Prove that the equation x2(a2 + b2) + 2x(ac + bd) + (c2 + d2)= 0 has

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

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

Answer»

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.