2021 AMC Fall 10A Problem 20

A monic quadratic $x^2 + px + q$ has two distinct real roots exactly when its discriminant $p^2 - 4q$ is positive. We need both discriminants to be not positive:
$$b^2 \le 4c \qquad\text{and}\qquad c^2 \le 4b.$$
(Equality is allowed: a double root is a single solution, not two distinct ones.)

Bound. Chain the inequalities: $b^4 \le 16c^2 \le 64b$, so $b^3 \le 64$ and $b \le 4$. By symmetry $c \le 4$ as well.

Enumerate. For each $b$ from $1$ to $4$, the conditions are $c \ge b^2/4$ and $c \le 2\sqrt{b}$:

• $b = 1$: $c \ge \tfrac14$ and $c \le 2$, so $c \in \{1, 2\}$.
• $b = 2$: $c \ge 1$ and $c \le \sqrt8 \approx 2.8$, so $c \in \{1, 2\}$.
• $b = 3$: $c \ge \tfrac94$ and $c \le \sqrt{12} \approx 3.5$, so $c = 3$.
• $b = 4$: $c \ge 4$ and $c \le 4$, so $c = 4$.

The pairs are $(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)$: six in all.

The answer is $\boxed{\textbf{(B)}\ 6}$.

"Neither has two distinct real solutions" translates directly into two discriminant inequalities. Because each variable is bounded by a function of the other ($b^2 \le 4c$, $c^2 \le 4b$), feeding one into the other produces an absolute bound, after which the search space is tiny. Read the wording carefully: "two distinct" means the boundary case (discriminant zero) counts as acceptable.

Since the conditions are symmetric, assume $b \le c$ first. Then $c^2 \le 4b \le 4c$ gives $c \le 4$, and listing $(b, c)$ with $b \le c \le 4$ and $b^2 \le 4c$, $c^2 \le 4b$ gives $(1,1), (1,2), (2,2), (3,3), (4,4)$. The one pair with $b \ne c$ contributes its mirror $(2,1)$, for a total of $6$.

• Requiring strictly negative discriminants (no real roots), which wrongly discards the double-root pairs (1,2), (2,1) and (4,4) and leaves only 3 pairs.
• Counting unordered pairs (getting $5$) or forgetting to bound $b$ and $c$ and missing $(4,4)$.