Reciprocal equation in quadratic equation Free JEE Math Tutorial

Reciprocal Equations

Key Definitions

  1. Reciprocal Equation: A polynomial equation f(x) = 0 where, if α is a root, 1/α is also a root.
  2. Reciprocal Polynomial of Type I: A polynomial where coefficients equidistant from the beginning and end are equal (aᵣ = aₙ₋ᵣ).
  3. Reciprocal Polynomial of Type II: A polynomial where coefficients equidistant from the beginning and end are equal in magnitude but opposite in sign (aᵣ = -aₙ₋ᵣ).
  4. Standard Form: The reduced form of a reciprocal equation of even degree 2m obtained by dividing by xᵐ and substituting y = x + 1/x.

Theory and Concepts

1. Properties of Reciprocal Equations

If f(x) = 0 is a reciprocal equation of degree n:

  • Type II equations: x = 1 is always a root (sum of coefficients is zero).
  • Type I equations (odd degree): x = -1 is always a root.
  • Type II equations (even degree): Both x = 1 and x = -1 are roots.

2. The Transformation Method (Even Degree Type I)

Common case: ax⁴ + bx³ + cx² + bx + a = 0.

  1. Divide by x²: a(x² + 1/x²) + b(x + 1/x) + c = 0.
  2. Substitute y = x + 1/x. Note: x² + 1/x² = y² – 2.
  3. Solve the quadratic: a(y² – 2) + by + c = 0.
  4. Solve x² – yx + 1 = 0 for each value of y.

3. Range Constraint

For real x, the substitution y = x + 1/x implies |y| ≥ 2.

Figure 9.1: Graph of y = x + 1/x showing the excluded region (-2, 2)

Solved Examples

Solved Example 1.1

Problem: Solve the equation: x⁴ – 10x³ + 26x² – 10x + 1 = 0.

Solution

Divide by x²: (x² + 1/x²) – 10(x + 1/x) + 26 = 0.
Let y = x + 1/x. Then (y² – 2) – 10y + 26 = 0 ⇒ y² – 10y + 24 = 0.
(y – 4)(y – 6) = 0 ⇒ y = 4 or y = 6.
If y = 4: x² – 4x + 1 = 0 ⇒ x = 2 ± √3.
If y = 6: x² – 6x + 1 = 0 ⇒ x = 3 ± 2√2.

Solved Example 1.2

Problem: Find the real roots of 6x⁴ – 35x³ + 62x² – 35x + 6 = 0.

Solution

6(y² – 2) – 35y + 62 = 0 ⇒ 6y² – 35y + 50 = 0.
y = [35 ± √(1225 – 1200)] / 12 = [35 ± 5] / 12 ⇒ y = 10/3, 5/2.
For y = 10/3: x = 3, 1/3.
For y = 5/2: x = 2, 1/2.

Solved Example 1.3

Problem: Show that x = -1 is a root of x⁵ + 2x⁴ + 3x³ + 3x² + 2x + 1 = 0.

Solution

f(-1) = (-1)⁵ + 2(-1)⁴ + 3(-1)³ + 3(-1)² + 2(-1) + 1
= -1 + 2 – 3 + 3 – 2 + 1 = 0.
Since f(-1) = 0, x = -1 is a root.

Solved Example 1.4

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

Solution

(x² + 1/x²) + (x + 1/x) + 1 = 0 ⇒ y² + y – 1 = 0.
y = (-1 ± √5) / 2.
Since both values of |y| < 2, all roots are non-real.

Solved Example 1.5

Problem: Solve 2x⁴ + x³ – 6x² + x + 2 = 0.

Solution

2(y² – 2) + y – 6 = 0 ⇒ 2y² + y – 10 = 0 ⇒ (2y + 5)(y – 2) = 0.
y = 2 ⇒ x = 1, 1.
y = -5/2 ⇒ x = -2, -1/2.

Solved Example 1.6

Problem: Condition for ax⁴ + bx³ + cx² + bx + a = 0 to have real roots.

Solution

The auxiliary quadratic ay² + by + (c – 2a) = 0 must have roots y such that |y| ≥ 2.

Solved Example 1.7

Problem: Solve x⁴ – 5x² + 1 = 0.

Solution

x² + 1/x² = 5 ⇒ y² – 2 = 5 ⇒ y = ±√7.
x² ∓ √7x + 1 = 0 ⇒ x = (±√7 ± √3) / 2.

Solved Example 1.8

Problem: Solve 12x⁴ – 56x³ + 89x² – 56x + 12 = 0.

Solution

12(y² – 2) – 56y + 89 = 0 ⇒ 12y² – 56y + 65 = 0.
y = 13/6, 5/2.
Roots: {3/2, 2/3, 2, 1/2}.

Solved Example 1.9

Problem: Transform ax⁶ + bx⁵ + cx⁴ + dx³ + cx² + bx + a = 0 into a cubic.

Solution

Divide by x³: a(x³ + 1/x³) + b(x² + 1/x²) + c(x + 1/x) + d = 0.
Using y = x + 1/x:
ay³ + by² + (c – 3a)y + (d – 2b) = 0.

Solved Example 1.10

Problem: Solve |x + 1/x| = 2 cos θ for real x.

Solution

Since |x + 1/x| ≥ 2 and 2 cos θ ≤ 2, equality holds only if cos θ = ±1.

  • If cos θ = 1: x = 1
  • If cos θ = -1: x = -1

Concept Application Exercise 1.1

  1. Solve x⁴ – 4x³ + 5x² – 4x + 1 = 0.

  2. Find the real roots of x⁴ + 1 = 3(x³ + x).

  3. Solve 2x⁴ – 9x³ + 14x² – 9x + 2 = 0.

  4. Solve the equation x⁵ – 1 = 0 using reciprocal equation properties.

  5. Determine the range of a for which x⁴ + ax³ + (2a+1)x² + ax + 1 = 0 has four real roots.

  6. Solve x⁴ – 5/2x³ + 2x² – 5/2x + 1 = 0.

  7. Find the number of real roots of x⁴ + 2x³ – x² + 2x + 1 = 0.

  8. If α is a root of x² – x + 1 = 0, show it satisfies x⁴ – x² + 1 = 0.

  9. Solve 6x⁴ + 5x³ – 38x² + 5x + 6 = 0.

  10. Solve the type II equation: x⁴ + 2x³ – 2x – 1 = 0.