solve
Asked by | 13th Jan, 2010, 09:58: PM
Expert Answer:
If the two roots are equal then,
B2-4AC = 0
(b-c)2 - 4(a-b)(c-a) =
b2-2bc+c2-4ac+4a2+4bc-4ab = 0
b2+2bc+c2-4ac+4a2-4ab = 0
b2+2bc+c2 = 4a(c-a+b)
(b+c)2 = 4a(c-a+b)
(b+c)2 = 4a(c+b - a)
Let's put b+c = 2a
(2a)2 = 4a(2a - a)
4a2 = 4a2
LHS = RHS
Hence 2a = b+c.
regards,
Team,
TopperLearning.
Answered by | 15th Jan, 2010, 09:57: AM
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