Difference between revisions of "2008 AMC 10B Problems/Problem 9"
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<math>\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ \frac ba\qquad\mathrm{(D)}\ \frac{2b}a\qquad\mathrm{(E)}\ \sqrt{2b-a}</math> | <math>\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ \frac ba\qquad\mathrm{(D)}\ \frac{2b}a\qquad\mathrm{(E)}\ \sqrt{2b-a}</math> | ||
− | ==Solution== | + | ==Solution 1== |
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>. | ||
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==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}} |
Revision as of 21:36, 14 January 2016
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 , 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.