If the roots of the equation (a — b)x² + (b — c) x + (c — a) = 0 are equal, prove that 2a = b + c.

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.