2021 Fall AMC 12A Problems/Problem 17

Problem

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?

$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$

Solution 1

If a quadratic equation does not have two distinct real solutions, then its discriminant must be $\le0$ . So, $b^2-4c\le0$ and $c^2-4b\le0$ . By inspection, there are $\boxed{\textbf{(B) } 6}$ ordered pairs of positive integers that fulfill these criteria: $(1,1)$ , $(1,2)$ , $(2,1)$ , $(2,2)$ , $(3,3)$ , and $(4,4)$ .

Solution 2

We need to solve the following system of inequalities: $\left\{ \begin{array}{ll} b^2 - 4 c \leq 0 \\ c^2 - 4 b \leq 0 \end{array} \right..$

Feasible solutions are in the region formed between two parabolas $b^2 - 4 c = 0$ and $c^2 - 4 b = 0$ .

Define $f \left( b \right) = \frac{b^2}{4}$ and $g \left( b \right) = 2 \sqrt{b}$ . Therefore, all feasible solutions are in the region formed between the graphs of these two functions.

For $b = 1$ , $f \left( b \right) = \frac{1}{4}$ and $g \left( b \right) = 2$ . Hence, the feasible $c$ are 1, 2.

For $b = 2$ , $f \left( b \right) = 1$ and $g \left( b \right) = 2 \sqrt{2}$ . Hence, the feasible $c$ are 1, 2.

For $b = 3$ , $f \left( b \right) = \frac{9}{4}$ and $g \left( b \right) = 2 \sqrt{3}$ . Hence, the feasible $c$ is 3.

For $b = 4$ , $f \left( b \right) = 4$ and $g \left( b \right) = 4$ . Hence, the feasible $c$ is 4.

For $b > 4$ , $f \left( b \right) > g \left( b \right)$ . Hence, there is no feasible $c$ .

Putting all cases together, the correct answer is $\boxed{\textbf{(B) }6}$ .

~Steven Chen (www.professorchenedu.com)

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2021 Fall AMC 12A Problems/Problem 17

Problem

Solution 1

Solution 2