Difference between revisions of "1970 AHSME Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | Let's imagine this as a quadratic equation. To find the minimum or maximum value, we always need to find the vertex of the quadratic equation. The vertex of the quadratic is <math>\frac{-b}{2a}</math> in <math>ax^2+bx+c=0</math>. Then to find the output, or the y value of the quadratic, we plug the vertex "x" value back into the equation. In this quadratic, a=1, b=8, and c=0. So the "x" value is <math>\frac{-b}{2a} \Rightarrow \frac{-8}{2} = -4</math>. Plugging it back into <math>x^2 + 8x</math>, we get <math>16-32 = -16 \Rightarrow</math> <math>\fbox{B}</math> |
== See also == | == See also == | ||
| − | {{AHSME box|year=1970|num-b=5|num-a=7}} | + | {{AHSME 35p box|year=1970|num-b=5|num-a=7}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 03:15, 12 March 2017
Problem
The smallest value of 
for real values of 
is
Solution
Let's imagine this as a quadratic equation. To find the minimum or maximum value, we always need to find the vertex of the quadratic equation. The vertex of the quadratic is 
in 
. Then to find the output, or the y value of the quadratic, we plug the vertex "x" value back into the equation. In this quadratic, a=1, b=8, and c=0. So the "x" value is 
. Plugging it back into 
, we get 
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
