Difference between revisions of "2001 AMC 12 Problems/Problem 19"
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== Solution == | == Solution == | ||
| − | We are given <math>c=2</math>. So the product of the roots is <math>-c = -2</math> by [[Vieta's formulas]]. These also tell us that <math>\frac{-a}{3}</math> is the average of the zeros, so <math>\frac{-a}3=-2 \implies a = 6</math>. We are also given that the sum of the coefficients is <math>-2</math>, so <math>1+6+b+2 = -2 \implies b=-11</math>. So the answer is <math>\ | + | We are given <math>c=2</math>. So the product of the roots is <math>-c = -2</math> by [[Vieta's formulas]]. These also tell us that <math>\frac{-a}{3}</math> is the average of the zeros, so <math>\frac{-a}3=-2 \implies a = 6</math>. We are also given that the sum of the coefficients is <math>-2</math>, so <math>1+6+b+2 = -2 \implies b=-11</math>. So the answer is <math>\fbox{A}</math>. |
== See Also == | == See Also == | ||
Revision as of 19:55, 13 June 2015
Problem
The polynomial 
has the property that the average of its zeros, the product of its zeros, and the sum of its coefficients are all equal. The 
-intercept of the graph of 
is 2. What is 
?
Solution
We are given 
. So the product of the roots is 
by Vieta's formulas. These also tell us that 
is the average of the zeros, so 
. We are also given that the sum of the coefficients is 
, so 
. So the answer is 
.
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
