China Math Olympiad Problem That Is Very Difficult.
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- čas přidán 9. 09. 2024
- Here is today's video, title; "China Math Olympiad Problem That Is Very Difficult" , enjoy it please.
Kindly leave a comment from this video in the comments section on how to solve the case two of this math problem please.
Thanks.
x^5+x^4+1=(x^2+x+1)(x^3-x+1)=0 , x^2+x+1=0 (simple complex conjugates). Also, x^3-x+1=0 (Cordano's formula) to obtain irrational root and complex conjugates. This is known to high school students, let alone to students preparing for maths Olympiad. 1) There is no available general formula for polynomials or degree 5 or higher 2) Factorisation 3) Application of Cordano's formula or Newton's method for numerical solution.
Please explain how you could use Cordano's formula to solve x³-x+1=0. There is no x² in the equation to apply Cordano's formula.
@@ASAPSquatterRemoval Cardano's formula does not need the x^2 term for application in solving the cubic equation. If there's x^2 term, then depress it using linear substitution y=x-a/3b.
@@user-hz5ne2rl5e Very interesting, haven't used Cordano's formula in a while. I'm assuming there is a typo in your statement and you meant to say "if there isn't a x² term", correct?
For equation x^3-x+1=0 make the substitution x=y+1/(3y) and everything will be fine
спасибо) только как догадаться до такой замены... а то автор показал задачу... и потом показал как он не может ее решить
мда... странное дело, но получается не хорошо... такая замена удивительным образом сводит это уравнение к квадратному с положительным дискриминантом... а это два действительных корня... что не соответствует действительности... эта функция абсолютно точно имеет один действительный корень и два комплексных, в чем можно убедиться, исследовав ее с помощью производной, а так же построив график
The video gave me exactly what I was looking for, a generalized method for handling a quintic equation.
Thank you for the clearly expounded, and written, explanation.
You are most welcome sir, and thank you for watching and dropping an encouraging comment.
Much love sir...💕💕💖💖♥️
So something that doesnt exist?
Excellent !!!!
You did well on the part you know my master. Well done 👍👍
Thanks for the accolade sir.
Isn't it easier to substitute x=cos(∆)+i(sin∆) and solve for Sin(5∆)+Sin(4∆)=0. This gives all the roots, works for this particular equation.
Ok, I will try that out sir.
Thanks for this special hint.
Maximum respect sir...🙋🙋🙋
This substitution will not give all solutions since not all roots lie on the unit circle
@@anatolyalikhanov9012But can't it give the imaginary solutions too?
X^4+2X^3+4X^2+8X+16 = 0 this is one of the difficult one as well
Awesome explanation 😊
x^3-x+1=0 -> x(x+1)(x-1)=-1. This implies that the real value root exist in the range x
So this formula is technically for Scipioni Del Ferro. Cardano and Tartaglia only got credit because one of them published after Del Ferro passed away after he had already gave the formula to his student. To solve a cubic in general you first must depress the cube. However in this specific case where, x^3-x+1=0 we already have a depressed cubic since there is no x^2 term. You could also use Lagrange Resolvent. If you decide to use the cubic formula you will get 1 real root and after synthetic or polynomial long division you will get a quadratic factor which will yield two more solutions which are conjugates of each other. Thus we have 4 non real solutions which come in 2 pairs of conjugates and 1 real solution and you can see this by Descartes Rule of sign. By Descartes rule there are no positive real roots and if you try x=0 it does not satisfy the equation now we know there are no non-negative real roots. And after trying the negative case you will see that there is only 1 sign change from -x^5 to x^4 therefore there is only 1 negative real solution and since the fundamental theorem of algebra implies a Quintic equation must have 5 solutions and there is only 1 real solution the rest must be complex and have to come in conjugate pairs. Note they only have to come in conjugate pairs in a polynomial with REAL COEFFICIENTS. If the coefficient or coefficients are not real then that does not have to be the case. You could have a repeated root that’s non real or not have a conjugate at all but two separate complex roots.
Thank you sir
Well solved brotha 😊
Why do you never specify the set in which you work? R or C? It must be said from the start.
Thank you! That's what I was looking for! Awesome! ❤
Awesome!! WOW
Thanks a million sir
Good Explained
Thanks and welcome
Nice handwriting.
Thanks sir
Putting x=omega or(omega)^2; then vanishing factor.
Use Cardano (1 real and 2 complex solutions):
x_1 = (-1/2 + (23/108)^0.5)^(1/3) + (-1/2 - (23/108)^0.5)^(1/3)
x_2 = (-1/2 + (sqrt{3}/2) . i) . (-1/2 + (23/108)^0.5)^(1/3) + (-1/2 - (sqrt{3}/2) . i) . (-1/2 - (23/108)^0.5)^(1/3)
x_3 = (-1/2 - (sqrt{3}/2) . i) . (-1/2 + (23/108)^0.5)^(1/3) + (-1/2 + (sqrt{3}/2) . i) . (-1/2 - (23/108)^0.5)^(1/3)
x3= 1.324..., ça sort d'où ?
x³ - x + 1 = 0
f(0)=1
f(-1)= 1>0
f(-2)= -5
Nice work through man.
А ещё 2 корня? И компьютерное решение - это несерьёзно.
Sure
X=-1
Tu as injecté une solution fausse pour l équation cubique tu aurais du résoudre par la formule de Cardan ou par factorisation ou changement de variable
x³ - x + 1 = 0
Cardano..
Ok sir, I will research on this Cardano method/approach and come up with a well detailed video on the problem in question.
Thanks for suggestion sir.
@@onlineMathsTV you are welcome. greetings from Brazil.
1 и 0'5 в о ответ тебе летит хим реакция.
thar's easy
(1 + x+ x^2) (1 - x + x^3) = 0
@TidakTerdefinisi There is no general method for solving a fifth-degree equation; if we have any chance, it is when we can find a factorization into a product of polynomials with rational coefficients. The equation x^5+x^4+1=0 , as can be easily checked, has no rational solutions because any solution would have to be a divisor of 1, so it is enough to check 1 and -1.
So, one factor is of degree 2 and the other is of degree 3.
We start by writing the general forms for two polynomials:
A(x) = A0 + A1*x + A2*x^2
B(x) = B0 + B1*x + B2*x^2 + B3*x^3
Next, we multiply these two polynomials:
(A0 + A1*x + A2*x^2) * (B0 + B1*x + B2*x^2 + B3*x^3)
Expanding this, we get:
A0*B0 + A0*B1*x + A0*B2*x^2 + A0*B3*x^3 + A1*B0*x + A1*B1*x^2 + A1*B2*x^3 + A1*B3*x^4 + A2*B0*x^2 + A2*B1*x^3 + A2*B2*x^4 + A2*B3*x^5
Now, combine like terms:
= A0*B0 + (A0*B1 + A1*B0)*x + (A0*B2 + A1*B1 + A2*B0)*x^2 + (A0*B3 + A1*B2 + A2*B1)*x^3 + (A1*B3 + A2*B2)*x^4 + A2*B3*x^5
We compare the coefficients with the equation x^5 + x^4 + 1 = 0:
A2*B3 = 1 (coefficient of x^5)
A1*B3 + A2*B2 = 1 (coefficient of x^4)
A0*B3 + A1*B2 + A2*B1 = 0 (coefficient of x^3)
A0*B2 + A1*B1 + A2*B0 = 0 (coefficient of x^2)
A0*B1 + A1*B0 = 0 (coefficient of x)
A0*B0 = 1 (constant term)
Now, we solve these equations to find integer solutions for the coefficients A0, A1, A2, B0, B1, B2, and B3.
We are looking for integer solutions.
From the first equation, we have A2*B3 = 1, which means A2 = 1 and B3 = 1 or A2 = -1 and B3 = -1. However, if A2 = -1 and B3 = -1, we can multiply each polynomial on the left by -1, which gives us A2*B3 = 1. Therefore, we can assume A2 = 1 and B3 = 1.
...
I know it is not nice but works... alternatively You can guess
@TidakTerdefinisi I don’t know if it’s sarcasm or a genuine thank you, because I can’t see my response. My first post was intended sarcasm because I don’t like tasks arranged using “reverse engineering.” Knowing the distribution, you can easily multiply and “pretend to be smart” by writing the calculation from the end to the beginning.
Nevertheless, in this case, the task can be relatively easily solved by looking for the factorization of a 5th-degree polynomial into two polynomials with integer coefficients of degrees two and three, by writing them in general form, multiplying, and comparing coefficients. We get a system of 6 equations which, thanks to the large number of zero and equal to 1 coefficients, is not as difficult to solve as it seems at first glance.
can you solve x^3+2x=100?
С помощью компьютера, как автор видео, за 1 секунду.
I will solve it make a video of it in no distant time sir.
x^3= - 2x +100
(a+b)^3=3ab(a+b)+ a^3+b^3
if we substitute x=(a+b) we get
x^3=3ab*x+(a^3+b^3)
so we have to find a and b
a^3+b^3=100
3ab= - 2
so
27*(a^3)*(b^3) =- 8
You can now solve wquations
uv=-8/27
u+v=100
a^3=u
b^3=v
x=a+b
solution does not looks nice:
x = (2 (225 + sqrt(50631)))^(1/3)/3^(2/3) - 2^(2/3)/(3 (225 + sqrt(50631)))^(1/3)
x = -(225 + sqrt(50631))^(1/3)/6^(2/3) + 1/(6 (225 + sqrt(50631)))^(1/3) + i (-(225 + sqrt(50631))^(1/3)/(2^(2/3) 3^(1/6)) - 3^(1/6)/(2 (225 + sqrt(50631)))^(1/3))
x = -(225 + sqrt(50631))^(1/3)/6^(2/3) + 1/(6 (225 + sqrt(50631)))^(1/3) + i ((225 + sqrt(50631))^(1/3)/(2^(2/3) 3^(1/6)) + 3^(1/6)/(2 (225 + sqrt(50631)))^(1/3))
@TidakTerdefinisi only one of them is real.
There are meny ways to get aproximate this real solution
You can try to divide x^3+2x-100 by ( x-r), then solve quadratic equation You will get.
Sum of roots in your equation is 0 so real part of complex root is r/2 (r is real root)
But the polynomial should have 5 roots, not only 3.
x4 = 066 + 0,56i
x5 = 0,66 - 0,56i
Yes, you are very right sir. I had issues solving for the 4th and 5th roots hence I stop at the 3rd root.
Please, sir, can you kindly share with OnlineMathsTV the approach you applied in solving for the 4th and 5th roots here? This platform will be forever be grateful for that.
Thanking you in advance sir...🙋🙋🙏🙏🙏💕💖
@@onlineMathsTV
x^3-x-+1=0
x^3=x-1
(a+b)^3=3ab(a+b)+ a^3+b^3
if we substitute x=(a+b) we get
x^3=3ab*x+(a^3+b^3)
so we have to find a and b
a^3+b^3=-1
3ab=1
so
27*(a^3)*(b^3)=1
and You can find a^3 and b^3 and a+b as solutions
It is how to apply Cardano method
remember u^3 = a has 3 solution for real a one is real othrt 2 are complex
2x3 faktoring , After a few tries... , (x^3-x+1)(x^2+x+1)=0 , test , x^5-x^3+x^2 +x^4-x^2+x +x^3-x+1 , -x^3+x^3 +x^2-x^2 +x-x =0 ,
-> x^5+x^4+1=0 , OK , x^3-x+1=0 , solu. , x^3-px+q=0 , p=-1 , q=1 ,
Thanks for the hint to this math puzzle sir.
I will work on it.
You are A Bad One
You should apply a => sign after each step, without which steps don't connected with each other. And this is a blunder my friend. 😢😢😢😢