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: $\pm 3i$.
$f(x) = x^2 +bx+ 9;\;g(x) = x^2 +ax+c$
The sum of the roots of $f(x)$ is $s=-b$ and their product is $p=9$
The sum the roots of $g(x)$ is $-a$ and their product is $c$
So we must have $a=-b;\;c=9$
$f(x)=x^2+bx+9;\;g(x)=x^2-bx+9$
$h(x)=f(x)+g(x)=2x^2+18=0\to x=\pm 3i$
