2019 AMC 12B Problems/Problem 21

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AMC 12 Problems and Solutions

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