Difference between revisions of "2008 AMC 10B Problems/Problem 9"
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==Solution 2== | ==Solution 2== | ||
| − | We know that for an equation <math>ax^2 + bx + c = 0</math>, the sum of the roots is <math>-b/a</math>. This means that the sum of the roots for <math>ax^2 - 2ax + b = 0</math> is <math>\frac{2a}{a}=2</math>. The average is the sum of the two roots divided by two, so the average is <math>\frac22 = 1</math>. | + | We know that for an equation <math>ax^2 + bx + c = 0</math>, the sum of the roots is <math>-b/a</math>. This means that the sum of the roots for <math>ax^2 - 2ax + b = 0</math> is <math>\frac{2a}{a}=2</math>. The average is the sum of the two roots divided by two, so the average is <math>\frac22 = 1 \Rightarrow \boxed{A}</math>. |
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2008|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:34, 7 June 2021
Contents
Problem
A quadratic equation 
has two real solutions. What is the average of these two solutions?
Solution 1
Dividing both sides by 
, we get 
. By Vieta's formulas, the sum of the roots is 
, therefore their average is 
.
Solution 2
We know that for an equation 
, the sum of the roots is 
. This means that the sum of the roots for is 
. The average is the sum of the two roots divided by two, so the average is 
.
See also
| 2008 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
