The Quartic Formula (Ferrari's Method)

After finding the cubic formula using Cardano's method, the next step is to find the quartic formula using Ferrari's method. It is quite similar and builds on the ideas and concepts of Cardano's method with an extra step to form two perfect square quadratics.

Our journey for polynomial solution formulas in radicals ends here though since there is no formula for the quintic and higher degrees using elementary functions, but the methods to solve the quadratic, cubic, and quartic equation were definitely ingenious and worth the ride.

Timecodes:
0:00 - Intro and statement
0:35 - Formula derivation (choose λ s.t. p + 2λ ≠ 0 in the rare case)**
21:13 - Worked example
32:32 - Alternative steps in solving
34:26 - Extra on the discriminant
42:29 - Closing and unsolvability of the quintic

Sources and other tidbits:

[MichaelMaths_]

I would say that it makes sense to use exact formulas for for the irreducible quadratic, cubic, the quartic might be pushing it a little, but for quintics and higher degrees it's just better to approximate using something like Newton's method, even for most that are solvable by radicals.

[pmccarthy]

Nice! Seeing a treatment of Ferrari's method in English. Most of what I've seen have been videos in Hindi.

[bobbyheffley]

The calculus shortcuts for finding p, q, and r provide an alternative to finding these values by algebra.

[carstenmeyer]

Great explanation of _Ferrari's Method_ !


There is an alternative method to solve the quartic without *⨦(y^2 + p + 𝜆)^2* . Consider the factorization of a general quartic where *ak, bk in ℝ* . Notice *bk* are _not_ squared to include both real and complex solutions:

*P(x) = [ (x + a1)^2 + b1 ] * [ (x + a2)^2 + b2 ]*

To depress it into *Q(y)*, we substitute *x = y - (a1 + a2) / 2* . Notice only *ak* change in the factorization above:

*Q(y) = [ (y - 𝝰)^2 + b1 ] * [ (y + 𝝰)^2 + b2 ] | 𝝰 := (a2 - a1) / 2*
*= y^4 + oy^2 + py + q*

Expanding the factorization, we may compare coefficients:

*o = b1 + b2 - 2𝝰^2*
*p = 2𝝰 * (b1 - b2)*
*q = (𝝰^2 + b1) * (𝝰^2 + b2)*

Let's use the first two equations to solve for *bk* depending on *𝝰* :

*b1 = 𝝰^2 + o / 2 + p / (4𝝰)*
*b2 = 𝝰^2 + o / 2 - p / (4𝝰)*

We insert both into the last coefficient *q* . Notice the product simplifies into a difference of squares:

*q = (2𝝰^2 + o / 2)^2 - p^2 / (4𝝰)^2*

Getting rid of the denominator, we are left with a cubic equation in *𝝰^2* . After solving that cubic, we are done!

[rafiihsanalfathin]

Amazing video, i wish you make more videos on obsecure math topics like this because the material isn’t very accesible anywhere else. And also aww no quintic video, but i wish you upload a video covering that transformation/hypergeometric thing

[nasrullahhusnan]

The equation is y⁴+py²+qy+r=0
The solution of y is given, but with no r; instead there is lambda. What is lambda in terms of p, q and r?
It is mentioned that p, q, and r may easily obtained from derivatives of a certain function f(.). But no mention is made what f(.) is.

[trix]

the most impressive thing is that this method was created by a car 👏👏👏👏

[lerat]

This video is helpful. But I did not succeed to use it for x^4+x-1=0 for instance. The cubic expression should be x^3+4x-1=0. Which I do not find with the simplified formulas given in the vidéo. My question is how do you get the expression of the four (4) roots with radicals for this specific équation ?

[shibam]

Solving a quartic equation using this formula
x^4-10x^2-20x-16=0
lambda=7, 9+3i, 9-3i(using your formula for calculating lambda)
If we put all the values into the formula
We get x=2, -4, 1+i, 1-i
But this isn't the value of x
x values are -2, 4, -1+i, -1-i
So why the formula is giving the wrong roots plz explain

[darshansinghrahal]

It is beyond my understanding why p, q and r are taken derivatives of different orders of the given function Prof. Kinde enough to answer my question

[holyshit]

In Ferrari method it is not nessesary to depress quartic, we can simply complete the square to get rid of cubic term

In the method of undetermined coefficients

there also is no need for depressing quartic but you will probably not avoid substitution

After comparing coefficients you will get system of equations and to solve it
I suggest substitution new variable for whatever you have in the denominator

There is also method which is generalization of the method for cubic and here you have to depress quartic