EXTREME quintic equation! (very tiring) 

I remember how to derive the formula for the cubic ( x^3 + a x^2 + bx + c = 0) by remembering these three steps: 1) Eliminate the quadratic term by a substitution x = y - a/3 (so that new cubic in y has no y^2 term). 2) Eliminate the fractions and then make a substitution absorbing the lead coefficient (it will be u = 3y). 3) Write u as the sum of two cube roots (so u = s^(1/3) + t^(1/3) ). (I don't really "remember" step 2 so much as I just "do it", because it's so natural; it's crying out to be done. But Step 1, and especially Step 3, I need to commit to memory. Step 3 is really the main one for me at least. I've derived this so many times that I now do remember the derivation without trouble.) With that setup, the solution is basically forced onto you by the algebra. You'll be forced into some algebra showing that you need to find s and t that simultaneously solve 2 equations: one for s+t in terms of the given coefficients, and another for st in terms of the given coefficients. But that means that s and t are the solutions to the quadratic: z^2 - (s+t) z + (st) = 0... and again, you know the values of s+t and st in terms of the original given a, b, and c. It works out like this: x^3 + a x^2 + bx + c = 0 Step 1: Let y = x + a/3, so that x = y - a/3. Then (y - a/3)^3 + a (y - a/3)^2 + b(y - a/3) + c = 0 [ y^3 - a y^2 + (a^2/3) y - a^3/27 ] + a [ y^2 - (2a/3) y + a^2/9 ] + b [ y - a/3 ] + c = 0 y^3 + ( a^2/3 - 2a^2/3 + b ) y + ( - a^3/27 + a^3/9 - ab/3 + c ) = 0 y^3 + ( - a^2/3 + b ) y + ( 2 a^3/27 - ab/3 + c ) = 0. Have a cubic in y with no quadratic term. Step 2: Eliminate the fractions and then make a substitution absorbing the lead coefficient (you can already see the coefficient expressions appearing in the Lagrange Resolvent) Multiply both sides by 27: 27 y^3 + ( - 9 a^2 + 27 b ) y + ( 2 a^3 - 9 ab + 27 c ) = 0 (3y)^3 - 3 ( a^2 - 3 b ) (3y) + ( 2 a^3 - 9 ab + 27 c ) = 0 u^3 - 3 ( a^2 - 3 b ) u + ( 2 a^3 - 9 ab + 27 c ) = 0, where u = 3y. Let P = a^2 - 3 b, and Q = 2 a^3 - 9 ab + 27 c, so P and Q are known values determined by the original cubic equation. The original cubic equation in x is now the cubic equation in u given by: u^3 - 3 P u + Q = 0. Step 3: Suppose s and t are values (possibly complex) such that u = s^(1/3) + t^(1/3). Then u^3 = [ s^(1/3) + t^(1/3) ]^3 = s^(3/3) + 3 s^(2/3) t^(1/3) + 3 s^(1/3) t^(2/3) + t^(3/3) = s + 3 s^(2/3) t^(1/3) + 3 s^(1/3) t^(2/3) + t = s + 3 [ s^(2/3) t^(1/3) + s^(1/3) t^(2/3) ] + t = s + 3 s^(1/3) t^(1/3) [ s^(1/3) + t^(1/3) ] + t = s + 3 s^(1/3) t^(1/3) [ u ] + t = 3 [ s^(1/3) t^(1/3) ] u + [ s + t ] = 3 [ st ]^(1/3) u + [ s + t ]. Thus u = s^(1/3) + t^(1/3) implies that u^3 - 3 [ st ]^(1/3) u - [ s + t ] = 0. That's a universal formula, but now make it apply to the specific equation u^3 - 3 P u + Q = 0, where P and Q are known. That requires that - 3 [ st ]^(1/3) = - 3 P, and - [ s + t ] = Q, and so that requires: st = P^3 and s+t = -Q. Thus s and t are the two solutions to the quadratic: z^2 - (s+t) z + (st) = 0, so here meaning z^2 + Q z + P^3 = 0. And that's the desired formula. Tracing it back in terms of x, get: x = y - a/3 = u/3 - a/3 = [ - a + s^(1/3) + t^(1/3) ] / 3, where s and t are the two solutions to the quadratic: z^2 + Q z + P^3 = 0, where P = a^2 - 3 b, and Q = 2 a^3 - 9 ab + 27 c.

Any Asian parent: "Don't accept sweets from strangers, it might be drugs" This guy: "Don't accept quintic equations from strangers, it might be unfactorable"

Solving degree 4 + equations through brute force is a real tedious task, good job.

And remember kids : Do not take quintic equation from a stranger ! 😜

I hate that I have to keep turning down people that try to hand me quintics when I’m walking on the sidewalk. Those people are so annoying.

You know shit gets real when he has to pull out the blue pen

My semester is indeed almost done :) I'm in grad school, and taking advanced algebra this semester. Next year I'll be starting my research for my master's thesis. But in my class we actually covered the insolvability of the general quintic a few weeks ago, so this video relates to what I'm learning about in my class!

in 2012 I finished my mathematics studies during my 2 years of intensive preparation in the so-called ``classes preparatoires`` in Morocco, which is a French education system that focuses on maths and physics for 2 successive years followed by a national competition to get a place in engineering schools. I graduated as an engineer and worked for many years and I am still enjoying your videos tackling some difficult calculus problems, and when you finish and find the solution it is like a stress-relieving process for me. Thank you for the great job, which is way better than watching the shitty social media content that has no point in our modern age.

I am reaching 12th Grade in September so I am preparing for math and physics already,your videos are very enjoyable and informative,keep it up! Sending love from Greece

台灣人路過 雖然大學不是念理工科系學到比較深的數學，但本身對代數和離散數學discrete math領域挺有興趣了解的 給個支持🎉

The complex solutions are straightforward, though. In your formule 1/3 (-a + ³√z1 + ³√z2) you apply a small change: 1/3 (-a + ω ³√z1 + 1/ω ³√z2) and 1/3 (-a + 1/ω ³√z1 + ω ³√z2) where ω is a complex cubic root of unity (ω³ = 1, ω = -1/2 + i√3/2) It's really the same recipe for Cardano's formula.

Why would I even attempt to solve a stranger's quintic equation? 🤔

The good "news" is that noone asks a mathematician "Yes, but why have you tried this (method/idea etc.) ?" as soon as it works. The winner is always right. Bravo for having created this working example !

Since you asked how I'm doing. I've been a subscriber for a long time. Back when I started at community college years ago your calculus 2 videos were very helpful in my studying. After that I would watch your videos periodically. Six years after initially subscribing I'm about to graduate with a degree in computer science. I'll continue to watch your videos even though I don't need calculus 2 videos to help me study because I like the content you make. Keep up the good work.

Please, please, please! Teach us everything about the Lagrange resolvent. Are there resolvents for polynomial equations of other powers? If there are, please give examples. How did Lagrange come up with this formula? Wikipedia gives a general formula for Lagrange resolvents (sum from i=0 to n-1 X sub i omega ^i), where omega is a primitive nth root of unity. How does that apply here, or in polynomial equations of other powers? I haven't seen anything like that on RU-vid. Thanks a lot!

Not as EXTREME PATIENCE as the extreme algebra question, but still extreme patience.

It’s fair to say the mechanics you explained are fantastic. For me. It’s substitution of trigonometric identity that is REQUIRED in order to come to a simplified expression. And that is where I stopped there’s a lot of thought in this video. Thank you it’s grand.

Surprisingly the most mind blowing part of this for me was the 65^2 trick

OK I used a root-finder. But the equation fits to phi=0.618... phi^2 = - phi + 1 phi^3 = 2 phi - 1 phi^4 = -3 phi + 2 phi^5 = 5 phi - 3 etc The coefficients are alternating, offset terms of the Fibonacci sequence. One to store in the memory bank EDIT: You could apply this idea to any equation of the type x^5 = Ax + B. I.e put x^2 = ax + b x^3 = ax^2 + bx = (a^2+b) x + ab ... x^5 = (a^4 + 3 a^2 b + b^2) x + ab(a^2 + 2 b) In this case A = a^4 + 3 a^2 b + b^2 = 5 [1] B = ab (a^2 + 2 b) = -3 [2] Solve [1] as a quadratic in a^2 a^2 = (-3 +/- sqrt(5b^2+20))/2 To get rational a^2 guess b^2=1 a^2 must be the positive root i.e. a^2=1 Solve [1] as a quadratic in b b = (-3*a^2 +/- sqrt(5 a^4+20))/2 b = (-3 +/- 5)/2 We must have b^2=1 so b = 1 Substitute in [2] 3a = -3 a = -1 The quadratic factor is x^2 - ax - b = x^2 + x - 1

I tried to prove langrange resolvent based on the construction of cubic formula and the three solutions would be (-a+(w^j)*³√z1+(w^(3-j))*³√z2)/3 Where j=0,1,2 and w=(1+i√3)/2

Hello! I am thinking about the "If a stranger gives you this quintic equsion....", and it gives giggles because it sounds like "Don't try this at home" xD. Regardless of me not being a student at any university, i still find your video entertaining

With a little linear algebra there is a general solution for a system of n quadratics in n unknowns. What you do is that for matrices A and B, you have x†Ax + Bx + C = 0 for a symmetric A where x is the vector of variables and x† is its transpose. Since A is symmetric you can choose a basis to diagonalize A, then (xS^(-1))† D (Sx) + BSx + C = 0. Then you can complete the square and take the square root. In finding the square root of D each eigenvalue has two square roots and so there are 2^n solutions.

The .618 is the power of root, Roots can be transformed in powers by raising the radicand to the power of a fraction, in which the numerator is the exponent (usually -1) of the radicand and the denominator will be the index of the radical

11:43 “Wow, D is pretty big”

There is an exam at my grad school which we have to take to start the research stage of our Ph.D. There is almost always a polynomial on there and they ask us if it can be factored using whole numbers. Usually there is a trick that makes it quick, but I was doing one of the older tests and one test writer gave one that had to be solved like this. It was very mean.

11:45 “Wow. D is big.” -BPRP 2022

He always have a t-shirt about the video he's making

It's not as precise or fun but you can also graph y=x^5 and y=5x-3 and find the intersection points. The x-coordinates of those points are the solutions.

Could you make a video about solving the quintic with elliptic integrals? That would be cool.

Another great video. Around 24 minutes shouldn't the quadratic have a 2(-1)=-2 in the denominator?

No need to solve for exact decimal to decimal complex solution But, you can numerically calculate the value of 'r' and then form the last quadratic which could be solved further. Will land up on an approximate complex solution.

its also nice you can just use iteration / newton rapson to get a non exact answer in a few seconds :)

Son did you see that guy up over. He's more than just crazy.

Well said 😂 "I dont know if you like extreme sports, but i like extreme math questions." Story of my life

8 months late, but... The final quadratic can be solved in terms of r to give x = (1 - r)/2 +/- i sqrt(3 r^2 - 2 r + 7)/2. Letting w1=z1^(1/3), w2=z2^(1/3) (where z1 and z2 are the solutions of the resolvent, (65 +/- 15 sqrt(21))/2), we have r = (1 + w1 + w2)/3, and the solution expands to x = (2 - w1 - w2 +/- i sqrt(3 w1^2 + 3 w2^2 + 30))/6. Substituting w1 and w2 into this expression, expanding, and then rationalizing denominators, gives: x = (4 - (260 + 60 sqrt(21))^(1/3) - (260 - 60 sqrt(21))^(1/3) +/- i sqrt(6 (17900 + 3900 sqrt(21))^(1/3) + 6 (17900 - 3900 sqrt(21))^(1/3) + 120))/12 Numerically, this is about x = - 0.13784110182549249453 +/- i*1.52731225088662939067, and substituting either numerical approximation for the root into x^3 - x^2 + 2 x - 3 gives a value that is equal to zero for 18 decimal places (as does substituting it into x^5 - 5*x + 3).

I could not follow this myriad of computations if I hadn't studied cubic and quartic equations for decades. I started being fascinated by these algebraic equations when I was about 15. The next year, we learned Newton's method at school and tried to solve x^3 - 15x - 30 = 0. The other pupils used their electronic calculators and wrote down a series of approximations, converging to about 4.634. I took my notes on Cardan's formula and got x = cubrt(25) + cubrt(5), which produced the same aproximated value on my calculator. Unfortunately, the teacher called another pupil to präsentiert his solution, and I was still too shy at that time. Imagine the reaction of y teacher and class if I had gone to the blackboard and written down the exact solution in terms of radicals!

good job , and I didn't know the lagrange resolvent but do remember cardano's formula for the cubic , rather you than me though

Why didn't you use simon's favorite method from AOPS?

I like to walk around the bad part of town in a shady hat and a big jacket and offer strangers unfactorable quintic equations

We got this lil thing in Italy called "Ruffini's Rule", we're taught this during algebra classes but everyone tends to forget it (cause it's tedious). I'm pretty sure that would've allowed us to find some real solutions from the beginning. What do you think about this? Cheers, Thomas! 💗

Me encantó este problema; además, maravilloso el rostro de buena persona de este matemático. Gracias.

Well you see, suppose for integers a, b, p, and q, if x^2-ax-b divides x^5-px-q, then a divides q^2. Because of the small margins, I won't leave a proof.

Another form by brute force, suppose that q(x) is a quadratic factor of f(x)=x^5-5*x+3, then f(0)=3, f(1)=-1. We can assume that q is monic, q(x)=x^2+b*x+c. Therefore q(0) divides 3 and q(1) divides -1. We take many finitely cases for q(x). This method proves that factoring polynomials (with integers coeficients) is equivalent to factor integer numbers. But this method is pure brute force.

I like watching you cause I wanna go back to school eventually, don't want to forget everything in the mean time

It will be very nice if you solve question paper of J.E.E mains and advance which is tricky and popular

A good starting point would be 0.6 since 0.6^5 is close to 0 (it is about 7/90), and -5 (0.6) is 3, so the we would have (almost 0) - 3 + 3 = 0 which is "almost" true. From there, I just used a computer to check values close to 0.6 and I got about 0.618 as the final answer. He basically has the answer on his shirt but with 1 added to it. 0.618034 is even more accurate. The answer to this "hard" problem can be easily approximated simply by inspection. Why even bother with problems like this when you can use wolframalpha and/or a computer to find the answer?

Sir, where you from? You sound so smart that i couldn't help myself that i instantly became your new subscriber from India 🧘 We need more teachers like you.

Your explanation, as always, was very clear..

In a^3+b^3+c^3=∆ then ; √∆ = a+b+c Where a,b,c are selected values A:-1 b:-2 c:-3 ≈ From ｉｎｄｉａｎ

When will I do math like this? I'm freshman in highschool

Sheesh bro lemme sleep. Ngl, I really find your vids so satisfying to watch that it made me appreciate more the innate beauty of Math. God bless you man!!! Much love from ph!

Only some quintic and higher degree equations can be solved in radicals, in contrast to linear, quadratic, cubic, and quartic equations.

In what subject of math would you learn the sorcery to solve such an equation?

I imagine that Lagrange's strategy was that if r is a root of the cubic, then (x-r) is a factor and he divided this from the cubic to see how the coefficients of the remaining quadratic must be related to the root and the original a, b, and c.

What is this for? What class requires solving quintics by hand?

watched till the end so cool lol

Can you use the rational zeros theorem for this?

I’m one of the people watching this just for fun! Really interesting stuff here. Thank you for making this 🙏🏻

isnt z = (65 + or - sqrt(4725)) / -2, since a = -1 then 2 x a = -2 not 2

May be a dumb suggestion. But could you not take the fifth root of x^5, 5x, 3 and the 0. And then do the pq formula?

How can you be sure there are no other possible values for A,B… other than the ones you found? (For example in the complex realm)

if SQRT(27) - SQRT(12) = SQRT(3), how would you figure out what it would be if we replace the SQRT(27) with a SQRT(29)?

can you find the indefinite integral f(x) = x! on (0, infinity)?

Hard to imagine! You have done it. Salute you 👏

For degree 5 it's very unusual to try and no result I think with this result it was very easy to seperate this equation in two equation and draw them and the cross point will be the roots Y1 =x5 easily can draw it with some points such as 0, 1 , 2 , 3,-1 ,-2 ,-3 Y2 = kx+ d line The cross points are equation roots , Or it was better ; You make x^5-1 + 3x -3 -2=0 (X^5-1)=( x-1) ( ........ ) Divide ( X^5 -1) to (x-1) therefore you can Make x-1 factor ...

I don't know why RU-vid showed me this old video today, but I watched it. Nice work. But (sadly?) Wolfram Alpha gives both real and complex solutions in either approximate or exact forms in about two seconds.

nếu phương trình bậc 3 trên có cả 3 nghiệm là số thực thì sao ạ Tại phương pháp trên em thấy chỉ tìm được 1 nghiệm thôi

Sir in which university do you teach

Sir can I ask you one question ..Is positive infinite and negative infinite are same point ?? If on real axis positive infinite and negative infinite are two different point then distance between them should be 2*Infinite now this became meaningless or simply we can say that a person is standing on infinite will treat infinite as zero so for that person positive infinite and negative is positive zero and negative zero and positive infinite is positive zero and negative infinite is negative so zero and infinite not a number rather than symbolic representation ??

The answers are proportionally perfect (golden ratio)👌

he would be a cool teacher

"If a stranger on the street just hands you a random quintic equation..." - who doesn´t know about this daily challenge in life?

How ur maths is so good n what u should do for that.

You: doing some out of the world maths We: still adding numbers of 5 to 6 digits in our heads for 10min ultimately getting wrong answer and the checking other topper students answers. XD XD XD

Quintic formula😵

You could have use the quadratic formula.... b^2-4ac

I love extreme sports but what I love way more is extreme math problems.

Yup, watched it for fun, very interesting, I'm not at that level of math and honestly, I dont need to, not yet, so it was all just fun and interesting

11:44 sorry

I'm doing pretty well, thanks

I love how we invented 5 different variables to solve for our one variable "x"

15:24 yep, just for fun!

I'm watching this for fun. I should be working on literary theory right now.

What if I do X3*x2-5x+3=0 And removing that x3 Solving the equation x2-5x+3=0

Random indian mc donalds worker solving depressed octics within 20 mins : 🗿

dont they teach ruffini in the usa?

nice and informative video thanks for sharing this video

I've never seen this formula to solve cubics, I don't know if it's possible but I would like to see where it comes from.

Can this be solved using Rational Zeros Theorem and synthetic division ?

It's 2am and this blew my f*cking mind for some reason

I wanna be that weirdo who asks quintic equations to people on the street.

I will try the other way to do this exam .

I was expecting all the solutions!😭

Even better, use cubic discriminant to determine the type of roots we will get :)

Z has a -2 at the bottom of the fraction, not a positive 2

Imagine having a channel called BLACKpenREDpen and doing math shinanigans with a blue pen. Still awesome :).

Watching video for fun. Left school decades ago.

For the complex solutions can't one just take the complex cube roots of z1 and z2?

Gotta love the Pokeball in every video

I learned something today. Don't accept quintic polynomials from strangers.