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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