(a-b) + (b-c) +(c-a) = 0 prove that 2a=b+c
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a-b+b-c+c-a
Here +a and -a , +b and -b ,+c -c get cancelled which is equal to 0
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Given question should be
(a-b)x^2 + (b-c)x + (c - a) = 0
Since the roots are equal b^2 - 4ac = 0 That is (b - c)^2 - 4(a - b)(c - a) = 0
b^2 - 2bc + c^2 - 4(ac - a^2 - bc + ab) = 0 b^2 - 2bc + c^2 - 4ac + 4a^2 + 4bc - 4ab = 0
b^2 + 2bc + c^2 - 4ac + 4a^2 - 4ab = 0
(- 2a)^2 + b^2 + c^2 + 2(- 2a)b + 2bc+ 2c(- 2a) = 0
(- 2a + b + c)^2 = 0 - 2a + b + c = 0 Therefore, 2a = b + c
(a-b)x^2 + (b-c)x + (c - a) = 0
Since the roots are equal b^2 - 4ac = 0 That is (b - c)^2 - 4(a - b)(c - a) = 0
b^2 - 2bc + c^2 - 4(ac - a^2 - bc + ab) = 0 b^2 - 2bc + c^2 - 4ac + 4a^2 + 4bc - 4ab = 0
b^2 + 2bc + c^2 - 4ac + 4a^2 - 4ab = 0
(- 2a)^2 + b^2 + c^2 + 2(- 2a)b + 2bc+ 2c(- 2a) = 0
(- 2a + b + c)^2 = 0 - 2a + b + c = 0 Therefore, 2a = b + c
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