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Difference between revisions of "Vieta's Formulas"
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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>.
 
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>.
 
Also, <math>-b/a = p + q, c/a = p \cdot q</math>.
 
Also, <math>-b/a = p + q, c/a = p \cdot q</math>.
 +
 +
Provide links to problems that use vieta formular:
 +
Examples:
 +
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23
  
 
==Proving Vieta's Formula==
 
==Proving Vieta's Formula==

Revision as of 14:59, 6 August 2021

Vieta's Formulas, otherwise called Viète's Laws, are a set of equations relating the roots and the coefficients of polynomials.

Introduction

Vieta's Formulas were discovered by the French mathematician François Viète. 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 $x^2+ax+b=0$ with solutions $p$ and $q$, then we know that we can factor it as:

$x^2+ax+b=(x-p)(x-q)$

(Note that the first term is $x^2$, not $ax^2$.) Using the distributive property to expand the right side we now have

$x^2+ax+b=x^2-(p+q)x+pq$

Vieta's Formulas are often used when finding the sum and products of the roots of a quadratic in the form $ax^2 + bx +c$ with roots $r_1$ and $r_2.$ They state that:

\[r_1 + r_2 = -\frac{b}{a}\] and \[r_1 \cdot r_2 = \frac{c}{a}.\]

We know that two polynomials are equal if and only if their coefficients are equal, so $x^2+ax+b=x^2-(p+q)x+pq$ means that $a=-(p+q)$ and $b=pq$. 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 $x$ term.

A similar set of relations for cubics can be found by expanding $x^3+ax^2+bx+c=(x-p)(x-q)(x-r)$.

We can state Vieta's formulas more rigorously and generally. Let $P(x)$ be a polynomial of degree $n$, so $P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0$, where the coefficient of $x^{i}$ is ${a}_i$ and $a_n \neq 0$. As a consequence of the Fundamental Theorem of Algebra, we can also write $P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)$, where ${r}_i$ are the roots of $P(x)$. We thus have that

$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).$

Expanding out the right-hand side gives us

$a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.$

The coefficient of $x^k$ in this expression will be the $(n-k)$-th elementary symmetric sum of the $r_i$.

We now have two different expressions for $P(x)$. These must be equal. However, the only way for two polynomials to be equal for all values of $x$ is for each of their corresponding coefficients to be equal. So, starting with the coefficient of $x^n$, we see that

$a_n = a_n$
$a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)$
$a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)$
$\vdots$
$a_0 = (-1)^n a_n r_1r_2\cdots r_n$

More commonly, these are written with the roots on one side and the $a_i$ on the other (this can be arrived at by dividing both sides of all the equations by $a_n$).

If we denote $\sigma_k$ as the $k$-th elementary symmetric sum, then we can write those formulas more compactly as $\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$, for $1\le k\le {n}$. Also, $-b/a = p + q, c/a = p \cdot q$.

Provide links to problems that use vieta formular: Examples: https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23

Proving Vieta's Formula

Basic proof: This has already been proved earlier, but I will explain it more. If we have $x^2+ax+b=(x-p)(x-q)$, the roots are $p$ and $q$. Now expanding the left side, we get: $x^2+ax+b=x^2-qx-px+pq$. Factor out an $x$ on the right hand side and we get: $x^2+ax+b=x^2-x(p+q)+pq$ Looking at the two sides, we can quickly see that the coefficient $a$ is equal to $-(p+q)$. $p+q$ is the actual sum of roots, however. Therefore, it makes sense that $p+q= \frac{-b}{a}$. The same proof can be given for $pq=\frac{c}{a}$.

Note: If you do not understand why we must divide by $a$, try rewriting the original equation as $ax^2+bx+c=(x-p)(x-q)$ SuperJJ

General Form

For a polynomial of the form $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ with roots $r_1,r_2,r_3,...r_n$, Vieta's formulas state that:

\begin{align*} s_1&= & r_1+r_2+r_3&+\cdots+r_n & &=-\frac{a_{n-1}}{a_n} \\ 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} \\ 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} \\ & & &\vdots & & \\ s_n&= & r_1r_2r_3&\cdots r_n & &=(-1)^n\frac{a_0}{a_n}.\\ \end{align*}

These formulas are widely used in competitions, and it is best to remember that when the $n$ roots are taken in groups of $k$ (i.e. $r_1+r_2+r_3...+r_n$ is taken in groups of 1 and $r_1r_2r_3...r_n$ is taken in groups of $n$), this is equivalent to $(-1)^k\frac{a_{n-k}}{a_n}$.

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