Difference between revisions of "Vieta's Formulas"
m (→Proving Vieta's Formula) |
m (→Introduction) |
||
Line 3: | Line 3: | ||
== Introduction == | == Introduction == | ||
− | Vieta's Formulas were discovered by the French mathematician [[François Viète]]. | + | 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 <math>x^2+ax+b=0</math> with solutions <math>p</math> and <math>q</math>, then we know that we can factor it as: | 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 <math>x^2+ax+b=0</math> with solutions <math>p</math> and <math>q</math>, then we know that we can factor it as: | ||
Line 43: | Line 43: | ||
If we denote <math>\sigma_k</math> as the <math>k</math>-th elementary 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>. | If we denote <math>\sigma_k</math> as the <math>k</math>-th elementary 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>. | ||
Also, <math>-b/a = p + q, c/a = p \cdot q</math>. | Also, <math>-b/a = p + q, c/a = p \cdot q</math>. | ||
− | |||
==Proving Vieta's Formula== | ==Proving Vieta's Formula== | ||
When is this going to come out? | When is this going to come out? |
Revision as of 05:54, 13 October 2020
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:
![$x^2+ax+b=(x-p)(x-q)$](http://latex.artofproblemsolving.com/2/0/7/20778e61a5565cccb66e9bf26674af6b92e8ed08.png)
(Note that the first term is , not
.) Using the distributive property to expand the right side we now have
![$x^2+ax+b=x^2-(p+q)x+pq$](http://latex.artofproblemsolving.com/f/e/9/fe9ef40e2e81d7ac64a56d496bce78113e21220e.png)
Vieta's Formulas are often used when finding the sum and products of the roots of a quadratic in the form with roots
and
They state that:
and
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 formulas 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
![$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).$](http://latex.artofproblemsolving.com/4/4/b/44b0375f3fff54d4440ac27ed32a29ea26fa6f10.png)
Expanding out the right-hand side gives us
![$a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.$](http://latex.artofproblemsolving.com/4/6/e/46e7e70100dd310b8f3cb9d5edeaed8c7fbf771a.png)
The coefficient of in this expression will be the
-th elementary 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
![$a_n = a_n$](http://latex.artofproblemsolving.com/b/0/b/b0bdcbc9e71069ce32a5b6c5a3137b1832620d2c.png)
![$a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)$](http://latex.artofproblemsolving.com/e/9/8/e98bba0737992471e152a486d4e12fc9d38f8803.png)
![$a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)$](http://latex.artofproblemsolving.com/e/5/7/e5734e66af5754c9f39706aafe68f3a0eb2bff8c.png)
![$\vdots$](http://latex.artofproblemsolving.com/6/9/c/69cdf7ec84ff918cbe7f21acf7543bf1bd243080.png)
![$a_0 = (-1)^n a_n r_1r_2\cdots r_n$](http://latex.artofproblemsolving.com/1/3/f/13f11a26327cb547c656475c515683b949d4823a.png)
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 elementary symmetric sum, then we can write those formulas more compactly as
, for
.
Also,
.
Proving Vieta's Formula
When is this going to come out?