### Instruction: Let $f(x) = x^2 + bx + 9$ and $g(x) = x^2 + dx + e.$  If $f(x) = 0$ has roots $r$ and $s,$ and $g(x) = 0$ has roots $-r$ and $-s,$ compute the two roots of $f(x) + g(x) = 0.$.### Response: To find the two roots of $f(x) + g(x) = 0,$ I need to first find the values of $b, d,$ and $e$ in terms of $r$ and $s.$
I know that $f(x) = 0$ has roots $r$ and $s,$ so by the factor theorem, I can write $f(x) = (x - r)(x - s).$
Expanding this gives $f(x) = x^2 - (r + s)x + rs.$
Comparing this with $f(x) = x^2 + bx + 9,$ I can see that $b = -(r + s)$ and $9 = rs.$
Similarly, I know that $g(x) = 0$ has roots $-r$ and $-s,$ so I can write $g(x) = (x + r)(x + s).$
Expanding this gives $g(x) = x^2 + (r + s)x + rs.$
Comparing this with $g(x) = x^2 + dx + e,$ I can see that $d = r + s$ and $e = rs.$
Now I have $f(x) = x^2 - (r + s)x + 9$ and $g(x) = x^2 + (r + s)x + 9.$
Adding them together gives $f(x) + g(x) = 2x^2 + 18.$
To find the roots of $f(x) + g(x) = 0,$ I need to solve $2x^2 + 18 = 0.$
This is a simple quadratic equation, which I can divide by $2$ to get $x^2 + 9 = 0.$
I can use the quadratic formula to find the solutions of $x^2 + 9 = 0.$
They are $x = \frac{-0 \pm \sqrt{-36}}{2} = \pm 3i.$
Therefore, the two roots of $f(x) + g(x) = 0$ are $\pm 3i.$