# Proved that cubic equation w/ real coefficients always has 2 complex conjugate roots but that's clearly not the case.

Given a cubic polynomial

$$x^3 + c_2 x^2 + c_1 x + c_0 = 0$$

We know that the coefficients can be factored into different functions of the roots, for example, given roots $$r_1$$, $$r_2$$, and $$r_3$$:

$$c_0 = -r_1 r_2 r_3, \\c_1 = r_1r_2 + r_2r_3 + r_1 r_3 \\ c_2 = -(r_1 + r_2 + r_3)$$

For cubic equations with real coefficients I know that there is at least one real root. I can use the imaginary part of the $$c_2$$ equation to then give me that:

$$r_{2i} = - r_{3i }$$

because $$c_2$$ and $$r_1$$ are real. However, this confuses me when I get to the $$c_1$$ equation and check:

$$\text{Im}[c_1] = 0 = \text{Im} [ r_1 ( r_2 + r_3 ) + r_2 r_3 ] = \text{Im}[ r_2 r_3] = \text{Im}[r_{2r}r_{3r} + r_{2i}r_{3i} + i(r_{2r} r_{3i}+ r_{3r}r_{2i})]$$

Since $$r_{3i} = - r_{2i}$$, this would seem to imply that $$r_{2r} = r_{3r}$$, the roots are complex conjugates of each other. I know that this is the case when the other two roots are complex, but it's obviously not true when there are 3 real roots. Where am I going wrong here?

• Remember that the real numbers are a subgroup of the complex numbers, with the imaginary part of a real number being $0$. So a real number can be written as $a + bi$, where $b = 0$ Jan 18 '19 at 4:54

Since $$r_{2i} = -r_{3i}$$, this would seem to imply that $$r_{2r} = r_{3r}$$
Yes, except when $$r_{2i} = r_{3i}=0$$, in which case $$r_{2r}$$ and $$r_{3r}$$ may assume arbitrary values, thus making $$r_2$$ and $$r_3$$ arbitrary real values, corresponding to the case of three real roots.
$$\mbox{Im}[r_{2r}r_{3r}+r_{2i}r_{3i} + i(r_{2r}r_{3i} + r_{2i}r_{3r})] = r_{2r}r_{3i}+r_{2i}r_{3r} = r_{3i}(r_{2r} - r_{3r}) = 0$$
This means that $$r_{2r} = r_{3r}$$ only when $$r_{3i} \neq 0$$, or that $$r_3$$ is strictly complex. In that case, $$r_2$$ is also strictly complex, and $$r_2$$ and $$r_3$$ are complex conjugates of each other. If $$r_{3i} = 0$$, then $$r_3$$ is real, forcing $$r_2$$ to be real as well, but not related to $$r_3$$ in any way.