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 $f$.

Question

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.

Answer 1

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 $f$.

Answer 2

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...

Answer 3

$$ 0=f(r)=r^2+17r+a=r^2+17r-a=g(-r)=0, $$ and thus $a=0$.

This implies that the roots of $f$ are $0$ and $-17$, while the roots of $g$ are $0$ and $17$.

Answer 4

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) \ . $