Solve the equation $h(x) = f(x) + g(x) = 0$, $f(x)$ and $g(x)$ having roots that are negatives of each other. Let $f(x) = x^2 +bx+ 9$ and let $g(x) = x^2 +ax+c, a, b, c ∈ R$. The roots of $f(x) = 0$ and
$g(x) = 0$ are negatives of each other. If $h(x) = f(x)+g(x)$, then solve the equation $h(x) = 0$.
I'm not sure how to solve this at all, maybe Vieta's formulas for quadratics can help.

One can certainly use Vieta's formulas, but I'll give a self-contained solution. (My argument essentially reproves them though.)
Suppose $r_1, r_2$ are the two roots of $f$. By the Fundamental Theorem of Algebra (and using the fact that $f$ is monic--i.e., the leading coefficient of $x^2$ is $1$), we may write
$$f(x) = x^2+bx+9 = (x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2.$$
Since $-r_1$ and $-r_2$ are the roots of $g$ and $g$ is also monic, we have
$$g(x) = x^2+ax+c = (x+r_1)(x+r_2) = x^2+(r_1+r_2)x+r_1r_2.$$
Comparing these two equations, we see $c=9$ and $a=-b$. Thus
$$h(x) = f(x)+g(x) = (x^2+bx+9)+(x^2-bx+9) = 2x^2+18.$$
So $h$ has two imaginary roots: $\pm 3i$.
