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.