1955 AHSME Problems/Problem 39: Difference between revisions
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==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== | ==Solution== | ||
Latest revision as of 12:26, 19 July 2021
Problem
If 
, then if the least possible value of 
is zero 
is equal to:
Solution
The least possible value of is given at the coordinate of the vertex. The 
- coordinate is given by
![\[\frac{-p}{(2)(1)} = \frac{-p}{2}\]](http://latex.artofproblemsolving.com/b/8/2/b82345ef393204f0d40274e191064b70a4295c2c.png)
Plugging this into the quadratic, we get
![\[y = \frac{p^2}{4} - \frac{p^2}{2} + q\]](http://latex.artofproblemsolving.com/6/a/2/6a2afa567a387b10458f22e8fc82be8049ae82f9.png)
![\[0 = \frac{p^2}{4} - \frac{2p^2}{4} + q\]](http://latex.artofproblemsolving.com/a/7/5/a759167c387574d4a12f3635947c8284350ca532.png)
![\[0 = \frac{-p^2}{4} + q\]](http://latex.artofproblemsolving.com/9/e/5/9e5bca1b0f9c9eebe8b449b05c38e8bdd37c31f0.png)
![\[q = \frac{p^2}{4} = \boxed{B}\]](http://latex.artofproblemsolving.com/3/5/4/354bd18bcb66eab365a35361dd29404c5dccc2ed.png)
~JustinLee2017
