Since the equation

(a² + b²) x² + 2 (ac + bd) x + c² + d² = 0 has no real root

D < 0

b² – 4ac < 0

b² < 4ac

Here a = (a² + b²), b = 2 (ab + bd), c = c² + d²

4(ac + bd)² – 4 (a² + b²)(c² + d²) < 0

4a²c² + 4b²d² + 8abcd – 4(a²c² + b²c² + a²d² + b²d²) < 0

– 4(a²d² + b²c² – 2abcd) < 0

– 4(ad + bc)²< 0

∴ d is always negative

And ad bc