Difference between revisions of "2003 AMC 12B Problems/Problem 20"
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<cmath>f(x) - g(x) = ax(x-1)(x+1) = ax^3 - ax</cmath> | <cmath>f(x) - g(x) = ax(x-1)(x+1) = ax^3 - ax</cmath> | ||
implies by <math>x^2</math> coefficient, <math>b + 2 = 0, b = -2 \Rightarrow \mathrm{(B)}</math>. | implies by <math>x^2</math> coefficient, <math>b + 2 = 0, b = -2 \Rightarrow \mathrm{(B)}</math>. | ||
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+ | ==Solution 4== | ||
+ | The roots of this equation are <math>-1, 1, \text{ and } x</math>, letting <math>x</math> be the root not shown in the graph. By Vieta, we know that <math>-1+1+x=x=-\frac{b}{a}</math> and <math>-1\cdot 1\cdot x=-x=-\frac{d}{a}</math>. Therefore, <math>x=\frac{d}{a}</math>. Setting the two equations for <math>x</math> equal to each other, <math>\frac{d}{a}=-\frac{b}{a}</math>. We know that the y-intercept of the polynomial is <math>d</math>, so <math>d=2</math>. Plugging in for <math>d</math>, <math>\frac{2}{a}=-\frac{b}{a}</math>. | ||
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+ | Therefore, <math>b=-2 \Rightarrow \boxed{B}</math> | ||
== See also == | == See also == |
Revision as of 15:18, 23 October 2017
Problem
Part of the graph of is shown. What is ?
Solution
Solution 1
Since
It follows that . Also, , so .
Solution 2
Two of the roots of are , and we let the third one be . Then Notice that , so .
Solution 3
Notice that if , then vanishes at and so implies by coefficient, .
Solution 4
The roots of this equation are , letting be the root not shown in the graph. By Vieta, we know that and . Therefore, . Setting the two equations for equal to each other, . We know that the y-intercept of the polynomial is , so . Plugging in for , .
Therefore,
See also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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.