# Number of solutions to an equation

Hello guys I have a simple question to ask. For example I have the equation :

$$x^n + x^{n-1} + x^{n-2} + ... + 1 = 0$$

I read somewhere that the number of solutions to an equation is given by the biggest power in the equation. So in the equation above, there should be $n$ solutions. Is this correct ? If it is, can anyone provide a proof?

• are you interested in purely real solutions? – oldrinb Jan 23 '15 at 20:19
• Sometimes, there are so-called "multiple" roots. For example, $1$ is a "double root" of $x^2 - 2x + 1 = 0$, and there are no other roots. So the theorem, called the "fundamental theorem of algebra" is only valid if roots are taken into account with their multiplicities. Also, the theorem doesn't work if you only want real solutions; you need to allow complex roots. The fundamental theorem of algebra is a difficult theorem to prove. In the specific example you gave, however, the $n$ roots are easy to find. They're the complex numbers $e^{2\pi k i/(n+1)}$ for $k = 1, 2, \dots, n$. That's... – user208259 Jan 23 '15 at 20:41
• ...because if you multiply both sides of the equation by $x-1$, you get $x^{n+1} - 1 = 0$. So the roots are all $(n+1)$st complex roots of unity besides $1$. If you're only interested in real roots, the answer is that there are none of $n$ is even, and just $x=-1$ if $n$ is odd. – user208259 Jan 23 '15 at 20:43

• Working over a field different from $\mathbb{C}$ the situation could be totally different. – Nicky Hekster Jan 23 '15 at 20:19