A Polynomial Equation | x^5=(x-1)^5

[@WahranRai]

A classic approach is to use complex numbers leading to the resolution of the nth roots of unity. Equation gives (it is obvious that x =1 or x=0 are not solutions ) : ( (x-1)/x)^5 = 1 ---> z^5 = 1 with z = (x-1)/x ---> z = e^(i*2kπ/5) with k= 0,1,2,3,4 and x = 1/(1-z) gives the 4 solutions for x (z=1 is removed)

[@SyberMath]

Nice!

[@FirstnameLastname-hg5gt]

This is exactly what I was thinking while watching the video. I was expecting that this one would be the second way to solve the equation when a second method was announced. After watching the video I moved to the comments to write about that, but I saw that the solution already existed from you. Nice job!.

[@wing_butterfly]

I am definitely a beginner in the complex aspect, so it will be appreciated if you can please tell me why z=that kind of stuff? Thanks a lot

[@Ronelel]

Just take the fifth root of both sides (we can do that since the power is odd) - (if the power were 2 for instance, we'll put an absolute value on both sides after taking the square root). Then, we'll be remaining with the following equation: x=x-1 and after subtracting an x from both side we will get 0=-1, which is a false statement and therefore there are no real solutions.

[@wychan7574]

You cannot do this kind of power reductions for complex values of x.

[@ilias-4252]

You still have 4 roots to find though that doesnt help much

[@kinshuksinghania4289]

There are 4 complex solutions because while taking the fifth root you only considered the real fifth root of unity and forgot the other four complex roots

[@pauselab5569]

Yes but there is 5 solutions to the equation so you need to check them all. The balance method doesn’t work well when there is multiple solutions(because you are always going to find a single solution). The reverse function for ^5 is found with the unit circle

[@nickcampbell3812]

You aren't wrong, but your answer is incomplete, partially because the question is incomplete. The question never specified what field you are solving over. If the question was to find all real solutions, then you would be correct, however, it seems like they want to find all solutions, which includes complex solutions.

[@MrLidless]

There are no real solutions for the following reason: 1 x^5 is strictly increasing for all x 2 The two graphs would be exactly the same, but transposed L to R by one 3 Therefore they will never cross, and therefore no real solution.

[@tontonbeber4555]

But 4 complex solutions ...

[@inkerilain]

I'm from an imaginary universe. They do cross here.

[@ugurcansayan]

Reality is more complex than real numbers

[@randomhuman9479]

x^5 is NOT strictly increasing for all x in real space. What if x is negative real number? Or a fraction less than 1?

[@miikey_lol]

@@randomhuman9479 Still strictly increasing

[@JensenPlaysMC]

Just do 1 = [(x-1)/x]^5 , then solve for z^5 = 1, [re^(i○)]^5 = e^i2pi compare coefficients on left and right. r = 1, ○=2/5pi*n. Then (x-1)/x = e^(i2/5pi*n) and thus x = 1/[1-e^(i2/5pi*n)] for n from 0 to 4, but 0 is undefined hence from 1 to 4.

[@federicoquaranta824]

You can divide both sides by x^5 since x=0 is not a solution, then you have (x-1)^5/x^5=1 which is also ((x-1)/x)^5=1. Then just take the 5th complex root. One of the solutions will be missing (since the original polynomial is a “false” 5th grade) and the other 4 solutions are the one in the video (which make sense because it is a 4th grade polynomial with real coefficients, so for the Fundamental Theorem of Algebra it has 4 complex roots, two by two paired with the conjugate.

[@rajeshbuya]

I was probably waiting for a 3rd method. Knowing we have only complex solutions, we could do a SUBSTITUTION (your fav has become my fav too). Substitute x with r ( cosA + i.sinA) and try to find possible values of r and A. Works?

[@SyberMath]

Great point!

[@armacham]

Another method, and I'm not saying it is a good one, is factoring from 1:11, instead of dividing by x^2 you can just try to factor it into squares. Multiply both sides by -1 first so it looks like a normal polynomial (biggest coefficient is positive) to get: 0 = 5x^4 - 10x^3 + 10x^2 - 5x + 1 0 = 5*x^2 *(x-1)^2 +5x^2 - 5x + 1 0 = 5*x^2 *(x-1)^2 +x^2 + 4x^2 - 4x - x + 1 0 = 5*x^2 *(x-1)^2 +(x-2)^2 + x^2 - x 0.25 = 5*x^2 *(x-1)^2 +(x-2)^2 + x^2 - x + 0.25 0.25 = 5*x^2 *(x-1)^2 +(x-2)^2 + (x-0.5)^2 from here, in order for there to be a solution, all 3 parts must sum to exactly 0.25, so we can say: any solution MUST have all 3 parts less than 0.25 and it's not that hard to show that there is no overlap between regions where ALL 3 of those parts are less than 0.25 the first part, 5x^2(x-1)^2 is only less than 0.25 inside a range smaller than (-1.2, -0.8) and (-0.2, 0.2) But the second part, (x-2)^2 is only less than 0.25 inside a range smaller than (1, 3) Since there there is no x-value where all 3 parts are less than 0.25 at the same time, you can confidently say there are no solutions

[@Minskeeeee]

my polar method: first off, x^5 is a monotonically increasing function, so we know that there's no real solution thinking in terms of the polar form of x as r*exp(i*theta), we know that subtracting 1 from x will give us another complex number with its own radius and angle. we know from our equation that these radii must be equal, which results in real(x) = 1/2 since angles multiply when you take powers, we know that the angles of x^5 and (x-1)^5 must differ by a multiple of 2pi. we also know that subtracting 1 would increase the angle in the upper half of the plane, and decrease the angle in the lower half. I only need to solve in the upper half of the complex plane, since we are solving a real polynomial, and the solutions will be conjugate pairs. this, combined with the above imply that the difference in angles of x and (x-1) should be either 2pi/5 or 4pi/5 (other multiples don't fit in the upper part of the plane). symmetry gives us that the angle of x is either 3pi/10 or pi/10. geometry gives us that the imaginary part of x is half the tangent of this angle. bringing it all together, I get that x is either (1 [+/-] i*tan(3pi/10))/2 or (1 [+/-] i*tan(pi/10))/2

[@Hobbitangle]

About the symmetry. The initial equation may be transformed to symmetric form by simple substitution x=y+½, what leads to equation (y+½)^5=(y-½)^5 after expanding the five lever bynom all the odd power coefficients cancel and the equation gets biquadratic: 5•y^4/2+10•y^2/8+1/32=0, which is simply solvable.

[@WikiBidoz]

Which is exactly method #2, why rewriting it?

[@vladimirkaplun5774]

a^5-b^5=(a-b)(a^4+...+b^4)=0. Second brackets is an homogeneous equation in (x/(x-1)+(x-1)/x).

[@haakoflo]

You can also solve this geometrically. From basic properties of real and complex numbers, we see that x must be complex, that the real component is 1/2. (the r component of x and x-1 must be identical, since polar component radius (r) is multiplicative under complex number multiplicaton) This means that a solution must be such that in polar coordinates, 5*theta must equal 5*(pi-theta) (modulo 2pi, where theta < pi/2). From symmetry, that meanst that 5*theta must be on the imaginary axis. Solutions in the first quadrant are theta = pi/10 and theta = 3*pi/10. The complex conjugates would also be solutions. That means that the imaginary components would be +- 1/2 * cot(pi/10) and +-1/2 * cot(3*pi/10). Now you could go on proving that thes cot() values are sqrt((5 +- 2*sqrt(5)/5), just look it up, or just leave the expression as is, with the cot() function. In fact, I prefer the latter, since I believe the geometric solution is more illuminating than the arithmetic one, especially so if illustrated with a drawing.

[@MrElCharpi]

Easiest way to prove that there is no real solution ? Case disjunction. - x and x-1 trivially cannot be both positive or both negative - x and x-1 trivially cannot be one positive and one negative - neither x nor x-1 can be 0.

[@FractalMannequin]

Third method: - Notice it's a 4th degree polynomial equation and exclude x=1. - Turn it to (x/(x-1))⁵ = 1, hence x/(x-1) Is a fifth root of 1, call it z. - From x/(x-1) = z get x = z/(z-1). - z can be any fifth root of 1 except 1 and we're done.

[@PremjitTalwar]

Why not take the 5th root of both sides! And quickly one realizes there are no real solutions.

[@Wolf-if1bt]

took the same substitution z=x-1/2 to get ((2z+1)/(2z-1))^5=1 thus (by taking the roots of 1) : z=2(w+1)/(w-1) with w=e(ik2pi/5)-1 for k=1 to 4

[@ivaneiring4732]

Вообще-то, нет решений для некомплексных чисел. Т.к. рассматриваем две функции: исходную f(x) = x^5 и вторую f(x-1) = (x-1)^5. Они никогда не пересекутся, следовательно, нет решений.

[@musicsubicandcebu1774]

For x = 1/2, for odd powers, the curves get closer and closer together. I tried up to power 25 and ran out of calculator.

[@DrAndyShick]

I'm glad I skipped to the end of the video to confirm what I already knew. Odd exponent, so x would have to equal x-1. If it were an even exponent, it would be easy. x=1/2

[@Qermaq]

x = 1/2 is as close as you can get, the values of (1/5)^5 and (-1/5)^2 differ by only 1/16.

[@zunaidparker]

Syber, I think you could benefit from writing out your script before hand. This was an interesting problem but presented in a way that was difficult to "get" what you were trying to demonstrate. If I were to provide some concrete ideas: 1. Start by explaining there are no real roots (taking 5th roots both sides) 2. Then, once that is out of the way, proceed with the Z-substitution as you did, explaining what and why you're doing as you go. Actually complete the work and show us the 4 solutions in complex conjugate pairs. 3. Then show the graphs to complete the video. The way it's presented now feels very haphazard - you show there is no real solution as a tangent midway through abandoning Method 1. Then when you do Method 2 which is clearly on the right track, you abandon the algebra right when you're about to solve the entire problem. It feels very unsatisfying - like, if you were just gonna show us a Wolfram Alpha solution and a Desmos graph instead of doing the problem fully, then what was the purpose of you making this video? I hope this doesn't come across too harsh. I'm just trying to make the videos more structured in a way that we know what we are supposed to take away from each one.

[@SyberMath]

Good feedback! Thanks. I can be all over the place sometimes! (Did I say sometimes? 😂)

[@zunaidparker]

@@SyberMath indeed🤣 And I'm someone that really needs structure to be able to follow along otherwise I get lost!

[@ВалерийХарченко-ш5д]

1:20 Error: right side does not change sign after reducing x^5 Ошибка: правая часть не меняет знак после сокращения х^5.

[@matufika]

Bro you forgot that it doesn't matter. If you put everything from the right side to the left one (by additions and subtractions), they do change sign. Also, if you multiply (or divide) the whole equation by -1 (which is an equivalent transformation), the signs are changed. Now, as a=b and b=a mean the same thing, the newly written equation is totally correct again.

[@alessiomazza5166]

If u cancel out the exponent u'll have x=x-1, x goes away, 0=1 so it's inpossible in R

[@28aminoacids]

Let c be a complex fifth root of 1. Then x=c(x-1) => x(1-c) = -c => x = c/(c-1) Now just replace c with some complex roots.

[@ИльхамАбдуллаев-ь6й]

Hello Syber Math.It's very interesting equation.May you solve this equation 2sinx+3tgx=5?

[@hoangphat3090]

you solve it very complicated. I solve by some step step 1 i use newton's binomial to develop (x-1)5 step 2 I will derivative the equation and prove the equation has no solution by y=0 doent cut y=-5x4+10x3 -10x2+5x-1 hope you see my comment i from vietnam

[@lightyagami1752]

x^5 = (x-1)^5 implies x = z(x-1) where z is a fifth root of unity. Reject z = 1 as this gives x = x-1, so there are no real solutions. The remainder are all complex solutions with z = exp(2nπi/5), n = 1,2,3,4.

[@haakoflo]

This would work if the exponent was 6, but not so with 5, since exp(2pi/5) - 1 is not exp(4pi/5) (unlike exp(2pi/6) -1 which is equal to exp(4pi/6) ). Instead you need to look for z-values where z^5 ends up on the imaginary axis.

[@lightyagami1752]

@@haakoflo My solutions are all fine. Maybe you misunderstood because I didn't show all the work. x = z/(z-1) where z = exp(2πi/5) or z = exp(4πi/5) or z = exp(6πi/5) or z = exp(8πi/5). Calculate x for each possible z, those x values are your solutions.

[@haakoflo]

@@lightyagami1752 You're right, I misread your first post as giving z as the solution, not z/(z-1)

[@lightyagami1752]

@@haakoflo Yeah, I was going for brevity, I'm doing this in my head straight to my phone lol. Cheers.

[@asparkdeity8717]

It’s actually a fourth degree polynomial with all complex roots, as there can’t be any real ones taking the fifth root inverse operator to both sides yields no solutions

[@viktor-kolyadenko]

x^5 - y^5 = (x-y)(x^4+x^3*y+x^2*y^2 + x*y^3+y^4). The 4th degree equation can already be solved in theory :) Yep, it's much easier.

[@kmsbean]

anyone find a way to factor 5x4-10x3+10x2-5x+1 = 0 ? I could google the quartic formula but I already know I don't want to have to use it

[@prabhagupta6871]

Take fifth root on both sides 🤯

[@mariocaroselli9665]

La chiacchierata è sminuita e controllata si esprime un servizio ?

[@raminrasouli7565]

How about imaginary Solutions?

[@martinrosol7719]

Math is the most accurate science. Also math: so this looks kinda symmetrical, so maybe we divide it by x^2...

[@Chriib]

One would think that A^C=B^C would imply that A=B but that is not the case here.

[@poppylikecats]

(-1)^2 = (1)^2 but -1 =/= 1

[@moeberry8226]

It would only imply A=B if the statement was always true to begin with. For example if it were (2x)^5=(x-1)^5 then we can say x=-1 but it still doesn’t mean 2x=x-1 for all x. It just means if there is a solution then it’s for some value or values of x. It won’t always be for all values of x. The difference is when A=B for all x that is called an identity.

[@sametkar6565]

where are u from i am really curios

[@samarthchohan106]

Those solutions will give me nightmares for some time

[@ardiris2715]

Avoid engineering then. (:

[@samarthchohan106]

@@ardiris2715 whenever i(square root of -1) gets involved,then the solution becomes scary

[@ardiris2715]

@@samarthchohan106 That is when the fun begins. Getting those i's to collapse to a real. (:

[@samarthchohan106]

@@ardiris2715 fear of the unknown

[@darkknightmax4818]

why do we can't log() everything ?

[@jenssletteberg3974]

All good, but how can there not be five roots to a quintic equation? Which root is double?

[@SyberMath]

This is not a quintic. x^5 cancels out

[@jenssletteberg3974]

@@SyberMath Indeed. Easy to miss, because it looks so very 5th degree, like the 5th order version of the problem. Reminds you that an equation can "fake" being more than it is.

[@framexor1186]

f(t)=t^5, then f(x)=f(x-1), but we know that f’(t)>0. In this case we can say that equation is working only when x=x-1, 0=-1. So we cant have solution in real numbers ever.

[@mcwulf25]

I could see these were two parallel curves, increasing apart from a stationary point. So there's no real solution.

[@alphalunamare]

neat!

[@notgeorge5697]

Both sides is on the power of 5 so you can delete that, and u have x=x-1 but then u have x-x=1 which gives us 0=1. That’s not possible so there aren’t real solutions.

[@SyberMath]

right

[@davidmagdalena8149]

a^5 = b^5 => a=b, so x=x-1, which isn't possible, therefore there are no real solutions

[@giuseppemalaguti435]

L'equazione di 4 grado dà 4 soluzioni che risultano da: x^2-x+1/2-sqrt(1/20)=0 e x^2-x+1/2+sqrt(1/20)=0

[@vaarmendel1657]

Yes I can

[@hans-joachimdreher2287]

In the set of real numbers: Powers with the same exponent lead to x = x - 1. No real solution.

[@virendraverma5855]

Binomial expansion 👍

[@YTSparty]

My simple solution "There is no real number solution" QED

[@SyberMath]

hmm 😁

[@anthonytesla8382]

X⁵ = (x - 1)⁵ X = X - 1 0 = -1 Therefore no real solutions

[@barakathaider6333]

👍

[@adeelalqershl7831]

Newton's polynomail

[@jeriesakroush7763]

Why not 5th root both sides then go from there. Youll get 0=-1 which is a false statement then you know there’s no solution

[@SyberMath]

Right!

[@davidbrown8763]

Exactly! The fact that the x's cancel on both sides confirms that the equation does not exist. It is impossible to "solve for x".when x has disappeared!.

[@CombustibleL3mon]

I did it by the second method :)

[@MuenchnerSaiten]

"The mapping: x->x^5 is bijective. This implies that f(x) is not equal to f(y) if x is not equal to y. (Note: represents not equal). Since x-1 and x are always not equal, (x-1)^5 can never be equal to x^5, the equation has no solution. Why is x^5 bijective? The derivative is 5x^4, so the graph is only increasing. x^5 also tends towards positive or negative infinity at the edges.

[@eddietime1811]

Expand the polynomial and use Quartic equation … NEXT

[@hovnovamjepomymjmenu4710]

That "z" is no "zii", it is "zet".

[@SyberMath]

To zee or not to zee! 😜😁

[@АндрейАнцышкин]

Не может быть такого❗ это всё равно что х=х-1.

[@antoningarcic471]

No x satisfies this equation!

[@ناصرالعتيبي-ك1و3ب]

X=0

[@matheushenriquegamer4914]

4

[@nol2521]

yes 4

[@choreboy3906]

Nope.

[@FM-yq8yfXYZ]

The equation is wrong as 1=2 is wrong. Why all this analysis?

[@SyberMath]

Right

[@archilarkania7203]

There is no complex fifth root either, because it's odd power and complex roots comes in pairs. One ,,root" doesn't exist, because fifth powers cancel out each other naturally so they're not part of the equation.

[@MrArcan10]

how to know that Z should be exactly like in method 2? It is very smart but not obvious for average people.

[@SyberMath]

Z is in the middle (it’s the average) so we’re taking advantage of symmetry