If the roots of the equation (a-b)x^{2} + (b-c) x+ (c - a)= 0 are equal. Prove that 2a=b+c.
Ans: (a-b)x^{2} + (b-c) x+ (c - a) = 0
T.P 2a = b + c
B^{2} - 4AC = 0
(b-c)^{2} - [4(a-b) (c - a)] = 0
b^{2}-2bc + c^{2} - [4(ac-a^{2} - bc + ab)] = 0
⇒ b^{2}-2bc + c^{2} - 4ac + 4a^{2} + 4bc - 4ab = 0
⇒ b^{2}+ 2bc + c^{2} + 4a2 - 4ac - 4ab= 0
⇒ (b + c - 2a)^{2} = 0
⇒ b + c = 2a