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 | + | 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>. |
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+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/czk6OsfrbxQ?t=678 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=qtlXAxj4y2Y | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC12 box|year=2001|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Latest revision as of 21:48, 18 August 2023
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 .
Video Solution by OmegaLearn
https://youtu.be/czk6OsfrbxQ?t=678
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=qtlXAxj4y2Y
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 |
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