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 $\boxed{B}$

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]

2019 AMC 12B (Problems • Answer Key • Resources)
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2019 AMC 12B Problems/Problem 21

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Solution 1

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