2021 AMC 10A Fall Problem 20

Question

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

Use the discriminant condition for each quadratic

The inequalities are

b^2\le4c

and

c^2\le4b

A quadratic fails to have two distinct real solutions exactly when its discriminant is nonpositive. Thus we need

b^2-4c\le0\qquad c^2-4b\le0,

or

b^2\le4c

and

c^2\le4b.

From

b^2\le4c,

we get

b^4\le16c^2.

Combining this with

c^2\le4b

gives

b^4\le64b,

so

b\le4.

Now check

b=1,2,3,4.

The inequalities give respectively

\begin{gathered} (c=1,2),\quad (c=1,2), \\ \quad (c=3),\quad (c=4). \end{gathered}

There are

2+2+1+1=6

ordered pairs.

Thus, B is the correct answer.

Problem 20 in other years