The Quartic Formula (Ferrari's Method)

[@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.

[@bobbyheffley4955]

The easiest equations to solve have at least one rational root.

[@pmccarthy001]

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

[@rafiihsanalfathin9479]

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

[@MichaelMaths_]

Due to the sheer amount of additional computation needed to jump up to solve the general quintic from the general quartic, it is highly impractical to use an analytical method. But I still have plans for a future video with an example where you can get a series solution for a reduced form of a quintic!

[@rafiihsanalfathin9479]

@@MichaelMaths_ i know, but it would be cool to know the idea behind it tho :v, anyways goodluck for your future videos

[@MichaelMaths_]

@@rafiihsanalfathin9479 Ok I eventually came back to this and found one solution method I more or less understood, so here you go (and to other people interested in this): kzbin.info/www/bejne/iXe7gpJvZ5WXatU. Tread with caution ⚠

[@carstenmeyer7786]

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!

[@bobbyheffley4955]

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

[@em_zon2643]

Unique! Thank you!

[@lerat72]

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 ?

[@MichaelMaths_]

I think you may have calculated the coefficients for the cubic in λ incorrectly. We have λ³ + 5/2 p λ² + (2 p^2 - r) λ + (p^3/2 - p r/2 - q^2/8) = 0 with (p,q,r) = (0,1,-1), so λ³ + λ - 1/8 = 0. We can take the real solution of this equation as λ and then substitute it in to the formula x = (± √(p + 2 λ) ± √(-3 p - 2 λ ∓ 2 q/√(p + 2 λ)))/2, separating the roots by the 4 permutations of signs shown in the video. The full expressions in radicals are too long to write here, but you can just plug everything into the formula in a symbolic computation engine like Mathematica to get them. I also checked their numerical values and compared them with the output of the numerical equation solver, which all matched {0.248126 + 1.03398 i, 0.724492, 0.248126 - 1.03398 i, -1.22074}

[@lerat72]

​@@MichaelMaths_ Thank you very much for your help. This cubic formula gives 0,12313308608386 as a real root; which leads to -1,2207441 (one of the four roots of the quartic formula) when using x = (± √(p + 2 λ) ± √(-3 p - 2 λ ∓ 2 q/√(p + 2 λ)))/2. However I found in "La Petite encyclopédie des mathématiques" (archive.com), page 110 that the auxiliary cubic formula is to be of the form z^3 + 2pz^2 + (p^2 - 4r)z -q^2 = 0 which leads to z^3+4z-1. This leads to a real root of approx. 0,246266172167723 wich comes from : x = ∛t1+∛t2 with t1= (1/18) * (9+✓849) and t2= (1/18) * (9 -✓849); From that it comes "z" is a solution of 𝑥^4 + 𝑥 −1= 0: with x= ½ (√ √ 4z^2 + 16 -z - √z) ≈ 0,724491959000516 x= -½ (√ √ 4z^2 + 16 -z + √z) ≈ -1,22074408460576. Which also works and gives a nice solution with radicals for these 2 real solutions. Can you see why the L. Ferrari method differs here ? And where these 2 expressions : x1= -1/2*(√(√(4z^2+16)-z) +√z) x2= 1/2*(√(√(4z^2+16)-z) -√z) are coming from ? Thank you again for your help.

[@lerat72]

@@MichaelMaths_ Thank you very much for your help. This cubic formula gives 0,12313308608386 as a real root; which leads to -1,2207441 (one of the four roots of the quartic formula) when using x = (± √(p + 2 λ) ± √(-3 p - 2 λ ∓ 2 q/√(p + 2 λ)))/2. However I found in "La Petite encyclopédie des mathématiques" (archive.com), page 110 that the auxiliary cubic formula is to be of the form z^3 + 2pz^2 + (p^2 - 4r)z -q^2 = 0 which leads to z^3+4z-1. This leads to a real root of approx. 0,246266172167723 wich comes from : x = ∛t1+∛t2 with t1= (1/18) * (9+✓849) and t2= (1/18) * (9 -✓849); From that it comes "z" is a solution of 𝑥^4 + 𝑥 −1= 0: with x1= ½ (√ √ 4z^2 + 16 -z - √z) ≈ 0,724491959000516 x2= -½ (√ √ 4z^2 + 16 -z + √z) ≈ -1,22074408460576. Which also works and gives a nice solution with radicals for these 2 real solutions. Can you see why the L. Ferrari method differs here ? And where these 2 expressions (x1,x2) are coming from ? Thank you again for your help.

[@lerat72]

@MichaelMaths_ Thank you very much for your help. This cubic formula gives 0,12313308608386 as a real root; which leads to -1,2207441 (one of the four (4) roots of the quartic formula) when using x = (± √(p + 2 λ) ± √(-3 p - 2 λ ∓ 2 q/√(p + 2 λ)))/2. However I found in "La Petite encyclopédie des mathématiques" (archive.com), page 110 that the auxiliary cubic formula is to be of the form: z^3 + 2pz^2 + (p^2 - 4r)z -q^2 = 0, which leads to: z^3+4z-1. This leads to a real root of approx. 0,246266172167723 which comes from: z = ∛t1 + ∛t2 where: t1= (1/18) * (9+✓849) and t2= (1/18) * (9 -✓849). From that it comes "z" is a solution of the auxiliary formula (z^3+4z-1= 0) to 𝑥^4 + 𝑥 −1= 0: with: x1= ½ (√ √ 4z^2 + 16 -z - √z) ≈ 0,724491959000516 x2= -½ (√ √ 4z^2 + 16 -z + √z) ≈ -1,22074408460576. Which also works and gives a nice solution with radicals for these 2 real solutions. Can you see why the L. Ferrari method differs here ? image006.png X1= ½ * image008.png X2= -½ * And where these 2 expressions (x1 and x2) are coming from ? Thank you again for your help.

[@lerat72]

@@MichaelMaths_Thank you for your message. I tried to send a detailed answer which failed to go trough. How can I send it to you ? Best. Arnaud

[@trix609]

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

[@TerryFerrellmathematics]

You are awesome!

[@nasrullahhusnan2289]

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.

[@shibam4182]

Lamda is an extra term from which we can get the cubic in the form of lamda then putting back to the equation we can get the value of y

[@MichaelMaths_]

Yeah lambda is just the solution of the cubic to complete the square, I didn't write it out in terms of all the coefficients because the general form takes up a lot of space. Also the function f(x) for getting the depressed quartic is just the initial quartic itself (not set equal to 0), I mentioned this in the cubic formula video.

[@detectiveandspynovels7140]

♥️♥️♥️♥️♥️♥️♥️

[@shibam4182]

Where is 'r' in the formula

[@MichaelMaths_]

It may not seem to appear in the factoring of the perfect square, but remember that lambda (which solves the previous cubic) does include r in the equation’s coefficients.

[@shibam4182]

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

[@MichaelMaths_]

I just tried it myself and I got the correct roots. Maybe there was a small algebra mistake somewhere or you didn’t use the correct plus-minus sign combinations at the end, happens to me sometimes when using it.

[@shibam4182]

​​​@@MichaelMaths_I have checked more than 5 times and found nothing wrong in solving Q-x^4-10x^2-20x-16=0 For calculating lambda we have the formula -8λ^3-20pλ^2+(8r-16p^2)λ+(4pr+q^2-4p^3) Put all the values we will get -8λ^3+200λ^2-1728λ+5040=0 So λ=7,9+3i,9-3i put λ=7 so x=(+/-√p+2λ +/-√-3p-2λ+/-2q/√p+2λ)/2 Putting all the values we get x=(+/-√4 +/-√30-14+/-40/√4)/2 x=(+/-2 +/-√16+/-20)/2 So x1=(2+√16-20)/2 x2=(2-√16-20)/2 x3=(-2-√16+20)/2 x4=(-2+√16+20)/2 So we will get x=1+i,1-i,-4,2 What's the mistake here?

[@MichaelMaths_]

​@@shibam4182 The first and last double signs are normally dependent and opposites of each other, the middle one is independent. However, if you distribute the -1 from 2q being negative, the third double sign should be matching the first in this case.

[@shibam4182]

​But -/+ and+/-are same so both should be right This is going over my head 😩 Can you please solve it in comment section

[@MichaelMaths_]

@@shibam4182 The depedence comes from at what point the roots were extracted in the derivation of the formula. The final solutions should come from the following expressions: (2 + √(16 + 20)) / 2 = 4 (-2 + √(16 - 20)) / 2 = -1 + i (2 - √(16 + 20)) / 2 = -2 (-2 - √(16 - 20)) / 2 = -1 - i

[@alissoneuclides3407]

Very Good!

[@holyshit922]

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 (x^2+px+q)(x^2+rx+s)=x^4+a_{3}x^3+a_{2}x^{2}+a_{1}x+a_{0} there also is no need for depressing quartic but you will probably not avoid substitution (x^2+px+q)(x^2+rx+s)=x^4+a_{3}x^3+a_{2}x^{2}+a_{1}x+a_{0} 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

[@appybane8481]

I think we need to reduce the equation first ,then factor as (x^2+tx+u)(x^2-tx+v)

[@darshansinghrahal2549]

You cannot even solve by completing the square of equation X⁴-10X³ +35X² - 50X + 24 = 0

[@darshansinghrahal2549]

Because no value of lemda will satisfy the expression

[@darshansinghrahal2549]

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

[@MichaelMaths_]

This basically comes from expanding as a Taylor polynomial at x = -b/(4a) as the the coefficient of the (x + b/(4a))^2 term depends on f’’(-b/(4a)), which is 0 at that point (and substitute x + b/(4a) by y).

[@glairepaglinawan5421]

16:05 why 21/4?

[@MichaelMaths_]

I'm assuming you are referring to the choice of lambda in the worked example. This was determined using rational root theorem and all of the other roots are irrational, so picking a rational root if at least one is present will the best choice for simplifying the process of completing the square on the right side afterwards.

[@glairepaglinawan5421]

Oh wow....now i get! Thank you!!!

[@gnet9397]

dislike

[@briannguyen4570]

hater