As a high school student this is the first problem from your channel that I can solve hahaha, thanks for the help, each day growing stronger in mathematics because of you
But that wouldn't give us the solution x=0. Of course, we could say that multiplicative cancellations equal to 0. But some people, mainly children, won't understand.
@@hasanjakir360 Yeah well, what I mean is that we're talking about a 28th degree equation while it's the exact equivalent of a 21st degree one. You could as well multiply both sides by x^1000 and say we're solving a 1028th degree equation. But that sounds just weird to me.
x^7 = { 0, 1, 1 ± √3) } So, there are 4 real roots and 6 times of that complex conjugate (3 pairs of complex reciprocals). w^n = e ^ i(2*π*n/7) where n = { 0, 1, ..., 6 }. Fun note - 7 of the roots are equal to 0 (1 real 0 and 6 complex conjugate zeros).
Take the four real solutions and multiply the set by the seven complex roots of unity and there you get 28 solutions (7 of them are degenerate, namely the ones with magnitude 0)
2 x power7 + x power 28 = 3 x power 21 ( bring one x power 21 to the left , take 2 x power 7 to the right) X power 28 - x power 21 = 2 x power 21 - 2 x power 7 X power 21 ( x power 7 - 1 ) = 2xpower 7 (x power 14 - 1) ( x po7 gets cancelled) x power 14( x power 7 - 1 ) = 2 (x power7+1)( x power7 - 1 ) xpower 7 - 1 gets cancelled xpower 14 = 2 ( x power 7 + 1) xpower 14 - 2 x power7 - 2 = 0 Let y = x power7 y square - 2 y - 2 = 0 Apply quadratic formula y = 2 + or - square root of ( 4+8) divided by 2 = 2 + or - root 12 divided by 2 ( root 12 = 2 root 3) Take 2 outside and cancel 2 ( 1 + or - root 3) by 2 Hence y = 1 + or - root 3 x power 7 = 1+ or - root3 x = 7 th root or 1 + or - root 3
@@SyberMath sure! I think odd roots of negative numbers always have one negative (and real) number as a solution (and, of course, other complex answers). Only even roots of negative numbers have no real solutions.
I believe all four real solutions each belong to a family of solutions in the complex plane except for 0, which has neither a real part nor a complex part. Does this mean 0 is a septuple root (a root with multiplicity 7)?
Why diddn't you consider the other 5 root of unity? I mean x^7=1-> x=e^(i2n Pi/7) with n in {0,1,2,3,4,5,6}, giving 7 solution instead of just 2 Sale thing for 1+- sqrt(3)
I solved this problem for X=1 2x^7+x^28 = 3x²¹ Taking x^7 common on both sides x^7(2+x⁴) = x^7 (3x³) Cancelling of x^7 on both sides 2 + x⁴ = 3x³ -3x³ + 2 + x⁴ =0 As we know if the sum of a,b,c = 0 Then X=1 -3+2+1=0 So X= 1
Let y = x^7 y^4-3y^3+2y=0 =>y=0=>x=0 y^3-3y^2+2=0=>y=1=>x=1 (y-1)(y^2-2y -2)=0 y=(2+-sqrt(12))/2 y= 1+-sqrt(3) x= (1+-sqrt(3)) ^1/7 Now i can watch the video😂😂😂😂
@@SyberMath Well, what makes French smooth is what makes it difficult: the pronunciation. So, his name is like giving a kiss (mwah) but with a extra v at the end.
@@angelmendez-rivera351 ok so you are the one with whom I got into an argument about whether 0 is a natural number or not in Michael Penn's channel. I still say 0 is not a natural number
@@The_Math_Enthusiast You can say that all you want, but you will never manage to change the mathematical consensus, or the ISO 80000-2, both of which very plainly say that you are wrong. So I will leave it at that.
Many math geeks say that higher degree polynomials are almost unsolvable, what we see in this video ;-) The best in all of that is "Octovigintic" word ;-)
@@SyberMath Whenever there is nothing, the world functions how it should, hopelessness and despair is the only way to move forward, thus 0 is the only un-trivial answer for this equation.
Why is it unvigintic if there’s a 28th degree
It should be octovigintic! 😂
@@SyberMath Hahaha awesome 🤣
@@Happy_Abe Don't you love those names? 😂
@@SyberMath Solving a centillion degree equation next!😀
@@EtemKaya Türk gibi konuşuyor değil mi? Daha önce sormuştum ama cevap alamamıştım.
As a high school student this is the first problem from your channel that I can solve hahaha, thanks for the help, each day growing stronger in mathematics because of you
Oh, thanks! Glad to hear that! 😊
I found it satisfying to follow you through this, but also weird that we didn't just divide by x^7 as first passage
You're right! We could've
But that wouldn't give us the solution x=0.
Of course, we could say that multiplicative cancellations equal to 0. But some people, mainly children, won't understand.
Dividing by x^7 would have been incorrect, since you lose solutions doing this.
@@angelmendez-rivera351 but you could state that first solution (which is actually obvious) and then divide looking for the other solutions
@@hasanjakir360
Yeah well, what I mean is that we're talking about a 28th degree equation while it's the exact equivalent of a 21st degree one.
You could as well multiply both sides by x^1000 and say we're solving a 1028th degree equation.
But that sounds just weird to me.
x^7 - 1 being a divisor of
2x^7 +x^(28) = 3x^(21), one gets
x^7 (x^(21) - 3*x^(14 ) +2) = 0
or x^7 (x^(21) - x^(14)
- 2*x^(14)+ 2*x^7 - 2*x^7 +2) = 0.
or x^7 ( x^7-1)(x^14 - 2*x^7 -2) = 0
Now x^14 - 2*x^7 -2= 0 implies
(x^7-1)^2 -3 = 0
or (x^7 -1-√3)(x^7 -1+√3)= 0
Hereby x^7= 0, 1, 1-√3,1 + √3 completes the solution set for x^7
or x^7 = 0, 1
Next on SyberMath: Solving an duotrigintic Polynomial equation
No, we godda do a centic first! 😁🙃
Please solve a centillion polynomial equation next
Solve: x^32-x^31 = 0
x^31*(x-1)= 0
x = 0 and x = 1 are the only real or complex solution.
Done.
😬
x^7 = { 0, 1, 1 ± √3) } So, there are 4 real roots and 6 times of that complex conjugate (3 pairs of complex reciprocals). w^n = e ^ i(2*π*n/7) where n = { 0, 1, ..., 6 }. Fun note - 7 of the roots are equal to 0 (1 real 0 and 6 complex conjugate zeros).
That's right! It's octovigintic after all! 😁
fun fact : De Moivre is pronounced kind of like duh maw-ah-vr-uh (with a French r)
10:24
There would be a total of 28 roots to this equation
Yes, including the repeated roots
Awesome video Syber! Love seeing your channel grow
Thank you! Same here. You make good videos! 😊
@@SyberMath Thank you so much
@@MathElite You're very welcome!
It's funny that you would get 28 roots (multiple roots in some cases) if you wanted to work in complex world
Yes it is!
Take the four real solutions and multiply the set by the seven complex roots of unity and there you get 28 solutions (7 of them are degenerate, namely the ones with magnitude 0)
Putting the "fun" in the fundamental theorem of algebra
@@morbidmanatee5550 When I read degenerate, I thought you're talking about me. I'm definitely not sober.
@@AnnXYZ666 lol. No in mathematics and physics it means an identical solution :)
fancy way of solving a cubic and finding seventh roots of unity
Thanks
2 x power7 + x power 28 = 3 x power 21 ( bring one x power 21 to the left , take 2 x power 7 to the right)
X power 28 - x power 21 = 2 x power 21 - 2 x power 7
X power 21 ( x power 7 - 1 ) = 2xpower 7 (x power 14 - 1)
( x po7 gets cancelled)
x power 14( x power 7 - 1 )
= 2 (x power7+1)( x power7 - 1 )
xpower 7 - 1 gets cancelled
xpower 14 = 2 ( x power 7 + 1)
xpower 14 - 2 x power7 - 2 = 0
Let y = x power7
y square - 2 y - 2 = 0
Apply quadratic formula
y = 2 + or - square root of ( 4+8) divided by 2
= 2 + or - root 12 divided by 2 ( root 12 = 2 root 3)
Take 2 outside and cancel
2 ( 1 + or - root 3) by 2
Hence y = 1 + or - root 3
x power 7 = 1+ or - root3
x = 7 th root or 1 + or - root 3
Another great explanation!
Glad you liked it!
What a peace you get when u solve this type of problem without seeing the solution
Absolutely!
@@SyberMath I have a que for you such that :x^5+x^4+1=0 then find real values of x .
Hey, wait! Isn't the solution of the form " ⁷√(1 - √3) " also complex? Cubic root of 3 is larger than 1!
7th root of a negative number is a negative real number. Think about 7th root of -128
@@SyberMath sure! I think odd roots of negative numbers always have one negative (and real) number as a solution (and, of course, other complex answers). Only even roots of negative numbers have no real solutions.
Amazing !
Thank you!
Excellent!
Many thanks!
I believe all four real solutions each belong to a family of solutions in the complex plane except for 0, which has neither a real part nor a complex part. Does this mean 0 is a septuple root (a root with multiplicity 7)?
7:35 shouldn't it be +2 instead of -2?
No, -2(u+1)=-2u-2
@@SyberMath ah i see i somehow tought you were multiplying -2 and (u-1)
4:06 i believe his name is spelled more like "de muavr"
Hey, firstly you could take x^7 common from both sides, then it would be easy to solve, and there no need to put x^7 = u
Why diddn't you consider the other 5 root of unity?
I mean x^7=1-> x=e^(i2n Pi/7) with n in {0,1,2,3,4,5,6}, giving 7 solution instead of just 2
Sale thing for 1+- sqrt(3)
Yeah, I just showed what one of them is going to look like. Too lazy to consider all 28 roots! 🤣
I solved this problem for X=1
2x^7+x^28 = 3x²¹
Taking x^7 common on both sides
x^7(2+x⁴) = x^7 (3x³)
Cancelling of x^7 on both sides
2 + x⁴ = 3x³
-3x³ + 2 + x⁴ =0
As we know if the sum of a,b,c = 0
Then X=1
-3+2+1=0
So X= 1
Nice!
I look at it and I see 1 and 0 instantly
Nice!
Nice one👏👏
Thank you!
@@SyberMath damn you're so good.....
Thanks!
2:09
You said that 1 of the solutions is 0, but in fact - according to the fundamental theorem of algebra - 7 of the are 0 lol
PS. nice video
Yes, coinciding roots
Thank you!
@@SyberMath Who has birthday? 😆😆
First of all at first glimpse 1 and 0 are 2 roots
Shri Dharacharya formulae
May be some where you find
Shreedhara Acharya's formula
What is it?
@@SyberMath google it bro plz and made a video on this
@@SyberMath it is basically the quadratic formula .... In india it is commonly referred to as shreedharacharya formulae
Finally one I can solve:
x=1,x=0,x=[1±sqrt(3)]^(1/7)
I got the solution to it with my Texas Instruments Nspire CX CAS calculator. Thanks to you for a good problem and T.I. for the lazy mans solution
Oh, you're welcome!
If we go by the logic of complex roots of 1,then any number n can be expressed as n x 1 and we'll get infinite complex roots!
Good solution
Thank you!
why degrees are in french?????
My goodness !!
In which class will we have to read such problems?
I would say high school level but it really depends on the curriculum
Amazinng problem !!!! 0:56 to youuuu or you toooo 😁😊
Happy Birthday!!!!
🍰🙃😁
@@SyberMath 🤣🤣🤣🎊🎊🎉
damn i love our videos....................................................................................
Can u try like some good problems from tough exams like jee,Putnam,etc
I will take a look!
Yes , that's a great idea
4:30 this is why complexe numbers are dumb
Haha! I agree! 😁
"1"
The answer is 1 by observation.
Güzel video
Tesekkurler!
I like to watch your channel.👍
Glad to hear that
0:55 😀😀
Happy Birthday!!!
🎂😁
Damn I like your videos,............................
x=1!!!!!!!!!!! Sorry I should have been patient.
Np 😁
What does S.S. mean?
Solution set
@@SyberMath ahh ty, we always use that fancy L for "Lösungsmenge" which is German for area of solution (probably not rlly I translated 1 to 1)
I can't understand it perfectly.
Because i am 6th grade and not use English for our main language
And yet you still watch it! 👍👏
Lol crazy but method to the madness
Crazy is our middle name! 😁
@@SyberMath Yes! Lol
Euler = oil + R
Definitely! That's how you say it!
Let y = x^7
y^4-3y^3+2y=0 =>y=0=>x=0
y^3-3y^2+2=0=>y=1=>x=1
(y-1)(y^2-2y -2)=0
y=(2+-sqrt(12))/2
y= 1+-sqrt(3)
x= (1+-sqrt(3)) ^1/7
Now i can watch the video😂😂😂😂
de Moivre is pronounced as [de MWAHV]
That's really hard! 😁
@@SyberMath Well, what makes French smooth is what makes it difficult: the pronunciation. So, his name is like giving a kiss (mwah) but with a extra v at the end.
@@XAE-yc9rr Very affectionate! 😁
جيد جدا شرح رياضيات الله يوفقك فقط ممكن سؤال اي مرحله هذه مواضيع تدرس please the exm for any class sorryEnglish not very good
You're welcome! Thanks for the good wishes! This is more like high school/math competition level
@@SyberMath thank you very mach
Noice
X = 1
That's it?
OMG wow
😊
X can be 1 and 0
You can make that X^7=y and resolve this equation....
x = 1 or 0
No other solutions?
In fact, seventh root of (1-sqrt(3)) is not real, it is complex one. Therefore, may be omitted from the solution set.
7.
No
Why are you so obsessed with higher degree polynomials?
I don't know
Because they are fun. Being able to solve higher-degree polynomial equations is also an important skill in mathematics.
@@angelmendez-rivera351 ok so you are the one with whom I got into an argument about whether 0 is a natural number or not in Michael Penn's channel. I still say 0 is not a natural number
@@The_Math_Enthusiast You can say that all you want, but you will never manage to change the mathematical consensus, or the ISO 80000-2, both of which very plainly say that you are wrong. So I will leave it at that.
Many math geeks say that higher degree polynomials are almost unsolvable, what we see in this video ;-) The best in all of that is "Octovigintic" word ;-)
0
Is that the only one?
@@SyberMath Whenever there is nothing, the world functions how it should, hopelessness and despair is the only way to move forward, thus 0 is the only un-trivial answer for this equation.
1,0
That's right!
Totally not obvious substitution
Really?
It was obvious