Question

(a^2+b^2)x^2 + 2 (ac+bd)x + (c^2+d^2) = 0 if ad is nt equal to bc prove it​

Answer 1

Answer:

This quadratic in x has no real solutions when ad ≠ bc.

Explanation:

Recall that a quadratic

Ax² + Bx + C = 0

has real roots only if the discriminant B²-4AC ≥ 0.

For the given quadratic, the discriminant is

[ 2 ( ac + bd ) ]² - 4 ( a² + b² ) ( c² + d² )

= 4 [ ( ac + bd )² - ( a² + b² ) ( c² + d² ) ]

= 4 [ a²c² + b²d² + 2abcd - a²c² - a²d² - b²c² - b²d² ]

= 4 [ 2abcd - a²d² - b²c² ]

= -4 [ a²d² + b²c² - 2abcd ]

= -4 ( ad - bc )²

< 0,  when ad ≠ bc.

Since the discriminant is negative when ad ≠ bc, the quadratic has no real roots.

Incidentally, if ad = bc, the discriminant is equal to 0, so there is exactly one real root in that case.