Difference between revisions of "2003 AMC 12B Problems/Problem 20"
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\qquad\mathrm{(E)}\ 4</math> | \qquad\mathrm{(E)}\ 4</math> | ||
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
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Two of the roots of <math>f(x) = 0</math> are <math>\pm 1</math>, and we let the third one be <math>n</math>. Then | Two of the roots of <math>f(x) = 0</math> are <math>\pm 1</math>, and we let the third one be <math>n</math>. Then | ||
<cmath>a(x-1)(x+1)(x-n) = ax^3-anx^2-ax+an = ax^3 + bx^2 + cx + d = 0</cmath> | <cmath>a(x-1)(x+1)(x-n) = ax^3-anx^2-ax+an = ax^3 + bx^2 + cx + d = 0</cmath> | ||
− | Notice that <math>f(0) = d = an = 2</math>, so <math>b = -an = -2</math>. | + | Notice that <math>f(0) = d = an = 2</math>, so <math>b = -an = -2 \Rightarrow \mathrm{(B)}</math>. |
+ | |||
+ | === Solution 3 === | ||
+ | Notice that if <math>g(x) = 2 - 2x^2</math>, then <math>f - g</math> vanishes at <math>x = -1, 0, 1</math> and so | ||
+ | <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>. | ||
+ | |||
+ | ===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>. | ||
+ | |||
+ | Therefore, <math>b=-2 \Rightarrow \boxed{B}</math> | ||
+ | |||
+ | ===Solution 5=== | ||
+ | From the graph, we have <math>f(0)=2</math> so <math>d=2</math>. Also from the graph, we have <math>f(1)=a+b+c+2=0</math>. But we also have from the graph <math>f(-1)=-a+b-c+2=0</math>. Summing <math>f(1)+f(2)</math> we get <math>2b+4=0</math> so <math>b = -2 \Rightarrow \mathrm{(B)}</math>. | ||
+ | |||
+ | Solution by franzliszt | ||
== See also == | == See also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:01, 23 November 2020
Contents
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,
Solution 5
From the graph, we have so
. Also from the graph, we have
. But we also have from the graph
. Summing
we get
so
.
Solution by franzliszt
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.