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

We know, D = b² – 4ac

Thus,

D = [2(ac + bd)²] – 4(a² + b²)(c² + d²)

= [4(a²c² + b²d² + 2abcd)] – 4(a² + b²)(c² + d²)

= 4[(a²c² + b²d² + 2abcd) – (a²c² + a²d² + b²c² + b²d²)]

= 4[a²c² + b²d² + 2abcd – a²c² – a²d² – b²c² – b²d²]

= 4[2abcd – b²c² – a²d²]

= –4[a²d² + b²c² – 2abcd]

= –4[ad – bc]²

But we know that ad ≠ bc

Therefore, 

(ad – bc) ≠ 0

⇒ (ad – bc)² > 0

⇒ –4(ad – bc)² < 0

⇒ D < 0

Hence, the given equation has no real roots.