500 years of NOT teaching THE CUBIC FORMULA. What is it they think you can't handle?

[@boomjonggol5757]

My dad: the main reason math is such a feared subject for many students is the teachers don't really master it. Those who master it are paid better elsewhere.

[@martinxXsuto]

I don't see much point in mastering the cubic formula when numerical methods exist.

[@nonamemike7149]

the rules regarding rearranging formulas AND the rules regarding simplifying . "Verifying Trig Identities Examples" & " Simplifying Trigonometric Expressions (Using Identities)" are just two videos not very good though but shows you what i'm referring to. When i was doing these in college for trig. each one required a page to a page and a half to write down. those videos have problems you can solve way too easily and RUNNING IN TO QUESTIONS REGARDING WHAT YOU CAN AND CAN NOT DO (and finding the answer by asking your professor questions like "can I?" "why not?). "IS THE WHOLE POINT.

[@YodaWhat]

@@huizilin65 - This is GOLD!!! "ignorance reinforces itself and sells this prejudice as proof." I see it almost everywhere, and what it has done to discourse and politics in USA is utterly sickening.

[@megauser8512]

Sad but true 😓

[@yash1152]

@@huizilin65 (ignore this:) #save something about ignorance which i didnt understand quite well supported with a link

[@vedant519]

The shirt is about square roots when he explains quadratics and cube roots when he explains cubic equations.

[@Matiburon04]

came to the comments for this

[@Mathologer]

Wasn't sure whether anybody would notice :)

[@eduardosuela7291]

In fact it looks like a cubic root.

[@AndreasDelleske]

The T-shirt has a projection of cubic roots on two dimensions.

[@eduardosuela7291]

Next thing would be fractal roots

[@leonard3468]

0:07 Fun fact: The usual term for the quadratic formula in Germany is "midnight formula" (in German "Mitternachtsformel") because we say "when the teacher wakes you at midnight, you should be able to know this formula" which is exactly what he said at the beginning

[@leonard3468]

@ Sure, can be. But in my school in Unterfranken (en. Lower Franconia) we used to always say "Mitternachtsformel". I only know the term "abc-Formel" from the Internet.

[@ericsbuds]

nice! thanks!

[@GrashalmTuts]

It's called Mitternachtsformel everywhere in Switzerland, as far as I know.

[@flippert0]

I tried this once on my son and he was able to recite it instantly (at midnight). So they really learn this by heart.

[@flippert0]

@ He survived it ;-)

[@sunilsoni3310]

Quadratic formula- Simple and easy to memorize Cubic formula- Kinda ugly, but not enough to be totally ignored Quartic formula- Way uglier than the cubic formula, not worth taking a look at Quintic formula- So ugly that it doesn't even exist

[@Mathologer]

Hmm, as far as I and a lot of other people with a mathematical soul are concerned the cubic formula for reduced cubic equation x^3+px+q=0 is one of the most beautiful equations in mathematics :)

[@matthewmatics6928]

There are actually several formulas for the Quintic equation, but they are the quartic formula factorial in terms of complexity, and use elliptic functions.

[@russcrawford3310]

@@Mathologer - Navier-Stokes in spherical ... there's an equation I'd put my wedding ring on ...

[@mattgsm]

I guess quintics are like my dad

[@karolakkolo123]

Quintic formula (and higher) exist, but they are not in terms of elementary functions. They use bring radicals and elliptic functions

[@Mathologer]

Another long one that I’ve been meaning to make for a long time. A video in which the mission is to rediscover the famous cubic formula in as motivated a way as possible. Also, another stepping stone towards a video about Galois theory in the hopefully not too distant future. Let me know how well this one worked for you :)

[@keylanoslokj1806]

the tshirt rocks and is woke af. it encodes the secret of life and the universe. for those who know: 1/Φ^-3

[@umitozusta6610]

Wow

[@ronak42]

You are amazing!!! Thanks a LOT for such great videos!! ❤️

[@ruinenlust_]

Great video!

[@victoramezcua4713]

As mentioned before, a brilliant video, a poem I can't wait for the Galois Theory video

[@MrDudoProudfoot]

In italian we still use the very common expression "fare un terzo grado" (literally "to make a third degree") to refer to an aggressive interrogation like the ones you can get from the police. It comes right from 500 years ago when the mathematical duels that Mathloger was referring to were popular. But nobody knows the reason for the expression anymore!

[@xyz.ijk.]

Thank you for sharing that! I often wondered where that came from, as if there were a first-degree or second-degree as well.

[@riccardodipietro4362]

Oh God, I always thought it came from the legal system

[@Jesin00]

I've heard "stop giving me the third degree" in English too!

[@runeodin7237]

Pretty interesting - in Danish you can also talk about a "third degree interrogation", but I had never realized it had anything with mathematics to do.

[@steveurquell3031]

@@runeodin7237 It could be from third degree burns, i.e. the most severe type

[@veloxsouth]

I noticed that completing the square and pascal's triangle were related in middle school and tried to go one further than my algebra class and derive the cubic formula and failed over and over. Every math teacher I had for the next 8 years could not or would not attempt to solve it or point me in the right direction. Finally, I asked my calculus II professor and he smiled and whispered excitedly "Oh! Cardano method!" and left me to enjoy. Seeing this video reminded me of that pursuit and the joy of finally solving it.

[@RadicalCaveman]

Cardano passed it down secretly through many generations to your calculus professor.

[@RockBrentwood]

I asked my math teacher in my Junior high school year, how to solve x³ + x + 1 = 0 and he said you need Calculus for that. So, I went to the bookstore, bought a $5 book on Calculus, carefully studied all the sections and worked out all the exercises over the weekend and came back. "Ok. I finished learning Calculus. But it didn't do me any good, because there was nothing in there on how to solve x³ + x + 1 = 0". So about 12 months later, I asked the college professor of the 3rd semester Calculus course, that I was in, in my first semester of college, how to solve x³ + x + 1 = 0 and he said to substitute z - 1/(3z) for x. Oh. Ok.

[@sonubanyal773]

[@pedroteran5885]

👍

[@SimonDoesmath]

They teach it in precalc however some algebra books will include it (McDougal doesn't but Blitzer does)

[@barissannan2731]

What makes this video even more beautiful, and many other similar endeavours, is that: they teach you all those wonderful things WITHOUT DEMANDING ANYTHING IN RETURN. So lovely.

[@slightlygruff]

But he receives a better understanding of the subject and lots of smart kids around him who might one day save his life from aging. That's a lot

[@BloodSprite-tan]

they kind of demand that you watch them though, not sure why you wouldn't though.

[@therealgadielsepulveda]

Don't give him ideas...

[@melbourneopera]

They have. Which is to expect you to solve more difficult questions.

[@dominusfons4455]

Ad revenue xD

[@tansoon8257]

Wow, I've watched this video four times over the past 1+ year and each time I understand the cubic formula better, seriously! Especially since I learnt about complex numbers last year only. Amazing video, one of my favourites on all of KZbin!

[@geekygurl8949]

‘Complex numbers were way beyond their imagination’

[@dansman1729]

Has anyone done Mathologer's homework for this video? They're at 18:35 and 23:18.

[@99bits46]

yea Bombelli imagined the unimaginable and called it imaginary

[@irrelevant_noob]

ArtiniTM Well the first one is easy enough: the local extremes are solutions for df = 0, so 3x²+p=0, x = ± √(-p/3). Plugging these in, the values for them will be x³+px+q, so ± (-p/3)√(-p/3) ± p √(-p/3) + q = q ± (2p/3)√(-p/3), and from this the half-drop is (2p/3)√(-p/3) = √(-4p³/27). Which fits nicely with what comes next in the video, since comparing this to q, it indeed is equivalent to comparing (p/3)³+(q/2)² with 0. As for the second one, my long-trained "sense" for these things tells me that the other combos would not respect the requirements that uv = -p/3 and u³+v³=-q. Good luck double-checking. ;)

[@psychedelicfungi]

I can't believe that one sailed past me first time!

[@akshataggarwal4002]

The joke is complex numbers are also called imaginary numbers,you dum dum.

[@isaacchatfield9366]

For the homework at 18:38 (figuring out the discriminant for the cubic): cubic equation -> x^3 + px + q = 0 - first derivative -> 3x^2 + p = 0 3x^2 = -p x = +-sqrt(-p/3) These are the extrema (x values) - [+-sqrt(-p/3)]^3 + p[+-sqrt(-p/3)] + q = 0 This gives us the y values for the extrema - [sqrt(-p/3)]^3 + p[sqrt(-p/3)] = green (difference in height between the inflection point and the 2 extrema) - | q | = absolute value of q - | q | > green -> 1 solution | q | < green -> 3 solutions | q | = green -> 2 solutions - q^2 > green^2 -> 1 solution q^2 < green^2 -> 3 solutions q^2 = green^2 -> 2 solutions - q^2 > {[sqrt(-p/3)]^3 + p[sqrt(-p/3)]}^2 -> 1 solution q^2 < {[sqrt(-p/3)]^3 + p[sqrt(-p/3)]}^2 -> 3 solutions q^2 = {[sqrt(-p/3)]^3 + p[sqrt(-p/3)]}^2 -> 2 solutions - (q/2)^2 + (p/3)^3 > 0 -> 1 solution (q/2)^2 + (p/3)^3 < 0 -> 3 solutions (q/2)^2 + (p/3)^3 = 0 -> 2 solutions Edit: I just realized that "green" can be simplified into {[sqrt(-p/3)]^2 + p}[sqrt(-p/3)] -> [(-p/3) + p] * [sqrt(-p/3)] -> [(-p/3) + (3p/3)] * [sqrt(-p/3)] -> (2p/3) * [sqrt(-p/3)] -> sqrt(-4p^3/27). This is easier to substitute in. - q^2 = green^2 -> 2 solutions q^2 = [sqrt(-4p^3/27)]^2 -> 2 solutions q^2 = -4p^3/27 -> 2 solutions q^2 + 4p^3/27 = 0 -> 2 solutions (q^2)/4 + p^3/27 = 0 -> 2 solutions (q/2)^2 + (p/3)^3 = 0 -> 2 solutions - which gives us the same final answer: (q/2)^2 + (p/3)^3 > 0 -> 1 solution (q/2)^2 + (p/3)^3 < 0 -> 3 solutions (q/2)^2 + (p/3)^3 = 0 -> 2 solutions

[@Svemirsky]

It's 4am, I'm watching math instead of sleeping and I'm also completely fascinated by it - If anyone told me 15 years ago that this would happen... :)

[@raph2550]

Haha same!

[@patrioticwhitemail9119]

The public education system isn't ment to teach you, it was made to condition bumpkins how to function socially in city factories. It has outlived it's usefulness as America's cities are no longer production centers. Now it needs to be remade to focus on getting kids to want to learn on their own, but unfortunately that won't happen. The tumorous bureaucracy and unions that have made it it's host will fight tooth and nail to stop anything that threatens their pension supply.

[@kumuthapriya.s5960]

1:23 am here ,

[@Janshevik]

@@patrioticwhitemail9119 the problem is that public education is teaching this from 7am and I'm definitely not in condition of learning anything when I can barely have my eyes open 😴

[@harrisonprocter8947]

Haha same here

[@photonjones5908]

I'm not sure what they think I can't handle; but I believe they are correct.

[@DennisJHarrisonJrHere]

Haha! Agreed :)

[@Killerlllnumll1]

It's kind of unfair to students in high school math, because to them, it seems like they are learning all of these disparate things not connected to each other. pi, complex numbers, each formula, it's like teaching the verbs of a language one conjugation at a time. It doesn't really start to come together until calculus. There's a lot of beauty in mathematics, but to a kid being taught in public school it's just boring worksheets, and "oh look, another standalone abstraction, what am I supposed to do with this?". In this video, you can clearly see how geometry, algebra, calculus, and every mathematical discipline are deeply intertwined. As a student, you only retroactively get the sense that you've built up to something bigger; that what you've learned actually has use and meaning. The most common question and complaint you hear in math classes is "how does this apply to the real world". This is tragic, because math is nothing but the real world in it's truest and least ambiguous form. The disconnect is disheartening, and it makes me wonder what educators are doing wrong.

[@hardlyb]

@@Killerlllnumll1 Most K-12 math teachers I've encountered (both as a student and as a parent) don't know much math. It's pretty hard to teach what you don't know. And if you are actually good at math, you have lots of options that pay better than teaching, so if you get fed up it's easy to leave. This sifting process leaves even fewer qualified teachers than it did when I was a kid. There are wonderful exceptions, of course, but I never had any of them in public school, and neither did my kids. (Despite this I became a mathematician - not sure how that worked.)

[@Metalhammer1993]

@Kevin Begley not showing them phyiscs/engineering will bore 99% of students, probably more to death. just is no compelling reason for pure maths to most of them. If the maths teachers face lights up, the students faces darken. that´s just how it is. If the teacher´s face even ights up with the more interesting pure maths parts. and that as i said is just a warning sign "now it gets boring" for most students. you´d be surprised how easy students pick up on that. And usually they get annoyed at the teacher´s enthusiasm. Most students respond MUCH better to practical application as long as it isn´t obviously constructed like Pythagorean theorem to put up a ladder. (no joke i saw that in my maths textbook in 8th grade) it´s much simpler to just say, that this will be important later, when we are talking about angles and trigonometric functions. Most kids accept that modern cutting tools for example use sine and cosine functions to graph out a wavy section they are meant to cut out. ANd kids are usually good at spotting patterns. That´s something you can do. but that is miles off from pure maths. which i´ll repeat will bore 99% of the students. Maths has its reputation as students least favorite subject shared with chemistry for the same three reasons. While physics, pretty much the linking bridge between the two doesn´t share their rep and usually is pretty reliably the favourite of the natural sciences after biology. Maths and chemistry are difficult, boring and useless. I´m not saying tha.t that is the typical student opinion. Chemistry is fun as long as you burn stuff. after that it gets less and less interesting, while physics is observable on a day to day basis. From why you need to hold yourself to something on the bus to why your cellphone runs out of battery so quickly. So in my opinion the best course of action is to prove the kids wrong on the useless count and show practical uses. That´s what they want to see and what gets them motivated to learn. Then you can slip in a bit of pure maths. But it´s pretty much hiding vegetables in pizza. You´re fucked if one of the students tastes them. Most kids just will not see anything but a bore in pure maths ,and if ythe teacher is a pure maths geek with absolute disregard for practical use, he won´t impart any share of his knowledge onto the students, sorry. Had teachers like tha.t not bad people. definitely highly knowledgeable. but pure maths until you throw up. just for the sake of maths and entirely annoyed if you DARED to ask for the usefulness of it. If you did not see it, you were in the wrong class. Too bad itr wasn´t my choice to be in that class. Maths is mandatory. When i started studying chemistry i literally was bummed out by all the maths. because it was maths. and maths could not possibly be useful. You were a criminal for asking for the usefulness of maths. And now we were supposed that thing, for which the very concept of usefulness was an insult, to actually get shit done? Syntax error my brain could not compute that. That´s the kind of student you get, if you try to drill them on pure maths on its own merit. No matter how you do it.

[@chemmandan9231]

Metalhammer1993 you sound like you never got to the point of understanding that chemistry IS physics or that math is the universal language that describes both of them. The cell phone battery example you gave is more chemistry than it is physics actually. Physics is just as much math as chemistry with plenty of useless practicality problems to be solved; ‘a power cable is joined by two 30 m tall towers. The cable makes a dip that forms a perfectly symmetrical parabola modeled by the equation y=x^2+4. How long is the cable?’ Dont get me wrong, I love chemistry math and physics but you really can’t go spewing a bunch of opinionated nonsense about why kids don’t like or understand them when clearly you don’t either. Hell, even biology presents some novel math problems when you’ve learned deep enough into the subject.

[@realdarthplagueis]

I remember I spent hours trying to develop a formula for the cubic in my high school years. I never managed it, but the problem was fun to work on (I had not read the solution in any textbooks)

[@AyanKhan-if3mm]

Me too but I used an approximation and simplified it like quadratic.

[@lawrencedoliveiro9104]

I tried doing the same. Not long after that I discovered computer programming. And I figured out how to do a first-order binary search between two points on the curve, one with positive Y and the other with negative Y, looking for the zero point. And I also figured out that the solutions for the (n+1)-degree polynomial, if they existed, lay between those for the n-degree polynomial which was its derivative. All this was before I actually got my hands on a computer for the first time.

[@AHeil1963]

Me too.

[@harishankar8575]

I first took an approximate root then Found y at it and divided (-y) by derivative at that approximate root to get even better approximation. Then we can repeat same on new approximation to get more accurate root. I also tried to develop cubic formula when i was 15, but was not able to. But i got some special cases If eq is of form x^3+ ax +b And a is much greater than b (say |a|>|10b|) then one root is approximately = -b/a So same here.

[@MrSeezero]

I can turn the general cubic equation into a transformed equation that you can solve by simply completing the cube. Check out my 3 videos about this subject on my KZbin handle "MrSeeZero".

[@swampwiz]

A Mathologer video about Galois theory and the impossibility of the quintic would be highly welcome. This is really the last step in gaining a complete understanding of regular algebra.

[@sizur]

As always, Mathologer and 3Blue1Brown produce the best presentations in baby steps optimized for time. Love your amazing work Mathologer! Edit: adding Welch Labs and Think Twice to the list.

[@mathieup.corbeil894]

Absolutely not. I can vouch for 3B1B, but Math doesn’t excel with the visuals. You are undermining a lot of others. Search for Welch Labs and Think Twice. I, by no means, am trying to attack Mathologer im any way, but Mathologer explains in his own way, so does 3B1B, who, while less rigorous, is understandable much more easily. I mention visuals because, on KZbin, that’s what people are looking for.

[@doommaker4000]

@@mathieup.corbeil894 I agree 3B1B is always easy to understand while I have to slow down or rewatch parts of Mathologer videos.

[@rationalmind3567]

mathtutor dvd is also good

[@koenth2359]

@SeaweedWorker Wow. Great school, where & when was it?

[@Mr.Not_Sure]

Totally agree. They both create really nontrivial videos, comparing to the rest math youtubers.

[@luck3949]

Happy thought 1: Whenever I will have to solve cubic equation by hand I will be able to apply Cardano' formula, instead of guessing roots using Viète theorem. Yey! Happy thought 2: I've graduated from school 7 years ago, so I'll never need to solve cubic equation by hand. Double yey!

[@mohammadfahrurrozy8082]

This guy is a genius...

[@meiz1795]

Sad thought 1: It would be much shorter to use any other existing method than cardano because most of the times you get some ridiculous fractions fractions, unless the equation was specifically designed to be solved using this method.

[@alejrandom6592]

I watched this a few months ago and understood very little, now after a semester of calculus I can really appreciate the beauty of it. Great video Mathologer!

[@NazriB]

Lies again? Center Fold

[@theboombody]

Mathologer is great at delivering a whole LOT of unexpected good stuff in a short span of time.

[@Krieglocke]

In finland we have this in a upper secondary school (read as the grades you go through before uni/college) math textbook alongside with the quadratic formula. It wasn't mandatory to learn or memorize it but it was still interesting to read about it.

[@maxsch.6555]

Same here in Germany

[@tensor131]

Here in the UK it was touched on when teaching polynomial theory to further maths students when I started teaching (1970's) ... Alas the regimentation of maths syllabuses has confined much of the fascinating advanced school maths to the bin (where you'll find almost ALL geometry 😩)

[@dansman1729]

Has anyone done Mathologer's homework for this video? They're at 18:35 and 23:18.

[@TheCarpenterUnion]

I always felt like if I had learned more about the extremities and histories of our topics, that I'd understand them (and the need for them) much better... But alas, can't have any child left behind *eyeroll*

[@ghenulo]

I remember the quadratic formula being covered in college algebra, but I don't remember the formula. However, I do remember that once I dropped college algebra, I was able to catch up in my other subjects.

[@sergiomanuel2206]

He changes his t-shirt at 13:50!!!! From square to cube!!

[@SKO_PL]

Wow, nice catch! I didn't even notice that :o

[@anandsuralkar2947]

Lol

[@abramthiessen8749]

Wow. His shirts are radical.

[@chocolatechocochoco]

omfg

[@LMDAVE29]

Wow, I thought it was an optical illusion I just didn't notice before. Cool catch.

[@andrewstockton72]

I love the subtle shirt change from a tree with 2D roots to a tree with 3D roots!

[@ramanmann2793]

yeah me too

[@PC_Simo]

So do I 🌳² -> 🌳³ 😌.

[@ARavingLobster]

I didn't expect an appearance of the Inquisition in a video about cubics... but then again, no one ever does.

[@hugoestevesrj]

I don't have the knowledge to fully understand it, but it gave me a lot of insights. Thanks for sharing your knowledge with so much passion. Those videos seems to require a lot of work, and the quality is amazing.

[@guiorasokolovsky6351]

My detailed solution for fun #1 challenge The cubic function is (1) : y = x^3 + px + q It's deriviative : (2) : ydot = 3*x^2 + p The equation is : (3) : x^3 + px + q = 0 Let r be one representative of the three roots (4) : r = r1 or r2 or r3 r belongs both to the tangent line and the cubic function (5) : ydot =3*r^2 + p To make the general line function (6) : y = a*x + b To become the tangent line we substitute : (7) : a = ydot ; b = - ydot * r So the tangent line is (8) : y = ydot*x - ydot * r Or the tangent line is also (9) : y=ydot*( x - r) Let X be the x coordinte of the intersection point of the tangent line with the cubic function (10) : ydot*( X - r) = X^3 + pX + q Thus (11) : X^3 + pX + q - ydot*( X - r) = 0 Or (12) : X^3 + pX + q - (3*r^2 + p)*(X - r) = 0 And after px is cancelled (13) : X^3 + q + 3*r^3 - (3 * r^2) * X + p * r = 0 And writing it as a new cubic (14) : X^3 + [(-3 * r^2)]*X + q+ [ 3*r^3 + p * r ] = 0 Let Pn Qn (p new q new) be the p and q of the new cubic so (15) : X^3 + Pn * X + Qn = 0 The General formula for the 3 roots is ( see video at 23:17 ) [ (-Qn/2) - sqrt( (Qn/2)^2 + ( Pn/3)^3 ) ) ]^ (1/3) + [ (-Qn/2) + sqrt( (Qn/2)^2 + ( Pn/3) ^3) ) ]^(1/3) But this new cubic has only two real roots: first root is the intersection at coordinate X, second root where the two roots coinside to one tangency point .Note that the tangent line of the new cubic function coincides with the x axis, like in the video at 18:51 where the sqrt argument becomes zero And the formula is simplified to : (-Qn/2) ^(1/3) + (-Qn/2) ^(1/3) : 2* (-Qn/2)^(1/3) Each of the two cube roots has three solutions . one real and two conjugate complexes.All with same absolute value. If (-Qn) is negative the two complexes have +60 degrees and -60 degrees with x axis. thus each with real part half of the real solution (and opposite sign ! ) the same for positive (-Qn) with angles +120 degrees and --120 degrees The sum of the two halves is r since the tangency point (r,y=0) is on both cubic functions. And the sum of the full two real solutions is thus (-2*r) ! which is the X coordinate of the tangent line other intersection point with the original qubic function. Let's find the Y coordinate ! From (9) : y = ydot * (-2*r - r) From (2) : y = (3 * r^2 +p ) * (-3 * r ) Thus : y = (-9) * r^3 - 3 * r *p Thus : y/(-9) = r^3 + r * p / 3 Adding and subtracting (r * p + q) yields : y/(-9) = r^3 + r * p + q + r * p / 3 - r * p - q Since r zeroizes the cubic function : y/(-9) = r * p / 3 - r * p - q Thus : y/(-9) = -(2/3) * r * p - q Multiplying by (-9) gives the desired Y : Y = 6 * r * p + 9 * q We now refer to the thee point: : (X,Y) : (X1,Y1) ,(X2,Y2) ,(X3,Y3) and to the three roots : r : r1, r2, r3 And check the three points linearity : (X1 - X3) /( X2 - X3 ) VS (Y1 - Y3) /( Y2 - Y3 ) (15) (X1- X3 ) /(X2 - X3 ) = ( -2*r1 - (- 2*r3) ) / ( -2*r2 - (- 2*r3) ) (Y1- Y3 ) /(Y2 - Y3 ) = ( 6 * r1 * p - 6 * r3 * p) / ( 6 * r2 * p - 6 * r3 * p) (16) (Y1- Y3 ) /(Y2 - Y3 ) = -3 * p * ( (- 2*r1 - (-2*r3)) / (-3 * p*( - 2*r2 - (- 2*r3 ) ) Thus (15) = (16) q.e.d

[@CaptainCalculus]

“Cubic Nightmare” sounds like a great name for a maths heavy metal band

[@galacticbob1]

I love their hit single, "Mind your P's & Q's" off the album "Equal to Zero" 🎸🎶

[@boboften9952]

With "Sum" Band Members "Quest" , "Roots" , And The Additional Parts "The Figures" Making Up The Numbers . Intergers Watching Intensely From The Side Lines .

[@EvenStarLoveAnanda]

Sounds like something the Borg would teach us to fuck up our brains so they can take over our planet.

[@cygil1]

So Tesseract or Helix Nebula or Intervals, then?

[@AchtungBaby77]

I was thinking the same thing, but about "Cubical Conundrums" instead! 😂

[@flamingpaper7751]

You should have just flashed the entire generalized quartic formula on screen in the last frame of the video to show just how complicated it is

[@Lucky10279]

My gosh, I've seen it before and it's _insanely_ long! I can't imagine trying to actually use it to manually find the roots of polynomial. Give me Newton's method any day before that. That method can get tedious pretty quickly, but it's far simpler. If I had to solve a quartic by hand that couldn't be solved by factoring or substitution, I'd definitely choose Newton's method. Thank goodness for computers that can do all the tedious iterations of succ algorithms in seconds though. I love math, but not tedious calculations, that's for sure!

[@qwertyqwerty7881]

I actually had to use the scary formula once in the time in order to solve Taylor series approximation. Those were some intense months I spent dealing with it :-/

[@Lucky10279]

@@qwertyqwerty7881 Why would you need the quartic formula? And what do you mean by "solve" Taylor Series?

[@vitakyo982]

Here you are : www.dropbox.com/s/g710eosav1f40ht/EQUATION%20DU%20QUATRIEME%20DEGRE%2B.doc?dl=0

[@rylaczero3740]

I wonder if they are reprintable.

[@gvssen]

Thank you for this presentation as this brings my old memories back. I sat the examination to enter the University in 1976 April and was admitted to the University in 1977 June. In between I read old books of my father, mostly books of Algebra by English Authors and Cardan’s (not Cardano) solution for the reduced cubic equation was one of the chapters of a book, and in yet another of another was the use of trigonometry, substituting x=r Sin θ for the reduced cubic equation x^3 +p.x +q=0 such that r is so chosen that the equation takes the form for the solution of 3θ. Never in my wildest dreams I thought that I would have to use all this when in 1979, I was answering a paper in Engineering Mechanics in the University that required me to solve the reduced cubic equation. That was the time that I thanked my lucky stars in having recalled what I read a few years back and I still have those old books preserved for posterity. Probably formal education will not touch the solution of the cubic equation for the simple reason that there are no applications widely in use. But in my case it became handy two years later. A BIG THANK YOU ONCE AGAIN.

[@friendlyone2706]

We live in a 3-D world, not 2-D, there are lots of applications overly simplified into 2-D questions with 2-D answers. Which questions are over simplified? I don't know, but when someone figures it out, we will all look and think it should have been obvious.

[@tylergoerlich9494]

Can you tell me what those books were called? I’m trying to find one of the solutions with trigonometry so that I don’t have to delve into imaginary numbers land, and so if there’s a solution with trig in those books that would help a lot.

[@nikoladjuric9904]

@@tylergoerlich9494 x³+px+q=0, Replace x=y/k, where k is some number, y³/k³+py/k+q=0, multiply by y³+pk²y+qk³=0, Now we know cos(3t)=4cos³t-3cost So ratio of coefficients is 4/(-3)=-4/3 (cubic and linear term), So here 1/(pk²)=-4/3, p is known number, k²=-3/(4p). Notist that p need to be less than 0 to k be real nimber.You can choose for k any of +-_/(-3/(4p)). I ussualy do it with + sign, k=_/(3/(4p)), replace back, y³-3/4*y=-q*3_/3/(8p_/p) Multiply by 4, 4y³-3y=-3_/3*q/(2p_/p), Now put y=cos(z) 4cos³(z)-3cos(z)=cos(3z)= -3_/3*q/(2p_/p) And we get 3 solutions for z,one obvious z1=arccos(-3_/3*q/(2p_/p))/3 z2=z1+2pi/3 (3z2=3z1+2pi) z3=z1-2pi/3 (3z3=3z1-2pi) Replace back, x=y/k=cos(z)/k =cos(z)*2_/p / _/3

[@MrSigmaSharp]

In Iran we have learned the cubic equation solution formula although not in all schools and not nearly as beautiful and pleasant as you put it

[@hachikiina]

@FAT cat but its all lost in history...

[@lawrencedoliveiro9104]

@@hachikiina Not lost www.aljazeera.com/programmes/science-in-a-golden-age/

[@gregoryferber3231]

Five minutes to take attendance... Ten minutes to review homework... 37 Mathologer video... Three minutes of wrap up... Monday's lesson plan done...

[@ThePharphis]

5 mins on attendance is just inefficiency!

[@7636kei]

Wait, did the teaching plan actually involves the generic formula to solve cubic equation?

[@allentwowalktwo9946]

Know the script then

[@WanderingWolfe]

@@ThePharphis The students refuse to sit in assigned seats. ~30 unexpected variables slows the counting process. Or some of maths excuse...

[@davidhenningson4782]

I wish I had KZbin growing up... I didn't take school seriously until grade 11 when I finally got interested in the subject matter and went straight A. I'd probably be a PhD if the internet was around back then.

[@jppitol]

Fun fact: Brazil is probably the only country who calls the quadratic formula "Bhaskara's formula". Really don't know why, but is really unusual the use of the regular one

[@Schieman]

It's in the honour of Bhaskaracharya, a 12th century Indian mathematician...

[@brunojuarez1883]

In Argentina it's also called that as well😄

[@BlackDragon-tf6rv]

​@@brunojuarez1883Siempre le dije cuadrática o resolvente

[@zamundaaa776]

"Complex numbers waay beyond their imagination" see what you did there.

[@vaneakatok]

c'mon guys, that deserves to be seen in the top comments

[@神のユージン]

@@vaneakatok there is already the same comment in the top comments.

[@ArGyProductions]

@Zamundaaa El Capitan.

[@mz7315]

aaaaaaaaaaaaaaaaaayyyyyyyyyyyyyyyy

[@KendrixTermina]

"16th century math feuds" tho What sort of epic dark academic stuff is this

[@shreyassamirkolte2378]

Fun Fact: the guy who proved that there can be no general solution (a formula, so to say) for polynomials of degree greater than 4 was Evariste Galois, and he was killed in a duel too.

[@qui-gonnjinn8949]

Bruh two guys in the 1600s made up calculus independently and argued about it for years

[@shreyassamirkolte2378]

@@qui-gonnjinn8949 True 😂, but thankfully they didnt kill each other in a physical duel😂, otherwise math and physics would've missed the english guy's brilliance.

[@JSSTyger]

5,000 chalk boards lost their lives.

[@daxbruce3491]

Sounds like modern scientists wanting to be the "fist recogized" inversed

[@steven3958]

An example when the coefficients are complex. If you multiply out (x-i).(x-2i).(x-3i) you get x^3 - 6ix^2 - 11x + 6i = 0. So we have a = 1, b = -6i, c = -11, and d = 6i. When plugging these into the formula, you get [ (1/27)^1/2 ] ^1/3 + [ - (1/27)^1/2 ] ^1/3 + 2i as the solution(s). Obviously the solutions are the three complex values i , 2i, and 3i by construction. Therefore, since the expression [ (1/27)^1/2 ] ^1/3 + [ - (1/27)^1/2 ] ^1/3 is being added to 2i, we should derive the three values of zero, i and -i from that expression. [ (1/27)^1/2 ] ^1/3 + [ - (1/27)^1/2 ] ^1/3 adds to zero when we take the real cube roots of each addend because one is the positive cube root of [ (1/27)^1/2 ] and the other is the negative cube root of [ - (1/27)^1/2 ] and their sum cancels to zero. Interestingly when we take the complex cube roots of both [ (1/27)^1/2 ] and [ - (1/27)^1/2 ] we get 1/sqrt(3).cis120 and 1/sqrt(3).cis240 for the positive value, and 1/sqrt(3).cis60 and 1/sqrt(3).cis300 for the negative value. Now when you add together both complex numbers above the real axis, you get ( 1/2 + 1/2 ) i which equals i . Similarly by adding both complex numbers below the real axis you get -i , which is what we wanted. i. e. The three solutions are 2i - i = i, 2i + 0 = 2i , and 2i + i = 3i. My point or suggestion...... The cubic formula also works for complex coefficients. However the ( elegant ) way complex conjugates are added to give real numbers for real coefficients is replaced by the equally elegant addition of complex numbers which are reflections of each other along the vertical ( imaginary ) axis to give complex solutions. It certainly worked that way for the example given. The discriminant was positive in this example, yet there were three solutions. So the sign of the discriminant may not have the same bearing on the number of distinct solutions when the coefficients are complex. It is well known that the formula for quadratic equations works for both real and complex coefficients. It appears that the same statement is true for cubic equations, just by tweaking the trigonometry used in the complex plane. Also the sign of the discriminant in the quadratic formula only affects the number of real solutions for real coefficients, but is not relevant when dealing with complex coefficients, and I think we see the same principle for cubic equations at work in this example.

[@DISCONIUM]

that -b/3a is just the math equation equivalent to "& knuckles"

[@Vandarte_translator]

This is some valuable information

[@tophattaco9052]

What does this mean

[@war_reimon8343]

@@tophattaco9052 it is a Video-game meme. To make fun of ridiculous naming like sonic & knuckles game. It can also be complemented with "with Dante from devil may cry series" or "new funky mode".

[@simohayha6031]

@@war_reimon8343 ridiculous naming of exoplanets

[@dansman1729]

Has anyone done Mathologer's homework for this video? They're at 18:35 and 23:18.

[@isaac10231]

That completing the square part blew my mind.

[@tetsi0815]

You've probably kinda seen this in high school math but generally teachers leave out the visual part and only do it algebraically, which is very unintuitive for most of the students so they forget it very quickly and only learn the formula. :-(

[@easymathematik]

@@tetsi0815 If one understands the simple relation (a+b)² = a² + 2ab + b² then one is able to understand "completeting the square" without intuition. Intuition is usseful but not nessecary.

[@ConorBreakell]

your completing the square explanation was the most clear and concise one I've ever seen!

[@berndhutschenreuther8342]

Hi, I learned it in school in the GDR, in an extended math class in the begin of the 1970ths. It was rather interesting and fascinating.

[@192ali1]

Fascinating. I am really impressed. You explained it to the comprehension of a junior High school level student. I was already aware of all these tricks you applied but never accord to me to put them together the way you did. I thank you for your excellent demonstration.

[@joellogan2420]

I found the cubic formula to be highly valuable in construction of houses. The problem was that this equation was infinitely more complicated when using American/English units especially on a hot summer day when my brain was likely to misfire. I solved the problem with a calculator that calculated the cubic formula in inches and feet. My houses were very square because I could lay out a foundation on a slope with great accuracy.

[@Player-pj9kt]

For what reason do u use the cubic formula when building houses?

[@aks8403]

Uhm what about using a mathematical tool like Wolfram alpha

[@clobre_]

​@@aks8403you never always had a phone with you

[@saltyyf1802]

High school me:"Damn you teacher, why not just tell us the formula for these equations" **SEES THE FORMULA* Me:"Ok yeah nah I'm good"

[@somatia350]

Abdallah ٰ i always thought rational root theorem was easier

[@Lucky10279]

@@somatia350 The problem is that it only works for finding the rational root. The _vast_ majority of cubics won't have any rational roots. That's not to say that the RRT is useless, just that it's not good enough on its own. In real life though, you'd be more likely to have a computer give an approximate answer obtained using some numerical method, such as Newton's method using derivatives. It can still be interesting learning about the symbolic methods though. I love finding unexpected patterns in mathematics. :)

[@rayniac211]

@@Lucky10279 I'm sorry I don't quite understand what you're saying. What kind of a cubic equation does not have a single rational root? How do you make one so the curve never crosses the x-axis? It sounds impossible to me.

[@gtahk-vy6io]

@@rayniac211 rational ≠ real

[@randomnobody660]

@@rayniac211 you might be confusing "real" with "rational". A cubic equation always has at least 1 real root. It's often not rational. (so sqrt(3) is real but not rational, and (1+i) is not real and um...i don't know if it's rational or not tbo?) unless it's a question constructed to be solved via rrt (which is likely every time outside of when you are learning rrt) you are very unlikely to have rational root.

[@denisdaly1708]

Alot of work gone into this video. Engaging, informative, and visually stunning, aestethic.

[@josephsilver9162]

I apparently came to this party a little late (by a few years). But wanted to share my joy at finding this video and persevering through it a little at a time until it ALL SUNK IN! I now have an excellent understanding of solving even general cubic equations - which frankly I never dreamed of. What a wonderful video and a delight once I was able to comprehend it completely. Many thanks. Your site is fabulous!

[@robertgumpi7235]

Mr. Mathologer. Thank you for your good work. I understand every single thought, every single word in every single second in this video and it is an extreme pleasure to follow the steps of these old math geniuses. The perfectness and beauty of math amazes me for years now. And it never stops. In math we see ... god.

[@Mathologer]

That's great :)

[@bt_18]

7:18 "..and a cameo appearance of the Inquisition" *_NOBODY EXPECTS THE SPANISH INQUISITION!_*

[@GRBtutorials]

Except they were Italian, so it should be the Italian Inquisition.

[@notsmoothie]

@@GRBtutorials if you were Italian you wouldn't expect the Spanish inquisition either...

[@melkiorwiseman5234]

Fun Fact: The Spanish Inquisition had to give 30 days notice of charges in order to allow the accused to prepare a defense.

[@BatkoNashBandera774]

@@GRBtutorials If you remember history, southern Italy used to be Spanish for a time.

[@jamesalexander7540]

Thanks for the chuckle.

[@DroneMTL]

To come back to the choice of root (+/-) when solving for (u+v), I have thought a lot about it and read some of the comments, but I still couldn't get my head around it. But I ended up coming with a simpler example to use as an analogy. Let's say you want to solve these two equations: x^2 + y^2 = 8 x + y = 0 In this case you could also say that x and y are indisguishable, and choose to start with solving x by substituting y = -x like so: x^2 + (-x)^2 = 4 2x^2 = 8 x = +/- 2 So, here we have a choice between +2 or -2 for x. y is indistinguishable from x, so y can also be +2 or -2. Does that mean we can have 4 combinations of x and y? No! The confusion arise from the false sense of freedom, which is to start solving again for y by substituting for x. y is not indistinguishable from x once we chose to solve for x or the other way around. So the other root belongs to y exclusively. And it can also be found by using the second equation, x=-y which is why solving for x decides the fate for y or vice versa.

[@guillermo7871]

I think +/+ and -/- dont work because when he multiplied by V cube, he added extraneous solutions to the final equation that solved for u and v, but didnt solve for x

[@pauselab5569]

that's what I told myself. if you pick a negative then the other is positive and vice versa

[@mikumikuareka]

Oh my god, thank you for this video, it literally changed my life. I had never been taught how to solve cubic equations algebraically but had a lot of situations where I had to but I couldn't. Simply because I had no idea how. I'm going to write down all the formulas occurred in the video and learn them all. I wish I saw this video before entering university. Thank you very much.

[@ivansincic7304]

7:22 nobody expects the Spanish Inquisition.

[@aditya95sriram]

Nooooo!! I was gonna post that. Never mind -_-

[@macarc985]

@@aditya95sriram Same haha.

[@aDifferentJT]

Technically it’s an Italian inquisition

[@FogToo]

They did give a 30 day notice so you could prepare your defense.

[@michaelsommers2356]

@@FogToo It also gave you time to prepare a list of your enemies to inform on.

[@xyz.ijk.]

Two years later this is still just as enjoyable and informative.

[@jcb3393]

16:03 - The fact that the x-transpose shift completely eliminated the x^2 term just blew my mind! I had to rewind that quite a few times!

[@david203]

Eliminating the x^2 term creates the symmetry that makes cubic equations easier to solve in closed form.

[@Ottmar555]

I believe it's a Tschirnhaus transformation.

[@johnchessant3012]

I'm so glad to have joined the Polynomial Brotherhood.

[@ArGyProductions]

incredulous!

[@Anenome5]

Your first mission: assassinate the number 3.

[@wolfsden6479]

​@@Anenome5 , but the first recite the secret poem.

[@Anenome5]

@@wolfsden6479 No, first you must understand why 6 was afraid of 7, because 7, 8, 9.

[@alexgottlieb7363]

What a lovely derivation of the cubic formula! A hueristic and satisfying approach: just rearranging the identity for (u+v)^3 to match the form of the depressed cubic. Cardano's formula has never seemed easy to me, until now. Thank you, Mathologer!

[@claytonl724]

I once asked if there existed a cubic formula...many years ago and was told unequivocally no. And now I learn that I was lied to and that it has existed for 500 years. Awesome!

[@99bits46]

he saved you the agony of using it

[@jismeraiverhoeven]

not many would say it was awesome they where being lied to

[@carultch]

Either he didn't know there was a cubic formula and didn't want to admit his ignorance, or he didn't want to open an endless conversation to try to explain something that is far beyond the scope of the class.

[@QuinsonHonQBB123XX]

21:34 An alternative route you can take: cube both sides of the green equation to get u^3*v^3=-(p^3)/27. Now, you realise you have found the sum and product of u^3 and v^3. Therefore, you can write a quadratic equation with roots u^3 and v^3, which then you can use to solve for x. In this case, the quadratic equation is x^2+q+(p^3)/27=0. Solve the equation to get the values of u^3 and v^3, thus finding u and v, thus allowing you to solve for x.

[@meowington2747]

I like how your shirt changes once you start explaining the cubic part! The video was also very interesting and good!

[@dedstrock]

Me: *uses the cubic formula instead of gauss' method* Teacher: Wait, that's illegal

[@samegawa_sharkskin]

what is gauss method and what question?

[@Begeru]

@@samegawa_sharkskin Matrices/linear algebra

[@clashingallthetime2655]

I hate this sentence 😂

[@AlexandrBorschchev]

Uh ok

[@marioulrichb.a.5607]

Are those conjugates under the radical?

[@Metalcape]

I still remember that in 2nd grade of high school our math teacher mentioned the existence of the cubic formula, and her comment was just: "I'm not going to teach it to you, because you're not going to remember it".

[@pennyoflaherty1345]

Metalcape Id like 2no if the teacher understood completely / incompletely not knowing how they might relay to a group with their teaching standards? Levels have fallen significantly not only with teaching ethics but understanding too!💡

[@allentwowalktwo9946]

I was just shown a square with a circle in it and told it was important. All we had then we're a book of logarims and slide rules which are next to useless

[@spdcrzy]

That's a bad teacher.

[@friendlyone2706]

If we were taught only what we were expected to remember....History class would be a LOT shorter.

[@bruj2444]

@Willie Reber don't fail lol

[@carultch]

I came up with a poem of my own for the cubic formula: x^3 + n*x = m x = cbrt(m/2 + sqrt(D)) + cbrt(m/2 - sqrt(D)) D = (m/2)^2 + (n/3)^3 When x cubed and x times n, Are added and equal to m. The values of x, The goals of our quest, Here's how to calculate them. Cube roots to add, Square roots they had, Both of a term we'll call D. Square half of m, Cube third of n, Add together and see. Half of m, adds to the root, First with a sign of plus. Its little brother, Is just like the other, Except with a sign of minus. Cube rooting time, of both the brothers, Add up the roots with glee. We found our first x, But where is the next? I know there have to be three. With help from DeMoivre, Who's theorem, we love ya, There's cube roots all over the plane Yes, they're complex, But do not perplex, A new kind of numbers we gain.

[@tipoima]

"What is it they think you can't handle?" This. This is what I can't handle

[@NaughtyOddity]

This video was truly amazing! I loved the animations, the multitude of angles from which you highlighted this topic (history, investigation of the quadratics to prepare for the cubics, analogies from geometry, illustrations of the calculus ideas, ...) and the amount of depth you go into. You also did a great job of not overwhelming potentially less well-versed viewers by unfolding additional layers of depth one by one - like mentioning the complex roots towards the end of the explanation, eliminating the x² term of the cubic but only giving the explanation later, once the basic ideas have been outlined, and so on. I am very much looking forward to your video about galois theory including the proof of why equivalents for degree 5 or higher do not exist. Keep up the amazing work. Greetings from a Math student at Bonn, Germany

[@Mathologer]

:)

[@gamemaster.......]

You saved my day! The formula for cubic equations is nowhere to be found online

[@andrewtychinin3308]

Hooray, can't wait for Mathologerized video-proof of Abel-Ruffini theorem

[@13chomp4]

“You’ve all done this a million times” Why is it that algebra is the thing that loses me? Why can’t I read a complex equation like a sentence describing a function? I want to learn how to visualize complex math, and I’m getting stuck at this very simple idea of visualizing algebraic balancing of a quadratic. 11 minutes in, and I feel like imagining a cube will make me froth at the mouth and keel over from exertion. Maybe I need to just get a degree in math in my free time or something, because I really want to do mental math some crazy stuff.

[@13chomp4]

Like these dudes were passing notes around Florence with crazy symbology, knowing what to do without having had some disaffected coach or underpaid teacher scream formulas and scribble cowboys and donuts and shit.

[@fredrosse]

To visualize math, at least in the engineering world (I am a Mechanical Engineer), I plot functions in EXCEL, and several variables can be examined. For example, say Y = A*X^alpha + B. Plot this function with a range of alpha, say alpha = 0.1, 0.2, 0.3,.....1.6. Then another plot varying B, etc. Then you can get a "feel" for the function, and, at least in my profession, that can lead to a better understanding. The example I have given here is very trivial, but the idea is illustrated.

[@DavidDiLillo]

@@fredrosse Amazing advice! Testing different models.

[@gibbogle]

Imagine how difficult is was when mathematical theorems and proofs were expressed with words, not with symbols.

[@13chomp4]

@@gibbogle I feel like I might better understand it if I knew how people were first forming these expressions with language rather than abstracting them with symbology.

[@Jenghis-Kh4n]

Very informative and satisfying, and invokes the flux of curiosity! Very good video !Literally took 11+ pages of notes watching this

[@diegoparodi3854]

Oh my god, I can't wait for the Galois theory video

[@peppermann]

These maths videos really are the best out there. Superbly animated and narrated with cheeky excitement! Bravo, Mathologer! 👏

[@ricardolichtler3195]

Dear professor, I really love this video. Could you make a video explaining the trigonometric approach to solve cubic equations?

[@mohammedal-haddad2652]

Watching this video was like living an adventure. Thank you very much.

[@Singularidade]

Really Amazing! this geometric demonstration (complete the square or cube) gave me an intuition simple and elegant

[@craigwillford5113]

Once again you have blown me away with the presentation to help simplify understanding. Thank you very much for all your time that this had to have taken!

[@mfrdbigolin]

A outstanding and in-depth video of a essential element of mathematics.

[@aquilazyy1125]

2:03 “It’d be good if we can get rid of all those b” Me: Heh surely you can’t that’s.. “And it turns out we can.” Me:”Wut”

[@protoman1365]

25:09 - this value is 4, just not simplified. I will prove this, albeit it would be hard to decipher what it is if you didn’t know the value was 4. First, I will set the monstrosity of cuberoot(20 - squareroot(392)) equal to a, cuberoot(20 + squareroot(392)) equal to b, and their sum equal to x for the sake of simplicity. If we cube both sides, we get a^3 + 3ab(a+b) + b^3 = x. But since a + b = x, this means that a^3 + 3abx + b^3 = x. Next, if we do a^3 + b^3, we get 20 - squareroot(392) + 20 + squareroot(392). This simplifies nicely to 40. Now, if we do ab, we are multiplying two values of the form (x+y) and (x-y). This means that the product is cuberoot(400-392), or cuberoot(8), which is 2. Now, we have 40 + 6x = x^3, which in other words is the basic equation we solved for. However, since we deduced this from simple algebra, this means that the equation having a solution of x = 4 means that our nasty radical does in fact equal 4.

[@benheideveld4617]

A triumph! About as far as a simple algebrain can be stretched in a non-reduceable way. More involved stuff is higher math. So this video represents the outer envelope of how far simple algebra can take you... Fascinating!

[@noahtaul]

25:10 Yeah, inside the cube root is (2+sqrt(2))^3 and (2-sqrt(2))^3, just check by expanding; so we take the cube roots and add and we get 4.

[@Mathologer]

Cool :) Maybe also share your method for finding these roots. I had a section on this in the original draft of the video but then ended up cutting this for better flow. Would be nice if somebody talked about this a bit :)

[@algebranograzie1396]

Thanks. I couldn't see it at all. I wonder, though, is there a way to "see" that (20+14sqrt(2)) is a cube or do we just have to know it?

[@MichaelRothwell1]

You assume that 20+14√2=(a+b√2)^3 for some nice rational a & b. Then expand and simplify RHS. Then it is clear that also 20-14√2=(a-b√2)^3. By equating terms with √2 and equating terms without √2 you get 2 equations in a & b. A bit of trial and error gives the result. I haven't worked through the details of this particular case, but have for another similar one, x^3+6x-20=0, for which x=2 is clearly a solution. Applying Cardano's formula leads to cbrt(10+6√3)+cbrt(10-6√3)=2. Here one finds, using the method described, that cbrt(10+6√3)=1+√3, cbrt(10-6√3)=1-√3, and indeed (1+√3) + (1-√3) = 2. I once set as a competition question the evaluation as a rational value the expression cbrt(10+6√3)+cbrt(10-6√3) (without technology, of course) :)

[@heaslyben]

Thank you! Thank you for showing this nice visual motivation for our "preprocessing" variable changes. It's so much more fun, insightful, and sticky than "depressing" the polynomial, just 'cause you're supposed to.

[@hpalpha7323]

I do remember touching on the cubic formula briefly in high school, but like so much of what we're taught in high school it never came up again

[@phiefer3]

I also vaguely remember being shown it in high school as well. But it was more of a "this also exists, but it's generally too 'complex' to be very useful"

[@xxaidanxxsniperz6404]

It's much easier to take the long polynomial look at the factors for the constant, and divide by the opposite sign of the constants factor and get 1 zero. Den continue until u get a quadratic and then use the quadratic formula to get the other two zeros. Takes a little longer but no need to memorize the long cubic formula.

[@xiaokangzhang3077]

For 18:41 you could also multiply the y values of the stationary points to obtain a polarity that will tell you whether this cubic has 1,2,or 3 solutions :)

[@the_eternal_student]

Amazing video! Congratulations for tackling this obscure but famous topic.

[@h4c_18]

The reason why only the +- pair works is that u*v=-p/3, and that's the only pair that accomplishes that.

[@leobidussi5039]

I guess the deaper reason is that if the signs are both equal, and therefore u=v, the p and q are no longer independent functions of u and v. This would mean that p and q are linked and knowing p would fix q as well. This is absurd since p and q are your initial values for the cubic equation and they are arbitrary, no connection between them actually exists.

[@harrymills2770]

Good stuff. Good production values. Almost makes me want to teach high-school algebra. They've got a whole year. You could definitely give this stuff some time.

[@SamueleCastiglioni]

i can't believe such a perfect video exists... going into detail but still understandable. thank you very much

[@saxbend]

I remember at school being explicitly told there is a technique for cubics, but no formula. This video title came as a massive surprise.

[@natan9065]

There's a formula for quartics and I believe quintics too

[@coopergates9680]

@@natan9065 34:40

[@dreamingforward]

That is freaking amazing. You can greatly simplify the cubic equation, of course, by defining some temporary variables for the common parenthetical expressions.

[@klepikovmd]

This is the best explanation of Cardano formula on KZbin!!! 👍👍👍

[@carultch]

Did you enjoy the subtitles from Михаил Диденко?

[@swankitydankity297]

Amazing video and wonderful animations. Your content is always quite the treat to watch !

[@John-jc3ty]

mathematicians: math is beatiful and perfect and makes sense me: 19:56 mathematicians: WE DONT TALK ABOUT THAT

[@onemadscientist7305]

Well, yeah, I mean there are two solutions, they're just the same solution. What ?

[@rossjennings4755]

Me: 25:08

[@igorface09]

@@onemadscientist7305 Aren't there actually three equal solutions though? There can never be an odd number of complex solutions.

[@igorface09]

@UCaeIYyDocCt8sPIJ8S5wJKQ But as far as I know, a cubic polynomial with real coefficients can only have either 1 or 3 real solutions (yes, because for each non-real root, its conjugate will also be a root, so they will always come in pairs). In the case of x^3=0, I'd argue there are 3 equal real solutions instead of only 2.

[@wilddogspam]

@@igorface09 yeah, you're right. The discriminant is zero if *at least* two solutions are equal. In this case all three solutions are zero in the factorization (x - 0)(x - 0)(x - 0).

[@pierre-marcshinkaretzky8851]

Excellent! Best explanation I found . History and analytics aims are very well explained. 20/20

[@wymarsane7305]

"Now it'd be good to get rid of all those bees..." Please don't, we need them and they're already endangered.

[@Mathologer]

:)

[@ravikantpatil3398]

Very good explanation as well as fun, delight and much more. ..... Thank a lot

[@rocklight4111]

Thanks for this, much love man. You've got the way to make these sweet things even sweeter

[@lordbertox4056]

6:20 me, italian speaking person, enjoing his funny pronunce, until I remember he knows maths better than I'll ever do.

[@riccardotrocano]

Lord Bertox però non abbiamo avuto bisogno della traduzione, un punto per noi

[@jimmyh2137]

@@riccardotrocano se riesci a capire l'italiano antico di quel testo tanto di cappello, la maggior parte degli italiani non ne sarebbe capace.

[@escu5054]

However, he keeps his german accent.

[@MichaelPrice917]

It would be helpful to mention that cqrt (20 +/- sqrt(392)) = 2 +/- sqrt(2), An amazingly simple result, IMO! And says why the answer, at 25:00, in the video really is 4.

[@henrikljungstrand2036]

Interesting. Let's check it out. (2+sqrt(2))³ = 8+2sqrt(2)+3*2sqrt(2)*(2+sqrt(2)) = 8+2sqrt(2)+6*2sqrt(2)+6*2 = 20 + 14sqrt(2). And indeed 392 = 196*2 = 14²*2 so sqrt(392) = 14sqrt(2). (2-sqrt(2))³ = 8-2sqrt(2)+3*2(-sqrt(2))*(2-sqrt(2)) = 8-2sqrt(2)-6*2sqrt(2)+6*2 = 20 - 14sqrt(2), the conjugate of the other term. So one solution is 2-sqrt(2)+2+sqrt(2) = 4. Another solution is (-½+½sqrt(-3))*(2-sqrt(2)) + (-½-½sqrt(-3))*(2+sqrt(2)) = -1+sqrt(-3)+½sqrt(2)-½sqrt(-6) + -1-sqrt(-3)-½sqrt(2)-½sqrt(-6) = -2 - sqrt(-6). And the third solution is (-½-½sqrt(-3))*(2-sqrt(2)) + (-½+½sqrt(-3))*(2+sqrt(2)) = -1-sqrt(-3)+½sqrt(2)+½sqrt(-6) + -1+sqrt(-3)-½sqrt(2)+½sqrt(-6) = -2 + sqrt(-6). Where -2-sqrt(-6) and -2+sqrt(-6) are conjugate pairs in sqrt(-6). Remember that conjugate pairs work even for something like 1-sqrt(2) and 1+sqrt(2), there is an (rational) automorphism swapping the two irrational roots even in real quadratic field extensions, not just in imaginary quadratic field extensions, these field extensions forming planes with (2d) Minkowsky metric instead of Euclidean metric, having hyperbolas of constant radius rather than circles of constant radius, yet without introducing any zero divisors. In most cubic field extensions (of the rational numbers) there are "conjugate" automorphisms of order 3 swapping the 3 roots as well, these are given by multiplying the two similar cube root terms in the sum, by conjugate primitive cube roots of 1, this we may call a cubic conjugation. In our specific example this fails to be an automorphism since 4 is not irrational enough, in fact it is rational, and the extension is merely a quadratic one since the polynomial of our equation splits over the rational numbers into a product of one first degree polynomial and one (irreducible) second degree polynomial. In most cubic field extensions we thus have six automorphisms, the trivial one, three of order 2 and two of order 3, forming the non-commutative group Sym(3). In some cubic field extensions though, the proposed order 2 conjugations fail to be automorphisms since the number under the square root sign happens to be a perfect square of a rational number, in this case we only get automorphisms cycling the three cubic conjugate roots, not automorphisms flipping the inner quadratic conjugates, thus our automorphisms form the commutative group Alt(3). In this case the cubic extension is a real extension, and all three roots have real rather than complex places, despite using the complex primitive cube roots of 1 to cycle these cubic conjugate roots of our equation. This field extension is really 3 dimensional nonetheless, not 1 dimensional, and contains (real) irrational numbers with similar caveats for calculations as (real or imaginary) irrational numbers produced by irreducible quadratic polynomials, it also is a Galois field, containing all roots of the cubic polynomial. In the general case the resulting field extension (having automorphism group Sym(2)) is 3 dimensional over the rational numbers, but is not a Galois field, containing only one root of our cubic polynomial, its Galois closure however contains all three roots, and is a 6 dimensional field extension of the rational numbers. I suppose it is possible to define a cubic metric for cubic field extensions, at least those whose Galois group (of any polynomial yielding them) is Alt(3), similar to the quadratic Euclidean and Minkowsky metrics for quadratic field extensions. This is work for the future however.

[@AntonBourbon]

I understand the final answer ³√(20+√392) = 2+√2, and it is easy to check it, but I'm wondering what the *solution* may be. I tried to fiddle with a³+b³=(a+b)(a²-ab+b²), it is easy to see that ab = ³√(20+√392) * ³√(20-√392) = 2, but I'm still getting nowhere as I can't calculate a²+b² . Any hints? Thanks! (Sorry, I can't read through all the 3000+ comments)

[@NadiehFan]

@@AntonBourbon Let ³√(20 + √392) = a ³√(20 − √392) = b You've already noted that we have ab = 2 and you can also easily see that we have a³ + b³ = 40 Now note that we have the identity (a + b)³ = (a³ + b³) + 3ab(a + b) so we have (a + b)³ = 40 + 6(a + b) This is really nothing but a cubic equation in (a + b). If we set a + b = x we have x³ = 40 + 6x or x³ − 6x − 40 = 0 Now, we could of course try to solve this cubic using the formulae named after Cardano, but then you end up with the exact same two nested cubic roots ³√(20+√392) and ³√(20−√392) you were trying to evaluate in the first place, so that is not the way to go. But what you can do is try to find a rational solution of this cubic in another way. The rational root theorem guarantees that any rational solutions of this equation - if they exist - must be integers and that these must divide the constant term 40. So, we only need to try the divisors of 40 which are 1, 2, 4, 5, 8, 10, 20, 40. Furthermore, we don't even need to try 1 and 2 because x³ − 6x = x(x² − 6) is evidently negative for x = 1 and x = 2. Also, x = −1 and x = −2 are not solutions and we dont need to try any other negative integers because x³ − 6x = x(x² − 6) is negative for x < −3. So, we try x = 4, and, sure enough, this is a solution of the equation since 4³ − 6·4 − 40 = 64 − 24 − 40 = 0. Are there any other divisors of 40 we should try? No, because x³ − 6x = x(x² − 6) is strictly increasing for x > 4 which means that the equation cannot have any real solutions larger than 4. Now, what do we do next? Since x = 4 is a solution, the factoring theorem tells us that (x − 4) is a factor of x³ − 6x − 40 so let's factor that out: x³ − 6x − 40 = 0 x²(x − 4) + 4x² − 6x − 40 = 0 x²(x − 4) + 4x(x − 4) + 16x − 6x − 40 = 0 x²(x − 4) + 4x(x − 4) + 10x − 40 = 0 x²(x − 4) + 4x(x − 4) + 10(x − 4) = 0 (x − 4)(x² + 4x + 10) = 0 A product can be zero only if (at least) one of its factors is itself zero so we have x = 4 or x² + 4x + 10 = 0 The other two solutions of the cubic equation are the solutions of the quadratic equation, but you can easily verify that this quadratic has no real solutions. Since x² + 4x + 4 = (x + 2)² we can write x² + 4x + 10 = 0 as (x + 2)² + 6 = 0 or (x + 2)² = −6 The square of a real number is never negative, so this equation cannot have any real solutions. But we know that x = a + b must be real because a = ³√(20+√392) and b = ³√(20-√392) are both real. Therefore, we must have x = 4. So, now we have a + b = 4 and ab = 2 This is a system of two equations in the two unknowns a and b which we can solve. How? One approach would be to consider a and b as the solutions of a quadratic equation t² − 4t + 2 = 0 which we can easily set up because we know that the sum of the solutions of a quadratic is equal to the negative of the coefficient of the linear term while the product is equal to the constant term. But I won't do that and show you another approach. Using the identity (a − b)² = (a + b)² − 4ab we have (a − b)² = 4² − 4·2 = 16 − 8 = 8 Now, we also know that a > b since a = ³√(20+√392) and b = ³√(20−√392). Therefore, a−b must be positive. So, we find that a − b = √8 = 2√2 From a + b = 4 and a − b = 2√2 we get 2a = (a + b) + (a − b) = 4 + √2 so a = 2 + √2 and 2b = (a + b) − (a − b) = 4 − √2 so b = 2 − √2 and we have derived and proved that ³√(20 + √392) = 2 + √2 and ³√(20 − √392) = 2 − √2 and so ³√(20 + √392) + ³√(20 − √392) = 4 Perhaps you feel a bit disappointed that denesting these nested cube roots required solving the very equation which gave rise to these nested cube roots in the first place, but that's just the way it is. It is impossible to denest a nested cube root by solving only quadratic equations (which appears to have been your idea, since you wanted to find a² and b² first).

[@AntonBourbon]

@@NadiehFan Thank you so much. To be honest, I'll only have time in a couple of days to properly read, understand and appreciate all this. Will update this (or write another) comment then.

[@jgone4856]

This channel is amazing. Great math explanations + humor

[@podemosurss8316]

23:23 For equation simmetry, after all we came from taking two differents u, v and making (u+v)^3. If they were to be equal, then this would just be (2u)^3

[@ursulapainter5787]

The great Leibniz would be pleased with you! Good work and nice music, too! Thank you very much!

[@peakperformancetrain]

Ursula Painter ? Did that author who wrote Candide , not going to name him, try to cut off Leibniz from being credited as the father of calculus? He who I will still not name was a friend of Issac Newton and had much to gain if Newton received this credit as well. That author hated and tried to mocks Leibniz’s theological concept of best possible worlds. That author had a creepy life.

[@petermhart]

Brilliant video! Another fun fact about cubics (with three real roots) is that the tangent to the curve where the x value is exactly half-way between the first two intercepts actually passes through the third intercept of the curve with the x-axis. In other words for a cubic with equation a(x-x1)(x-x2)(x-x3) the line tangential to the curve where x = (x1+x2)/2 intercepts the x-axis at x = x3.

[@afrasiabsleem7198]

I had been trying to find out the solution of the cubic equation for months and didn't look through internet just supposing I should be able to do that easily and took it as a challenge... After tiring efforts I m here and can imagine how close I was... :) explanation in this video is excellent... I loved each and every part of this video... (Y)

[@sudeshnaguhaneogi5486]

28:00 If there is only one quadratic formula, excluding Muller's method, then there is a big reason to why there should not be more than one cubic formula. Cardano's formula is already complex. - See what I did in the last sentence? 28:55, yeah but not EVERYbody knows trigonometry. - Quadratic Formula No.2 - x = (-p/2) +/- sqrt((p/2)^2 - q) 31:22, did Bombelli himself discover this?

[@captainri_2062]

Thank youuuu for this! Your way of teaching makes me inspired to learn the fun in learning by heart what you are teaching. Your audience can see and feel how you enjoy what you do. So it's very infectious. 💖

[@prathamsrivastava1367]

Sir, this is one of the best video on maths. i have sent this to other people also. this should be in curriculum, or at least as extra fun chapter in high school mathematics. My request, please also, make an awesome video on quartic equation (x^4).