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Vieta's formulas

Revision as of 14:59, 5 November 2021 by Etmetalakret (talk | contribs)

In algebra, Vieta's formulas are a set of formulas that relate the coefficients of a polynomial to its roots.

(WIP)

Statement

Let $P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ be any polynomial with complex coefficients with roots $r_1, r_2, \ldots , r_n$, and let $s_n$ be the elementary symmetric polynomial of the roots with degree $n$. Vietas formulas then state that \[s_1 = r_1 + r_2 + \cdots + r_n = - \frac{a_{n-1}}{a_n}\] \[s_2 = r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n = \frac{a_{n-2}}{a_n}\] \[\vdots\] \[s_n = r_1r_2r_3 \cdots r_n = (-1)^n \frac{a_0}{a_n}.\] This can be compactly written as if $j$ is any integer such that $0<j<n$, then $s_j = (-1)^j \frac{a_{n-j}}{a_n}$.

Proof

Let all terms be defined as above. By the factor theorem, $P(x) = a_n (x-r_1)(x-r_2) \cdots (x-r_n)$; we will then prove these formulas by expanding this polynomial. When expanding this polynomial, every term is generated by $n$ choices whether to include $x$ or $-r_{n-j}$ from any factor $(x-r_{n-j})$.

Consider all the expanded terms of $P(x)$ with degree $j$; they are formed by choosing $j$ of the negative roots, then by making the remaining $n-j$ choices $x$. Thus, every term is equal to a product of $j$ of the negative roots multiplied by $x_{n-j}$. If one factors out $(-1^{j})x_{n-j}$, we are left with the $j$th elementary symmetric polynomial of the roots. Thus, when expanding this product, the coefficient of $x_{n-j}$ is equal to $(-1)^j a_n s_j$. However, we defined the coefficient of $x_{n-j}$ to be $a_{n-j}$. Thus, $(-1)^j a_n s_j$ = a_{n-j}$, or$$s_j = (-1)^j \frac{a_{n-j}}{a_n}$, as required. $\square$

Provide links to problems that use vieta formulas: Examples: https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_21

Proving Vieta's Formula

Basic proof: This has already been proved earlier, but I will explain it more. If we have $x^2+ax+b=(x-p)(x-q)$, the roots are $p$ and $q$. Now expanding the left side, we get: $x^2+ax+b=x^2-qx-px+pq$. Factor out an $x$ on the right hand side and we get: $x^2+ax+b=x^2-x(p+q)+pq$ Looking at the two sides, we can quickly see that the coefficient $a$ is equal to $-(p+q)$. $p+q$ is the actual sum of roots, however. Therefore, it makes sense that $p+q= \frac{-b}{a}$. The same proof can be given for $pq=\frac{c}{a}$.

Note: If you do not understand why we must divide by $a$, try rewriting the original equation as $ax^2+bx+c=(x-p)(x-q)$

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