Relation Between Roots and Coefficients Free JEE Math Tutorial

Relation Between Roots and Coefficients

Key Definitions

1. Elementary Symmetric Functions: For a polynomial equation, the sums and products of roots taken k at a time are known as elementary symmetric functions.
2. Symmetric Expressions: An expression in roots α, β, γ, … is symmetric if interchanging any two roots does not change the value of the expression (e.g., α² + β²).
3. Vieta’s Formulas: These are the formulas that relate the coefficients of a polynomial to the sums and products of its roots.
4. Newton’s Sums: A specialized recurrence relation used to find the sum of powers of roots, denoted as Sₙ = αⁿ + βⁿ + …

Theory and Concepts

1. Quadratic Equations

If α and β are the roots of ax² + bx + c = 0, then:

2. Cubic Equations

If α, β, γ are the roots of ax³ + bx² + cx + d = 0, then:

• Σ α = α + β + γ = -b/a
• Σ αβ = αβ + βγ + γα = c/a
• αβγ = -d/a

3. Higher Degree Polynomials

For the general polynomial aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ = 0:

• Sum of roots taken one at a time: S₁ = -aₙ₋₁/aₙ
• Sum of roots taken two at a time: S₂ = aₙ₋₂/aₙ
• Product of roots: Sₙ = (-1)ⁿ (a₀/aₙ)

4. Newton’s Theorem for Sum of Powers

If α, β are roots of ax² + bx + c = 0 and Vₙ = αⁿ + βⁿ, then for n ≥ 2:

Solved Examples

Solved Example 1.1

If α, β are roots of ax² + bx + c = 0, find the value of 1/α² + 1/β².

Solution

We know α + β = -b/a and αβ = c/a.

Substituting values:

Solved Example 1.2

If the roots of x³ – 12x² + 39x – 28 = 0 are in arithmetic progression, find them.

Solution

Let the roots be A-d, A, A+d.
Sum of roots: (A-d) + A + (A+d) = 12 ⇒ 3A = 12 ⇒ A = 4.
Product of roots: (A-d)A(A+d) = 28 ⇒ 4(16 – d²) = 28.
16 – d² = 7 ⇒ d² = 9 ⇒ d = ± 3.
Roots are {4-3, 4, 4+3} = {1, 4, 7}.

Solved Example 1.3

Find the value of k if one root of x² – 5x + k = 0 is 2 more than the other.

Solution

Let roots be α and α+2.
Sum of roots: α + (α+2) = 5 ⇒ 2α = 3 ⇒ α = 1.5.
Product of roots: α(α+2) = k.
k = (1.5)(3.5) = 5.25.

Solved Example 1.4

If α, β are roots of x² – x – 1 = 0, find the value of α⁵ + β⁵.

Solution

Let Vₙ = αⁿ + βⁿ. From Newton’s Theorem: Vₙ – Vₙ₋₁ – Vₙ₋₂ = 0.
V₀ = 2, V₁ = 1.
V₂ = V₁ + V₀ = 1 + 2 = 3.
V₃ = V₂ + V₁ = 3 + 1 = 4.
V₄ = V₃ + V₂ = 4 + 3 = 7.
V₅ = V₄ + V₃ = 7 + 4 = 11.

Solved Example 1.5

Find the sum of the squares of the roots of xⁿ + nx – 1 = 0.

Solution

Sum of roots S₁ = 0 (since coefficient of xⁿ⁻¹ is 0).
Sum of roots taken two at a time S₂ = 0 (for n > 2).
Σ α² = (Σ α)² – 2Σ αβ = 0² – 2(0) = 0.

Solved Example 1.6

If α, β, γ are roots of x³ + px + q = 0, find Σ α³.

Solution

α + β + γ = 0.
Using the identity: α³ + β³ + γ³ – 3αβγ = (α+β+γ)(α² + β² + γ² – αβ – βγ – γα).
Since Σ α = 0, we have Σ α³ = 3αβγ.
From Vieta: αβγ = -q. Therefore, Σ α³ = -3q.

Solved Example 1.7

If the product of two roots of x³ – 3x² + 4x – k = 0 is 2, find k.

Solution

Let roots be α, β, γ. Given αβ = 2.
From αβγ = k ⇒ 2γ = k ⇒ γ = k/2.
Since γ is a root: (k/2)³ – 3(k/2)² + 4(k/2) – k = 0.
k³/8 – 3k²/4 + k = 0 ⇒ k(k² – 6k + 8) = 0.
Possible values: k = 0, 2, 4.

Solved Example 1.8

Roots of x² – 6x + a = 0 and x² – 24x + b = 0 form a G.P. Find a, b.

Solution

Let roots be r, rk, rk², rk³.
r(1+k) = 6 and rk²(1+k) = 24.
Dividing: k² = 4 ⇒ k = 2 (assuming increasing G.P.).
r(3) = 6 ⇒ r = 2. Roots: 2, 4, 8, 16.
a = 2 × 4 = 8; b = 8 × 16 = 128.

Solved Example 1.9

Find a such that the sum of squares of roots of x² – (a-2)x – a – 1 = 0 is minimum.

Solution

S = (α+β)² – 2αβ = (a-2)² – 2(-a-1) = a² – 2a + 6.
S = (a-1)² + 5. Minimum at a = 1.

Solved Example 1.10

If roots of x² – 10cx – 11d = 0 are a, b and roots of x² – 10ax – 11b = 0 are c, d, find a+b+c+d.

Solution

Summing equations of roots and coefficients yields the relation a+b+c+d = 1210.