Question

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

Answer 1

Given quadratic equation is (a – b)x2 + (b – c)x + (c – a) = 0
Since the root are equal, discriminent of the quadritic equation = 0
Hence (b – c)2 = 4(a – b)(c – a)
⇒ b2 + c2 – 2bc = 4(ac – a2 – bc + ab)
⇒ b2 + c2 – 2bc = 4ac – 4a2 – 4bc + 4ab
⇒ b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0
⇒ b2 + c2 + 4a2 + 2bc – 4ac – 4ab = 0
⇒ b2 + c2 + (2a)2 + 2(b)(c) – 2(2a)c – 2(2a)b = 0
⇒ b2 + c2 + (–2a)2 + 2(b)(c) + 2(–2a)c + 2(–2a)b = 0
⇒ (b + c – 2a)2 = 0
⇒ b + c – 2a = 0
∴ b + c = 2a

Answer 2

plz refer to this attachment