Difference between revisions of "1955 AHSME Problems/Problem 39"
(Wrote a Solution for this problem, since it didn't have one.) |
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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== | ||
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 | ||
− | <cmath>\frac{-p}{2}</cmath> | + | <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> | ||
− | < | + | <cmath>0 = \frac{-p^2}{4} + q</cmath> |
<cmath>q = \frac{p^2}{4} = \boxed{B}</cmath> | <cmath>q = \frac{p^2}{4} = \boxed{B}</cmath> | ||
~JustinLee2017 | ~JustinLee2017 |
Latest revision as of 13:26, 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
Plugging this into the quadratic, we get
~JustinLee2017