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Difference between revisions of "2001 AMC 12 Problems/Problem 19"

(New, more inefficient solution)
 
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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>\fbox{A}</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>.
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== Solution 2 (inefficient) ==
 
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If you don't know Vieta's formulas, you can go through the desperate route. Let <math>p(x) = (x-r_1)(x-r_2)(x-r_3)</math>. After distributing, we find that <math>p(x) = x^3 - (r_1+r_2+r_3)x^2 + (r_1r_2+r_2r_3+r_1r_3)x - r_1r_2r_3</math>. We know that the last term (<math>c</math>) is <math>2</math>, so <math>r_1r_2r_3</math> must be <math>-2</math>. We know from our equation that <math>b = r_1r_2+r_2r_3+r_1r_3</math>. The problem gives that the average of the roots is equal to their product (their product is <math>-2</math>), so the sum of the roots is hence <math>-2\cdot3 = -6</math> (giving us <math>a = 6</math>, since a is the sum of the roots times <math>-1</math>). Finally, we know that the sum of the coefficients is the same as the product of the roots, so the sum of the coefficients is <math>-2</math>. Thus <math>1 + a + b + c = -2</math>, so <math>1 + 6 + b + 2 = -2</math>, so <math>b + 9 = -2</math> and <math>b = -11</math>. The answer is then <math>\fbox{A}</math>.
 
  
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==

Latest revision as of 21:48, 18 August 2023

Problem

The polynomial $p(x) = x^3+ax^2+bx+c$ has the property that the average of its zeros, the product of its zeros, and the sum of its coefficients are all equal. The $y$-intercept of the graph of $y=p(x)$ is 2. What is $b$?

$(\mathrm{A})\ -11 \qquad (\mathrm{B})\ -10 \qquad (\mathrm{C})\ -9 \qquad (\mathrm{D})\ 1 \qquad (\mathrm{E})\ 5$

Solution

We are given $c=2$. So the product of the roots is $-c = -2$ by Vieta's formulas. These also tell us that $\frac{-a}{3}$ is the average of the zeros, so $\frac{-a}3=-2 \implies a = 6$. We are also given that the sum of the coefficients is $-2$, so $1+6+b+2 = -2 \implies b=-11$. So the answer is $\fbox{A}$.

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
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All AMC 12 Problems and Solutions

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