Question

Answer this question..... Number 17​

Answer 1

For no real roots, discriminant must be less than 0.

We have to prove D < 0

D = B² - 4 A C

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

= 4a²c² + 4b²d² + 8abcd - 4 (a²c² +a²d² + b²c² + b²d²)

= 4a²c² + 4b²d² + 8abcd - 4a²c² - 4a²d² - 4b²c² -4b²d²

= 8abcd - 4a²d² - 4b²c²

= 8abcd - (2ad)² - (2bc)²

= - ((2ad)² +(2bc)² - 8abcd)

= - (2ad - 2bc)²

= -4 (ad-bc)²

Since ad≠bc given, so Discriminant can't be zero.

and since (ad-bc)² is always positive, multiplying it by -4 will always yield a negative answer.

As discriminant is less than 0, there are no real roots.

Answer 2

hope it helps..........