2021 Fall AMC 10A Problems/Problem 20

Problem

How many ordered pairs of positive integers $(b,c)$ exist where both $x^2+bx+c=0$ and $x^2+cx+b=0$ do not have distinct, real solutions?

$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 12 \qquad$

Solution

A quadratic equation does not have real solutions if and only if the discriminant is nonpositive. We conclude that:

Since $x^2+bx+c=0$ does not have real solutions, we have $b^2\leq 4c.$

Since $x^2+cx+b=0$ does not have real solutions, we have $c^2\leq 4b.$

Squaring the first inequality, we get $b^4\leq 16c^2.$ Multiplying the second inequality by $16,$ we get $16c^2\leq 64b.$ Combining these results, we get $b^4\leq 16c^2\leq 64b.$ Note that:

If $b=1,$ then $1\leq 16c^2\leq 64,$ from which $c=1,2.$

If $b=2,$ then $16\leq 16c^2\leq 128,$ from which $c=1,2.$

If $b=3,$ then $81\leq 16c^2\leq 192,$ from which $c=3.$

If $b=4,$ then $256\leq 16c^2\leq 256,$ from which $c=4.$

Together, there are

~MRENTHUSIASM

Solution 1(Oversimplified but risky)

We want both $x^2+bx+c$ to be $1$ value or imaginary and $x^2+cx+b$ to be $1$ value or imaginary. $x^2+4x+4$ is one such case since $\sqrt {b^2-4ac}$ is $0$ . Also, $x^2+3x+3, x^2+2x+2, x^2+x+1$ are always imaginary for both b and c. We also have $x^2+x+2$ along with $x^2+2x+1$ since the latter has one solution, while the first one is imaginary. Therefore, we have $6$ total ordered pairs of integers, which is $\boxed {(B) 6}$

~Arcticturn

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2021 Fall AMC 10A Problems/Problem 20

Contents

Problem

Solution

Solution 1(Oversimplified but risky)

See Also