Question

If ad ≠ bc, then prove that the equation (a² + b²)x² + 2 (ac + bd)x + (c² + d²) = 0 has no real roots.

Answer 1

The given equation is :

(a² + b²)x² + 2(ac + bd)x + (c² + d²) = 0

Here, A = (a² + b²), B = 2(ac + bd) and C = (c² + d²)

⇒ B² – 4AC = 0

⇒ 12(ac + bd)]² – 4 × (a² + b²) × (c² + d²) = 0

⇒ 4(a²c² + b²d² + 2abcd) – 4(a²c² + a²d² + b²c² + b²d²) = 0

⇒ 4a²c² + 4b²d² + 8abcd – 4a²c² – 4a²c² – 4b²c² – 4b²d² = 0

⇒ – 4a²d² – 4b²c² + 8abcd = 0

⇒ – 4 (a²d² + b²c² – 2abcd) = 0

⇒ – 4(ad – bc)² = 0

Since ad ≠ bc = 0 Therefore D < 0

Hence, the equation has no real roots.