# Condition for real roots of depressed cubic equation

My book sets out to prove that three normals drawn to any parabola $$y^2=4ax$$ from a given point $$(h,k)$$ are real when $$h>2a$$. It gives the following proof:

Proof: When normals are real, then all the three roots of $$am^3+m(2a-h)+k=0$$ are real and in that case,

$$m_1^2+m_2^2+m_3^2>0\\\Rightarrow (m_1+m_2+m_3)^2-2(m_1m_2+m_2m_3+m_3m_1)>0\\\Rightarrow (0)^2-\frac{2(2a-h)}{a}>0\\\Rightarrow h>2a$$

I don't see how $$m_1^2+m_2^2+m_3^2>0$$ is the condition for real roots of $$m$$. I tried solving few depressed cubic equations with imaginary roots, and found that the above condition is satisfied (at least for the ones I tried), but I was unable to find any proof online.

So how do I prove this condition?

• What do you mean by a normal is real? Feb 13, 2022 at 7:42
• @PaulFrost Normal whose slope is not a complex number. Feb 13, 2022 at 9:02

My book sets out to prove that three normals drawn to any parabola $$y^2=4ax$$ from a given point $$(h,k)$$ are real when $$h>2a$$. It gives the following proof:

The quoted proof appears to prove the opposite implication i.e. if the three normals are real then $$\,h \gt 2a\,$$, and also seems to assume that $$\,a \gt 0\,$$.

The statement that the proof actually proves is that when the three normals are real then the relation $$h>2a$$ is satisfied.

I don't see how $$m_1^2+m_2^2+m_3^2>0$$ is the condition for real roots of $$m$$.

$$m_1^2+m_2^2+m_3^2 \ge 0$$ is a necessary condition for the three roots to be real, but it is not a sufficient condition, for example $$\,(-2)^2 + (1+i)^2 + (1-i)^2 = 4\ge 0\,$$ where $$\,-2, 1+i, 1-i\,$$ are the roots of the depressed cubic $$\,m^3-2m+4\,$$.

However, the proof only requires the forward implication three real roots $$\,\implies m_1^2+m_2^2+m_3^2 \ge 0\,$$, so being a necessary condition is enough for the purpose of the proof.

• The condition $m_1^2+m_2^2+m_3^2 > 0$ is explicitly true only for depressed cubic equation. I think I mentioned that in the question post. Feb 13, 2022 at 7:23
• @AlphaDelta The cubic being depressed or not does not change anything here. 1) $\,m_1^2+m_2^2+m_3^2 \ge 0\,$ is a necessary condition for $\,m_1,m_2,m_3 \in \mathbb R\,$ regardless of whether $\,m_1+m_2+m_3=0\,$ (depressed cubic) or not. 2) The condition is not sufficient for the roots to be real, even if the cubic is constrained to be a depressed cubic, see the modified example in my answer.
– dxiv
Feb 13, 2022 at 7:30
• Yes you are correct. It appears I was misreading the statement. Thank you for clearing my doubt. Feb 13, 2022 at 9:28

Your question is when all three roots of $$am^3+m(2a-h)+k=0$$ are real. It is well-known that the roots of a cubic equation can be determined by the Cardano formula. For a survey see my answer to Is there really analytic solution to cubic equation? You will see that the expression $$R = \frac{k^2}{4a^2} + \left(\frac{2a-h}{a}\right)^3$$ allows to decide the question.

1. If $$R > 0$$, then you get one real and two non-real complex-conjugate roots.

2. If $$R = 0$$, then you get two real roots (one of multiplicity $$1$$, the other of multiplicity $$2$$).

3. If $$R < 0$$, then you get three real roots.

I do not know whether case 2 has significance for you. Noting that $$y^2=4ax$$ only yields a parabola for $$a > 0$$, it is clear that case 3 requires $$2a - h < 0$$, i.e. $$h > 2a .$$ This is a necessary condition, but obviously not a sufficient condition for the existence of three real roots.

The condition $$m_1^2 + m_2^2 + m_3^2 > 0$$ is of course satisfied if all roots $$m_i$$ are real. But it is not a sufficient condition. In case 1 above we have one real root $$m_1$$ and two roots $$m_2 = u + iv$$, $$m_3 = u -iv$$ with $$v > 0$$. We get $$m_1^2 + m_2^2 + m_3^2 = m_1^2 + 2(u^2 - v^2)$$ which may very well be positive (e.g if $$u^2 > v^2$$).