de Moivre quintic formula

Рет қаралды 107,106

Күн бұрын

We will explore the de Moivre quintic formula. This is just a formula for a special case of the quintic equation in the form of x^5+5ax^3+5a^2x+b=0. By Galois theory, there's no formula that solves all the quintic equations. But don't be upset, because here's a nice formula for a special quintic equation.
Quintic equation zh.m.wikipedia.org/zh/%E4%BA%...
how to use Pascal's triangle to expand binomials • the easy way to expand...
0:00 a quintic formula? finally!
9:54 how to get that quintic form
28:46 other insane quintic cases
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Пікірлер: 190

@blackpenredpen Жыл бұрын

Try this quintic equation next: kzfaq.info/get/bejne/fdV3prl908ycgp8.html

@taektok3152 Жыл бұрын

Damn..!!!!!!, i cant hold my mother finger to not subscribe this channel.. So i follow you sir.. Im so sorry.. 🥺🥺🥺🥺🥺

@debashishkumarbehera501 Жыл бұрын

Can u find the solution of the equation given by sinx+e^x=0

@tetraktys2786 Жыл бұрын

Please, can you give us dificult equations, you makes me feel good at math So can you give us the solution of this equation? w(ln(10/x))=-w(-ln(10*(10^(1/x)))) Just kidding, only I want the equation, please

@leonardobarrera2816 Жыл бұрын

@@tetraktys2786 Tetraktys, I was solving, and I didn't get the answer, thanks

@mumujibirb Жыл бұрын

I also found one for a calculator, but it uses newton's method

@kummer45 Жыл бұрын

I am a mathematician. I can safely say that my respect to you increases exponentially. We need a detailed derivation for the Bashkara formula or quadratic, the cubic and the quartic polynomial equation with real or complex coefficients. Many people and lots of books don't give the steps for the derivation of these formulas. I found cubic and quartic in ordinary differential equations. THEY DO HAPPEN THERE and of course with those weird Lagrangians and Hamiltonians. I've seen your channel lately seeing an increment in difficulty tackling hard problems. Please continue doing so. You deserve getting the Patreon and and more subscribers. The algorithm of KZfaq must be benevolent with you. Teachers makes the difference. You are one of them.

@hoogreen Жыл бұрын

@oreos2653 Жыл бұрын

bro wtf i thought this stopped at quadratics bro where the 3rd and 4th and 5th exponent come from

@chuashanganluciennhps9992 Жыл бұрын

@@oreos2653 there is more to math than you think :)

@Mystery_Biscuits Жыл бұрын

11:03 I think the minus being there has a more important role in general. When looking for solutions, we need to make sure that the range of the substitution still completely covers the desired range of answers. For example, the substitution x= t+1/t has the range (-infty,-2] union [2,infty). This means, if the solution to the equation lies inside (-2,2), there does not exist a t that can find it. This is why the minus version works better. x = t -1/t has the whole real line as it's range so the problem of potentially missing a solution is removed.

@willie333b Жыл бұрын

Hmm imaginary numbers?

@Mystery_Biscuits Жыл бұрын

@@willie333b well, yeah of course this won’t find all the solutions but this video was just interested in finding the real solution to certain quintics…

@78anurag Жыл бұрын

Niels Henrik Abel crying rn

@rohanmehta8536 3 ай бұрын

He proved the unsolvability of the general quintic,not for all of them.

@rohanganapathy8 Жыл бұрын

Well nowadays since there are so many ways to find solutions to these big equations without actually solving them but with computers, no one has ever tried anything such a way. This is actually really fun! This is why I love mathematics, we do the impossible!!

@obinator9065 Жыл бұрын

Now do the quartic equation, but with the formula for general solutions.

@fantiscious Жыл бұрын

I don't think mathologer provides the formula, but he does provide the method

@kummer45 Жыл бұрын

@@fantiscious The topic BEGS for the long derivation. However the quadratic and cubic should be done first and the roots of unity for z^3=1 should be discussed first. There is some complex number algebra needed in such adventure. Binomial theorem plays a big role too completing the square, the cube and the quartic. Many books evade this because the algebra goes heavy on the details, moreover when Galois Theory enters when we confront the quantic The whole thing becomes a tour the force in Abstract Algebra. Besides the theory is extremely beautiful.

@NintendoGamer789 Жыл бұрын

Wiki page covers it pretty well, I was just too lazy to look for the details when I first searched this up in like 2021 lol

@shibam4182 11 ай бұрын

ax⁴+bx³+cx²+dx+e=0(a≠0) Substitute x=y-b/4a It will convert to y⁴+py²+qy+r=0(after dividing by a) Then y=(±√p+2λ±√-3p-2λ±2q/√p+2λ)/2 (where p+2λ≠0) And for the value of λ we have to solve a cubic equation in λ term 8λ³+20pλ²+(16p²-8r)λ+(4p³-4pr-q²)=0 Also if q is positive then {+()+√()-()}/2 {+()-√()-()}/2 {-()+√()+()}/2 {-()-√()+()}/2 & If q is negative then {+()+√()+()}/2 {+()-√()+()}/2 {-()+√()-()}/2 {-()-√()-()}/2 In this way the equation gives 4 roots

@johnchessant3012 Жыл бұрын

That's really cool! The x = t + 1/t trick works in reverse when you're trying to find the 5th roots of unity. x^4 + x^3 + x^2 + x + 1 = 0 -> x^2 + x + 1 + 1/x + 1/x^2 = 0 and now we set t = x + 1/x to get t^2 + t - 1 = 0. Solve for t using quadratic formula, then use the result to solve for x using quadratic formula again.

@DefenderTerrarian Жыл бұрын

I just started Calc 1, and your videos have let me get a leg up on the competition, so to speak. Starting college two years early (I graduated out of tenth grade) I felt that I would be, how to say..., a little dumber than the other students in Calc 1, but thanks to you I am more familiar with Calculus. Keep up the good work.

@chaiotic Жыл бұрын

This is so good! There's something about polynomial equations that's just so neat, like cool puzzles

@rihankota2021 Жыл бұрын

literally Awesome ur one of the best calculus teacher u have seen ur explanation is awesome👍🏻🥰...

@Peter_1986 Жыл бұрын

No matter how much I practise math, blackpenredpen always manages to bring up new challenges.

@DavidvanDeijk Жыл бұрын

Nice how you structured the video, with first the plain example and next the advanced with every thing matching the location on the board, very visual like blackpenredpen is famous for.

@trollme.trollmehard.9524 Жыл бұрын

I honestly watch these discoveries with a little apprehension. I would love to drop that and just focus on helping where my own skills / research lies, so would appreciate knowing how people manage this better.

@ntth74 Жыл бұрын

8:48 the imaginary roots can be find by mutiply the t value with 5 fifth-root of 1: e^0/5 (which is the value used in the video); e^i(2pi/5); e^i(-2pi/5); e^i(4pi/5); e^i(-4pi/5)

@Migeters Жыл бұрын

My guy I love the videos best calculus channel on yt

@letstalksciencewithshashwa9527 Жыл бұрын

Bprp : quintic equation Yt captions : green tea equation

@wetwillyis_1881 Жыл бұрын

I think someone needs to make you and intro, with music, that makes you seem like the final boss of a math game.

@charliearmour1628 Жыл бұрын

Great, very entertaining. I loved it.

@Chill---- Жыл бұрын

A very smooth solution indeed.(for the general case)

@afuyeas9914 Жыл бұрын

If the equation is of the form ax^5 + bx^3 + cx + d = 0 it will be solvable using this formula if and only if b^2-5*a*c = 0; the quintics that satisfy this condition are known as "De Moivre's quintics" and notice how the form of the quintic in the video satisfy this relationship. The reason why the formula ends up looking very much like the cubic formula is that it can be derived using the same idea. Let x = u + v, if the quintic satisfies the initial condition stated above then you can impose a condition on the product u*v so that you're only left with u^5+v^5 = - d. Take the fifth power of the product and you get a quadratic that is then easily solved. It has to be noted there are clear limitations with the nature of the roots of De Moivre's quintics. If what is inside the square root (let's call it the discriminant) is a real number then you only get 1 real root and 2 pairs of complex conjugates, if the discrimant is negative then you always add two complex conjugates to get your five roots so you end up with 5 real roots. Finally if the discrimant is 0 then you can show that the equation has 1 simple root and 2 distinct double roots, all real if the coefficients are real. In conclusion such equations will never have 3 real roots and a pair of conjugate real roots because the formula given is incapable of expressing such a case, nor will you ever get 1 double root, 1 simple root and 1 pair of complex conjugate roots with this formula even though there are quintics that have such roots. I really like De Moivre's quintics because even though there are very specific cases of quintics they still give you some hints about the general quintic not being solvable in radicals because the limitations of radical expressions are very clear when solving them.

@acelm8437 11 күн бұрын

It's nice that b^2-5ac resembles the discriminant for quadratic equations!

@tonyhaddad1394 Жыл бұрын

Great strategie

@scottleung9587 Жыл бұрын

Wow, I wish this worked for all cases of quintic functions - but this is a good start!

@LuigiElettrico Жыл бұрын

At 9:51 he says: "Keep all this in mind, because this was a very nice demonstration and now of course I will have to erase the board aaaaand..." 21:07 - "Extra B (?) Tada (?), that's very nice eh? And now I just have to erase the board" More hidden messages please :D And of course best math material as always!

@spencerswaggington Жыл бұрын

i love your videos!

@Monster-nn.007 Жыл бұрын

U did it man👌✨

@pepebriguglio6125 7 ай бұрын

I see why you chose the third formula. First and second options are very complicated. The first one is fascinating though! All powers of x are allowed to have non-zero coefficients, and still there is only restrictions on two of those coefficients: those of x and x². That is indeed impressive 🙏

@Dan_Tyme Жыл бұрын

Love the videos! Where did you get that identities poster behind you?

@stevemonkey6666 Жыл бұрын

👍 this is a great mathematical journey

@johnedwards4394 Күн бұрын

Great topic. Rarely discussed.

@angelespinosa906 Жыл бұрын

Muy interesante!

@quanticansh Жыл бұрын

instead of using general solutions, can I make an infinite series inverse function using the Lagrange's inversion theoram, to get the accurate roots of any degree polynomial equation I guess?

@astecheee1519 Жыл бұрын

Just use "inspect graph" on Desmos. Duh.

@shapshooter7769 Жыл бұрын

@@astecheee1519 But it's not Lagrange, it's most likely Newton-Rhapson

@niltondasilva1645 Жыл бұрын

Maravilha! Eu gosto desses cálculos DIFÍCEIS da equação de grau 5. 👏👏👏👏👏👏

@tobybartels8426 Жыл бұрын

Once you have one real solution, the other solutions are trivial. Because you can factor out x−r (where r is the root that you found) from the original polynomial to get a quartic polynomial, then apply the quartic formula to that.

@shobhitnegi6756 Жыл бұрын

it work on my pc thx bro vеry much

@dav1gamer287 9 ай бұрын

Please do this for the other formulas

@harshrajsinhsarvaiya3024 Жыл бұрын

Nice T-shirt

@iyziejane Жыл бұрын

What's the minimal set of special functions to add to arithmetic + radicals to allow for a quintic formula?

@gigantopithecus8254 8 ай бұрын

elliptic functions

@Eli-kz3bw Жыл бұрын

That c/a at 16:27 seems more sinister than at first glance. Since the coefficients are guaranteed to be rational numbers, the c/a seems to force you into having a rational constant as well. To me though it seems that c only needs to be a real constant. Do I have all that right?

@Eric-uf7dw Жыл бұрын

Next do the general quartic formula :)

@vishalmishra3046 Жыл бұрын

Any general Cubic equation can be changed to x^3 + 3 m x = 2 n. Then, x = (n + √D)^(1/3) + (n - √D)^(1/3) where cubic discriminant D = n^2 + m^3. To solve Quintic equation, let's change x to be sum of 5th root (instead of 3rd root) and change D to quintic discriminant D = n^2 + m^5. Therefore, IF { x = (n + √D)^(1/5) + (n - √D)^(1/5) } THEN { x^5 + 5 m x (x^2 + m) = 2 n } So, transform your Quintic equation to the form, x^5 + 5 m x (x^2 + m) = 2 n to get x = (n + √D)^(1/5) + (n - √D)^(1/5) where quintic discriminant D = n^2 + m^5.

@ErezMarom Жыл бұрын

9:47 impressing people is the best reason for inventing a formula.

@abu-karz Жыл бұрын

hi bprp Pls take a close pic of that information chart of that unit circle behind you and give it in the community tab

@kinshuksinghania4289 Жыл бұрын

This is great even though it’s just a special case

@user-gs3wx7ew9p Жыл бұрын

Hello thank you very much for it. Can you also try 100 trigonometric equations? It will be great, if you decide to solve it

@kamalrihani9609 Жыл бұрын

The coefficients are numbers that have a precise relationship, so this solution is for some specific five-degree equations and not all equations. De Galois is right again. Thans for you.

@kaddaderrer6152 Жыл бұрын

Fantastic

@ardizzle06 Жыл бұрын

i too want the quintic in my life

@antoniogeronimocabral126 Жыл бұрын

how about solving this equation?: 64+log[2](x)=x

@Abacadaba702 Жыл бұрын

Merch idea: cat and indeterminate forms as a hoodie?????

@himalperera3007 Жыл бұрын

Why integration of 1/(x.lnx) by parts gives 0=1

@thereaper3745 Жыл бұрын

I have a question : Can we solve this equation ? a^x + bx + c = 0 Please share it

@aayushkc5824 Жыл бұрын

yes x = (-b W((a^(-c/b) log(a))/b) - c log(a))/(b log(a))

@shophaune2298 Жыл бұрын

a^x + bx + c = 0 e^(x ln a) + bx + c = 0 e^(x ln a) = -bx - c (-bx -c)e^(-x ln a) = 1 (-bx -c)e^(-x ln a)a^(-c/b) = a^(-c/b) (-bx -c)e^(-x ln a)e^(-c/b ln a) = a^(-c/b) (-bx -c)e^(-x ln a -c/b ln a) = a^(-c/b) (-x -c/b)e^(ln a (-x -c/b)) = a^(-c/b)/b ln a (-x -c/b)e^(ln a (-x -c/b)) = a^(-c/b) ln a /b ln a (-x -c/b) = W(a^(-c/b) ln a /b) where W is the product logarithm (-x -c/b) = W(a^(-c/b) ln a /b)/ ln a x + c/b = -W(a^(-c/b) ln a /b)/ ln a bx + c = -bW(a^(-c/b) ln a /b)/ ln a bx = (-bW(a^(-c/b) ln a /b) - c ln a)/ ln a x = (-bW(a^(-c/b) ln a /b) - c ln a)/ (b ln a) when a = e, this simplifies to: x = -W( e^(-c/b)/b) - c / b

@alexwang982 Жыл бұрын

It’s solvable with the W function

@shophaune2298 Жыл бұрын

@@artophile7777 Yes.

@That_One_Guy... Жыл бұрын

That's solvable using lambert W function : a^x = -bx-c a^x/b = -x - c/b 1/b = -(x + c/b) * a^(-x) a^(-c/b) / b = -(x + c/b) * a^( -(x + c/b)) ln(a) * a^(-c/b) / b = -ln(a) * (x + c/b) * e^(-ln(a) * (x + c/b))

@nanamacapagal8342 11 ай бұрын

Just realized that's how the cubic formula was derived x^3 + 3px + 2q = 0 becomes a little more obvious it resembles the form if p = b/a and q = c/(2a) we can also solve more general cubics: for Ax^3 + Bx^2 + Cx + D = 0, make the substitution x = u - B/(3A), then after simplifying, divide through by A. You're back to u^3 + 3pu + 2q = 0 The basic cubic formula carries the rest from there, and the only thing left to do is to undo the substitution

@samehwaleed4872 10 ай бұрын

I can with ax^5 +bx^4 +cx^3 +dx^2 +ex+f=0 But f =retaion ship between a, b, c, d, e So can you put 5 exchange no exchange a, b, c only

@fabiosantucci6628 Жыл бұрын

6:51 look the bottom on the right

@giuseppemalaguti435 Жыл бұрын

al min 23 circa manca la radice 5 a denominatore(red)

@trix609 6 ай бұрын

Can you do the ferrari's fouth formula?

@ruud9767 Жыл бұрын

I can't believe I was here at this historic moment!

@tomctutor Жыл бұрын

Just thinking out loud here! What if we have the general quintic(omial) with integer coefficients: f(x)= ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0 then make a linear transform x=mt+n such that f(t) = a(m,n)t^5 + (0)t^4 + c(m,n)t^3 + d(m,n)t^2 + e(m,n)t + f(m,n) = 0 (choose m,n to make b(m,n)=0) Now make another linear transform for t= ps + q f(s) = a(p,q)s^5 + c(p,q)s^3 +(0)s^2 + e(p,q)s +f(p,q) = 0 (choose p,q to make d(p,q) = 0) Now make a final linear transform for s =ju + v f(u) = a(j,k)u^5 + c(j,k)u^3 + e(j,k)u + f(j,k) = 0 (choosing j,k so that e(j,k)/c(j,k) are in the correct ratio to match your fornula?) By the Fundamental Theorem any quintic must have a real solution, so this procedure is guaranteed to get that solution at least? I am not 100% sure this would work but maybe some programmer out there can write a program to check out my idea! 😁

@bjornfeuerbacher5514 Жыл бұрын

"Now make another linear transform for t= ps + q f(s) = a(p,q)s^5 + c(p,q)s^3 +(0)s^2 + e(p,q)s +f(p,q) = 0 (choose p,q to make d(p,q) = 0)" The problem is already in this step: You have to insert t = ps + q into a(m,n)t^5. If you choose p,q to make d(p,q) = 0, this will result in a non-vanishing term with s^4.

@tomctutor Жыл бұрын

@@bjornfeuerbacher5514 I see what you mean, if you t-transform out one s^n term then another spurious t^m term will resurface again! Seems to be a no win scenario, ah at least you can always get rid of one of the lower order terms! As I said just an idea. But thanks for pointing out the logical flaw.🤔

@Ninja20704 Жыл бұрын

Is there a formula for a general quartic equation? I know that for quintic and higher degree polynomials we can’t always get exact answers, but I forgot if its true for quartics

@d.l.7416 Жыл бұрын

There is one but its very long

@Ninja20704 Жыл бұрын

@@d.l.7416 Thank you. Even the general cubic formula already looks incredibly long and confusing for me

@Brian-pf6yb Жыл бұрын

28:37 look at top middle

@dharunrahul1700 Жыл бұрын

Plot twist : THE QUINTIC EQUATION FORMULA WAS FOUND BY STEVE ( bprp)

@infinity-yy3qn Жыл бұрын

find the range of f(x) f(x) = cosx[sinx + √(sin²x + sin²z)] here z is constant value sir please solve this problem

@PennyLapin Жыл бұрын

The hilarity of seeing what the English captions interpret "quintic" as each time

@trollme.trollmehard.9524 Жыл бұрын

Ahh! They're everywhere! :)

@ferre-jv3rp Жыл бұрын

Can u solve this equation please? Ln (sin(x)) = log(cos(x)) Greetings from Argentina

@AliHassan-hb1bn Жыл бұрын

I would say integral of secx is ln of its derivative. Thank you

@blackpenredpen Жыл бұрын

@-rachmaninoff Жыл бұрын

Can you solve e^3x + 5x^2 - e^2x = 2 ? :)

@VinTheFox Жыл бұрын

Seems like c/a is expressed as a ratio so that 0 can't be allowed. Otherwise the formula would be incomplete, because the equation can have multiple real solutions.

@dad-ms8mz Жыл бұрын

I guess ur the most productive person .

@ahmadmazbouh Жыл бұрын

5:55 i would use (-1)

@PrudentialViews Жыл бұрын

YAY!

@Predaking4ever Жыл бұрын

Awesome

@paolopalmisano827 Жыл бұрын

I can do this for polinomials of degree bigger

@SuperYoonHo Жыл бұрын

Awesome!!! I uploaded the link:)

@minimath5882 Жыл бұрын

Math finds a way!

@militantpacifist4087 Жыл бұрын

4:33 Fly wanted to be in the video.

@bruh____784 Жыл бұрын

√36 (possibly)

@Mono_Autophobic Жыл бұрын

He's obsessed with quintic equations

@blackpenredpen Жыл бұрын

Who isn’t? 😆

@phenixorbitall3917 Жыл бұрын

Magic

@sabasmoreno6705 Жыл бұрын

Congratulations

@mansoormohamedali122 Жыл бұрын

Hello Mr. BpRp. I have a question that has been keeping me all night these past few days. Hope you can help. Question: Let have any area on earth ( a city or a region). That we can calculate or know its area. We have n (a natural number) number of fixed spot we can put on the area.How can we calculate the minimal distance we can place n number of fixed spot such that we have the same distance between any point on the edge of the boundary to the next fixed spot or from fixed spot to fixed spot. For example: if we have n =1, so the only logical place to put the fixed point is at the center. From the the center ( fixed spot) to any point in the boundary are the same distance. What is the reasoning in the case n = 2,or n = 3, … n = inf?

@Ceratops17 Жыл бұрын

Okay so the first question is do you want your set (region) do be convex, this means every connection between points is also in the set itself. Also these points you mentioned just have the same distance to the center if your set is a ball. (If I misunderstood you, please correct me.) Consider a ellipses for example, we need a closed set M to also have the points on the boundary B, you can’t find a point x where the distance from any point on the boundary is the the same for all of them so to write it mathematically, no x (element) M: d(x,b)=d(x,c) for b,c (element) B and x (not element) bc (I mean the straight line connecting b and c here)

@awildstevey Жыл бұрын

Y’all know the know the meme where newton and like hawking are holding back Einstein? We need one but with Abel and Galois

@blackpenredpen Жыл бұрын

😂

@chrisleon27 Жыл бұрын

I shall make more contribution in the future if I keep my word

@evionlast Жыл бұрын

I'm gonna check that Chinese Wikipedia often, I don't know Chinese so... wish me luck 🤞

@blackpenredpen Жыл бұрын

I actually didn’t really read the description. I just read the formula 😆

@tsurutajunichiro6250 Жыл бұрын

hello, I want to present to you a problem from the moroccan math olympiad, here it is find all solutions for x and y in: x^4 + y^4 + 2 = 4xy hope you can solve it, Good Luck

@sendai-shimin 7 ай бұрын

Please challenge also the equation "型式2" on this Wikipedia.

@bean751-n2y Жыл бұрын

Would that mean that this formula can be generalized to any quintic that can be transformed into the quintic form mentioned at the start? If so, what properties does this particular class of quintics obey?

@social6332 Жыл бұрын

wow good.

@Magic73805 Жыл бұрын

Sir, I humbly request you to start a new series for mathematics. There’s an exam of maths in india. This exam is conducted by UPSC. Actually this exam is also known as maths optional in UPSC. I’m an aspirant and preparing for this maths optional exam. If possible please start this series so that we can get some help from you sir.

@yamanktail6573 Жыл бұрын

Chines guys are too powerful ☠️

@schizoframia4874 Жыл бұрын

The quintish formula 🙃

@84nsxtile55 Жыл бұрын

The equation looks like the cubic formula.

@identityelement7729 Жыл бұрын

Could someone solve any quintic equation if this person would have an countable infinite amount of such special case formulaes? One formulae is not enough I know but more than One?

@leonardobarrera2816 Жыл бұрын

Amazing, I will put you doble like

@chessematics Жыл бұрын

I became very very very upset after learning that angle trisection is impossible.

@blackpenredpen Жыл бұрын

Doubling a cube is also impossible… 😔

@chessematics Жыл бұрын

@@blackpenredpen yeah. But i was 7th grade and literally working my bowels out to find a way of trisecting an angle. Just for the record, i had no idea how formal constructions or proofs work. I just wanted to brute force it. That was 2019 February. In 202 i found a way of pretty closely approximating ANY rational multiple (i.e. multiplying the angle with ANY rational number). But it was still just a close approximation and of course, i must admit with all my grief, an angle in general can't be trisected. I got to know the doubling of cube and squaring of circle at the at the same time.

@urvpatel2003 Жыл бұрын

Sir how to become mathematician

@chrisleon27 Жыл бұрын

Hello from China

@blackpenredpen Жыл бұрын

Greetings! Thank you for your super chat!

@prodromoskonstandas155 Жыл бұрын

Shouldn't t have 10 roots ? Because it's a tenth degree ?

@That_One_Guy... Жыл бұрын

As BPRP said in the video, the number of solution in t variable is actually only 1/2 of the degree of t because the other half is the conjugate of the other one (so there won't actually be 10 instead it will just be 5), and the other 4 required solution is hidden away and to find them all i suspect you either : 1) Probably gonna need to use trig function to discover it (just like in cubic function). 2) Multiply the first solution by the fifth root of unity (correct me if i'm wrong on this one because i just recently read it from wikipedia about cardano's formula). These root of unity are all of the solution of x^5 = 1.