????

Asked by nmishra320 | 22nd Jan, 2010, 05:43: PM

Expert Answer:

4x2-2(a2+b2)x+a2b2 = 0

A = 4, B = -2(a2+b2), C = a2b2

x = {-B±(B2-4AC)}/2A

= {2(a2+b2) ± (4(a2+b2)2 - 4x4xa2b2)}/8

={2(a2+b2) ±2((a2+b2)2 - 4a2b2)}/8

= {(a2+b2) ±((a2+b2)2 - 4a2b2)}/4

={(a2+b2) ±(a2-b2)}/4

= a2/2 or b2/2

Hence the zeros are, a2/2 and  b2/2.

Regards,

Team,

TopperLearning.

Answered by  | 22nd Jan, 2010, 08:50: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.