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.
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
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😁
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.
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.
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!!
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
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
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
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.
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
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)
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.
❤️🙏
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.
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.
How can we prove Vieta's Formula?? That would be an interesting video! Greetings from Hellas!
Indeed.
Greetings from Heavenas!
Super useful theorem in Algebra Problems
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.
Lmao, I was taught this in yesterday's class and I couldn't understand it, and here you are!
Thank you very much Sir
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.
Muy buena explicación en el video, esa forma de expresar el conocimiento es muy buena
So, is there a way we can use these results to derive the roots of the cubic equation?
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.
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
Could someone explain/calculate what the values are of a,b,c ,which are used in the second example?
never been this early man! love your videos! your enthusiasm and knowledge always brighten up my day lol
You should do things on quaternions instead of just complex numbers.
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
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
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
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
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.
yayyy the wait is overrr
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
Thx bro i needed this
Day 1 of asking to put on a birthday cap while solving question…..hope we get it.
It is the Girard's formula, isn't?
I know the relations but I did not know they are called Vieta's formula
Interesting! I never studied Vieta Formula mathematics for polynomial roots finding.
Aren't these simply known as Gerard identities? That's how I learned in high school.
I knew that formula and method but not the name *Vieta's formula* thanks for that😊😊
Merci Newton.
This sounds the same as equations of Girard…….
❤
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)