it is given that (a-b)x^2 + (b-c)x + (c – a) = 0

=> d=b^2 – 4ac

it is given that a = (a-b), b= (b-c), c = (c-a)

Now, d = (b -c)^2 – 4 * (a-b)(c-a) = 0

by using identity (a -b)^2 = a^2 + b^2 – 2*a*b

=> b^2 + c^2 -2bc – 4[ac – a^2 -bc + ab] =0

=> b^2 + c^2 -2bc – 4ac +4a^2 + 4 bc – 4ab =0

=> (2a^2) + (b^2) + (c^2) + 2(-2a)(c) + 2(b)(c) + 2(-2a)(b) =0

by using formula

=> a^2 + b^2 + c^2 + 2ab + 2bc +2ca = (a + b+ c)^2

=> -2a + b + c = 0

=> b + c = 2a (proved)

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