How to prove the cubic formula (from scratch)

[@blackpenredpen]

Here’s a similar formula for quintic. de Moivre quintic formula kzbin.info/www/bejne/gmm6Z2tqpah7r8U

[@748813592415]

I thought quintic was impossible

[@Adventurin_hobbit]

Sir where do you learn that?

[@appybane8481]

​@@748813592415It's a special one.

[@alerikaisattera1465]

@@748813592415 Possible, but requires special functions

[@klepikovmd]

​@@748813592415 it is, but only as a universal solution. Special cases and numerical solution still possible.

[@edwardhuang5885]

Finally, a handy formula I can use when I come across x^3 + 3x^2 + 3x + 1 = 0

[@blackpenredpen]

😂

[@suddeneevee9441]

The equivalent of using a gaming PC to calculate '1+1=?'

[@Harbo1003]

-1

[@Amoeby]

This is a good one ngl.

[@spoon_s3]

It is just (x+1)^3=0 right?

[@alexeton86]

I don't speak English much but this guy explains it so simply and in such detail that I understand everything. Very interesting.

[@blackpenredpen]

Thank you. I actually added some translated captions. Check out the CC and let me know if your language is there. If not, then let me know and I will see what I can do.

[@kokulanselvakumaran]

​@@blackpenredpen there aren't tamil subtitles.

[@BigganFan]

Please add Bangla ​@@blackpenredpen

[@user-hp6pj4sy3k]

Finally a real derivation of the cubic formula. It's honestly not terribly difficult to follow, just tedious.

[@matte14-70]

NOW THE QUARTIC FORMULA 😁🙏🏼

[@saraxum9773]

*formulas

[@fantiscious]

Mathologer has a video on it, it's really tedious tho since it requires solving a cubic equation, and then a quadratic equation kzbin.info/www/bejne/hF6uiYaqqtWqqcU

[@Pacvalham]

-What about a generalization for any degree?- I have received more than enough replies telling me this is impossible.

[@DavidSilva-dq5cm]

​@@Pacvalham only until 3 actually

[@mr.nobody.01]

Will be so cool 😂

[@Awesome-ct7vr]

Tip: In ax³+bx²+cx+d=0 Before going to the formula try this. If a+b+c+d=0 , x=1 If (a+c)-(b+d)=0 , x= -1 Afterwards you can factor it into a quadratic easily. Ex. 2x³+5x²+2x-1=0 Test1: 2+5+2-1=0 ---> false. x≠1 Test2: (2+2)-(5-1)=0 4-4=0 ---> true. x= -1 Now factor: with solution (x=-1) 2-0=2 ,-1×2= -2 , 5-2=3 , -1×3=-3 , 2-3=-1 , (-1)×(-1)=1 , -1+1=0. Getting 0 means x=-1 is valid solution.(if not then its incorrect solution) Now take all numbers from the addition and subtraction results in order. 2,3,-1,0 and now you get a quadratic. 2x²+3x-1=0 x²+3/2x=1/2 x²+3/2x+9/16=1/2+9/16 (x+3/4)²=17/16 x=-3/4+-sqrt17/4. X[1,2,3]={-1 , (-3-sqrt17)/4 , (sqrt17-3)/4}

[@thedictator1454]

Thank

[@dhruvveersinghtanwar3336]

Thank you brother 🙏🙏🙏

[@laprankster3264]

If x doesn’t equal ∓1, then you can still use the rational root theorem. If you do find a rational root r, then you can simply factor the cubic into (x-r) and a quadratic.

[@Peter_1986]

I love learning derivations for things; it feels a lot better to have the skill to derive as many formulas as possible, instead of having to look them up in tables.

[@FNShadow008]

Same!

[@Fire_Axus]

your feelings are irrational

[@sriprasadjoshi3036]

Finally a great masterpiece, was very curious about this cubic formula...

[@azzteke]

Masterpiece? No, simple algebra.

[@fantiscious]

@@azzteke bro solves quintic equations 💀

[@General12th]

@@azzteke Are you trying to brag?

[@That_One_Guy...]

@@azzteke who even are you little kid

[@sriprasadjoshi3036]

@@azzteke Are you serious??? Like are you quantum physics student or 13 y/o kid commenting here like a genius???

[@Wmann]

This was a blast to watch. I watched the whole thing while not feeling bored, because the process was just so adventurous and fun. Can’t believe the dedication put into the presentation of this formula

[@blackpenredpen]

Thank you!

[@Jonny_XD_]

I love how you finally talked about the p-q-formular. In germany we students only learn this formular to solve quadratic equations. We were never taught abc-formular. It always confused me why every youtuber uses a different formular. I thought the p-q-formular would never be used somewhere else in the world because of that.😅

[@tollspiller2043]

that's really strange. I am here in the swiss school system which you would suspect not to be that different, and we only shortly talked about the p-q stuff, but also mainly use the main quadratic formula. A main problem may be that the p-q formula is only viable if the coefficient of the x^2 term is 1, which in most cases isn't the case, so it's quite strange because they basically teach you how to solve a special case

[@Jonny_XD_]

@@tollspiller2043 About the schools system: In germany the school system is managed by each state independently. It wouldn't wounder me if the school system in switzerland is different. About the coefficient needing to be 1: The teachers in my school told us just to divide by the coefficient of the x^2 term. I really didn't like it but I really didn't know math that well back then so I just did what they told me.🤷🤷

[@_trm3847]

​@@tollspiller2043 you just divide by the coefficient in front of x^2 to achieve the form for the pq-formula. In some cases that can be faster than using the abc-formula but eh

[@TyphAle99]

This probably only happens up to the 10th grade, in the upper school we were only taught the quadratic formula, bc the pq one tends to be inconvenient due to the ugly numbers you get when you divide with a

[@Jonny_XD_]

@@TyphAle99 Yeah... about that... We are expected to solve quadratic equations by factoring at that point... we never used any formular to solve equations after 10th grade. Wich is better... I think?

[@user-tz7ki8tu6d]

I tried deriving it myself but I got stuck at the complex analysis part so thank you for explaining it. it makes more sense now 😊

[@LexxusTheSpark]

You can also do it very fast by completing the square ^^

[@Oliver-wv4bd]

@@LexxusTheSpark In which part do you mean?

[@LexxusTheSpark]

@@Oliver-wv4bd the part where he derives the pq-formula.

[@fantiscious]

You can use a rhyme to remember the quadratic formula. This needs a whole song though 😵

[@fantiscious]

And the quartic formula needs a whole album

[@haaansolo8568]

7:15 'and this is then actually just a quadratic formula in terms of z^3' Thanks to watching you for years, I am actually able to follow ❤

[@ThetaMaths]

Now WE need a video for the quartic formula

[@DendrocnideMoroides]

and also the quintic formula, yes it does exist you just need to use Elliptic functions

[@r.w.emersonii3501]

@@DendrocnideMoroides I don't know what elliptic functions are, but I think I like them!

[@maxwelljohnson8076]

Hey BPRP, I just wanted to thank you for posting this video. I know this was a ton of work and took a lot of planning, and I really appreciate the effort you put into here. It definitely did not go unnoticed over here. You’ve been one of my favorite math educators ever sense I discovered your channel. I have been subscribed for almost a year now and in that time you have taught me so much. Even more, I have been trying to solve the Cubic formula for a couple months, and had recently given up thinking it was too hard and you come up and post a clear proof explaining how and why each step works. You have inspired me to think critically about problems and also kindled in me a love for math that will affect me in my college studies coming up. You are the best. Keep up the great work!

[@blackpenredpen]

I am very happy to hear this. Thank you and best wishes to you!

[@blackpenredpen]

39:57 you’re welcome 😃

[@k_wl]

thank you

[@aditya-u7s9v]

I love your videos Please do some more IMO or international Olympiad problems @blackpenredpen

[@asheep7797]

oh my gaah

[@drpeyam]

Whoa!!!

[@boltez6507]

I love the energy you have while teaching ,you really do like what you are doing thats cool

[@blackpenredpen]

Thank you!

[@zeroo8756]

@@blackpenredpen Please make a video about range of 1/tan(x) wolfram alpha say X€R but another say no x€R-(0) now who’s true and why??????!!

[@yacovaaa9042]

Please go quartic equations

[@swarley2500]

😈

[@Wmann]

he’s gonna need a bigger whiteboard for that

[@Ninja20704]

True. I dont’t even think he can fit the formula alone on the board, let alone the derivation.

[@ramzidaous5324]

Je suis un prof de math et j'adore votre méthode par laquelle vous expliquez des concepts mathématiques. Bravo ❤

[@JasonKwann]

Give a respect to those who made these valuable captions

[@howdoi_yt]

i had so much fun watching this video!! all this time ive been wondering how people get the cubic equation. i'm in 7th grade and i have just been a math fanatic since june 2022 and i have improved a lot since then. it got to the point that i understand calculus and how to do those limits, l'h rule, that type of thing. basically what im trying to say is i enjoyed this video a lot and it felt like i dived into the math rabbithole.

[@jakobthomsen1595]

Fantastic! I was wondering how the cubic formula can be derived. Thanks for explaining!

[@beanzthumbz]

Gotta pay respect to this man who dragged himself through algebraic perdition for you guys

[@blackpenredpen]

lol thanks!

[@lemondrop9506]

due to some complicated equation formula, my attention was diverted to the first 100 or 200 numbers of e and the stacks of markers at the back. (but this is very useful, thx for this video…it helped me alot)

[@omkartikekar6016]

Pls give this man an infinitely large board so that he dont need to rub every time 😭😭 great explanation sir

[@Fire_Axus]

stop getting so emotional

[@totor6553]

Do quatric formula next

[@wolffire8819]

I love watching math videos that I have no hope of fully understanding!

[@Velorant2]

I never get bored watching this guy, instead it inspires me to do more maths and calculas hard maths etc its relaxing and asmr

[@chhromms.8138]

how about deriving Lagrange's Resolvent next (for cubic equation case)?

[@jopaodomingo408]

bro you just legit helped me with one of the biggest obstacles in my thesis THANK YOU SO MUCH

[@serae4060]

17:45 what is the white rectangle that seems to be on the camera?

[@paulbuchinger4585]

This is something that I wanted to know for years!

[@treythompson342]

Had to go into blackpenredpenbluepen mode for this one!

[@r.w.emersonii3501]

Your ebullient attitude turns mathematics into a joyful pursuit. The approach you demonstrated for solving the quadratic was a revelation to me, and it set the stage for the cubic solution perfectly. Use an additional variable, "k", to eliminate one of the powers: What an ingenious idea! Yesterday, I was reading about Feynmann integration: There too, an additional variable is introduced. Is this a conceptual parallel? Anyway, I suspect that you wrote "2" beneath the "q" deliberately, to test your viewers. I saw it immediately, and I hoped that you would see it too -- which you did, fortunately!

[@djttv]

When taking the sqrt of something, we always remember to put +/- recognizing that there are 2 roots. Call them +/- r. I suppose to be completely thorough, we should do similarly when taking cube roots of a number. Using omega (w), where w1=1, w2=(1+i*sqrt(3))/2, w3=(1-i*sqrt(3))/2, the roots of a real number # would be r*w1, r*w2, r*w3. Where r^3 = # My question is: in intermediate steps such as at about 30:00 in the video, the cube root of 27 was just put down as 3, then the 3's canceled out. How do we justify only using the real root of 27 (3*w1)? Why would we not include 3*w2 and 3*w3 at that step to be more thorough?

[@Happy_Abe]

I have the same question, why not consider the complex roots there too?

[@沈博智-x5y]

yeah. it gets a bit hairy when trying to find the other cube roots (not the principal root). i guess in his case, he is trying to just focus on the principal root first (principal cube root of a real number is real) then deal with the other cube roots later with w2 and w3. however, it does indeed get hairy when p is complex... like cbrt(p^3) = p is not necesarily the principal root i believe if p is complex. and also it gets hairy when he brings the cube roots together in the denominator cbrt(z1)cbrt(z2) = cbrt(z1z2) but we know that sqrt(z1z2) is not necesarily the same as sqrt(z1)sqrt(z2) if we consider principal roots only. It all seems okay when p and q are real... but yeah, it's super hairy and tricky there to justify which cuberoot to use. so i dont quite have an answer either...

[@Happy_Abe]

@@沈博智-x5y I know how one can take such cube roots using the nth root of unity. My question is more so how it doesn’t contradict there only being 3 complex solutions to the cubic. When taking the cube root of 27 and getting three different answers earlier and combining that with our 3 final solutions it appears like we should get more solutions.

[@沈博智-x5y]

@@Happy_Abe yeah, this is why you just take the principal root and instead of using the cube roots of unity method by choosing a different branch of the cube root, you use the fact from cube roots of unity and multiply by w2 ans w3 which is equivalent. this way you dont have to really think about multiple solns nested within multiple solns. start with principal branch, worry about the others later

[@Happy_Abe]

@@沈博智-x5y of course one can do this, but why does doing this not miss out on some of the solutions. Clearly, there are only 3 solutions to cubics, but the question is why don’t those other roots lead to other solutions. I’m guessing if one goes through the very long and tedious algebra they will end up being the same solutions anyway, but it still makes me curious

[@mohan153doshi]

Absolutely awesome, enjoyed every moment of this video. BPRP, you rock!😀

[@bruhh489]

i understood like maybe a third of the video but i still watched the whole thing 😂

[@loekvanderzijde1701]

The proof I know is different (at least for the real solution, I don't remember if I used the same method for the complex ones as well) Once you have x³+px+q, Let x=u+v, with u≥v and 3uv=-p So we have (u+v)³+(u+v)p+q=0 Expanding: u³+v³+3uv(u+v)+(u+v)p+q=0 u³+v³+(u+v)(3uv+p)+q=0 (u³+v³+q)+[(u+v)(3uv+p)]=0 Since we chose u and v such that 3uv=-p, the second half becomes 0, leaving us with: u³+v³+q=0 u³+v³=-q Combined with the condition 3uv=-p we get a symmetric system of in equationa in two variables (I'll use two { parentheses but just pretend it's a single one): { u³+v³=-q { uv=-p/3 We can raise the second equation to the third power to solve the system in the unknowns u³ and v³. { u³+v³=-q { u³v³=-p³/27 Let u³=i, v³=j { i+j=-q { ij=-p³/27 Notice that the solutions i,j to this system of equations are the solutions to the quadratic equation z²+qz-p³/27=0 where, using the quadratic formula, z1,2= (-q±√(q²+4p³/27))/2 By going back to all substitutions we made you can have the formula for x (I don't want to write that monster)

[@matheusalmeidadamata]

Great!

[@lamaspacos]

Yes, is an amazing classic solution.

[@thuglife1219]

My worst nightmare came true: Getting such recommendations

[@alienbroccoli8296]

I treat his long videos about concepts i dont know about as movie night(with popcorn!)

[@HighSchoolMathsTKP]

Excellent I liked the explanation for finding the value of k which no body has tried to address

[@Exquisitetutorials]

Wow! Perfect way to derive the quadratic formula

[@dracokinerek]

Why do the cubic equations never have p and q when they repeatedly appear

[@ukdavepianoman]

Very well explained. The different colour pens really help viewers follow the terms too. It's not difficult to solve once you know the trick but there is an awful lot of algebra. I could sense your relief and enjoyment getting to the final answers. If you do a video on the quartic..."You're gonna need a bigger board".

[@brian554xx]

this is what it feels like when i try to plan tomorrow when way too sleepy.

[@stephenphilbrook6239]

Love your channel. Super helpful. I am doing Laplace Transform of discontinuous forcing functions and your videos helped out alot.

[@Bjowolf2]

Great 😉 Mathologer has another really nice video about this concept called "500 Years of not Teaching the Cubic Formula" 😊

[@Michael-sb8jf]

That and the veritasium video

[@Bjowolf2]

@@Michael-sb8jf YES, I wanted to mention that one as well, but couldn't remember the name 😂

[@Peter_1986]

I always use these kinds of long, overwhelming formulas as excuses to write programming code in MatLab (one of my personal favourite programming languages); there is something satisfying about writing a program file, and then being able to have that program file calculate things for me.

[@epsilia3611]

About the substitution y = z + k/z, there is something weird happening. First of all, if we suppose that k>0 in the function f(z) = z + k/z, it doesn't offer us a valid substitution. Indeed, in that case, the minimum value on the positive side will be (k+1)*sqrt(k), and the maximum value on the negative side will be (k-1)*sqrt(k). If we treat the y=0 case apart from this substitution, we have a big problem in the case of k>0, since there is plenty of values which are not reached by the substitution. More precisely, the open interval ( (k-1)*sqrt(k) , (k+1)*sqrt(k) ) isn't contained in the image of f... What does that mean ? It means that for k>0, the substitution doesn't reach some values we were considering and not neglecting at the first place, y = k*sqrt(k) being one of them for example. What if k

[@xenumi]

This is such a beautiful thing.

[@DoomsMusic614]

Bro's smile is the best thing in the world.

[@codingforest7442]

I used this formula with x^3 -x^2 -x -1 = 0, couldn't get a precise value, I only get an approximate value by 10^-9. I don't know why? I appreciate your help.

[@peterchan6082]

Finally, at long last, you've presented us with the derivation of the legendary Cardano Formula (or was it due to Cardano at all?)

[@ridefast0]

I asked Mathematica to solve a general quartic, and it gave me the formula for that - though I wouldn't want to write it by hand!

[@physicsmylife]

Bring part 3 of 100 Intrigral please

[@user-hp6pj4sy3k]

I hope we can get another video of BPRP deriving the quartic formula. Will almost certainly be a (very) long video, but I would watch every minute. I have a feeling he's already working on it though.

[@garfield36788]

Can you solve for x? (1/2)x+(1/2+x)x^2+(1/2+2x)x^3+(1/2+3x)x^4=(1/2+4x)x^5 I'm curious because it's a quintuple but it's apparently possible.

[@shanathered5910]

and now for the quartic formula/j

[@averagejojofan8649]

Quartic formula is much easier

[@pawel_maslanka]

WHOOOA I WAS ALSO THINKING ABOUT DOING THAT but I realized I'm not that good at all the stuff with cubics so I didn't get far

[@la1m1e]

Today I witnessed the most complicated way to solve quadratic equation

[@wqltr1822]

Yesss sin of 10 degrees. I remember a video you made where you showed sin of 10 degrees was a root of a polynomial with irrational roots. So i can see how the formula would help

[@angelosterizakis7635]

You are AMAZING. You explain best why the only solution to a cubic equation is cube root(S + (sqrt T)) + cube root(S-(sqrt T)) which you obtain by substituting y= z+ k/z. However, my question is how you are led to this substitution by the Vieta formulae.

[@redsurfer_255]

can you please explain me that how the heck we get transcendental roots of cubic with rational or irrational coefficients even tho thats not happening in answer, also we know that sin3x = (3sinx - sin^3x)/4, putting x = 10 degrees we get depressed cubic with rational coefficients , hence we can find value of sin 10 , which would be one of the roots of the cubic

[@tonio9351]

Please do ax⁴+bx³+cx²+dx+e=0 for us 🙏🙏🙏

[@PW-qi1gi]

I want the quartic formula!

[@ElGordoFreeman]

Nice demonstration, but, what happens when all the roots are real? what happens with the imaginary roots?

[@blackpenredpen]

they will turn out real, i.e. the complex parts will cancel out

[@stormswindy3013]

someone get this guy a bigger board

[@vishalmishra3046]

Any cubic equation can be turned into the form - { x^3 + 3 m x = 2 n } and then solved as - x = sum of 2 cube-roots = (n + d)^(1/3) + (n - d)^(1/3) where d^2 = D = n^2 + m^3 i.e. D = cubic discriminant and it's square root is +/- d. *Very easy to remember* right ?

[@tank2256]

do one for the quartic formula

[@GicaKontraglobalismului]

Blackpenredpen, I have always regarded this as too impressive to try, but this time I followed to the very end! Thank you! Now you have to plug the solutions you found into the original equation to see if they verify it... However, I think you should have taken a=1 when started with the third degree equation - it would have simplified it a little without any reduction of generality...

[@blackpenredpen]

Thank you! : )

[@person1082]

i used to do this proof so many times i memorized the final result

[@monkee3613]

Hello, do you mind explaining how the graph of y = x^y works

[@saravanarajeswaran2626]

After seeing your video, i tried to derive the formula and i did it!!!!!! now i can flex it to my friends!!!

[@jasimmathsandphysics]

Thanks

[@evanfortunato2382]

This is grueling, man. I love it.

[@gcewing]

39:59 I'd hate to have to play that piece of music!

[@armanavagyan1876]

PROF thanks as always UR second of none)

[@MrConverse]

27:25, *(q^2)/4

[@MrConverse]

31:20, you fixed it!

[@serae4060]

I also got so destracted by looking at that typo all the time until he fixed it😂

[@DestroyerWolfFenrir]

I derived that p q for formula of the cubic equation when I was in 9th grade and pissed that there was one available for the quadratic but not cubic in the book...then I lost the page where I solved it and forgot about it...😂😂

[@jakehobrath7721]

Man I was staring at that q^2/2 for what felt like a half-hour

[@carlopaternoster5878]

So, for the quadratic it was k = -p/2, with p = a/b, and in total p = -b/(2a). For the cubic, it is k = -a/(3b) Is it for the quartic k = -a/(4b)? Then, if the general quintic is not soluble, what happens with k = -a/(5b)?

[@hussainahmad27]

We substitute y = z + k/z because if we substitute y = z + k, then from y³ term, we get two terms similar, since we want to get rid of linear terms, like 3z²k and 3zk², and from py term, we get pz. We cannot take 3zk² and pz to find k as k, like 3zk² + pz = 0. Then k² = - pz / 3z, then k² = - p / 3. But here is a problem, we cannot take k² as it may not work in both terms ( ALSO WE ARE LEAVING THE 3Z²K TERM BEHIND, the squared term). Therefore we take 3z²k + pz = 0, then k = - p / 3z. If in 1st situation y = z + k / z, k = - p/3 and in 2nd situation y = z + k, k = - p / 3a. The substitution, y = z + k is more easy rather than the other one.

[@dhananjaysawant4646]

I memorized this set as well as the quartic ones

[@MC_Transport]

Good. Now derive the quartic formula.

[@EduardoViruenaSilva]

It was epic! I have a doubt. You made: y= z - p / (3z) but, what happens if z=0? Suppose: x^2 + 2 =0, this yields to p=0, q= 2. z= cbrt( -q/2 + sqrt( q^2/4 + p^3/3 ) ) = cbrt (-2/2 + sqrt( 4/4) ) = 0. Now, the formulae work: x1= cbrt(2) x2= w1 cbrt(2) x3= w2 cbrt(2) but they make me feel itchy

[@andrewchen802]

@blackpenredpen, a lot of us probably have calc exams coming up. Are you planning on doing any livestreams? We need your help reviewing!

[@blackpenredpen]

Prob no livestreams but definitely more practice problems. Please check out my “bprp calculus” channel.

[@kummer45]

Can you do the Quartic Formula without skipping steps?

[@apokalypthoapokalypsys9573]

At 1:55 why would (y+k)^2 equal 2yk? Let y=1, k=2. Now (1+2)^2=9 does not equal 2*1*2=4.

[@jusinocasino11]

The pq formula is what we use in sweden. When i saw the normal quadratic formula they use in the US i was confused but they are similar

[@hqTheToaster]

a_0 + a_1 x ^ a_2 + a_3 x ^ a_4 + a_5 x ^ a_6 ^ x + a_7 x ^ x ^ a_8 = 0 ; abs(a_2 - a_4) is more than or equal to 1 Just kidding. You are a good person, and I find it hard to apply such an equation irl. Good video.

[@excellent808]

Great I want the quintic formula

[@matheusfariassantos7490]

Hello, que legal video legendado em minha lingua nativa. Thanks! Your job is really nice!

[@hoanganhdao68]

Great work! 😁😁

[@JonDisnard]

I'm dumb, and so far only got freshman level algebra education. But for cubic polynomials I just use Vieta's formula, and ignore pretty much everything else. It like a simultaneous equation for r1 + r2 + r3 = …, and r1 * r2 * r3 = … Once the roots are factors it's pretty much done. But I like this cubic equation, but it's very lengthy. Good to know, but seems inefficient. I guess the quadratic equation vs formula is the same way?

[@elen1ap]

Mathmeticians, please learn that φ is pronounced fi and not phai and π is pronounced pi and not pai. I just picked a random video to tell this.

[@holyshit922]

I prefer subsitution y = u + v because we dont have to take care of division by zero and it is quite easy to generalize this substitution for quartic and Euler did it Substitution presented on the video doesn't work when p = 0

[@gamingwithessam3383]

I have this memorized. I did a study on it last year for fun :)

[@이름없음-y4f8l]

손이 아프시겠지만 혹시 사차공식도 가능한가요?

[@GatriAbdelkader]

Bravo je vais l'essayer sur un exemple