Descartes’ rule of signs pdf jee mains – Free JEE Math Tutorial

Descartes’ Rule of Signs

Key Definitions

  1. Variation of Sign: Occurs in a polynomial f(x) when two successive non-zero coefficients (in descending order) have opposite signs.
  2. Positive Real Root: A root x = α where α ∈ ℝ and α > 0.
  3. Negative Real Root: A root x = α where α ∈ ℝ and α < 0.
  4. Polynomial Bounds: This rule provides an upper bound on the number of real roots; it does not give the exact count but narrows possibilities.

Theory and Concepts

Descartes’ Rule of Signs is a technique to determine the maximum possible number of positive and negative real roots of a polynomial equation f(x) = aₙxⁿ + … + a₀ = 0.

1. Number of Positive Real Roots

The number of positive real roots is either equal to the number of sign changes (V) in f(x) or less than that by an even integer.
P ∈ {V, V-2, V-4, …}

2. Number of Negative Real Roots

The number of negative real roots is either equal to the number of sign changes (V’) in f(-x) or less than that by an even integer.
N ∈ {V’, V’-2, V’-4, …}

3. Analysis of Imaginary Roots

If the total degree of the polynomial is n and the total number of real roots is R, the minimum number of imaginary roots is n – R. Imaginary roots always occur in conjugate pairs.

Figure 13.1: Visualizing root distribution on the Cartesian plane.

Solved Examples

Solved Example 1.1

Problem: Determine the maximum possible number of positive and negative roots for f(x) = x³ + 3x + 1 = 0.

Solution

Signs of f(x): (+, +, +). V = 0. (No positive roots).
f(-x) = -x³ – 3x + 1. Signs: (-, -, +). V’ = 1.
Result: Exactly 1 negative root.

Solved Example 1.2

Problem: Examine the nature of roots of x⁷ – 3x⁴ + 2x³ – 1 = 0.

Solution

f(x) signs: (+ → – → + → -). V = 3. Positive roots: 3 or 1.
f(-x) = -x⁷ – 3x⁴ – 2x³ – 1. Signs: (-, -, -, -). V’ = 0.
Result: No negative roots.

Solved Example 1.3

Problem: Show that x⁴ + 2x² + 3x – 1 = 0 has exactly two imaginary roots.

Solution

f(x): (+, +, +, -) ⇒ V=1. (1 positive root).
f(-x) = x⁴ + 2x² – 3x – 1: (+, +, -, -) ⇒ V’=1. (1 negative root).
Total real roots = 2. Degree = 4.
Imaginary roots = 4 – 2 = 2.

Solved Example 1.4

Problem: Analyze x⁵ – 1 = 0.

Solution

f(x): (+, -) ⇒ 1 positive root (x=1).
f(-x) = -x⁵ – 1: (-, -) ⇒ 0 negative roots. 4 imaginary roots.

Solved Example 1.5

Problem: Discuss roots of x⁴ + 15x² + 7x – 11 = 0.

Solution

f(x): (+, +, +, -) ⇒ 1 pos root. f(-x): (+, +, -, -) ⇒ 1 neg root. 2 imaginary.

Solved Example 1.6

Problem: Find real roots of xⁿ – 1 = 0 where n is even.

Solution

V=1, V’=1. Total real roots = 2 (namely ±1).

Solved Example 1.7

Problem: How many imaginary roots does x⁴ + 3x – 10 = 0 have?

Solution

V=1, V’=1. Real = 2, Degree = 4. Imaginary = 2.

Solved Example 1.8

Problem: Does x⁶ + x⁴ + x² + 1 = 0 have any real roots?

Solution

V=0, V’=0. f(0) ≠ 0. All 6 roots are imaginary.

Solved Example 1.9

Problem: Find min imaginary roots of x⁹ – x⁵ + x⁴ + x² + 1 = 0.

Solution

V=2 (pos: 2 or 0). V’=1 (neg: 1). Max real = 3. Min imaginary = 6.

Solved Example 1.10

Problem: For f(x) = x³ – x² – x – 1 = 0, analyze the roots.

Solution

V=1 (1 pos). f(-x) = -x³ – x² + x – 1. V’=2 (2 or 0 neg).

Concept Application Exercise 1.1

  1. Find the maximum number of positive and negative roots of x⁴ + 3x³ – 5x² + 2x – 1 = 0.

  2. Prove that x⁶ – 3x² – x + 1 = 0 has at least two imaginary roots.

  3. Show that x⁵ + x³ + x = 0 has only one real root.

  4. Determine the nature of roots for 3x⁵ – 2x² – x + 1 = 0.

  5. Discuss the roots of the equation x⁴ + 1 = 0.

  6. How many real roots are possible for x⁷ + x⁵ – x³ + x – 1 = 0?

  7. Show that xⁿ + 1 = 0 has no real roots if n is even.

  8. For x³ + px + q = 0, if p, q > 0, prove all real roots must be negative.

  9. Find the number of imaginary roots of x⁴ + 2x² + 3 = 0.

  10. Analyze the roots of x⁵ + 2x⁴ + 3x³ + 4x² + 5x + 6 = 0.