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

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Solution

Given $(a - b) x^{2} + (b - c) x + (c - a) = 0$

Therefore, $D = (b - c)^{2} - 4 (a - b) (c - a)$

For real and equal roots, $D = 0$ .

Now,

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

$4 a^{2} + b^{2} + c^{2} - 4 a b + 2 b c - 4 a c = 0$ $[∵ (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + x z)]$

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

$- 2 a + b + c = 0$

$b + c = 2 a$

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

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