This is a very convoluted way of solving it and presenting the roots... Since x^5 = 1 all the roots need to be on the unit circle... And multiplication of complex numbers that are on the unit circle is just rotation its obvious that all the solutions are e^(i*2*pi*n/5) where n is a natural number...

@kappascopezz5122

Жыл бұрын

I had the same thoughts, but what's kind of nice about this way of solving it is that you don't need to know the values of sin and cos to be able to write the solution

@BKKGarrett

Жыл бұрын

I have no idea what you're talking about. No student who has gone through the standard Alg 2 - PreCalculus - AP Calc progression is going to know what you're talking about. That's why I love this guy's videos. He can make anyone who has had just Alg 2 (usually not even PreCal) understand how to find the solutions to interesting problems.

@kappascopezz5122

Жыл бұрын

​@@BKKGarrett The comment obviously isn't explaining Euler's formula, and I'd expect that to be because the comment was directed at the video's author, not some people defending them without making an attempt to read the comment. Knowing the calculation rules of the polar form of complex numbers as well as Euler's formula are very basic facts once the concept of complex numbers has been introduced, so the method being too complicated is a bad justification. If the video was made with the assumption that the viewer already knows complex numbers but somehow managed to skip the polar form, as seems to be your claim, then it just would've needed to teach you for 3 minutes and the video would've been finished in 5 minutes. Of course, that doesn't mean that there can't be a good reason for choosing such a bad method. For example, the video could be trying to show how you don't always need a good approach to solve the problem, or something. But in either case, they should've at least acknowledged the existence of an easier method before basically telling everyone they should do it the annoying way.

@justrandomthings8158

Жыл бұрын

@@kappascopezz5122 this is why mathematicians are generally considered to be antisocial. What an ignorant, hand wavy comment

@penteractgaming

Жыл бұрын

​​@@BKKGarrett algebra 2 students would be lucky to know that there are 5 roots let alone anything more than that. hell do they even know about complex numbers?

На множестве R действительных чисел, функция f(x)=x^5 возрастает, значение f(x)=1 принимаeт 1 раз, корень x=1 единственный. На множестве C комплексных чисел, уравнение x^5-1=0, 5 степени имеет 5 корней. Это 5 точек на окружности, с центром в (0;0) и радиуса r=1, образованные поворотом на угол a=360°/5. Спасибо.

@-wx-78-

Жыл бұрын

Не пугай иностранцев градусеями, решат что всё меряем водкой. 😉 2π/5 и все дела.

@AlexeyEvpalov

Жыл бұрын

@@-wx-78- На планшете (пи) набрать не могу, пишу (°) градусы. Лично не знакомы, пересекались в комментариях у Волкова.

@-wx-78-

Жыл бұрын

@@AlexeyEvpalov Добро пожаловать в тёплую компанию! Вроде с доп. клавиатурой андроиды умеют греческие, а я себе соорудил спец. драйвер клавы и повесил с помощью AutoHotKey на правый альт массу полезного (заодно и на левый, и вообще по всей клаве). Издержки способности программирования.

@AlexeyEvpalov

Жыл бұрын

@@-wx-78- Спасибо.

@nikolaymatveychuk6145

Жыл бұрын

О, русское комьюнити под английским видео. Прикольно. :)

Nice solution! My approach would be in complex numbers modulus-argument. As a root of unity, any solution of the 5th root of 1 must be of modulus. 5 times the argument must be a multiple of 2pi, so the argument of the solution could be 0, 2pi/5, 4pi/5, 6pi/5, or 8pi/5. The solution with argument 0 is obviously 1. For the rest we could use de Moivre’s theorem where the solution is cos(arg) + i sin(arg). The tricky part then becomes to apply trigonometric formulae to obtain the sine and cousins of 2pi/5 or 72 degrees. I believe this will also involves polynomial equations of degree 5.

@mattoucas869

Жыл бұрын

My solution is: X = 1 Lol

@andyrobertshaw9120

Жыл бұрын

@@mattoucas869 Really? That pun was intended :)

@camgere

Жыл бұрын

Complex numbers are the way to go. There are many videos on "nth roots of a complex number".

@chriswatson7965

Жыл бұрын

That is the standard method. As soon as you need to start looking for complex numbers in the solution you should use the techniques that we have in that area. I thought the solution provided was painful.

@oahuhawaii2141

Жыл бұрын

The sine of 18° and 54°, or the cosine of 72° and 36°, can be found with a quartic or cubic polynomial equation in x; the extraneous non-surd roots can be factored out by inspection to yield a quadratic equation with the solution. With the sine, it's easy to find the cosine, and vice versa.

interesting way to derive the 5th roots of unity without any complex numbers. Of course this doesn't work in general, 5 is one of a few special cases

many people mention complex numbers to solve this, but when i was doing this level of math at school, we wouldn't have started complex numbers at that point, but we would be doing algebra to this level. complex numbers, did come but after this. so solving using algebra is very useful to show it is possible, before moving to more advanced techniques.

Here is another solution: Since the roots are in the unit circle and except for x=1, you can write x^4+x^3+x^2+x+1=(x^2+ax+1)(x^2+bx+1), as the independent term is equal to the product of the roots, and their absolute value is equal to 1, the main term is 1 and we have four complex roots.

Клево. Лично я x^5=1 тупо на комлексной плоскости разрисовал. Потому как не силен в алгебре. А тут отлично корни сели в вершинах правильного пятиугольника. Забавно, что решая x^5+1=0 пятиугольник уже не такой красивый)

Literally did this in class today, what coincidence (or youtube just reads my mind and reccomends videos accordingly)

can't you just do x^5=1 =>x=1 ?

@sly6386

Жыл бұрын

Yes 💀 💀 💀 (Keep In mind, because it's ^5 it's positive one, if it was ^4, we would need to put +-1 because (-1)^4=1)

@suryavanapalli2536

Жыл бұрын

Degree of the polynomial equation decides the total no. of roots of an equation. So here, the power is '5' so the roots should be 5 roots (either imaginary or real).

@youssefchihab1613

Жыл бұрын

@@suryavanapalli2536 Thanks for clarifying! but you can still probably find all the numbers from there right ? (x^5=1)

@suryavanapalli2536

Жыл бұрын

That's what he did in the video... He found x-1=0 first and later the rest of 4 roots with it.

@Sky-pg6xy

Жыл бұрын

@@suryavanapalli2536De Moivres theorem

почему я в калькуляторе комплексных чисел возвожу результат в 5 степень и не получается единица? x^5 при любом найденном значении кроме 1 получается около 32.

You made an error for the 4 complex roots. The denominator should be 4 instead of 2. You can verify that the magnitude of each x must be 1. I used the polar form, and got the 5 solutions in the form of e^(i*θ) = cos(θ)+i*sin(θ). I happen to know the sine and cosine of all multiples of π/10 or 18°, so I substituted the relevant numeric expressions. x^5 - 1 = 0 x^5 = 1 = e^(i*2*π*k), k ∈ I x = e^(i*2*π*k/5) We have 5 unique solutions with k = 0, ±1, ±2 . I convert 2*π/5 to 72° for convenience in notation. k = 0: x = e^0 = 1 k = ±1: x = e^(±i*2*π/5) = cos(±72°)+i*sin(±72°) = cos(72°)±i*sin(72°) = sin(18°)±i*cos(18°) = [√5-1±i*√(10+2*√5)]/4 k = ±2: x = e^(±i*4*π/5) = cos(±144°)+i*sin(±144°) = cos(144°)±i*sin(144°) = -sin(54°)±i*cos(54°) = [-√5-1±i*√(10-2*√5)]/4 Notes: sin(18°) = cos(72°) = (√5-1)/4 sin(36°) = cos(54°) = √(10-2*√5)/4 sin(54°) = cos(36°) = (√5+1)/4 sin(72°) = cos(18°) = √(10+2*√5)/4

Subscribed… ❤️ from 🇮🇳

Как-то всё сложно. Проще это сделать по элементарному математическому закону, что: один в любой степени это один. Ну и просто: Х^5-1=0 Х^5=1 Х=1 Проверка: 1-1=0 1=1 1+0=1 И да, тут в комментариях говорят, что на месте Х может быть много числе, которые в 5 степени дают 1. Ну... я бы сказал что на месте Х бесконечное количество чисел, которые могут в пятой степени давать 1, так что не имеет никакого смысла решать вот так сложно, как это показано в видео. Сосредаточим ответы на другие задания коллеги, а не на задании, где и так понятно что ответов бесконечное число.

@lerarosalene

Жыл бұрын

Их не бесконечное количество. Их ровно 5.

@arjunkc3227

Жыл бұрын

How you come up with infinite numbers. Keep in mind the highest degree of the polynomial tells you the maximum number of solutions you can get. Since degree is 5, you should have five solutions or less (with multiplicity)

Just draw a pentagon at the complex plane starting from 1+0i point

@oahuhawaii2141

Жыл бұрын

Regular pentagon inscribed in the unit circle on the complex plane with 1+i*0 as a vertex. The 5 vertices are the fifth roots of 1.

I type the solution on algebra calc and didn't get these result, and looking for a why, I found a mistake. When you multiply or divide by x the entire equation, you inject more solution from the original equation, then your solution are right on x2+x+1+1/x+1/x2=0 but not from the x4+x3+x2+x+1=0

@oahuhawaii2141

Жыл бұрын

His denominator is 2, but it's supposed to be 4. The magnitude of each solution is 1, but his complex ones have magnitude 2.

Можно уточнить? Нужно найти x, или её составляющие? Почему все так горячо спорят и доказывают, что есть реальный простой ответ, а есть множество правильных?

Wouldn’t it be simpler to solve using polar coordinates in the imaginary plane. X^5 = 1 => x_k = cis(k*2*pi/5) where k is from 0 to 4?

@antenym8947

Жыл бұрын

Yes and the formula would be x_(n+1)=1^1/5(cos(0+n*2pi/5)+sin(0+n*2pi/5)i, nez & 0

@oahuhawaii2141

Жыл бұрын

The problem is you're left with sine and cosine functions, whereas it's possible to express the complex solutions with terms in the forms of (√5±1)/4 and √(10±2*√5)/4 .

Я не очень понимаю разве правильно не будет x^5-1=0 x^5=1 И исходя из обычной математической логики того что один в любой степени это один, ответ разве не один? Зачем так мудрить?

@Skibitskiy

Жыл бұрын

Затем чтоб найти все корни а не только еденицу.

@alexandermorozov2248

Жыл бұрын

«один в любой степени это один, ответ разве не один» - один ответ один, но изюминка в том, что есть и другие числа, которые в пятой степени тоже дают один :) То есть, если один в пятой степени один, то это не значит, что только один в пятой степени равно один :)

@user-no8te7rv7t

Жыл бұрын

Потому что любое уравнение имеет столько корней, уровнением какой степени оно и является. Если бы х был в 6й степени, уравнение имело бы 6 корней. Как то так)

@user-wi4vs6xc3z

Жыл бұрын

@@alexandermorozov2248 Ну, проблема ещё в том что под Х можно подставить ООООООООООчень много чего, что в 5 степени - 1 будет давать ноль. Например корень по основанию 5 из 1 в 5 степени:) так что можно просто сказать, что на месте Х может быть бесконечное количество числе, да и всём будет спокойнее и проще просто подставить 1, так как в условии, а в вернее его отсутствии ничего не сказано про то, сколько именно чисел Х должно быть. Так что, я бы просто сказал что Х в 5 степени = 1, так как по факту, так оно и есть (элементарная математика). Это проще и понятнее всего. Ну или простое доказательство: х^5-1=0 X^5= 1 - 0= 1 X=1

@ДедВездесущий

Жыл бұрын

Нет ни одного числа в мире кроме 1 которое дало бы в степени пять 1. Есть константа i которая придумана для таких заумных, нах не нужных решений. Да здравствует 2*2=5???

the number in the denominator for x2, x3, x4, x5 is 4, not 2

To be honest, I don't understand complex numbers, since they weren't part of school program and were only briefly touched in university progam, so I don't understand how "i" appeared in four other answers. So yeah, this video is interesting, but doesn't bring complete understanding. How does the graph of this function "y = x^5 -1" look like? If it touches y=0 line five times.

@JohnSmith-nx7zj

Жыл бұрын

You can’t graph complex numbers on an x-y plot because complex numbers are already a plane.

@Awesome-ct7vr

Жыл бұрын

Only real numbers can be plugged on an axis. This exact example of y=x⁵-1 would like a flat third degree function originating from y=-1 And would intersect x axis at x=1 Lets explain imaginary number: To explain it we first need to understand the base. I mean the very simple base. Axis of consecutive numbers like first grade. 1 2 3 4 5 6 7 8 9 ... •-|-|-|-|-|-|-|-|-|->... That is the first dimension of numbers we learn about as kids. After that. We learn that nothing can exist and is called 0 0 1 2 3 4 5 6 7 8 9 ... •-|-|-|-|-|-|-|-|-|-> ... And then we learn about negative numbers which exist in the opposite direction of the axis. -5 -4 -3 -2 -1 0 1 2 3 4 5 ...... Now if you look at this axis you can see that if it were to be a radi axis the negative numbers exist in perfect symmetry from their positive relative. In a 180° turn So 1 on the axis at 0° is +1 But 1 on the axis at 180° is -1 So for this demonstration we will use: 0° --> (+) 180° --> (-) Notice that as we keep spining our signs will meet again. 0° and 360° represent same angle on our axis. So 1 at (360°) is also +1 Now lets apply this to basic math. 1(0°) + 1(0°) meaning both going towards the right so we move 1 on the axis 0 --> 1 and + another movement towards 0° is 1-->2 Now you can guess it work the same the other way. Now lets try multiplication. 2 × 2 = 4 ---> 2(0°) × 2(0°) = 4(0°+0°) 2 × (-2) = -4 --> 2(0°) × 2(180°) = 4(0°+180°) You can see that in multiplication we add those degrees of the 2 numbers together and get a final degree for the result. Last test for multiplication and we go on to the actually intersting part about all of this which is the imaginary i that we waited for. (-2)×(-2)=4 obviously minus times minus is a + But look at that 2(180°)×2(180°)=4(360°) As we said earlier 360° is the same axis represntation as 0° there for a + Now when dealling with roots (which is where imaginary number come from) We get 2 answers per rooted number. Therefor we will split the angle into 2 1 positive and 1 negative number For example: Sqrt 9 is 3 and also -3 So Sqrt 9(0°) = 3(0°/2) But 0/2 is still 0 so where is our negative solution? Here is the magic. Again 360° is the same as 0° Sqrt 9(360°) = 3(360°/2) is 3(180°) Which is -3. At last lets see what happens when we sqrt a negative number. Sqrt 9(180°) = 3(180°/2) which is 3(90°) So what is 90°? What happened is a new dimension was created. A dimension of imaginary numbers 90° -|-3i -|-2i -1 -2 -|-i 1 2 180° 0°/360° -|- (-i) -|- (-2i) -|- (-3i) 270° So we got 3(90°) which is now we know 3i But what about the 2nd root? 9(180°) if we do another full 360 spin we are back to negative but at 540° So 9(180°) is also 9(540°) And root of 9(540°) is 3(540°/2) Which is 3(270°) which on the axis is the negative imaginary number. There for sqrt (-9) is 3i , -3i And thats how we get imaginary numbers

@sytrostormlord3275

Жыл бұрын

comples numbers just base on asumption that i^2 = -1 and that any number has it's real and complex part and could be presented as: a + b*i so X^2= -1 has no real solutions, but has 2 complex solutions: 0 + i and 0 - i With complex numbers every equation has exacly n solutions, where n is the highest power in equation. Some of them might be real, but don't have to. when solving ax^2 +bx +c = 0, if you got delta x^2 +x +1 =0 a=1,b=1,c=1 delta = b^2 -4ac = 1 -4 =-3 x1 = (-b -sqrt(delta))/2a x2 =(-b + sqrt(delta))/2a so x1 = (-1 - sqrt(-3))/2 = -1/2 - sqrt (-1 *3)/2 = -1/2 - sqrt(3) *i /2 x2 = -1/2 + sqrt(3) *i /2 Algebra on complex numbers is simple: X = a + bi Y = c +di X+Y = (a+c) + (b+d) *i X*Y = (a+bi) * (c +di) = (ac +adi +cbi +(bd * i^2)) and since i^2 = -1 it simplifies to (ac -bd) + (ad +cb)*i These are the basics. threat "i" in all calculation as additional variable and once its to the power of 2, substract it with i^2 =-1.

Denominator should be 4 not 2 when using quadratic formula. 2x^2+(1-sqrt5)x +2=0 X=-b+/-sqrt(b^2-4ac)÷2a

@omegajj

Жыл бұрын

no because a value is the coefficient and the coefficient is 1 so 2 X 1 = 2 so the denominator remains as 2

@empathy800

Жыл бұрын

@@omegajj no, the coefficient is 2.

@omegajj

Жыл бұрын

for your equation yes but for the video no

@general_bonapart6937

Жыл бұрын

It should be 4, otherwise the magnitude of x is not equal to 1

1, e^(2πi/5), e^(4πi/5), e^(6πi/5), e^(8πi/5) where i is the imaginary unit (i^2=-1). Thus, the solutions to X⁵-1=0 are: X=1, e^(2πi/5), e^(4πi/5), e^(6πi/5), e^(8πi/5) These solutions can be written in rectangular form as: X=1, cos(2π/5)+isin(2π/5), cos(4π/5)+isin(4π/5), cos(6π/5)+isin(6π/5), cos(8π/5)+isin(8π/5)

@fambam3679

Жыл бұрын

the q is asking for surd not exponential which is why that was doesn’t work well

Good resolution, it seems that x2, x3, x4 and x5 should be divided by 4 instead 2.

@oahuhawaii2141

Жыл бұрын

Yes, all his complex solutions are twice the correct ones.

Nice trick

When you expect the solutions to be in the complex domain, you should state it.

А ‘сольв солюшн’ разве не масло масляное?

@_Diana_S

Жыл бұрын

Так это не носитель языка говорит

what about x=e^(2πi/5) solution ?

@oahuhawaii2141

Жыл бұрын

You're missing 3 other complex solutions.

X^5-1=O= integral [5x-1]/O= 5x=x^5=5 =5/1=1 *5=5= 5 x=1 tai Daiktas sudarantis modelis yera lygus 5x =5 ir neziniomassis x=1

this probelm is so fantastic thanks it;s improve my Mathematics skill

@inq2605

Жыл бұрын

Focus on your grammar

@attran720f

Жыл бұрын

​@@inq2605 sure, it not our mother language

@inq2605

Жыл бұрын

@@attran720f where are you from?

@mememaster8050

Жыл бұрын

@@inq2605 grammar*

@inq2605

Жыл бұрын

@@mememaster8050 sry i misclicked it

Quite complicated

What about this solution: x = i^(4/5) ?

@drozdzuwa6669

Жыл бұрын

Because i^(4/5)=1 and she already found that solution in the first place. For comparison, if equation x-1=35 has solution x=36, then x=6^2 is the same solution.

@suryavanapalli2536

Жыл бұрын

​@@drozdzuwa6669 well explained...👏

I thought this was stats at first where x is the proportion lol

why you have done it in so complex way? x^^5 =1 is an equation for complex numbers. x^^5 = 1(cos(2pi) + i*sin(2pi)) taking fifth degree root from this means x = 1*[cos(2*k*pi/5) + i*sin(2*k*pi/5) ]. And roots looks easier than in your case no need for a lot of algebra transformation

@andyrobertshaw9120

Жыл бұрын

COMPLEX way? I like it!

@Mason11987

Жыл бұрын

"why the complex way" instead just introduce cos/sin and pi without any explanation of why, lol.

Unfortunately, perhaps, those who do not know imaginary and complex numbers believe that there is only one solution "1" for x³=1, x⁵=1, x⁷=1. ……. They know the solutions of x²=1 are 1,-1.

x = 1 ??

@thebeastlyiceking7416

Жыл бұрын

that's just one possible answer. There are infinite

@aristotle1337

Жыл бұрын

@@thebeastlyiceking7416 only 5 roots

@thebeastlyiceking7416

Жыл бұрын

@@aristotle1337 Yes only 5 roots but infinite answers. Extraneous solutions count

Why couldn't we do x^5=1?

@almaska82

Жыл бұрын

According to the fundamental theorem of the algebra of the roots of a polynomial, there must be 5, as well as its degree.

@Mason11987

Жыл бұрын

Because there are five values that when raised to the power of 5 become 1. Just like with x^2 = 4 you can't just say "it's 2". It is, but it's also -2.

@Fofodoido

Жыл бұрын

@@almaska82 ok, only one more question: why couldn't be five "ones"

@Fofodoido

Жыл бұрын

@@Mason11987 broo, ok, thank you so much guys 🙏

Why root over of -1 is undifined

@marcusgriese5966

Жыл бұрын

Only in N, Z, Q and R. Not in C. According to C, √-1 equals i.

@youssefchihab1613

Жыл бұрын

find a real number squared that equals -1 and tell me again( real numbers squared are always positive)

@marcusgriese5966

Жыл бұрын

@@youssefchihab1613 You are disregarding C.

@youssefchihab1613

Жыл бұрын

@@marcusgriese5966 I said real before number every single time bro

@dylwhs

Жыл бұрын

If this is because you don't know what complex numbers are, here is a quick intro. In the real number line, the square root of negative one (-1) is not defined because no number multiplied by itself gives -1. We therefore have to invent a way to represent it, and that is to say let √(-1) = i where i is the complex root of negative one. And i² = (-1). The complex root lies in an orthogonal direction to the real number line known as the imaginary number line, which form a complex number plane, such that complex numbers are represented as (x + iy) similar to say, on coordinate axis, but with real and imaginary components. When you multiply complex numbers, you multiply them as you would (a+b)(c+d) type algebraic expressions, for example: (p+iq)(u+iv) =pu +ipv+iqu+i²uv = pu +i(pv+qu)-uv, or gathering the real and imaginary parts =(pu-uv)-i(pv-qu) The imaginary part has i with its scalar value in the imaginary number line, and usually appears to the right.

Moreover, xⁿ−yⁿ is divisible by x−y. Golden ratio everywhere, and fifth roots of unity.

@AlexeyEvpalov

Жыл бұрын

Рад встрече на этом канале.

@-wx-78-

Жыл бұрын

@@AlexeyEvpalov Взаимно, хоть и не припомню где познакомились. Старенький ужо. 😉

x^5-1=0 => x^5 = 1 => x=1

Можно было схемой Горнера воспользоваться

@aristotle1337

Жыл бұрын

один раз Горнер, дальше по Феррари остальные корни найти просто

@user-vd8ze3kw1q

Жыл бұрын

@@aristotle1337 да

Nobody writes, so I write clearly. The four "correct" complex solutions are {(-1-√5)±i√(10-2√5)}/4, {(-1+√5)i√(10+2√5)}/4. If you are not convinced or think that these are wrong, please check them carefully.

@oahuhawaii2141

Жыл бұрын

You are wrong on 2 issues: 1) You write without using a multiplication symbol, such as "*", and expect adjacent entities without an operator has an implied multiply. This is bad technically, especially when typing out math problems and solutions, as they don't parse well in most computer languages. 2) Your latter 2 complex solutions are missing their "±" operator, indicating an implied multiplication operator.

@user-wu9hy4lt2w

Жыл бұрын

@@oahuhawaii2141 様(sama) Certainly, to leave out "±" in 2) was due to my carelessness. However, the omission of the multiplication symbol in 1) is what is taught at school. These videos of algebra assume that we will answer them by handwriting, so if people don't misunderstand when they see them, we need not to write multiplication symbols.

أتمرن للإمتحانات في الرياضيات (مهم للتلاميذ) : kzread.info/dash/bejne/pql6z5hvm7C1g6w.html

бля столько интересных комментов, здесь точно всего пять корней?

Itu nyangkut bil kompleks

I am seriously confused here X 'to the power 5' - 1 = 0 So X 'to the power 5' must = 1 So X must be 1 (1 x 1 x 1 x 1 x 1) - 1 = 0 1 - 1 = 0 ... there is no other answer that is just a number

@user-wu9hy4lt2w

Жыл бұрын

You need to learn imaginary and complex numbers from the beginning. Ask from math teachers or people who know algebra well, or get basic knowledge from books and the Internet.

I just solved in a fraction of second x is one.😟 Did i something wrong yo maths world .??

@user-wu9hy4lt2w

Жыл бұрын

If you are a junior high school student who has not studied imaginary and complex numbers, you are not wrong about the answer. However, if you are a high school student or older, please read other people's comments carefully, get information from books and the Internet, and think about whether it is really enough to answer that much.

Na minha mente: Número que -1 é 0, é 1, cabo.

Unfortunately, this is an easy problem for those who have knowledge of imaginary and complex numbers, but not for those who do not. If you don't understand the content of this video, you need to get the basic knowledge of it, otherwise it will not only be completely useless, but it may also hinder your future learning. You should watch with that in mind. In addition, it is four solutions of complex numbers, but the denominator is 2 due to a calculation error. The correct is 4. This can be checked by anyone, even a person without knowledge of imaginary and complex numbers, who knows the formula for the solution of quadratic equations.

X=1

х=1 так нельзя?

@user-rn8zi5rt7y

Жыл бұрын

Конечно можно... В видео просто от нечем заняться начали искать значения икса с учётом комплексных чисел. Даже интересно, что же он дальше не пошёл, и не начал искать иксы на множестве Кватернионов и т.д. Вообще это просто намеренное усложнение просто задачи

@_Diana_S

Жыл бұрын

Если искать ответ только в реальных числах, то можно и нужно. Просто давным-давно Гаусс доказал, что у полиномиального уравнения количество корней равно его степени. То есть если в уравнении стоит х в 5 степени, то должно быть 5 корней, а не один. Вот и стараются, вычисляют в комплексных числах ).

even without watching the video, i confidently say that x⁵ = 1

You MUST be wrong, since ChatGPT gives 1 and (-1 - sqrt(5)i)/2 (where i is the imaginary unit). I mean, Artificial Intelligence is bound to be right, don't you agree? Try it yourself ! 😂😂😂

@oahuhawaii2141

Жыл бұрын

It seems ChatGPT instead tried to solve for x in 2*x^3 + x = 3 . That's why self-driving cars still crash and kill people.

Trivial using 1 = e^i2npi. N=0-4 divide by 5 for all roots.

x^5 - 1 = 0 Plus one on both side x^5 = 1 The put root 5 x = 1

Where is it stated that you need to consider complex numbers? From the title only the solution is obvious ; x = 1

@oahuhawaii2141

Жыл бұрын

This is his video, and he calculated all 5 solutions.

You can do like this also x^5-1=0 x^5=1 x^5=1^5 => x=1

Simply take the -1 to the rhs It becomes x⁵=1 Therefore x =1 bruh

@oahuhawaii2141

Жыл бұрын

That's only 1 of 5 solutions. You get a 20% grade.

答え　1

@user-wu9hy4lt2w

Жыл бұрын

もし、あなたが虚数・複素数を学習していない日本の中学生なら、その答でも間違いではありません。しかし、高校生以上であれば、本当にそれだけの答で良いかどうか、他の人のコメントをよく読んだり、本やネットからの情報を得て、考えたうえで、コメントして下さい。 If you are a Japanese junior high school student who has not studied imaginary and complex numbers, you are not wrong about the answer. However, if you are a high school student or older, please read other people's comments carefully, get information from books and the Internet, and think about whether it is really enough to answer that much.

Ohhh, please... X⁵ - 1 = 0 X⁵ = 0 + 1 X⁵ = 1 X = 1 / 1⁵ X = 1 Done The simple, ever is most elegant.

@oahuhawaii2141

Жыл бұрын

You get a 20% score on this problem. You'll need to do bonus work to bring up your average to get a passing grade.

@diogovasconcellossilva969

Жыл бұрын

@@oahuhawaii2141 Bullshit. This is de correct answer, any other way to see this matter is a result of constant use of maryjuana. Learn this leasson stupid padawan: the simple way is ever, ever better. Got it?

Wouldn't this be simply: x^5 - 1=0 x^5 = 0 + 1 x^5=1 x*x*x*x*x = 1 so x =1.

@e4d578

Жыл бұрын

Yes, but the question is asking for all solutions, and while x =1 is correct for real numbers, there are other solutions for imaginary numbers (i).

@kobalt4083

Жыл бұрын

watch the video, there are complex solutions

@Todor81

Жыл бұрын

​@@e4d578Imaginary number is not a number. Real number is correct answer, and imaginary answer is only for imagination

@e4d578

Жыл бұрын

It's difficult to win a debate with a smart person....but absolutely impossible to win against a dumb person.

@kobalt4083

Жыл бұрын

@@Todor81 we know that. its easy to get the basic solutions, but this video is showing how to get ALL solutions, real and imaginary. also "i" has a value of sqrt(-1), so technically its not for our imagination. the imaginary solutions presented are correct answers, but the only difference is they dont have a real value.

X is equal to 1.

@oahuhawaii2141

Жыл бұрын

You get a 20% score on this problem. You'll need to do bonus work to bring up your average to get a passing grade.

1

Все заплутали. Х в 5 степені дорівнює 1. Значить х дорівнює корінь 5 степені з одиниці , дорівнює 1.

@oahuhawaii2141

Жыл бұрын

There are 4 more unique roots, which happen to be complex numbers.

Use sign of root

Da nije možda -1?

@oahuhawaii2141

Жыл бұрын

(-1)^5 = -1 So, x^5 - 1 ≠ 0 .

(X5-1= 0), (x = 1)

Just….x=1

Add 1 on both sides and boom you have the answer

@oahuhawaii2141

Жыл бұрын

No, you now have: x^5 = 1 Where are the 5 unique solutions for x?

Why do I love Physics so much but hate maths so bad

@_Diana_S

Жыл бұрын

Because physics' math makes sense and math's math - not always.

1)

x = 1😎

Dude… 1^# is always 1…

X5

Just Pentagon

That was an extremely CONVOLUTED method of solving!! Step1: x^5 = 1 Step2: x = 5th root of 1 Step3: x = 1 or -1 Done, END OF PROBLEM

@JohnSmith-nx7zj

Жыл бұрын

x = -1 isn’t a solution. And you missed the 4 complex solutions.

@shahbazahmad8972

Жыл бұрын

this equation has 5 roots

@kobalt4083

Жыл бұрын

-1 is not even a solution because a negative number raised to an odd power is negative... plus it doesnt give u complex solutions. you need to listen to the video before commenting

@Mason11987

Жыл бұрын

You're wrong in two ways. You really shouldn't be commenting on math videos, you obviously don't understand math.

В чем смысл комплексных чисел. Зачем 4 дополнительх корня. А если корней 30? Еще 25 хировыебаных, для тех кому мало.

1^5=1 | 1^5-1= 0

1 to the power anything is 1 therefore 1-1=0 😂

Please don’t solve the equation like this. For the love of god.

@oahuhawaii2141

Жыл бұрын

Why? This is an elegant way to find the 5 roots without using trig functions.

Trash dieses vid schwöre man kann nicht durch x teilen ya ariiiiiiiii

@marcusgriese5966

Жыл бұрын

Solange 0 als Wert für X ausgeschlossen werden kann bzw. ausgeschlossen wird, ist auch das Teilen durch X erlaubt.

Х=1 Другие четыре не изобразить числом, а только формулой, по которой только приближенно. Впрочем, производные покажут,что других корней нет Счастливо, если нужно, вычислит

Oh my goodness - all this is such a complicated way to get the answer. It took me about 5 seconds and I ain’t that smart : - take the 1 to the right hand side and the equation becomes “x to the power of five =1” - the answer *must* be x=1, because 1x1x1x1x1 = 1 (it couldn’t be anything else) All these complex calculations in the video are OK for educational purposes, I suppose. However I’m more interested in getting the job done in the most efficient and timely manner ...

@fabianbotello4917

Жыл бұрын

the point of this video is obtaining the complex solution, not the trivial real solution of 1.

@oneeyedrichmond

Жыл бұрын

You're missing the 4 other (complex number) solutions. There are 5 solutions given it's x to the power of 5. The video though is a complicated way of doing it. Simpler to use the fact that 1 = e^(2 k i pi) where k is an integer. -> x^5 = 1 = e^(2 k i pi) -> x = e^(2 k i pi/5) k=0 ---> x = e^(0) = 1 k=1 ---> x = e^(2 i pi/5) k=2 ---> x = e^(4 i pi/5) k=3 ---> x = e^(6 i pi/5) k=4 ---> x = e^(8 i pi/5) We stop here as: k=5 ---> x = e^(10 i pi/5) = e^(2 i pi) = 1 ... same solution as k=0 so not another unique solution. k=6,7, etc ... similarly are repeated non-unique solutions. So, the five unique solutions to x^5 = 1 are: x_1 = 1 x_2 = e^(2 i pi/5) = cos(2 pi/5) + i sin(2 pi/5) x_3 = e^(4 i pi/5) = cos(4 pi/5) + i sin(4 pi/5) = -cos(pi/5) + i sin(pi/5) x_4 = e^(6 i pi/5) = cos(6 pi/5) + i sin(6 pi/5) = -cos(pi/5) - i sin(pi/5) x_5 = e^(8 i pi/5) = cos(8 pi/5) + i sin(8 pi/5) = cos(2 pi/5) + i sin(2 pi/5) where cos(pi/5) = (1+Sqrt(5))/4 cos(2 pi/5) = (1-Sqrt(5))/4 sin(pi/5) = Sqrt(2)*Sqrt(5 - Sqrt(5))/4 sin(2 pi/5) = Sqrt(2)*Sqrt(5 + Sqrt(5))/4

@johnnydev9318

Жыл бұрын

@@oneeyedrichmond 😵‍💫😵‍💫😵‍💫 I’ll believe you !!! 😂😂😂

@Mason11987

Жыл бұрын

"Getting the job done" means getting the answer. Not a part of the answer. If you want to half-ass something you probably aren't doing a job that actually involves 5th roots anyway.

@johnnydev9318

Жыл бұрын

@@Mason11987 Yes, well - obviously I’m well below the intellectual level of some others here - because I’ve got no idea of what that all means.

Вообще-то ответ+-1

Решение только одно x=1. Все остальное математический онанизм

Х равен 1

В чем проблема сделать х = 1. 1 в любой степени будет 1. 1-1=0

X = 1

@kobalt4083

Жыл бұрын

The degree of the polynomial, 5, implies that there are 5 unique solutions.

X = 1

@kobalt4083

Жыл бұрын

The degree of the polynomial, 5, implies that there are 5 unique solutions.

@MSN539

Жыл бұрын

@@kobalt4083 X = 1