Q. 94.7( 11 Votes )

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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