????
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
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