Let $f(x)=x^2+17x+a$, $g(x)=x^2-17x-a$, $r$ a root of $f$ and $-r$ a root of $g$. Determine the roots of $f$. Let $f(x)=x^2+17x+a$ and $g(x)=x^2-17x-a$. Suppose $r$ is a root of $f$ and $-r$ is a root of $g$. Determine all roots of $f$.
From the descriptions, I can conclude that $f(x)-g(x)=2a$. But that doesn't help.
 A: Since $r$ is a root of $f$ and $-r$ is a root of $g$,
$$\begin{aligned}
0 &= r^2 + 17r + a\\
0 &= (-r)^2 - 17(-r) - a = r^2 + 17r - a
\end{aligned}$$
Combining both equations,
$$\begin{aligned}
0 &= 2r^2 + 34r
\end{aligned}$$
So
$$\begin{aligned}
0 &= 2r(r + 17)
\end{aligned}$$
which shows that $r = 0$ and $r = -17$ are possible roots of $f$.
A: Hints:
$$\begin{cases}r^2+17r+a=0\\r^2+17r-a=0\end{cases}\implies \begin{cases}r=0\;,\;\;or\\{}\\r=-17\end{cases}$$
Take it from here...
A: $$
0=f(r)=r^2+17r+a=r^2+17r-a=g(-r)=0,
$$
and thus $a=0$.
This implies that the roots of $f$ are $0$ and $-17$, while the roots of $g$ are $0$ and $17$. 
A: You can also see this from the quadratic formula.  For $ \ f(x) \ , $
$$ x_f \ = \ \frac{-17 \ \pm \sqrt{17^2 - 4a}}{2} \ , $$
while for $ \ g(x) \ , $
$$ x_g \ = \ \frac{17 \ \pm \sqrt{17^2 + 4a}}{2} \ . $$
In order to meet the stated condition, either both discriminants must be zero, which is not possible, or they must be equal, which requires $ \ a \ $ to be zero.  That situation produces the roots $ \ 0 \ \text{and} \ -17 \ $ for $ \ f(x) \ $ and $ \ 0 \ \text{and} \ +17 \ $ for $ \ g(x) \ . $
