Question

if the roots of the equation (a-b)x^2+(b-c)x+(c-a)=0 are equal proove that 2a=b+c​

Answer 1

Roots are equal. So discriminant is zero.

Therefore,

$(b-c)^2-(4(a-b)(c-a)) = 0 \\ \\ (b^2-2bc+c^2)-(4(ac-a^2-bc+ab)) = 0 \\ \\ (b^2-2bc+c^2)-(4(ab-bc+ac-a^2)) = 0 \\ \\ (b^2-2bc+c^2)-(4ab-4bc+4ac-4a^2)=0 \\ \\ b^2-2bc+c^2-4ab+4bc-4ac+4a^2=0 \\ \\ 4a^2+b^2+c^2-4ab+2bc-4ac=0 \\ \\ (2a-b-c)^2=0 \\ \\ 2a-b-c=0 \\ \\ 2a=b+c$

Hence proved!

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#adithyasajeevan

Answer 2

Answer:

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Step-by-step explanation: