Difference between revisions of "Vieta's formulas"
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− | + | In [[algebra]], '''Vieta's formulas''' are a set of results that relate the coefficients of a [[polynomial]] to its roots. In particular, it states that the [[elementary symmetric polynomial | elementary symmetric polynomials]] of its roots can be easily expressed as a ratio between two of the polynomial's coefficients. | |
+ | |||
+ | It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in many math contests/tournaments. | ||
+ | |||
+ | == Statement == | ||
+ | Let <math>P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0</math> be any polynomial with [[Complex number | complex]] coefficients with roots <math>r_1, r_2, \ldots , r_n</math>, and let <math>s_j</math> be the <math>j^{\text{th}}</math> elementary symmetric polynomial of the roots. | ||
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+ | Vieta’s formulas then state that <cmath>s_1 = r_1 + r_2 + \cdots + r_n = - \frac{a_{n-1}}{a_n}</cmath> <cmath>s_2 = r_1r_2 + r_2r_3 + \cdots + r_{n-1}r_n = \frac{a_{n-2}}{a_n}</cmath> <cmath>\vdots</cmath> <cmath>s_n = r_1r_2r_3 \cdots r_n = (-1)^n \frac{a_0}{a_n}.</cmath> This can be compactly summarized as <math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math> for some <math>j</math> such that <math>1 \leq j \leq n</math>. | ||
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+ | == Proof == | ||
+ | Let all terms be defined as above. By the [[factor theorem]], <math>P(x) = a_n (x-r_1)(x-r_2) \cdots (x-r_n)</math>. We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients. | ||
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+ | When expanding the factorization of <math>P(x)</math>, each term is generated by a series of <math>n</math> choices of whether to include <math>x</math> or the negative root <math>-r_{i}</math> from every factor <math>(x-r_{i})</math>. Consider all the expanded terms of the polynomial with degree <math>n-j</math>; they are formed by multiplying a choice of <math>j</math> negative roots, making the remaining <math>n-j</math> choices in the product <math>x</math>, and finally multiplying by the constant <math>a_n</math>. | ||
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+ | Note that adding together every multiplied choice of <math>j</math> negative roots yields <math>(-1)^j s_j</math>. Thus, when we expand <math>P(x)</math>, the coefficient of <math>x_{n-j}</math> is equal to <math>(-1)^j a_n s_j</math>. However, we defined the coefficient of <math>x^{n-j}</math> to be <math>a_{n-j}</math>. Thus, <math>(-1)^j a_n s_j = a_{n-j}</math>, or <math>s_j = (-1)^j a_{n-j}/a_n</math>, which completes the proof. <math>\square</math> | ||
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+ | == Problems == | ||
+ | Here are some problems with solutions that utilize Vieta's quadratic formulas: | ||
+ | |||
+ | |||
+ | === Introductory === | ||
+ | * [[2005 AMC 12B Problems/Problem 12 | 2005 AMC 12B Problem 12]] | ||
+ | * [[2007 AMC 12A Problems/Problem 21 | 2007 AMC 12A Problem 21]] | ||
+ | * [[2010 AMC 10A Problems/Problem 21 | 2010 AMC 10A Problem 21]] | ||
+ | * [[2003 AMC 10A Problems/Problem 18 | 2003 AMC 10A Problem 18]] | ||
+ | * [[2021 AMC 12A Problems/Problem 12 | 2021 AMC 12A Problem 12]] | ||
+ | |||
+ | === Intermediate === | ||
+ | * [[2017 AMC 12A Problems/Problem 23 | 2017 AMC 12A Problem 23]] | ||
+ | * [[2003 AIME II Problems/Problem 9 | 2003 AIME II Problem 9]] | ||
+ | * [[2008 AIME II Problems/Problem 7 | 2008 AIME II Problem 7]] | ||
+ | * [[2021 Fall AMC 12A Problems/Problem 23 | 2021 Fall AMC 12A Problem 23]] | ||
+ | |||
+ | == See also == | ||
+ | * [[Polynomial]] | ||
+ | |||
+ | [[Category:Algebra]] | ||
+ | [[Category:Polynomials]] | ||
+ | [[Category:Theorems]] |
Latest revision as of 20:24, 21 October 2024
In algebra, Vieta's formulas are a set of results that relate the coefficients of a polynomial to its roots. In particular, it states that the elementary symmetric polynomials of its roots can be easily expressed as a ratio between two of the polynomial's coefficients.
It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in many math contests/tournaments.
Statement
Let be any polynomial with complex coefficients with roots , and let be the elementary symmetric polynomial of the roots.
Vieta’s formulas then state that This can be compactly summarized as for some such that .
Proof
Let all terms be defined as above. By the factor theorem, . We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients.
When expanding the factorization of , each term is generated by a series of choices of whether to include or the negative root from every factor . Consider all the expanded terms of the polynomial with degree ; they are formed by multiplying a choice of negative roots, making the remaining choices in the product , and finally multiplying by the constant .
Note that adding together every multiplied choice of negative roots yields . Thus, when we expand , the coefficient of is equal to . However, we defined the coefficient of to be . Thus, , or , which completes the proof.
Problems
Here are some problems with solutions that utilize Vieta's quadratic formulas:
Introductory
- 2005 AMC 12B Problem 12
- 2007 AMC 12A Problem 21
- 2010 AMC 10A Problem 21
- 2003 AMC 10A Problem 18
- 2021 AMC 12A Problem 12