Difference between revisions of "Vieta's Formulas"

Revision as of 21:43, 9 March 2014

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 get

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

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 formula's 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 $k$ th 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 symmetric sum, then we can write those formulas more compactly as $\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$ , for $1\le k\le {n}$ .

Problems

Beginner

Let $r_1,r_2,$ and $r_3$ be the three roots of the cubic $x^3+3x^2+4x-4$ . Find the value of $r_1r_2+r_1r_3+r_2r_3$ .
Suppose the polynomial $5x^3+4x^2-8x+6$ has three real roots $a,b$ , and $c$ . Find the value of $a(1+b+c)+b(1+a+c)+c(1+a+b)$ .
Let $m$ and $n$ be the roots of the quadratic equation $4x^2 + 5x + 3 = 0$ . Find $(m + 7)(n + 7)$ .

Intermediate

Olympiad

External Links

Mathworld's Article

@@ Line 45: / Line 45: @@
 *Let <math>r_1,r_2,</math> and <math>r_3</math> be the three roots of the cubic <math>x^3+3x^2+4x-4</math>.  Find the value of <math>r_1r_2+r_1r_3+r_2r_3</math>.
 *Suppose the polynomial <math>5x^3+4x^2-8x+6</math> has three real roots <math>a,b</math>, and <math>c</math>.  Find the value of <math>a(1+b+c)+b(1+a+c)+c(1+a+b)</math>.
+*Let <math>m</math> and <math>n</math> be the roots of the quadratic equation <math>4x^2 + 5x + 3 = 0</math>. Find <math>(m + 7)(n + 7)</math>.
 ===Intermediate===

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