One small nuance that I tell my students all of the time: there are not 24 complex solutions, but instead there are 25, since the real solution is also complex. There is 1 real solution and 24 non-real solutions. As always, an excellent video! Congrats on 25k!!
Let's be careful both ways. Let's avoid gotchas. So one should understand what is meant by one real solution and 24 complex solutions. One routinely refers to "integer spin" and "half-integer spin", with the understanding that the half-integer means non-integer. I don't really like saying, "integer-plus-a-half spin".
Congratulations on another milestone! Hmm, as I had no intention to deal with 25th degree polynomial, I started by being creative this time and it did its job. : ) I won't spoil all the solution, it's your part. But one little tip for solvers: it might be helpful for you if you know where a function and its inverse meet.
@@SyberMathI believe that you can show that any continuous function will intersect its inverse either not at all (as with y=e^x) or on the diagonal. However, in the latter case, it may also intersect its inverse off the diagonal as well. (E.g., the function y=(x-10)^2 will intersect its inverse in four places, two on the diagonal and two off the diagonal.) However, you can also construct a *discontinuous* function that only intersects its inverse *off* the diagonal. For example, the function defined by y = {1, x=0} will intersect its inverse at 2 locations, both off the diagonal [at (-1,1) and (1,-1)].
Another great explanation, SyberMath! I actually figured out the real value of x in my head. I did not know that there are 24 complex solutions in this math problem. Thanks!
Good day. Congratulations on the subs. You got me as a new one. I love the way you describe the solutions and I also find your vocal tone very pleasing to listen to. And most importantly, I love your tempo of explanation. I think it is JUST RIGHT. Awesome!
This is late, but: you can rewrite the original equation as (x^5 -30)^5 -30 = x f is used in the video, so to avoid confusion let g(x) = x^5 - 30 We now have g(g(x)) = x But g(x) is an easy function to analyze, it is a simple translation of x->x^5, so it is a strictly increasing bijection, and it increases faster than x->x, so it will have a single point where g(x0)=x0, which we can either solve or guess is x=2, g(2)=2 and g(g(2))=g(2)=2. For yy, so again no possible solution. This, x=2 is the only real solution. As for complex solutions, I feel there should be a way to find them through the same functional analysis, but I'm afraid it's beyond my knowledge and skill :/
Hello sybermath congrats to you on getting 25k I learned a lot from your TH-cam channel including wlog assumptions using inequalities in smart ways and using Simon to solve fractional equations Thank you so much sybermath :D
(x^5-30)^5-x=30 Rearrange the equation as (x^5-32+2)^5-32-(x-2)=0 A=x^5-32 (A+2)^5-2^5-(x-2)=0 A{(A+2)^4+(A+2)^3*2+(A+2)^2*4+ +(A+2)*8+16}-(x-2)=0 Now x-2 is factor of A and so factor ofLHS.x=2
Congratulations for the milestone! I am looking forward for more geometry related problems ! In the equation you presented, the solution method could be similar for any number m that can be written as m=a^5-a, 30 being one case . Then the initial equation becomes (x^5-a^5+a)^5-a^5-(x-a)=0. If you factor the first two terms you get (x^5-a^5)(20th order pol. in x)-(x-a)=0 which again gives the real root a. This is essentially the same with what you presented, but factorization is applied to the first step. As a second remark any number m can be written as m=a^5-a since this equation has one real solution for the number a (positive or negative) as you showed by the graphic solution, therefore the initial equation has exactly one real solution for any real m.
Right from the start if you add oaks to both sides of the equation and take the 5th root on both sides you will realize that the functions are inverses of each other and therefore the only real solutions come from Y equals X and then you can solve from there.
Guessing x=2 is simple. We have a line y=x+30 and f(x)=(x^5-30)^5. It is not difficult to find its derivative and conclude that increasing function, slightly before x=2 it is zero, and it is increasing very fast... almost like a straght vertical line... Do you think it is acceptable to explain like that ? that x=2 is the only intersection ?
Here's why the quartic has no real roots. Rewrite 4x^2 as 2x^2+2x^2 so now you have 6 terms in decreasing order of exponents. The sum of the first 3 term is >=0, and last 3 terms is >0, both by the quadratic formula.
How do we demonstrate that x^4+2x^3+4x^2+8x+15=0 has no real solutions? With similar steps I concluded that x=2 is a solution, but I wasn’t able to show that it is the only solution.
u can easily see it because u are adding a bunch of terms and u have x⁴ as a leading term which will PROBABLY ALWAYS going to exceed other terms. for the whole proof u will need to express it as a sum of perfect squares.
@@dqrk0 Well, the leading term tells us about the limit to +infinity and -infinity, it doesn’t really tell us what happens “in between”. Anyway, I used your strategy and this is what I found: x^4+2x^3+4x^2+8x+15 = (x^2+x)^2+1/3(3x+4)^2+29/3, which is a sum of positive terms (actually, two are non-negative and one is positive), therefore the polynomial is always positive and has no real roots. Thanks a lot for your tip!!
@@policarpo4816 yes, i agree, its not really a proof, but its a really good sign when u have a bunch of terms that are just adding up, it USUALLY means it doesn't have any real roots. of course u need a rigorous proof of that, but u know that u are chasing ,,no real roots" as a result. great job, i am glad i have helped :)
@@MicheleCaine if there are no solutions there is nothing to divide the polynomial by (without a remainder). If you mean dividing x^5-x-30 by x-2, I used synthetic division, there is really no need for long division
Another solution: From the manipulation at 2:27 we know that x is a fixed point of f(f(t)), where: f(t) = t^5 - 30 But the succession a[n+1]=f(a[n]) is strictly increasing for initial condition a[0]>2 and strictly decreasing for a[0]
Let all 25 complex roots have absolute value 2. By Vieta's formulas, product of roots equals to -(-30^5-30), but it's absolute value equals to 2^25 by assumption. 30^5+30 is divided by 3, but 2^25 is not divisible by 3. Therefore, 30^5+30 ≠ 2^25. Contradiction.
I have a proof : We have f(x) = x^4 + 2x^3 + 4x^2 + 8x + 15 = 0 my claim is that its always > 0 f(x) can be writen as (X+(1/2))^4 + (5/2)x^2 + (15/2)x + 239/16 the first term is >= 0 (since 4 power) and the rest has negative descrimenent (and the coefficient of x^2 is positive) so the rest is always strictly bigger then 0 therefor f(x) is always > 0
It is difficult for me to see any point to these videos, since the "solutions" involve a huge amount of prior knowledge which cannot be explained or justified within the video. I would recommend abandoning this kind of effort.
One small nuance that I tell my students all of the time: there are not 24 complex solutions, but instead there are 25, since the real solution is also complex. There is 1 real solution and 24 non-real solutions.
As always, an excellent video! Congrats on 25k!!
That's right!
Yes. I also made the same mistake for infinitely many times!
Let's be careful both ways. Let's avoid gotchas. So one should understand what is meant by one real solution and 24 complex solutions.
One routinely refers to "integer spin" and "half-integer spin", with the understanding that the half-integer means non-integer. I don't really like saying, "integer-plus-a-half spin".
What do you mean the real solution is also complex?
@@squeezy8414 all real numbers are also complex number with zero imaginary part
Congrats on 25K man! You deserve it so much
I think 100K could come by the end of the year
I'll look forward to seeing the proper term for a 100th degree equation at that time!
definitely
Congrats, 25k, just awesome!!
@@brentwilson6692 So...how about an equation with 100 fraction mark and a radical?
No even more.
Btw, i can't see the join button
And i think it is only in the USA
Equation 1 uses the property that both functions are inverse of each other and satisfy f(x) = x.
Congratulations on another milestone! Hmm, as I had no intention to deal with 25th degree polynomial, I started by being creative this time and it did its job. : ) I won't spoil all the solution, it's your part. But one little tip for solvers: it might be helpful for you if you know where a function and its inverse meet.
Although you have to be careful. Some functions can also intersect their inverse off the diagonal.
Thank you! 💖
@@leickrobinson5186 That's interesting! You mean they will not intersect on the diagonal?
@@SyberMathI believe that you can show that any continuous function will intersect its inverse either not at all (as with y=e^x) or on the diagonal. However, in the latter case, it may also intersect its inverse off the diagonal as well. (E.g., the function y=(x-10)^2 will intersect its inverse in four places, two on the diagonal and two off the diagonal.)
However, you can also construct a *discontinuous* function that only intersects its inverse *off* the diagonal. For example, the function defined by
y = {1, x=0}
will intersect its inverse at 2 locations, both off the diagonal [at (-1,1) and (1,-1)].
@@leickrobinson5186 bro the functions u gave as example aren't even bijective how in the world are u talking about the graph of their inverses? 😂
Another great explanation, SyberMath! I actually figured out the real value of x in my head. I did not know that there are 24 complex solutions in this math problem. Thanks!
You're welcome! Thanks!
Good day. Congratulations on the subs. You got me as a new one. I love the way you describe the solutions and I also find your vocal tone very pleasing to listen to. And most importantly, I love your tempo of explanation. I think it is JUST RIGHT. Awesome!
Awesome, thank you and welcome aboard! 💖
I tried 2. Considering that you have to subtract x to get 30 (0 as ones digit), x should have the same ones digit as its 5th power.
This is late, but:
you can rewrite the original equation as (x^5 -30)^5 -30 = x
f is used in the video, so to avoid confusion let g(x) = x^5 - 30
We now have g(g(x)) = x
But g(x) is an easy function to analyze, it is a simple translation of x->x^5, so it is a strictly increasing bijection, and it increases faster than x->x, so it will have a single point where g(x0)=x0, which we can either solve or guess is x=2, g(2)=2 and g(g(2))=g(2)=2.
For yy, so again no possible solution.
This, x=2 is the only real solution.
As for complex solutions, I feel there should be a way to find them through the same functional analysis, but I'm afraid it's beyond my knowledge and skill :/
Wow! Already 25k! Congrants!
Thanks! 💖
Congratulations for 25k subscribers!!!
Thank you! 💖
Congrats on 25K subscribers! 👏🤯🎊🥳
Thank you! 💖
Congratulations for 25.2k !!
Thank you! 💖
Hello sybermath congrats to you on getting 25k I learned a lot from your TH-cam channel including wlog assumptions using inequalities in smart ways and using Simon to solve fractional equations
Thank you so much sybermath :D
Glad to hear that!
:D
Oh! No! We needed other 24 solutions not aproximations.
Congratulation mst sybermath for 25k sub , keep on !! ❤
You commented eallier today!
@@aashsyed1277 yes so what ? 😂
@@tonyhaddad1394 you don't comment this early!
@@aashsyed1277 yes bro it depend if im free or not
@@tonyhaddad1394 oh
(x^5-30)^5-x=30
Rearrange the equation as
(x^5-32+2)^5-32-(x-2)=0
A=x^5-32
(A+2)^5-2^5-(x-2)=0
A{(A+2)^4+(A+2)^3*2+(A+2)^2*4+
+(A+2)*8+16}-(x-2)=0
Now x-2 is factor of A and so factor
ofLHS.x=2
Congrats on 25K!!! 🎉
Thanks! 💖
I could see the X=2 solution at the start. The difficulty is proving it's the only solution. Nice problem 👍
Good point!
Congratulations for the milestone! I am looking forward for more geometry related problems ! In the equation you presented, the solution method could be similar for any number m that can be written as m=a^5-a, 30 being one case . Then the initial equation becomes (x^5-a^5+a)^5-a^5-(x-a)=0. If you factor the first two terms you get (x^5-a^5)(20th order pol. in x)-(x-a)=0 which again gives the real root a. This is essentially the same with what you presented, but factorization is applied to the first step. As a second remark any number m can be written as m=a^5-a since this equation has one real solution for the number a (positive or negative) as you showed by the graphic solution, therefore the initial equation has exactly one real solution for any real m.
Right from the start if you add oaks to both sides of the equation and take the 5th root on both sides you will realize that the functions are inverses of each other and therefore the only real solutions come from Y equals X and then you can solve from there.
Very nice!
I hope your channel has an exponential growth :D
congratulations on 25k,you deserve it brother :)
Thank you very much! 💖
Congraaats!!
Thank you! 💖
Man this guy is a champion at solving higher degree equations!
That's a strong term !! Quinviginticgigantic equation 😎
😅😅😅😅
😁
th-cam.com/video/JkaMSQNAxZ4/w-d-xo.html
Wow sybermath thanks so much!!!!!
Guessing x=2 is simple. We have a line y=x+30 and f(x)=(x^5-30)^5. It is not difficult to find its derivative and conclude that increasing function, slightly before x=2 it is zero, and it is increasing very fast... almost like a straght vertical line... Do you think it is acceptable to explain like that ? that x=2 is the only intersection ?
I agree. I was also sitting here thinking the same thing.
Sounds good to me
J’ai fait la même chose!
Is anybody else wondering about the logo
This was a wonderful video, and I definitely learned new stuff and had fun in the process. I suppose you could say that now, things are…five-by-five!
Awesome! Thanks! 💖
Please tell me how you got other 24 complex solutions 🥺 which method you used...can you make a video on it please
Here's why the quartic has no real roots. Rewrite 4x^2 as 2x^2+2x^2 so now you have 6 terms in decreasing order of exponents. The sum of the first 3 term is >=0, and last 3 terms is >0, both by the quadratic formula.
Congrats on your subscriber milestone! ... kilometerstone...
Thank you very much! 💖
Wow thanks for these joyful moments 😊😊☺️☺️😊😊
No problem! Thank you! 😊
How do we demonstrate that x^4+2x^3+4x^2+8x+15=0 has no real solutions? With similar steps I concluded that x=2 is a solution, but I wasn’t able to show that it is the only solution.
u can easily see it because u are adding a bunch of terms and u have x⁴ as a leading term which will PROBABLY ALWAYS going to exceed other terms. for the whole proof u will need to express it as a sum of perfect squares.
@@dqrk0 Well, the leading term tells us about the limit to +infinity and -infinity, it doesn’t really tell us what happens “in between”.
Anyway, I used your strategy and this is what I found: x^4+2x^3+4x^2+8x+15 = (x^2+x)^2+1/3(3x+4)^2+29/3, which is a sum of positive terms (actually, two are non-negative and one is positive), therefore the polynomial is always positive and has no real roots.
Thanks a lot for your tip!!
@@policarpo4816 yes, i agree, its not really a proof, but its a really good sign when u have a bunch of terms that are just adding up, it USUALLY means it doesn't have any real roots. of course u need a rigorous proof of that, but u know that u are chasing ,,no real roots" as a result. great job, i am glad i have helped :)
The hard way is to do long division in order to prove that the quotient has no real solutions
@@MicheleCaine if there are no solutions there is nothing to divide the polynomial by (without a remainder). If you mean dividing x^5-x-30 by x-2, I used synthetic division, there is really no need for long division
Congratulations
Thank you! 💖
Wish u get a million subs soon
💖 Thank you! 💖
I substituted x=2 r8 aftr seeing the eqn. 😎 my guess turned out correct!
Lol
To test x=1,x=2 find key x=2 it is very simple.
(x^5 - 30)^5 - x = 30 = 32 - 2
(x^5 - 30)^5 = 32 --> x^5 - 30 = 2 --> x^5 = 32 --> x = 2
answer: x = 2
amazing
Thank you!
Just try out nice numbers, such as x = ...
So how the heck do we divide out x-... from that polynomial?
I got x=2 as an answer by inspection
Nice!
Красивое решение.Спасибо.Привет из Баку.
Another solution:
From the manipulation at 2:27 we know that x is a fixed point of f(f(t)), where:
f(t) = t^5 - 30
But the succession a[n+1]=f(a[n]) is strictly increasing for initial condition a[0]>2 and strictly decreasing for a[0]
compi finds 25 solutions (24 complex, 1 real), here is the real solution:
*Solve[(x^5 - 30)^5 - x == 30, x, Reals]*
weird bot
@@archiebrew8184 lol by the way it isn't a bot but it looks like one a little
@@MathElite "it" = "bot" = "compi" ?
yee .
Yeet
Nice solution! I assume all the complex solutions have the same absolute value as the real one?
That's a good question! Why would that be the case?
@@SyberMath Because all solutions are written as
R(cosv+isinv) = z
IzI = sqrt(R^2(cos^2(v)+sin^2(v))) = R
Let all 25 complex roots have absolute value 2. By Vieta's formulas, product of roots equals to -(-30^5-30), but it's absolute value equals to 2^25 by assumption. 30^5+30 is divided by 3, but 2^25 is not divisible by 3. Therefore, 30^5+30 ≠ 2^25. Contradiction.
@@SyberMath Oh no, that was just a guess. I would love to see them mapped out and see if / why they make any kind of shape.
(x^5 - 30 )^5 - 30 = x ==> Let f ( x ) = x^5 - 30
so f ( f ( x ) ) = x
and as a result f ( x ) = f^( - 1 ) ( x )
I think you forgot the coefficient of the Pascal Triangle (1-4-6-4-1) in your demo....
That's awaome now supposedly someone asks to find all the four complex solution to the other factors then how to proceed.
Let y = x^5 -30 and we have system sym !
Only mathematicians can solve this
Why?
So what kind of an equation is (x^9-19680)^9 - x = 19680? I guess it would be an unoctogenic equation.
Probably unoctogintic 😁
👍
So why do yo know the second factor is complex solutions? Sir
Because the function is increasing, it can only have one intersection point with a horizontal line
Ok,also adway kumar pls respond to me,Its my bad that I made u feel bad,I dont mean to make u feel bad.
ở VN cái này gọi là hàm đặc trưng đấy mn ạ :))
X=2
11:06 but you didnt prove it (its another case ,not like (x^5+x =y^5+y)
I have a proof :
We have f(x) = x^4 + 2x^3 + 4x^2 + 8x + 15 = 0
my claim is that its always > 0
f(x) can be writen as
(X+(1/2))^4 + (5/2)x^2 + (15/2)x + 239/16
the first term is >= 0 (since 4 power)
and the rest has negative descrimenent (and the coefficient of x^2 is positive) so the rest is always strictly bigger then 0
therefor f(x) is always > 0
Nice!
@@SyberMath thank u 😍
You should say in the title real solutions.
Btw, how did Wolfram alpha solved a 25th degree polynomial cause there is no 25th degree solution.
It probably approximated
-30 -30 = 0?????
-30 - (-30) = 0
@@SyberMath ohhhh, srry i was thinkinf like 10 minutes that i dont figure how
thnxs a lot! i love your videos
I think the right spanish-latin word is quinvigesimic or quinvigesic not the way you've put it.
I think the right word is viginti quintic. I don't know why I used quinvigintic which is 10^78
en.wiktionary.org/wiki/quinvigintillion 😜
It is difficult for me to see any point to these videos, since the "solutions" involve a huge amount of prior knowledge which cannot be explained or justified within the video. I would recommend abandoning this kind of effort.
Congratulations for 25.2k !!