Cubic equations questions for JEE – Free JEE Math Tutorial

Cubic Equations

Key Definitions

  1. Standard Cubic Equation: An equation of the form ax³ + bx² + cx + d = 0, where a, b, c, d ∈ ℝ and a ≠ 0.
  2. Roots of a Cubic: The values (α, β, γ) that satisfy the equation. A cubic equation must have at least one real root because imaginary roots always occur in conjugate pairs.
  3. Monical Cubic: A cubic equation where the leading coefficient a = 1, represented as x³ + px² + qx + r = 0.
  4. Discriminant of a Cubic: For x³ + px + q = 0, the discriminant Δ = -(4p³ + 27q²) determines the nature of the roots.

Theory and Concepts

1. Relation Between Roots and Coefficients

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

  • Sum of roots: Σ α = α + β + γ = -b/a
  • Sum of roots taken two at a time: Σ αβ = αβ + βγ + γα = c/a
  • Product of roots: αβγ = -d/a

2. Transformation of Cubic Equations

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

  • kα, kβ, kγ: a(x/k)³ + b(x/k)² + c(x/k) + d = 0
  • 1/α, 1/β, 1/γ: dx³ + cx² + bx + a = 0
  • α-h, β-h, γ-h: a(x+h)³ + b(x+h)² + c(x+h) + d = 0

3. Removal of the Quadratic Term (x²)

To remove the x² term from ax³ + bx² + cx + d = 0, substitute x = y – b/3a. This results in the form y³ + Py + Q = 0.

4. Nature of Roots and Geometry

A cubic function can have three distinct real roots, repeated real roots, or one real and two imaginary roots.

Figure 11.1: Geometry of a cubic with three distinct real roots.

Solved Examples

Solved Example 1.1

Problem: If α, β, γ are the roots of x³ – 7x² + 14x – 8 = 0, find α² + β² + γ².

Solution

Σ α = 7, Σ αβ = 14.
Σ α² = (Σ α)² – 2Σ αβ = 7² – 2(14) = 49 – 28 = 21.

Solved Example 1.2

Problem: Find the condition that the roots of x³ – px² + qx – r = 0 are in A.P.

Solution

Let roots be a-d, a, a+d. Sum = 3a = p ⇒ a = p/3.
Substitute x = p/3 in the equation:
(p/3)³ – p(p/3)² + q(p/3) – r = 0 ⇒ 2p³ – 9pq + 27r = 0.

Solved Example 1.3

Problem: If α, β, γ are roots of x³ + qx + r = 0, find the equation whose roots are (β+γ), (γ+α), (α+β).

Solution

α+β+γ = 0 ⇒ β+γ = -α, etc. New roots are -α, -β, -γ.
Substitute x = -y:
(-y)³ + q(-y) + r = 0 ⇒ y³ + qy – r = 0.

Solved Example 1.4

Problem: Solve x³ – 3x² – 16x + 48 = 0 if the sum of two roots is zero.

Solution

Let roots be α, -α, β. Sum = β = 3.
Product = -3α² = -48 ⇒ α² = 16 ⇒ α = ±4.
Roots: {4, -4, 3}.

Solved Example 1.5

Problem: Find the sum of cubes of roots of x³ – px² + qx – r = 0.

Solution

Σ α³ – 3αβγ = (Σ α)(Σ α² – Σ αβ).
Σ α³ = p(p² – 2q – q) + 3r = p³ – 3pq + 3r.

Solved Example 1.6

Problem: Determine the number of real roots of x³ + 3x + 1 = 0.

Solution

f'(x) = 3x² + 3. Since f'(x) > 0 for all x, the function is strictly increasing. Thus, it has exactly 1 real root.

Solved Example 1.7

Problem: If roots of x³ – 6x² + 11x – 6 = 0 are α, β, γ, find Σ (1/αβ).

Solution

Σ (1/αβ) = (α+β+γ) / (αβγ) = 6/6 = 1.

Solved Example 1.8

Problem: Form a cubic equation whose roots are 1, 2, 3.

Solution

S₁ = 6, S₂ = 11, S₃ = 6.
Equation: x³ – 6x² + 11x – 6 = 0.

Solved Example 1.9

Problem: If α, β, γ are roots of x³ + px² + qx + r = 0, find Σ α²β.

Solution

(Σ α)(Σ αβ) = Σ α²β + 3αβγ
Σ α²β = ((-p)(q)) – 3(-r) = 3r – pq.

Solved Example 1.10

Problem: Solve x³ – 9x² + 23x – 15 = 0 if roots are in A.P.

Solution

3a = 9 ⇒ a = 3. Dividing original equation by (x-3) gives:
x² – 6x + 5 = 0 ⇒ (x-5)(x-1) = 0.
Roots: {1, 3, 5}.

Concept Application Exercise 1.1

  1. If α, β, γ are the roots of x³ – 6x² + 11x – 6 = 0, find the value of Σ α³.

  2. Find the condition that the roots of x³ + 3px² + 3qx + r = 0 are in G.P.

  3. If α, β, γ are roots of x³ + ax² + b = 0, find the value of the determinant involving α, β, γ.

  4. Solve x³ – 13x² + 15x + 189 = 0 given that one root exceeds another by 2.

  5. Find the number of real solutions of eˣ = x³.

  6. If the roots of x³ – 12x² + 39x – 28 = 0 are in A.P., find the roots.

  7. If α, β, γ are roots of x³ + px + q = 0, find α⁵ + β⁵ + γ⁵ in terms of p, q.

  8. Solve 4x³ – 24x² + 23x + 18 = 0 if roots are in A.P.

  9. If α, β, γ are roots of x³ – px² + qx – r = 0, find (α+β)(β+γ)(γ+α).

  10. Prove that any cubic equation with real coefficients must have at least one real root.