Why does the discriminant of a cubic polynomial being less than $0$ indicate complex roots?

The discriminant of any monic polynomial is the product $\prod_{i \neq j} (x_i - x_j)^2$ of the squares of the differences of the roots (in an algebraic closure, e.g. $C$). Cf. the Wikipedia article on this. Consequently, if the roots are all real and distinct, this must be positive.

(If the polynomial is not monic, the factor $a_0^{2n-2}$ is thrown in, for $a_0$ the leading coefficient and $n$ the degree; this is positive for a real polynomial.)

These implications are reached by considering the three, different cases for the roots $\{ r_1, r_2, r_3 \}$ of the polynomial: repeated root, all distinct real roots, or two complex roots and one real root.

When one of the roots is repeated, say $r_1$ and $r_2$, then it is clear that the discriminant is $0$ because the $r_1 - r_2$ term of the product is $0$.

When one root is a complex number $\rho = x+ yi$, then by the complex conjugate root theorem, $\overline{\rho} = x - yi$ is also a root. By the same theorem, the remaining third root must be real. Evaluating the product in the discriminant for this case,

$$ \begin{align*} (\rho - \overline{\rho})^2 (\rho - r_3)^2 (\overline{\rho} - r_3)^2 &= (2yi)^2 (x + yi - r^3)^2 (x - yi - r^3)^2 \\ &= -4y^2 [((x - r_3) + yi) ((x - r_3) - yi) ]^2 \\ &= -4y^2 ((x - r_3)^2 + y^2)^2 \end{align*} $$

which is less than or equal to $0$.

Finally, when all roots are real, the product is clearly positive.

Putting it all together, $\Delta$:

• less than $0$ implies that one root is complex;
• equal to $0$ implies that one root is repeated;
• greater than $0$ implies that all roots are distinct and real.