Sum of roots of any equation

Question

Is the sum of the roots of any equation of one term (e.g., $x^n=c$) always equal to zero? I found this out by having a look at the roots of any equations that have been solved till now. Does it hold true only for $n>1?$

$1.$ How to prove it mathematically

$2.$ What insight or intuition does it provide, if any, in the realm of abstract thinking or mathematics itself.

Answer 1

Consider $x-1=0$.

It has one root $x=1$ and it's not zero.

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There is a well-known theorem $($Vieta's theorem$)$ says the sum of root of polynomial

$$a_nx^n+a_{n-1}x^{n-1}+...a_1x+a_0$$

is equals to

$$-\frac{a_{n-1}}{a_n}$$ and not necessary zero. Probably all of your equations had $a_{n-1}=0$

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For your modified version, The $($complex$)$ roots of $x^n=c$ has an interesting form. They all are located on a circle and they sum to zero by symmetry for $n>1$. See the wikipedia page for more details.

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P.S: I've just found this video about roots of unity. I think you will probably like it:

Answer 2

The sum of the roots of any equation of one term (e.g., $x^n=c$) is always zero?

Yes if $n>1$, no if $n=1$. @Leila already gave a more general answer for polynomials, but for this specific form, it is more readily seen. Define $\omega\equiv e^{2\pi i/n}$. Then the sum of roots of $x^n=c$ is

$$\sqrt[n]{c}\left(1+\omega+\omega^2+\cdots+\omega^{n-1}\right)=\begin{cases}c&\text{if}\; n=1\\\sqrt[n]{c}\frac{\omega^n-1}{\omega-1}=0&\text{if}\;n>1\end{cases}.$$

What insight or intuition does it provide, if any, in the realm of abstract thinking or mathematics itself.

The roots of unity can be thought of as weights placed around the perimeter of a circular disc. When there is only 1 weight ($n=1$) then the weight can never be balanced around the center. When there is more than 1 weight ($n>1$) then by equally spacing the weights, as are the $n^\text{th}$ roots of unity, the center of gravity is always the center of the disc$-$in other words they sum to 0.

Answer 3

Let the polynomial $P(x)$ with roots $r_k$ such that $$\sum_{k=1}^n r_k=0.$$

Then $P(x-1)$ is also a polynomial, and it is such that the sum of its roots is $n$, which refutes the claim.

Answer 4

Since you asked for some insight in the realm of mathematics it worth noting that even though the term "roots of any equation" is much more general than "root of polynomials" (which is still more general than x^n=c) most of simple equations that brings to mind has some symmetric properties. One of the most common of these properties is that the sum of their roots are zero. Indeed, it's possible to be even/odd function which their roots are conjugated. Consider sin(x)=0 or cos(x)=1/2.

So an philosophical question may be arose. Is it something metaphysical in the nature or simply our mind have been attracted to this symmetries. Why you just consider x^n=c instead of x^2+x=0 ?