# Difference between revisions of "Vieta's Formulas"

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<center><math>x^2+ax+b=x^2-(p+q)x+pq</math></center> | <center><math>x^2+ax+b=x^2-(p+q)x+pq</math></center> | ||

− | We know that two | + | 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. |

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>. | 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>. |

## Revision as of 17:04, 26 December 2015

**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 with solutions and , then we know that we can factor it as

(Note that the first term is , not .) Using the distributive property to expand the right side we get

We know that two Polynomials are Equal if and only if their coefficieNts are equal, so means that and . 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 term.

A similar set of relations for cubics can be found by expanding .

We can state Vieta's formula's more rigorously and generally. Let be a polynomial of degree , so , where the coefficient of is and . As a consequence of the Fundamental Theorem of Algebra, we can also write , where are the roots of . We thus have that

Expanding out the right hand side gives us

The coefficient of in this expression will be the th symmetric sum of the .

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

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

If we denote as the th symmetric sum, then we can write those formulas more compactly as , for .

## Problems

- Let and be the three roots of the cubic . Find the value of . Yeah
- Suppose the polynomial has three real roots , and . Find the value of .
- Let and be the roots of the quadratic equation . Find .

### Intermediate

- Let , , and be positive real numbers with such that , , and . Find .
- (USAMTS 2010) Find such that if , , and are the roots of the cubic then
- (HMMT 2007) The complex numbers , , , and are the four distinct roots of the equation . Determine the unordered set

### Olympiad

[2008 AIME II] Let , , and be the three roots of the equation . Find .