Can a quadratic have non-zero coefficients when the roots are $k$ and $-k$?

Question

Can a polynomial like this:

$$ax^2+bx+c$$

have two opposite roots, without either $b$ or $a$ being equal to zero?

Answer 1

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, $\displaystyle(k+(-k))=\frac{-b}{a}$

$\implies\displaystyle\frac{-b}{a}=0$

$\implies b=0$

Answer 2

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 $r_1+r_2=-b/a$.

Answer 3

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 $0$.