Question

If (a² + b²) x² + 2 (ac + bd) x + c² + d² = 0 has no real roots, then 

A. ad = bc 

B. ab = cd 

C. ac = bd 

D. ad ≠ bc

Answer 1

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 \(\neq\) bc