Vieta's Formula
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- Опубликовано: 13 дек 2024
- This is, by far, the most important polynomial formula I have ever come across. It works for real and complex roots without fail. It basically interprets the expansion of the factors of a polynomial, no matter the degree. this is a must know for any math Olympian
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Trivia: Viète (or Vieta in Latin form) was a lawyer by profession, and did not bother to make this Formula publicized since he thought it was obvious or too basic.
Wow, I have never heard of Vieta Formula. Excellent presentation as usual.
It is used in quadratic of 11th
@@AyushGautam-gj6csAnd are they all Americans?😂
shut up
in my country we study Vieta's formula in middle school, along with discriminant. but as I remember, not in depth, because we only learn the -b/a formula for sum of roots and c/a for multiplication of roots
Yeah in skls they only teach for quadratics n cubics, not more than tht
U should be learning this in like grade 7
Long forgotten.
Thank you for reminding.
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Lmao, I was taught this in yesterday's class and I couldn't understand it, and here you are!
Thank you very much Sir
4:55 I've never seen inequalities under the summation sigma before. I tried to look it up but there is very little online about it. Great video. I enjoy your gentle humour.
Формулы Виета получаются очень просто. Надо привести многочлен к виду, когда коэффициент при неизвестном в высшей степени равен 1, потом многочлен представить в виде произведения разностей (X - Ri), раскрыть скобки и привести подобные относительно степеней X члены.
I’ve been watching your videos daily for several weeks now. I studied mathematics in university about 25-30 years ago. I never learned Vieta’s formula (or don’t recall it). I’m now questioning the quality of my education.
Super useful theorem in Algebra Problems
I Saw that Pattern when I tried to expand (x+a)(x+b)(x+c), then I did (x+a)(x+b)(x+c)(x+d), after that, I did (x+a)(x+b)(x+c)(x+d)(x+e)
After analasizing all the results I came to a formula, witch if u turn the other side to 0 and divede both sides by a_n, you get basicly the vieta's formula, I am so proud of myself😁
You just need to prove, by induction, that it is true for any polynomial with real coefficients.
never been this early man! love your videos! your enthusiasm and knowledge always brighten up my day lol
Way back when in the UK, I did 2 'O' levels and 2 'A' levels in school and higher school maths, plus a lot of maths during an electronic engineering degree and was never taught this!!
Yeah the igcse gce didnt teach us too
This formula is similar to the formula that we use for finding probability in a binomial distribution.
How can we prove Vieta's Formula?? That would be an interesting video! Greetings from Hellas!
Indeed.
Greetings from Heavenas!
Muy buena explicación en el video, esa forma de expresar el conocimiento es muy buena
Now I know where does root formula come from for quadratic equation. SOR= - b/a n POR =c/a are actually from vieta formula. Tq for showing me.
Not exactly. Supposing r1 and r2 are the roots, we have a(x-r1)(x-r2)=0 => ax²-a(r1+r2)x+ar1r2=0
Well, mathematicians knew about this rule for quadratics. Vieta’s Laws only generalized it for all polynomials of any nth degree.
So, is there a way we can use these results to derive the roots of the cubic equation?
Maybe series about symmetric polynomials
Then Vieta formulas can be put somewhere in such series
Symmetric polynomials are such polynomials that are invariant to permutation of variables
at any point are you planning to do a bunch of series on various topics from ground up like a 15 ep run on matrices or any
These are observational formulas so they r called Vieta’s Laws. Also, i think the second equation is wrong. It should be:
(r_1r_2 + r_1r_3 + … + r_1r_n) + (r_2r_3 + r_2r_4 + … + r_2r_n) + … + (r_{n-1}r_n) = a_{n-2} / a_n
Could you use this to solve for the roots by setting up a system of equation and solving for a, b, c or does that not work?
I don't think it helps in this case
Al presentar el tema, habría que decir que n es impar para que se cumplan tal como están escritas al principio, si se prueban para n un número par como un polinomio cuadrático, cuartico, etc,... no funcionarán
Can Vieta Formulas be used to numerically find the roots? It would seem plausible that a computer algorithm could use the formulas to iteratively narrow in on them.
You should do things on quaternions instead of just complex numbers.
Could someone explain/calculate what the values are of a,b,c ,which are used in the second example?
Thx bro i needed this
yayyy the wait is overrr
In school we only learned them for quadratic equations, I'm curious what is the connection between this and the algorithm for factoring quadratic polynomials, when you guess coefficients that if multiplied, are to be equal to a*c, but their sum is b. Those numbers should be opposite to the roots of a quadratic equation that could be guessed according Vieta's formula? Maybe you have a video, explaining how this algorithm of factorisation is proven , why it works. I feel like it's something to do with Vieta's formula
It is the Girard's formula, isn't?
Day 1 of asking to put on a birthday cap while solving question…..hope we get it.
Aren't these simply known as Gerard identities? That's how I learned in high school.
I know the relations but I did not know they are called Vieta's formula
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Interesting! I never studied Vieta Formula mathematics for polynomial roots finding.
I knew that formula and method but not the name *Vieta's formula* thanks for that😊😊
This sounds the same as equations of Girard…….
Merci Newton.
What ! a^2 +b^2 +c^2 =-23, not possible (not positive). I think that ab+ac+bc=-36. Else, good subject .
even though you did clarify that the values for a, b, and c would have to be positive to make the expression real, it is possible for it to be negative with complex solutions (as he said in the video)