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− | '''Vieta's Formulas''', otherwise called Viète's Laws, are a set of [[equation]]s relating the [[root]]s and the [[coefficient]]s of [[polynomial]]s.
| + | #REDIRECT[[Vieta's formulas]] |
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− | == Introduction ==
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− | Vieta's Formulas were discovered by the French mathematician [[François Viète]].
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− | Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics. If we have a quadratic <math>x^2+ax+b=0</math> with solutions <math>p</math> and <math>q</math>, then we know that we can factor it as
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− | <center><math>x^2+ax+b=(x-p)(x-q)</math></center>
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− | (Note that the first term is <math>x^2</math>, not <math>ax^2</math>.) Using the distributive property to expand the right side we get
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− | <center><math>x^2+ax+b=x^2-(p+q)x+pq</math></center>
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− | We know that two polynomials are equal if and only if their coefficients are equal, so <math>x^2+ax+b=x^2-(p+q)x+pq</math> means that <math>a=-(p+q)</math> and <math>b=pq</math>. In other words, the product of the roots is equal to the constant term, and the sum of the roots is the opposite of the coefficient of the <math>x</math> term.
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− | A similar set of relations for cubics can be found by expanding <math>x^3+ax^2+bx+c=(x-p)(x-q)(x-r)</math>.
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− | We can state Vieta's formula's more rigorously and generally. Let <math>P(x)</math> be a polynomial of degree <math>n</math>, so <math>P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0</math>,
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− | where the coefficient of <math>x^{i}</math> is <math>{a}_i</math> and <math>a_n \neq 0</math>. As a consequence of the [[Fundamental Theorem of Algebra]], we can also write <math>P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)</math>, where <math>{r}_i</math> are the roots of <math>P(x)</math>. We thus have that
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− | <center><math> a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).</math></center>
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− | Expanding out the right hand side gives us
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− | <math> a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.</math>
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− | The coefficient of <math> x^k </math> in this expression will be the <math>k </math>th [[symmetric sum]] of the <math>r_i</math>.
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− | We now have two different expressions for <math>P(x)</math>. These must be equal. However, the only way for two polynomials to be equal for all values of <math>x</math> is for each of their corresponding coefficients to be equal. So, starting with the coefficient of <math> x^n </math>, we see that
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− | <center><math>a_n = a_n</math></center>
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− | <center><math> a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)</math></center>
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− | <center><math> a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)</math></center>
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− | <center><math>\vdots</math></center>
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− | <center><math>a_0 = (-1)^n a_n r_1r_2\cdots r_n</math></center>
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− | More commonly, these are written with the roots on one side and the <math>a_i</math> on the other (this can be arrived at by dividing both sides of all the equations by <math>a_n</math>).
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− | If we denote <math>\sigma_k</math> as the <math>k</math>th symmetric sum, then we can write those formulas more compactly as <math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>1\le k\le {n}</math>.
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− | ==Problems==
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− | ===Intermediate===
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− | Let <math>a+b+c=12</math>, <math>a^2+b^2+c^2=50</math>, and <math>a^3+b^3+c^3=168</math>. Find <math>a,b,c</math>
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− | ===Olympiad===
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− | == See Also ==
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− | * [[Algebra]]
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− | * [[Polynomials]]
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− | * [[Newton's Sums]]
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− | == External Links ==
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− | *[http://mathworld.wolfram.com/VietasFormulas.html Mathworld's Article]
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− | [[Category:Elementary algebra]]
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