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Difference between revisions of "2019 AMC 12B Problems/Problem 21"

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<math>\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } \text{infinitely many}</math>
 
<math>\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } \text{infinitely many}</math>
  
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==Solution==
+
==Solution 1==
 +
First, if <math>r=s</math>, then <math>a=b=c</math>, equation <math>ax^2 + ax + a = 0</math> has no real roots.
 +
 
 +
Then there are three cases:
 +
 
 +
Case 1: <math>a=b=r</math>, <math>c=s</math>:
 +
 
 +
The equation becomes <math>ax^2+ax+c=0</math>, and by Vieta's theorem, we have <math>a+c=-1</math>, <math>ac = \frac{c}{a}</math> or <math>(a^2-1)c=0</math>
 +
if <math>c=0</math>, we get <math>a=-1</math>, equation <math>-x^2 - x = 0</math> has roots 0 and <math>-1</math>. <math>\boxed{1}</math>
 +
 
 +
if <math>c\ne 0</math>, then <math>a^2-1=0</math>,
 +
 
 +
if <math>a=1</math>, we have <math>c=-2</math>, and the equation is <math>x^2 + x - 2 = 0</math> <math>\boxed{2}</math>
 +
 
 +
if <math>a=-1</math>, we have <math>c=0</math>, same as <math>\boxed{1}</math>
 +
 
 +
Case 2: <math>a=c=r</math>, <math>b=s</math>:
 +
 
 +
The equation becomes <math>ax^2 + bx + a = 0</math>, by Vieta's theorem, we have <math>a+b = -\frac{b}{a}</math> and <math>ab = 1</math>. This can be rewritten as
 +
<math>a^3 + a + 1 = 0</math>, which has 1 real root of <math>a</math>. <math>\boxed{3}</math>
 +
 
 +
Case 3: <math>a=r</math>, <math>b=c=s</math>:
 +
 
 +
The equation becomes <math>a^x + bx+b=0</math>, by Vieta's theorem, we have <math>a+b = -\frac{b}{a}</math> and <math>ab = \frac{b}{a}</math>.
 +
<math>b</math> can not be 0, so we have <math>a=\frac{1}{a}</math>, or <math>a=\pm 1</math>.
 +
 
 +
if <math>a=1</math>, we have <math>1+b=-b</math>, and <math>b=-\frac{1}{2}</math>, the equation is <math>x^2 - \frac{1}{2}x - \frac{1}{2} = 0</math>, <math>\boxed{4}</math>
 +
 
 +
if <math>a=-1</math>, we have <math>1+b=b</math>, a contradiction.
 +
 
 +
The answer is <math>\boxed{B}</math>
  
 
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>.
 
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>.

Revision as of 21:00, 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 1

First, if $r=s$, then $a=b=c$, equation $ax^2 + ax + a = 0$ has no real roots.

Then there are three cases:

Case 1: $a=b=r$, $c=s$:

The equation becomes $ax^2+ax+c=0$, and by Vieta's theorem, we have $a+c=-1$, $ac = \frac{c}{a}$ or $(a^2-1)c=0$ if $c=0$, we get $a=-1$, equation $-x^2 - x = 0$ has roots 0 and $-1$. $\boxed{1}$

if $c\ne 0$, then $a^2-1=0$,

if $a=1$, we have $c=-2$, and the equation is $x^2 + x - 2 = 0$ $\boxed{2}$

if $a=-1$, we have $c=0$, same as $\boxed{1}$

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

The equation becomes $ax^2 + bx + a = 0$, by Vieta's theorem, we have $a+b = -\frac{b}{a}$ and $ab = 1$. This can be rewritten as $a^3 + a + 1 = 0$, which has 1 real root of $a$. $\boxed{3}$

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

The equation becomes $a^x + bx+b=0$, by Vieta's theorem, we have $a+b = -\frac{b}{a}$ and $ab = \frac{b}{a}$. $b$ can not be 0, so we have $a=\frac{1}{a}$, or $a=\pm 1$.

if $a=1$, we have $1+b=-b$, and $b=-\frac{1}{2}$, the equation is $x^2 - \frac{1}{2}x - \frac{1}{2} = 0$, $\boxed{4}$

if $a=-1$, we have $1+b=b$, a contradiction.

The answer is $\boxed{B}$

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
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