Problem==
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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:
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<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>
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==Solution==
 
==Solution==
  
 
The least possible value of <math>y</math> is given at the <math>y</math> coordinate of the vertex. The <math>x</math>- coordinate is given by  
 
The least possible value of <math>y</math> is given at the <math>y</math> coordinate of the vertex. The <math>x</math>- coordinate is given by  
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<cmath>\frac{-p}{2}</cmath>. Plugging this into the quadratic, we get
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<cmath>\frac{-p}{(2)(1)} = \frac{-p}{2}</cmath> Plugging this into the quadratic, we get
 
<cmath>y = \frac{p^2}{4} - \frac{p^2}{2} + q</cmath>
 
<cmath>y = \frac{p^2}{4} - \frac{p^2}{2} + q</cmath>
 
<cmath>0 = \frac{p^2}{4} - \frac{2p^2}{4} + q</cmath>
 
<cmath>0 = \frac{p^2}{4} - \frac{2p^2}{4} + q</cmath>

Latest revision as of 12:26, 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

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