Summer Problem Solving Marathon Solution #11

Let r be the solution of x² - 3x + c = 0 whose negative -r is a solution of x² + 3x - c = 0. Then we can substitute r for x in the first equation, and -r for x in the second. This yields:

r² - 3r + c = 0
r² - 3r - c = 0

Subtracting the second equation from the first, we obtain 2c = 0, so c = 0.

We want the solutions of x² - 3x + c = 0. Since c = 0, we thus want the solutions of x² - 3x = 0.

So we have x(x-3) = 0, and the solutions are x = 0 and x = 3.