Study Notes – Roots & Coefficients (Vieta’s Formulas)
Maths Olympiad – Algebra
Vieta’s formulas give a direct link between the coefficients of a monic (or any) polynomial and the symmetric sums of its roots.
They are the bread‑and‑butter of many Olympiad problems that involve roots, sums of powers, or inequalities.
1. Simple Explanation
For a polynomial
[ P(x)=a_nx^n+a_{n-1}x^{,n-1}+\dots +a_1x+a_0 \qquad (a_n\neq 0) ]
with roots (r_1,r_2,\dots ,r_n) (counting multiplicities) we can factor
[ P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n). ]
Expanding the product and comparing coefficients with the standard form yields the Vieta relations:
| Symmetric sum of roots | Coefficient |
|---|---|
| ( \displaystyle\sum_{i} r_i ) | (-\dfrac{a_{n-1}}{a_n}) |
| ( \displaystyle\sum_{i<j} r_i r_j ) | (\dfrac{a_{n-2}}{a_n}) |
| ( \displaystyle\sum_{i<j<k} r_i r_j r_k ) | (-\dfrac{a_{n-3}}{a_n}) |
| … | … |
| ( r_1 r_2 \dots r_n ) | ((-1)^n\dfrac{a_0}{a_n}) |
The signs alternate because each time a factor ((x-r_i)) contributes a (-r_i) when expanding the constant term.
2. Key Points
-
General Form
[ P(x)=a_n\prod_{i=1}^{n}(x-r_i) ] → ( \displaystyle\sum r_i=\frac{-a_{n-1}}{a_n}),
( \displaystyle\sum r_ir_j=\frac{a_{n-2}}{a_n}),
( \displaystyle\sum r_ir_jr_k=\frac{-a_{n-3}}{a_n}), … -
Monic Polynomials
If (a_n=1), the formulas simplify to signs only: [ \sum r_i = -a_{n-1},\quad \sum r_ir_j = a_{n-2},\quad \dots,\quad r_1\cdots r_n = (-1)^n a_0. ] -
Complex Conjugate Roots
For real‑coefficient polynomials, non‑real roots appear in conjugate pairs.
Example: if (a+bi) is a root, then (a-bi) is also a root. -
Symmetric Sums of Powers
Vieta gives first‑order symmetric sums.
Higher‑order sums (e.g., (\sum r_i^2)) can be expressed via Newton’s identities or by using Vieta on a derived polynomial. -
Applications in Olympiad
- Root bounds: compare sums/products to deduce inequalities.
- Constructing polynomials: given a set of symmetric conditions, find a polynomial with those roots.
- Eliminating variables: replace sums of roots with coefficients to reduce equations.
3. Worked Example 1 – Classic Cubic
Problem:
Find the sum of the roots, the sum of the products of the roots two‑at‑a‑time, and the product of the roots of
[
P(x)=x^3-6x^2+11x-6.
]
Solution:
-
Here (a_3=1,; a_2=-6,; a_1=11,; a_0=-6).
-
Apply Vieta:
- (\displaystyle\sum r_i = -\frac{a_2}{a_3}= -(-6)=6).
- (\displaystyle\sum_{i<j} r_ir_j = \frac{a_1}{a_3}=11).
- (r_1r_2r_3 = (-1)^3\frac{a_0}{a_3}= -(-6)=6).
-
Check: The polynomial factors as ((x-1)(x-2)(x-3)).
Sum (1+2+3=6), pairwise sum (1\cdot2+1\cdot3+2\cdot3=2+3+6=11), product (1\cdot2\cdot3=6).
✅ All matches.
4. Worked Example 2 – Using Vieta in a Problem
Problem (IMO‑style):
Let (P(x)=x^4+ax^3+bx^2+cx+d) have real coefficients.
Suppose the roots satisfy (r_1+r_2+r_3+r_4=0) and (r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=1).
Determine (d) (the constant term) given that (r_1r_2r_3r_4