Theorem of Vieta

The set of Vieta or root set of Vieta is a mathematical theorem from elementary algebra . It is named after the mathematician François Viète , who proved it in his posthumously published work “De aequationum recognitione et emendatione Tractatus duo” (“Two treatises on the investigation and improvement of equations”). The theorem makes a statement about the connection between the coefficients and the solutions of an algebraic equation .

statement

Let and be the coefficients of the quadratic equation ${\displaystyle p}$ ${\displaystyle q}$

${\displaystyle x^{2}+px+q=0}$

and and their solutions (roots). Then applies ${\displaystyle x_{1}}$ ${\displaystyle x_{2}}$

${\displaystyle {\begin{array}{r c l}x_{1}+x_{2}&=&-p\\x_{1}\cdot x_{2}&=&q\end{array}}}$

Examples

There are three important uses for the sentence:

• It can be used to construct quadratic equations for given solutions. For example, a quadratic equation for solutions 2 and 3 is . ${\displaystyle x^{2}-5x+6=0}$
• Systems of equations of the form

${\displaystyle {\begin{array}{r c l }x_{1}+x_{2}&=&-p\\x_{1}\cdot x_{2}&=&q\end{array}}}$

to solve. For example, the solutions and the system are the solutions of the associated quadratic equation . After solving formula arises , or , . ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle x+y=-(-5),x\cdot y=6}$ ${\displaystyle z^{2}-5z+6=0}$ ${\displaystyle x=2}$ ${\displaystyle y=3}$ ${\displaystyle x=3}$ ${\displaystyle y=2}$

• The theorem can help determine the solutions through trial and error: Is the quadratic equation

${\displaystyle x^{2}-7x+10=0}$

given, then for the zeros , the following must apply: ${\displaystyle x_{1}}$ ${\displaystyle x_{2}}$

${\displaystyle {\begin{aligned}x_{1}+x_{2}&=-(-7)=7\\x_{1}\cdot x_{2}&=10\,.\end{aligned}}}$

If we first look for integer zeros, the zeros must be divisors of 10, the sum of which is 7. The divisors of 10 are 2 and 5, or 1 and 10, or −2 and −5, or −1 and −10. 2 and 5 are actually zeros, there and is. ${\displaystyle 2+5=7}$ ${\displaystyle 2\cdot 5=10}$

proof

The sentence is obtained directly by multiplying the form of the zeros after the coefficient comparison :

${\displaystyle {\begin{aligned}x^{2}+px+q&=(x-x_{1})\cdot (x-x_{2})\\&=x^{2}-(x_{1}+x_{2})\cdot x+x_{1}\cdot x_{2}\end{aligned}}}$

and thus and . ${\displaystyle p=-(x_{1}+x_{2})}$ ${\displaystyle q=x_{1}\cdot x_{2}}$

Alternatively, the theorem follows from the pq formula : The following applies to the solutions of the equation ${\displaystyle x^{2}+px+q=0}$

${\displaystyle x_{1}=-{\frac {p}{2}}+{\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}}$ and ${\displaystyle x_{2}=-{\frac {p}{2}}-{\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}.}$

Adding the two equations gives:

${\displaystyle x_{1}+x_{2}=-{\frac {p}{2}}+{\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}+\left(-{\frac {p}{2}}-{\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}\right)=-p}$,

Multiplying results according to the third binomial formula:

${\displaystyle x_{1}\cdot x_{2}=\left(-{\frac {p}{2}}+{\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}\right)\cdot \left(-{\frac {p}{2}}-{\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}\right)=\left({\frac {p}{2}}\right)^{2}-\left(\left({\frac {p}{2}}\right)^{2}-q\right)=q}$.

generalization

Vieta's theorem on quadratic equations can be generalized to polynomial equations or polynomials of any degree. This generalization of Vieta's theorem is the basis for solving higher degree equations by polynomial division . According to the fundamental theorem of algebra :

Every (normalized) polynomial -th degree with coefficients in the complex numbers can be represented as the product of linear factors : ${\displaystyle n}$ ${\displaystyle n}$

${\displaystyle P(x)=x^{n}+a_{n-1}x^{n-1}+\dotsb +a_{2}x^{2}+a_{1}x^{1}+a_{0}=(x-x_{1})(x-x_{2})\dotsm (x-x_{n}).}$

${\displaystyle x_{1},x_{2},\dotsc ,x_{n}}$are the roots of the polynomial; even if all coefficients are real, the zeros can be complex. Not all have to be different. ${\displaystyle a_{0},a_{1},\dotsc ,a_{n-1}}$ ${\displaystyle x_{i}}$

Now the Vieta theorem results from multiplying and comparing coefficients :

${\displaystyle a_{n-j}=(-1)^{j}\sigma _{j},\quad j=1,\dotsc ,n}$,

in which

${\displaystyle \sigma _{k}=\sum _{1\leq i_{1}<i_{2}<\dotsb <i_{k}\leq n}x_{i_{1}}\cdots x_{i_{k}}}$

are the so-called elementary symmetric polynomials in to . For a fourth degree polynomial ${\displaystyle x_{1}}$ ${\displaystyle x_{n}}$

${\displaystyle P(x)=x^{4}+a_{3}\cdot x^{3}+a_{2}\cdot x^{2}+a_{1}\cdot x+a_{0}=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})}$

surrendered:

${\displaystyle {\begin{array}{r c l c l}-a_{3}&=&\sigma _{1}&=&x_{1}+x_{2}+x_{3}+x_{4}\\a_{2}&=&\sigma _{2}&=&x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}\\-a_{1}&=&\sigma _{3}&=&x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4}\\a_{0}&=&\sigma _{4}&=&x_{1}x_{2}x_{3}x_{4}\\\end{array}}}$

An important application of the theorem for and is the reduction of the cubic equation to a quadratic equation and the equation of the 4th degree to a cubic equation, the so-called cubic resolvent . ${\displaystyle n=2}$ ${\displaystyle n=3}$

In general, Vieta's theorem of roots also applies to polynomials with coefficients in other fields , as long as these are only algebraically closed .

literature

• Walter Gellert: Lexicon of Mathematics . Leipzig: Bibliographisches Institut, 1990, pp. 578, 200.

Web link

• Proof by direct invoice (for students) on Rudolf Brinkmann's website

Individual evidence

1. ^ Heinz-Wilhelm Alten : 4000 years of algebra. History, cultures, people . Springer, Berlin a. a. 2003, ISBN 3-540-43554-9 , pp. 268 .