Symmetric Functions of Roots
Key Definitions
- Symmetric Function: A function of the roots α, β, … is symmetric if its value remains unchanged when any two roots are interchanged. Example: f(α, β) = α² + β².
- Elementary Symmetric Functions: For a quadratic equation, these are the basic building blocks: e₁ = α + β and e₂ = αβ.
- Power Sums: The sum of the nᵗʰ powers of the roots, denoted by Sₙ = αⁿ + βⁿ.
- Newton’s Sums: A recursive relationship used to calculate high-power sums Sₙ using polynomial coefficients.
Theory and Concepts
1. Expressing Symmetric Functions in terms of Coefficients
Every symmetric polynomial of roots for ax² + bx + c = 0 can be written using (-b/a) and (c/a):
- α² + β² = (α + β)² – 2αβ
- α³ + β³ = (α + β)³ – 3αβ(α + β)
- (α – β)² = (α + β)² – 4αβ
- 1/α² + 1/β² = (α² + β²) / (αβ)²
2. Newton’s Theorem for Quadratic Equations
If α and β are roots of ax² + bx + c = 0 and Sₙ = αⁿ + βⁿ, then for n > 2:
This identity is a major shortcut for JEE Advanced problems involving high powers like S₁₀ or S₂₀₂₆.
3. Geometrical Insight into Root Symmetry
The roots of a quadratic are always symmetric about the vertical line passing through the vertex.
Figure 10.1: Symmetry of roots about the axis x = -b/2a
Solved Examples
Solved Example 1.1
Problem: If α, β are roots of x² – 5x + 6 = 0, find the value of α⁴ + β⁴.
Solution
α² + β² = (α+β)² – 2αβ = 25 – 12 = 13.
α⁴ + β⁴ = (α² + β²)² – 2(αβ)² = (13)² – 2(6)² = 169 – 72 = 97.
Solved Example 1.2
Problem: Let α, β be the roots of x² – 6x – 2 = 0. If aₙ = αⁿ – βⁿ, find (a₁₀ – 2a₈) / 2a₉.
Solution
For n=10: a₁₀ – 6a₉ – 2a₈ = 0 ⇒ a₁₀ – 2a₈ = 6a₉.
Thus, (6a₉) / (2a₉) = 3.
Solved Example 1.3
Problem: If α, β are roots of x² + px + q = 0, find the equation whose roots are α² and β².
Solution
New product P = (αβ)² = q².
Equation: x² – (p² – 2q)x + q² = 0.
Solved Example 1.4
Problem: Find the value of 1/α³ + 1/β³ for x² – 4x + 1 = 0.
Solution
1/α³ + 1/β³ = (α³ + β³) / (αβ)³ = [(α+β)³ – 3αβ(α+β)] / 1
= 4³ – 3(1)(4) = 64 – 12 = 52.
Solved Example 1.5
Problem: If α, β are roots of x² – x – 1 = 0, find S₅ = α⁵ + β⁵.
Solution
S₂ = 1 + 2 = 3; S₃ = 3 + 1 = 4;
S₄ = 4 + 3 = 7; S₅ = 7 + 4 = 11.
Solved Example 1.6
Problem: If α/β and β/α are the roots of ax² + bx + c = 0, find the relation between a, b, c.
Solution
Product of roots = (α/β) · (β/α) = 1. Thus, c/a = 1 ⇒ c = a.
Solved Example 1.7
Problem: Find (α – β)² if α, β are roots of 3x² – 5x + 1 = 0.
Solution
(α – β)² = (α + β)² – 4αβ = (25/9) – 4/3 = 13/9.
Result: 13/9.
Solved Example 1.8
Problem: Find the sum of roots of x² – |x| – 6 = 0.
Solution
Let |x| = y ≥ 0. y² – y – 6 = 0 ⇒ (y-3)(y+2) = 0.
y = 3 ⇒ |x| = 3 ⇒ x = ±3. Sum = 0.
Solved Example 1.9
Problem: If α, β are roots of ax² + bx + c = 0, find α²β + β²α.
Solution
α²β + β²α = αβ(α + β) = (c/a)(-b/a) = -bc/a².
Solved Example 1.10
Problem: Evaluate √α + √β for x² – 10x + 9 = 0.
Solution
K² = 10 + 2√(9) = 10 + 6 = 16.
K = 4.
Concept Application Exercise 1.1
If α, β are roots of x² – 7x + 1 = 0, find the value of α⁴ + β⁴.
Let α, β be roots of x² – 6x – 2 = 0. If Vₙ = αⁿ + βⁿ, find V₁₀ – 6V₉ – 2V₈.
Find the equation whose roots are 1/α² and 1/β² for ax² + bx + c = 0.
If α, β are roots of x² – ax + b = 0, find the value of α³ + β³.
Solve for x if the sum of squares of the roots of x² – (k-2)x – (k+1) = 0 is minimum.
If (α – β) = 5 and αβ = 6, form the quadratic equation.
Find the value of Σ(α/β) for the equation x² – 3x + 2 = 0.
If α² = 5α – 3 and β² = 5β – 3, find the equation whose roots are α/β and β/α.
If α, β are roots of x² + px + 1 = 0 and γ, δ are roots of x² + qx + 1 = 0, find (α-γ)(β-γ)(α+δ)(β+δ).
For the equation x² – 2x + 4 = 0, find the value of αⁿ + βⁿ using De Moivre’s Theorem.