# Root question

1. Nov 30, 2011

### Miike012

My friends teacher posted a sample problem saying find the real roots of x^6 + 1. Is this a trick question? All roots for this function are imaginary right?

2. Nov 30, 2011

### tjackson3

Yep!

Consider the fact that $x^6 + 1 = 0 \Rightarrow x^6 = -1 \Rightarrow x = (-1)^{1/6}$. If you write $-1 = e^{i(\pi + 2\pi n)}$, then this is simply $e^{i(\pi/6 + n\pi/3)} = \cos(\pi/6 + n\pi/3) + i\sin(\pi/6 + n\pi/3)$ for n = 0, 1, ..., 5. If there were any real roots, then the imaginary part of this would be zero; i.e., we would have $\pi/6 + n\pi/3 = m\pi$ for some integer m. But since there are only finitely many values of n, you can just do this manually. We have:

$\pi/6, \pi/2, 5\pi/6, 7\pi/6, 3\pi/2, 11\pi/6$

none of which would cancel the imaginary part.

3. Nov 30, 2011

### tjackson3

If you've taken calculus (and I assume you have if you're on this board), another way to see it would be to note that since this is an even degree polynomial, the limit as $x\rightarrow \pm\infty$ will be the same. In this case, it is positive infinity. Now find the minimum of this function by taking the derivative ($6x^5$) and setting it equal to zero. Then you can see that the only critical point is at x = 0, which corresponds to f(0) = 1 in the original function. You can perform a derivative test if you want to verify that it's a minimum, but graphically you can see that it is. Since the minimum is y = 1, there are no roots.

4. Nov 30, 2011

### HallsofIvy

Well, not "imaginary", but "complex". the only "imaginary" roots of $x^6+ 1= 0$ is are i and -i. The other for are non-real complex numbers.