Question

The equation x⁴ - 2x³ - 3x² + 4x - 1 = 0 has four distinct real rootsx₁, x₂, x₃, x₄ such that x₁ < x₂ < x₃ < x₄. Prove x₁ x₂ + x₁ x₃ + x₂ x₄ + x₃ x₄ = -3

Answer 1

Given equation is x⁴ - 2x³ - 3x² + 4x - 1 = 0

roots are x₁, x₂, x₃, x₄ such that x₁ < x₂ < x₃ < x₄.

x₁ x₂ + x₁x₃ + x₁x₄ + x₂x₃ + x₂x₄ + x₃x₄ = \(\frac{c}a = -3\)-----(i)

Given that the product of two roots is units.

Let x₁x₄ = 1

Also product of roots is x₁ x₂ x₃ x₄ = \(\frac{e}a=-1\)

⇒ x₂ x₃ = -1

(\(\because\) x₂ x₄ = 1 by assuming)

\(\therefore\) From (i), we get

x₁ x₂ + x₁ x₃ + 1 - 1 + x₂ x₄ + x₃ x₄ = -3

⇒ x₁ x₂ + x₁ x₃ + x₂ x₄ + x₃ x₄ = -3