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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.

Rodrigo de Azevedo
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Mathrix
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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.

José Carlos Santos
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  • 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. – Mathrix Nov 02 '19 at 09:37
  • It would be nice to explain why that polynomial is a good choice of discriminant. Also, do we divide away any unnecessary factors? – Allawonder Nov 02 '19 at 09:40
  • @M.Fatih I cannot answer to all that in a comment. I suggest that you take a look at this question. – José Carlos Santos Nov 02 '19 at 09:41
  • 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. – Lubin Apr 29 '22 at 21:41