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If the roots of the equation (a - b)x2 + (b - c) x + (c - a) = 0 are equal, prove that 2a = b + c. |
Answer» Since roots are equal ∴ d=0 (1) (a — b)x2 + (b — c) x + (c — a) = 0 d = b2 – 4ac d = (b–c)2 – 4 (a–b) (c–a) d = b2 + c2 – 2bc –4 [a (c – a) – b (c – a)] d = b2 + c2 – 2bc – 4 [ac – a2 – bc + ba] From (1), d = 0 ∴ Equation will be: 0 = b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ba b2 + c2 – (2a)2 2bc + 2c (–2a) + 2(–2a)b = 0 (b + c – 2a)2 = 0 (b + c – 2a) = 0 b + c = 2a Hence proved. |
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