Difference between revisions of "Vieta's Formulas"

Revision as of 13:41, 12 June 2011

Vieta's Formulas are a set of equations relating the roots and the coefficients of polynomials.

Introduction

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 $x^2+3x^2+4x-4$ . 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 $a(1+b+c)+b(1+a+c)+c(1+a+b)$ .

Intermediate

Olympiad

External Links

Mathworld's Article

@@ Line 37: / Line 37: @@
 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>.
+==Problems==
+===Beginner===
+*Let <math>r_1,r_2,</math> and <math>r_3</math> be the three roots of the cubic <math>x^2+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>.
+===Intermediate===
+===Olympiad===
 == See also ==

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