Biquadratic equations examples for JEE – Free JEE Math Tutorial

Biquadratic Equations

Key Definitions

  1. Biquadratic Equation: A polynomial equation of the fourth degree, typically of the form ax⁴ + bx³ + cx² + dx + e = 0, where a ≠ 0.
  2. Pure Biquadratic Equation: An equation containing only even powers of x, expressed as ax⁴ + cx² + e = 0.
  3. Standard Form: The monic form x⁴ + px³ + qx² + rx + s = 0, used to analyze relations between roots and coefficients.
  4. Ferrari’s Method: A classical algebraic method used to solve general biquadratic equations by completing squares to reduce them into two quadratic factors.

Theory and Concepts

1. Relation Between Roots and Coefficients

If α, β, γ, δ are the roots of ax⁴ + bx³ + cx² + dx + e = 0, then:

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

2. Reduction to Quadratic Form

For pure biquadratic equations ax⁴ + bx² + c = 0, substitute x² = t. The equation becomes at² + bt + c = 0. After finding t₁ and t₂, solve x = ±√t₁ and x = ±√t₂.

3. Special Forms (Reciprocal Equations)

Equations like ax⁴ + bx³ + cx² + bx + a = 0 are solved by dividing by x²:
a(x² + 1/x²) + b(x + 1/x) + c = 0.
Let z = x + 1/x, then x² + 1/x² = z² – 2.

4. Geometrical Interpretation

A biquadratic function can have up to four real roots, representing the intersection points with the x-axis.

Figure 12.1: A biquadratic function with four distinct real roots.

Solved Examples

Solved Example 1.1

Problem: Solve x⁴ – 5x² + 4 = 0.

Solution

Let x² = t. t² – 5t + 4 = 0 ⇒ (t-4)(t-1) = 0 ⇒ t = 4, 1.
x² = 4 ⇒ x = ±2; x² = 1 ⇒ x = ±1.
Roots: {±1, ±2}.

Solved Example 1.2

Problem: If α, β, γ, δ are roots of x⁴ – 10x³ + 35x² – 50x + 24 = 0, find Σ α².

Solution

Σ α = 10, Σ αβ = 35.
Σ α² = (Σ α)² – 2Σ αβ = 100 – 70 = 30.

Solved Example 1.3

Problem: Solve x⁴ + x³ – 4x² + x + 1 = 0.

Solution

Divide by x²: (x² + 1/x²) + (x + 1/x) – 4 = 0. Let z = x + 1/x.
(z² – 2) + z – 4 = 0 ⇒ z² + z – 6 = 0 ⇒ z = -3, 2.
If z = 2: x = 1, 1. If z = -3: x² + 3x + 1 = 0 ⇒ x = (-3 ± √5)/2.

Solved Example 1.4

Problem: Solve (x+1)(x+2)(x+3)(x+4) = 120.

Solution

Rearrange: (x²+5x+4)(x²+5x+6) = 120. Let y = x²+5x.
(y+4)(y+6) = 120 ⇒ y² + 10y – 96 = 0 ⇒ y = -16, 6.
y = 6 ⇒ x²+5x-6 = 0 ⇒ x = 1, -6. (y = -16 gives non-real roots).

Solved Example 1.5

Problem: Find the condition that roots of x⁴ + px³ + qx² + rx + s = 0 are in G.P.

Solution

Let roots be a/r³, a/r, ar, ar³. Product s = a⁴. The condition is r² = p²s.

Solved Example 1.6

Problem: If the product of two roots of x⁴ – 5x³ + 10x² + kx + 12 = 0 is 6, find k.

Solution

αβ = 6 ⇒ γδ = 2 (since αβγδ = 12).
Equation: (x² + px + 6)(x² + qx + 2) = 0.
Comparing coefficients, we find k = -14.

Solved Example 1.7

Problem: Determine the nature of roots of x⁴ + 4x – 1 = 0.

Solution

f'(x) = 4x³ + 4 = 0 ⇒ x = -1. f(-1) = -4. Since the minimum value is negative and f(x) → ∞, there are 2 real roots.

Solved Example 1.8

Problem: Solve x⁴ – 10x² + 9 = 0.

Solution

(x² – 9)(x² – 1) = 0 ⇒ x² = 9, 1. Roots: {±1, ±3}.

Solved Example 1.9

Problem: Form a biquadratic with rational coefficients having √2 + √3 as a root.

Solution

x = √2 + √3 ⇒ x² = 5 + 2√6 ⇒ x² – 5 = 2√6.
(x² – 5)² = 24 ⇒ x⁴ – 10x² + 1 = 0.

Solved Example 1.10

Problem: Solve x⁴ – 12x³ + 49x² – 78x + 40 = 0 if roots are in A.P.

Solution

4a = 12 ⇒ a = 3. Using product of roots, d = 1. Roots: {1, 2, 4, 5}.

Concept Application Exercise 1.1

  1. Solve x⁴ – 13x² + 36 = 0.

  2. If α, β, γ, δ are roots of x⁴ + px³ + qx² + rx + s = 0, find Σ α²β.

  3. Solve x² + 1/x² – 3(x – 1/x) – 2 = 0.

  4. Find the number of real roots of x⁴ – 8x² + 2 = 0.

  5. If two roots of x⁴ – 2x³ + 4x² + 6x – 21 = 0 are equal in magnitude but opposite in sign, solve the equation.

  6. Solve (x-1)⁴ + (x-5)⁴ = 82 by substituting y = x-3.

  7. If roots of x⁴ – 8x³ + 14x² + 8x – 15 = 0 are in A.P., find the roots.

  8. Solve the reciprocal equation 2x⁴ + x³ – 6x² + x + 2 = 0.

  9. Find the product of the non-real roots of x⁴ – 4x³ + 6x² – 4x + 5 = 0.

  10. If α is a root of x⁴ – x – 1 = 0, find the value of α⁴ – α.