# 2019 AMC 12B Problems/Problem 21

## 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 $\boxed{B}$

## Solution 2

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. This real root works for our value of $r$, so there is 1 polynomial in this case.

Case 4: $b = c = r$ and $a = s$

The polynomial will be in the form $sx^2 + rx + r$. By Vieta's formulas, $r + s = \frac{-r}{s}$ and $rs = \frac{r}{s}$. The second equation tells us that either $r = 0$ or $s^2 = 1$. If $r = 0$, $s = 0$, which causes a divide by 0 error. If $s = -1$, the first equation becomes $0 = 1$, a contradiction. If $s = 1$ then $r = \frac{-1}{2}$. There is 1 polynomial in this case.

Our answer is $\boxed{\mathbf{(B)}\ 4}$

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