# strange adventures in algebra

I’ve been catching up on my comic book reading recently, and saw a fun little math problem at the bottom of the first page of Tom King’s *Strange Adventures* #2. Mr. Terrific’s T-spheres are quizzing him on a little math problem:

In the equation \(x^2 + mx + n = 0\), \(m\) and \(n\) are integers. The only possible value of of \(x\) is \(-3\). What is the value of \(m\)?

Mr. Terrific correctly guesses, “six”, apparently nearly instantly. Why is this the answer?

This is a quadratic equation, and so it has either zero, one, or two real solutions. Recall the *quadratic formula*, i.e., a quadratic equation of the form \(a x^2 + bx + c = 0\) has solution(s):

In particular, we have a unique solution when \(b^2 - 4ac = 0\), and in this case, this solution is

\[x = -\frac{b}{2a}\]In Mr. Terrific’s equation, we have \(a = 1\), \(b = m\), and \(c = n\), so the unique solution is:

\[x = -\frac{m}{2\cdot 1} = -\frac{m}{2}\]Since the T-sphere says that \(x = -3\), we have

\[-3 = -\frac{m}{2}\] \[m = 6\]The T-sphere doesn’t ask Mr. Terrific the value of \(n\), but we can figure it out. Recall that \(b^2 - 4ac = 0\), or in this case:

\[m^2 - 4\cdot 1\cdot n = 0\] \[6^2 - 4n = 0\] \[4n = 36\] \[n = 9\]Also, the T-sphere didn’t really need to specify that \(m\) and \(n\) were integers. Indeed, these are the only *complex numbers* that work.

Geometrically, we’re looking for a parabola with an \(x\)-intercept of \(-3\), and so it must have equation

\(y = a(x + 3)^2 = a(x^2 + 6x + 9) = ax^2 + 6ax + 9a\),

but we know that \(a = 1\), and so we obtain the same result.