I am studying on cubic equations for an essay and I have reached the general formula for any cubic equation. However I didn't realise what is what while formulating it, like discriminant. Now, I am trying to obtain it just like how it is done in quadratic equations.

I know that the vertex point (or points where the concavity changes or what it is because I do not really know what is what exactly. I will call these vertex points.) is the average of the real roots in a quadratic function. So, for any cubic equation in the form of:

$$a x^3 + b x^2 + c x + d = 0$$

Our vertex point is $\frac{-b}{3a}$ if there are $3$ real roots, and $\frac{-b}{a}$ if there is $1$ real root, as the sum of real roots of a cubic equation is equal to $-\frac{b}{a}$.

The discriminant of a quadratic equation is obtained by replacing x with the vertex point, $\frac{-b}{2a}$. In this case I replaced x with $\frac{-b}{3a}$, for 3 real roots, and obtain something that I thought it is a point equidistant from all the real roots, but I don't know if it really is.

I would like to know if I am going wrong and the correct way to the discriminant of a cubic equation with explanation.

  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ Commented Nov 2, 2019 at 9:01
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    $\begingroup$ Thank you, I will. $\endgroup$
    – Mathrix
    Commented Nov 2, 2019 at 9:09
  • $\begingroup$ Your question raises many points. Could you please make it clear what are your questions exactly? For example, would you like to know what "Cubic Discriminant" is? How to calculate it? etc. If you have multiple questions, please number them. The concept of a point that is equidistant of all roots is interesting - Do you have a reference for this for my own learning? $\endgroup$
    – NoChance
    Commented Nov 2, 2019 at 10:33

1 Answer 1


You are going in the wrong direction. I don't know what is the vertex point of a cubic. However, it is true that the sum of all roots of that cubic (real or not) is $-\frac ba$.

Dealing with $-\frac b{3a}$ is useful because if you replace $x$ by $x-\frac b{3a}$ in your cubic, then you get a new cubic without a second degree monomial. Then, if you divide everything by $a$, you get a reduced cubic: $x^3+px+q$. And reduced cubics are easier to deal with than general ones.

On the other hand, you should keep in mind that the discriminant of your cubic is $\bigl((r_1-r_2)(r_1-r_3)(r_2-r_3)\bigr)^2$, where $r_1$, $r_2$, and $r_3$ are the roots of the cubic.

  • $\begingroup$ Thank you for your concern. Could you please explain how the discriminant is derived? I would like to learn how it is derived in order to understand how it determines the sort of roots. I also realised that the replacement, -b/(3a), is the second derivative of any cubic equation in the form of ax^3+bx^2+cx+d. So, I am also in search of any relationship between the second derivative and the quadratic term. $\endgroup$
    – Mathrix
    Commented Nov 2, 2019 at 9:37
  • $\begingroup$ It would be nice to explain why that polynomial is a good choice of discriminant. Also, do we divide away any unnecessary factors? $\endgroup$
    – Allawonder
    Commented Nov 2, 2019 at 9:40
  • $\begingroup$ @M.Fatih I cannot answer to all that in a comment. I suggest that you take a look at this question. $\endgroup$ Commented Nov 2, 2019 at 9:41
  • $\begingroup$ The question of what the discriminant really is has many answers. Perhaps the best one at the level you find yourself is that it tells when two of the $n$ roots (of a degree-$n$ polynomial) are equal. So you can see why the formula given by José Carlos Santos does the trick. $\endgroup$
    – Lubin
    Commented Apr 29, 2022 at 21:41

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