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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>ax^2+bx+c=(x-p)(x-q)</math></center>
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| − | Using the distributive property to expand the right side we now have
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| − | <center><math>x^2+ax+b=x^2-(p+q)x+pq</math></center>
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| − | Vieta's Formulas are often used when finding the sum and products of the roots of a quadratic in the form <math>ax^2 + bx +c</math> with roots <math>r_1</math> and <math>r_2.</math> They state that:
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| − | <cmath>r_1 + r_2 = -\frac{b}{a}</cmath> and <cmath>r_1 \cdot r_2 = \frac{c}{a}.</cmath>
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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 formulas 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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| − | <center><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></center>
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| − | The coefficient of <math> x^k </math> in this expression will be the <math>(n-k)</math>-th [[elementary 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 elementary 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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| − | Also, <math>-b/a = p + q, c/a = p \cdot q</math>.
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| − | Provide links to problems that use vieta formulas:
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| − | Examples:
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| − | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23
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| − | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_21
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| − | ==Proving Vieta's Formula==
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| − | Basic proof:
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| − | This has already been proved earlier, but I will explain it more.
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| − | If we have
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| − | <math>x^2+ax+b=(x-p)(x-q)</math>, the roots are <math>p</math> and <math>q</math>.
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| − | Now expanding the left side, we get: <math>x^2+ax+b=x^2-qx-px+pq</math>.
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| − | Factor out an <math>x</math> on the right hand side and we get: <math>x^2+ax+b=x^2-x(p+q)+pq</math>
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| − | Looking at the two sides, we can quickly see that the coefficient <math>a</math> is equal to <math>-(p+q)</math>. <math>p+q</math> is the actual sum of roots, however. Therefore, it makes sense that <math>p+q= \frac{-b}{a}</math>. The same proof can be given for <math>pq=\frac{c}{a}</math>.
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| − | Note: If you do not understand why we must divide by <math>a</math>, try rewriting the original equation as <math>ax^2+bx+c=(x-p)(x-q)</math>
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| − | SuperJJ
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| − | ==General Form==
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| − | For a polynomial of the form <math>f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0</math> with roots <math>r_1,r_2,r_3,...r_n</math>, Vieta's formulas state that:
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| − | <cmath>
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| − | \begin{align*}
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| − | s_1&= & r_1+r_2+r_3&+\cdots+r_n & &=-\frac{a_{n-1}}{a_n} \\
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| − | s_2&= & r_1r_2+r_1r_3+r_1r_4&+\cdots+r_{n-2}r_{n-1} & &=\phantom{-}\frac{a_{n-2}}{a_n} \\
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| − | s_3&= & r_1r_2r_3+r_1r_2r_4&+\cdots+r_{n-2}r_{n-1}r_n & &=-\frac{a_{n-3}}{a_n} \\
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| − | & & &\vdots & & \\
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| − | s_n&= & r_1r_2r_3&\cdots r_n & &=(-1)^n\frac{a_0}{a_n}.\\
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| − | \end{align*}
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| − | </cmath>
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| − | These formulas are widely used in competitions, and it is best to remember that when the <math>n</math> roots are taken in groups of <math>k</math> (i.e. <math>r_1+r_2+r_3...+r_n</math> is taken in groups of 1 and <math>r_1r_2r_3...r_n</math> is taken in groups of <math>n</math>), this is equivalent to <math>(-1)^k\frac{a_{n-k}}{a_n}</math>.
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| − | [[Category:Algebra]]
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| − | [[Category:Polynomials]]
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