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.
@@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?
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
@@alexnikola7520Alex, you are partly right and mostly wrong. The replacement x=y+1/(3y) does indeed reduce x^3-x+1=0 to a quadratic, in y^3. I agree, amazing. The solutions for the quadratic in y^3 does have a positive discriminant, and two solutions for y^3, both real and negative. Namely, y^3=-1/2 +/- sqrt(621)/54. Their (real) cube roots are negative, and both(!) lead to the same analytic solution for x (by construction). Plugging either of these into x=y+1/3y gives an analytic solution to the (real) root of x^3-x+1=0. The complex roots follow. One punch line: counting roots can be tricky. In this problem, there is no difficulty. Everything works.
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.
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.
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
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)
@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.
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
