2021 AMC 12A Fall Problem 17

Below is the professionally curated solution for Problem 17 of the 2021 AMC 12A Fall, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Fall solutions, or check the answer key.

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Concepts:quadratic inequality

Difficulty rating: 1910

17.

For how many ordered pairs $(b, c)$ of positive integers does neither $x^2 + bx + c = 0$ nor $x^2 + cx + b = 0$ have two distinct real solutions?

$4$

$6$

$8$

$12$

$16$

Solution:

Neither quadratic has two distinct real roots exactly when both discriminants are nonpositive: $b^2 \le 4c$ and $c^2 \le 4b.$

Multiplying gives $b^2c^2 \le 16bc,$ so $bc \le 16,$ forcing small values. Checking: $b = 1$ gives $c \in \{1,2\};$ $b = 2$ gives $c \in \{1,2\};$ $b = 3$ gives $c = 3;$ $b = 4$ gives $c = 4;$ and $b \ge 5$ gives none.

That is $(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)$ — $6$ ordered pairs.

Thus, the correct answer is B.

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