CEMC Fermat, 2016

Solution 1

Since $x=18$ is a solution to the equation $x^2+12x+c=0$, then $x=18$ satisfies this equation.

Thus, $18^2+12(18)+c=0$ and so $324+216 + c = 0$ or $c = -540$.

Therefore, the original equation becomes $x^2+12x-540=0$ or $(x-18)(x+30)=0$.

Therefore, the other solution is $x=-30$.

Solution 2

We use the fact that the sum of the roots of an equation of the form $x^2+bx+c=0$ is $-b$.

If the roots of the equation $x^2+12x+c=0$ are $18$ and $r$, then $18+r=-12$ or $r = -30$.

Therefore, the other solution is $x=-30$.