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Difference between revisions of "2019 AMC 12B Problems/Problem 21"
(Solution Work in Progress)
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==Solution==
 
==Solution==
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Because there are three coefficients and two roots, we need at least two elements in the set <math>\{a,b,c\}</math> to be equal to each other. It is possible that all three could be equal to each other. In the case that two elements in the set <math>\{a,b,c\}</math> are equal to each other, two of those elements will be equal to <math>r</math> and the third will be equal to <math>s</math>.
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Case 1: <math>r = s</math>
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We would need the polynomial <math>rx^2 + rx + r</math> to have a double root <math>r</math>. By inspection, there is no such polynomial, so there are no polynomials for this case.
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Case 2: <math>a = b = r</math> and <math>c = s</math>
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The polynomial will be in the form <math>rx^2 + rx + s</math>. By Vieta's formulas, <math>r + s = -1</math> and <math>rs = \frac{s}{r}</math>. The second equation tells us that either <math>r = 0</math> or <math>s^2 = 1</math>. Testing each possibility, we find the polynomials <math>-x^2 - x</math> and <math>x^2 + x - 2</math>, both of which work. There are 2 polynomials for this case.
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Case 3: <math>a = c = r</math> and <math>b = s</math>
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The polynomial will be in the form <math>rx^2 + sx + r</math>. By Vieta's formulas, <math>r + s = \frac{-s}{r}</math> and <math>rs = 1</math>. Through substitution, we get <math>r^3 + r + 1</math>. The function f(r) = <math>r^3 + r + 1</math> is a strictly increase function with one real root.
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[Work in Progress]
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=20|num-a=22}}
 
{{AMC12 box|year=2019|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:50, 14 February 2019

Problem

How many quadratic polynomials with real coefficients are there such that the set of roots equals the set of coefficients? (For clarification: If the polynomial is $ax^2+bx+c,a\neq 0,$ and the roots are $r$ and $s,$ then the requirement is that $\{a,b,c\}=\{r,s\}$.)

$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } \text{infinitely many}$

Solution

Because there are three coefficients and two roots, we need at least two elements in the set $\{a,b,c\}$ to be equal to each other. It is possible that all three could be equal to each other. In the case that two elements in the set $\{a,b,c\}$ are equal to each other, two of those elements will be equal to $r$ and the third will be equal to $s$.

Case 1: $r = s$

We would need the polynomial $rx^2 + rx + r$ to have a double root $r$. By inspection, there is no such polynomial, so there are no polynomials for this case.

Case 2: $a = b = r$ and $c = s$

The polynomial will be in the form $rx^2 + rx + s$. By Vieta's formulas, $r + s = -1$ and $rs = \frac{s}{r}$. The second equation tells us that either $r = 0$ or $s^2 = 1$. Testing each possibility, we find the polynomials $-x^2 - x$ and $x^2 + x - 2$, both of which work. There are 2 polynomials for this case.

Case 3: $a = c = r$ and $b = s$

The polynomial will be in the form $rx^2 + sx + r$. By Vieta's formulas, $r + s = \frac{-s}{r}$ and $rs = 1$. Through substitution, we get $r^3 + r + 1$. The function f(r) = $r^3 + r + 1$ is a strictly increase function with one real root.

[Work in Progress]

See Also

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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