Condition for real roots of depressed cubic equation

Question

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?

Answer 1

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.

Answer 2

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 $u^2 > v^2$).