Can a quadratic have non-zero coefficients when the roots are $k$ and $-k$? Can a polynomial like this:
$$ax^2+bx+c$$
have two opposite roots, without either $b$ or $a$ being equal to zero?
 A: Let, if possible, the roots be $k,-k$.
So, the quadratic polynomial must be $(x-k)(x+k)=x^2-k^2$.
So, $b$ must be equal to $0$.
Also, note that sum of roots $=\displaystyle\frac{-b}{a}$.
So, $\displaystyle(k+(-k))=\frac{-b}{a}$
$\implies\displaystyle\frac{-b}{a}=0$
$\implies b=0$
A: Vieta's formulas state that the (not necessarily distinct) roots $r_1,\dotsc,r_n$ of
$$
p(x)=a_n x^n+a_{n-1}x^{n-1}+\dotsb+a_1x+a_0
$$
are related to the coefficients $a_0,\dotsc,a_n$ by
\begin{align*}
r_1+\dotsb+r_n &= -\frac{a_{n-1}}{a_n} \\
(r_1r_2+r_1r_3+\dotsb+r_1r_n)+(r_2r_3+r_2r_4+\dotsb r_2r_n)+\dotsb+r_{n-1}r_n &= \frac{a_{n-2}}{a_n} \\
&\vdots\\
r_1r_2\dotsb r_n&=(-1)^n\frac{a_0}{a_n}
\end{align*}
In your case, applying the first of Vieta's formulas to the two roots $r_1$ and $r_2$ of
$$
p(x)=ax^2+bx+c
$$
gives the relation
$$
r_1+r_2=-\frac{b}{a}
$$
Thus $r_1=-r_2$ implies $-b/a=0$ so that $b=0$.
Of course, Vieta's formulas are complicated in general. However, your case $r_1+r_2=-b/a$ is quite simple to show. Note that your polynomial factors as
$$
ax^2+bx+c=a(x-r_1)(x-r_2)
$$
Expanding the right-hand side of this expression gives
$$
ax^2+bx+c=ax^2-a(r_1+r_2)x+ar_1r_2
$$
Comparing the coefficients of $x$ gives
$$
b=-a(r_1+r_2)
$$
Hence $r_1+r_2=-b/a$.
A: The other answerers have given good algebraic justifications for why you must have $b=0$, but here's a simple geometric one: we can see that parabolas are always symmetrical around their extreme point. If the roots are $k$ and $-k$, then the extreme point must be at $y=0$. Since setting $b$ is how you translate the parabola along the $x$-axis, it must still be $0$.
