We all can solve quadratic equations of the form $ax^2 + bx + c = 0$ where $a \neq 0$. Simple, isn't it. Even, we have also devised formulaes like:

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

for that general quadratic. Has anyone wondered about solving higher degree equations? Cubics (of degree 3), Quartics (of degree 4), Quintics (of degree 5), sextics ( of degree 6), etc.? Probably. We might have heard about Cardano's formula in Cubics, Ferrari's solution in quartics. Even, we might know about the Tschirnhausen Transformation and all to obtain depressed equations to solve the original. Hey Wait, we also have a solution for cubics using trigonometry.

We'll discuss here a way to solve cubics. One is the Traditional Cardano's Solution as we all know. Let's explore the rest.

- Lagrange's Resolvents ( A Brief Idea ) :

Guys, do you know about the Abel-Ruffini Theorem? Maybe. Forget that. I'm presenting you a statement, and if you don't know it, just accept it for the present moment.

Galois theory has established that formulas using a finite number of arithmetic operations and root extractions are impossible for general equations of degree greater than four.

What? Really? So, can't we generalize formulas for degree 5+ equations?

Joseph-Louis Lagrange (1736-1813) had something in store. What Lagrange realized was that to solve equations of **prime** degree $n$ with rational coefficients, one has to solve a resolvent equation of degree $n-1$ also with rational coefficients, which are now called Lagrange resolvents. Please remember that he is talking about prime degrees, like cubics, quintics, heptics, degrees - 11, 13, and so on.

Now here comes our Mathematics!

Using Lagrange's resolvents, to solve the cubic, one has to first solve a quadratic.

Given the general cubic, $x^3 + ax^2 + bx + c = 0$

it's resolvent equation is given by

$z^2 + (2a^3 - 9ab + 27c)z + (a^2-3b)^3 = 0$

such that the solution to the cubic is $x= \dfrac{-a + z_1^{1/3} + z_2^{1/3}}{3}$ where $z_1, z_2$ are roots of the resolvent.

One can note two things. First, the solution is analogous to the quadratic formula. And while Galois theory has established that formulas using a finite number of arithmetic operations and root extractions are impossible for general equations of degree greater than four, there are particular equations solvable as such. So the solution to the solvable quintic should be similar in form, namely,

$x = \dfrac{-a + z_1^{1/5} + z_2^{1/5} + z_3^{1/5} + z_4^{1/5} }{5}$

where $z_i$ are the roots of the Lagrange's quartic resolvent.

We can obtain similar solutions for solvable degrees - 7, 11, 13, 17, ..

Let's take an example:

Let the cubic be $x^3 - 5x^2 + x - 7 = 0$

Resolvent is $z^2 -394z + 22^3 = 0$

with roots $z_i = 197 \pm \sqrt{3129}$

Thus $x = \large{\boxed{ \dfrac{5 + \left(197 + \sqrt{3129} \right)^{1/3} + \left(197 - \sqrt{3129} \right)^{1/3}}{3} }} \approx 5.07475$

And we know how to obtain the rest two roots of a cubic if we are presented with one of the roots.

The important issue here is to find the Lagrange's Resolvents for other higher degrees. Can you think of some?

**In the subsequent issues, I'll be posting various other ways to solve a cubic like - Cardano's solution, Translation, Linear Fractional Transformations.**

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestAt last I will get rid of factorization and long division method.

Log in to reply

No.. Factorization and Division Method will be easy for cubic equations having simple rational/integral roots. My note is targeted towards those who are interested in finding the exact roots of any cubic given.

Log in to reply

Awesome! I previously thought that while solving cubic equations one root is to be guessed

Log in to reply

Nah! Well, this method is only one of the ways. I'll post various more ways in the future :)

Log in to reply

Waiting for those posts! :)

Log in to reply

Log in to reply

Thank you !

Log in to reply

Log in to reply

$x^3 - 5x^2 + x - 7 = 0$

Try your Vieta's ? :D

Log in to reply

Log in to reply