0

Some days ago I was solving a question which gave me a hard time. After doing some research I found out that it required the Discriminant of a Cubic Equation.

I looked up to the internet and I found out that the Discriminant of a Cubic Equation is $a^2b^2 + 18abc − 4b^3 − 4a^3c − 27c^2$.

So I was wondering if we have a general formula for finding the Discriminant an Equation of any degree.

I have been taught the discriminant of a Quadratic Equation where we prove it by Completing the Square method so I was wondering if there was any method to find/prove if for Cubic Equations.

So, the 2 main Question I want to ask are,

How can we find/prove the Discriminant of a Cubic Equation?

And,

Can we find the Discriminant of any equation of any degree?

Any help would be appreciated.

Rodrigo de Azevedo
  • 23,223
Toshu
  • 1,109
  • 14
  • 30
  • 3
    The discriminant can be determined for polynomials with arbitary degree. But if the degree exceeds $4$, the roots cannot be calculated in general by radicals (Galois). – Peter Nov 27 '19 at 14:02
  • Oh, is there any general formula for finding the discriminant till the degree 4? – Toshu Nov 27 '19 at 14:03
  • Enter poldisc ( a * x^4 + b * x^3 + c * x^2 + d * x + e ) in the pari/gp-calculator for the general formula for degree $4$, for example. – Peter Nov 27 '19 at 14:10
  • Note that the expression is very long already for degree $4$ – Peter Nov 27 '19 at 14:13
  • 1
    https://en.wikipedia.org/wiki/Discriminant – amd Nov 27 '19 at 18:41

1 Answers1

0

I think your formula for the discriminant of a cubic equation is wrong. This formula is independent of $d$. (Assuming that you considered the cubic equation as $ax^3+bx^2+cx+d$). The correct formula for the discriminant is given by $D = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd.$ Refer https://brilliant.org/wiki/cubic-discriminant/ for detailed proof.

Also, Have a look at a simiar question asked here: Proof of the cubic discriminant

A general formula for finding the discriminant of an equation of any degree $(P(x)=a_nx^n+a_{n−1}x^{n−1}+\cdots+a_1x+a_0$ having roots $x_1,x_2,\ldots,x_n)$ is given by:

$D = a_n^{2n-2} \prod_{1 \leq i < j \leq n}{(x_i-x_j)^2}.$

Moreover, you can refer to Galois Theory by Ian Stewart. The book has a nice treatment on the solution of polynomial equation in section $1.4$ of chapter $1$.

SARTHAK GUPTA
  • 1,501