Cubic Discriminant

Question

For a real cubic polynomial, this website states that if discriminant is 0 then all roots are still real. I don't see why it should be true. Can anyone explain?

Answer 1

The discriminant is $0$ iff the cubic has a double root. On the other hand, if a polynomial with real coefficients has a non-real complex root $z$, then it also has its conjugate $\bar z \ne z$ as a root. This excludes the possiblity that the double root of the cubic could be a non-real complex, since in that case the cubic would have $\ge 4$ roots counting multiplicities. And, once the cubic has a real double root, the third one must be real as well.

Answer 2

1. If any polynomial has a non-real root $z$, then it also must have the complex conjugate $z^*$ as a root. This includes multiplicities, so: $z$ is a non-real root with multiplicity $k$ if and only if $z^*$ is a non-real root with multiplicity $k$.

2. Consequently, if you count the number of non-real roots of a polynomial (including multiplicities), the result is always even: each non-real root paired up with its complex conjugate.

3. Now, every cubic has exactly three roots (including multiplicities). This means that it can have either 0 or 2 non-real roots.

4. In other words, it can have either 3 or 1 real roots.

$$\begin{array}{rr|l}\text{non-real}&\text{real}&\text{total}\\\hline 0&3&3\\\hline 2&1&3\end{array}$$

4. The discriminant of a cubic is zero if and only if the cubic has a double-root.

5. That double-root must be real. After all, if it were a non-real number $z$, that would necessarily imply that the roots of the cubic include $z, z, z^*, z^*$—that's too many roots for a cubic.

6. If your cubic has two real roots so far, and all cubics have 1 or 3 real roots, your cubic must have 3 real roots.

7. Hence every cubic with discriminant zero has a double root, which is a real valued double-root, which implies that the remaining root is also real. Hence all three roots are real.

Answer 3

Have a look at "A new approach to solving the cubic: Cardan's method revealed", which does a nice job of mapping the algebra of the solution of the cubic to the geometry of the cubic.

Take a look at figure 1, and mentally shift it down until one of the turning points of the cubic just touches the x-axis. Also pay attention to the geometric parameters $y_N$ and $h$ on that figure.

In footnote 10, on page 4 of that paper, Nickalls notes that the discriminant to which you linked

$$18abcd -4ac^3-27a^2d^2+b^2c^2-4b^3d = -27a^2\left(y_N^2 - h^2\right)$$

Since $$\left(y_N^2 - h^2\right) = \left(y_N + h\right)\left(y_N -h\right)$$ then if the discriminant is 0, then either $$y_N = h$$ or $$y_N = -h$$ or $$y_N = h = 0$$

The first two possibilities correspond to the scenario I asked you to mentally visualize earlier: one of the two turning points of the cubic just touches the x-axis, since the distance of the N point from the x-axis, $y_N$, is exactly the same distance as one of the turning points from the x-axis, $h$.

The final possibility corresponds to a degenerate case, where the cubic doesn't have distinct turning points and has a triple, real root.