Question

prove that the equation (open a square + b [close X square + 2) open AC + BD (close X + c square) + D square has no real root​

Answer 1

Step-by-step explanation:

a² + b²)x² -2( ab + cd)x +( c² + d²) = 0

roots are equal so, D = b² -4ac =0

{2(ab + cd)}² -4(a² +b²)(c² + d²) =0

4(ab+ cd)² -4(a² + b²)(c²+ d²) =0

( a²b²+c²d² +2abcd ) -a²c²-a²d²-b²d² -b²c² =0

-a²c² -b²d² + 2abcd =0

-( a²c² + b²d² -2abcd) =0

{(ac-bd)²} =0

ac -bd =0

ac = bd

a/b = d/c

hence proved /////