Question

Difference between the corresponding roots of x² + px + q = 0 and x² + qx + p = 0 is same and p ≠ q, then
1. q + p + 4 = 0
2. q – p – 4 = 0
3. p – q + 2 = 0
4. q + q – 4 = 0

Answer 1

Correct Answer - Option 1 : q + p + 4 = 0

Calculation:

Let α, β be roots of equation x² + qx + p = 0

∴ α + β = -q and αβ = p

Let γ, θ are roots of equation x² + px + q = 0

∴ γ + θ = -p and γθ = q

Now,

α – β = γ – θ

⇒ (α – β)² = (γ – θ)²

⇒ (α – β)² - 4αβ = (γ – θ)² - 4γθ

⇒ q² – 4p = p² – 4q

⇒ q² – p² = -4 (q - p)

⇒ (q - p)(q + p + 4) = 0

⇒ q + p + 4 = 0 (q ≠ p)

∴ If difference between the corresponding roots of x² + px + q = 0 and x² + qx + p = 0 is same and p ≠ q, then q + p + 4 = 0.