Prove that the equation x²(a² + b²) + 2x(ac + bd) + (c² + d²)= 0 has no real root, if ad ≠ bc.

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

d = b² – 4ac

d = (2ac + 2bd)² – 4 (a² + b²) (c² + d²)

d = 4a²c² + 4b²d² + 8abcd – 4 [a² (c² + d²) + b² (c²+d²)]

d = 4a²c² + 4b²d² + 8abcd – 4 [a²c² + a²d² + b²c² + b²d²]

d = 4a²c² + 4b²d² + 8abcd – 4a²c² – 4a²d² – 4b²c² – 4b² d²

d = 8abcd – 4a²d² – 4b²c²

d = 8abcd – 4(a²d² + b² c²)

d = –4 (a² d² + b²c² – 2abcd)

d = –4 [(ad + bc)²]

For ad ≠ bc

d= –4 × [value of (ad + bc)²]

∴ d is always negative

So, d < 0

The given equation has no real roots.