2008 AMC 10B Problems/Problem 9: Difference between revisions
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Dividing both sides by <math>a</math>, we get <math>x^2 - 2x + b/a = 0</math>. By Vieta's formulas, the sum of the roots is <math>2</math>, therefore their average is <math>1\Rightarrow \boxed{A}</math>. | Dividing both sides by <math>a</math>, we get <math>x^2 - 2x + b/a = 0</math>. By Vieta's formulas, the sum of the roots is <math>2</math>, therefore their average is <math>1\Rightarrow \boxed{A}</math>. | ||
== | ==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>2a/a</math>, or 2. The average is the sum of the two roots divided by two, so the average is <math>2/2 = 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>2a/a</math>, or 2. The average is the sum of the two roots divided by two, so the average is <math>2/2 = 1</math>. | ||
Revision as of 20:36, 14 January 2016
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 
, or 2. 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 | ||
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