If the roots of the quadratic equation $(a - b) x^{2} + (b - c) x + (c - a) = 0$ are equal, prove that $2 a = b + c$ .

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Solution

$(a - b) x^{2} + (b - c) x + (c - a) = 0 r o o t s a r e e q u a l$

$∴ △ = 0 i . e b^{2} - 4 a c = 0$

${(b - c)}^{2} - 4 (a - b) (c - a) = 0 b^{2} + c^{2} - 2 b c + 4 (a^{2} - a (b + c) + b c) = 0 4 a^{2} + b^{2} + c^{2} - 4 a b + 2 b c - 4 a c = 0 {(2 a - b - c)}^{2} = 0 ∴ 2 a - b - c = 0 2 a = b + c$

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