If the roots of (a – b)x2 + (b – c)x + (c – a) = 0 are real and

1.	If the roots of (a – b)x2 + (b – c)x + (c – a) = 0 are real and equal, then prove that b, a, c are in arithmetic progression.
Answer» (a – b)x² + (b – c)x + (c – a) = 0 A = (a – b), B = (b – c), C = (c – a) Δ = b² – 4ac = 0 ⇒ (b – c)² – 4(a – b)(c – a) ⇒ b² – 2bc + c² - 4 (ac – bc – a² + ab) ⇒ b² – 2bc + c² – 4ac + 4bc + 4a² – 4ab = 0 ⇒ 4a² + b² + c² + 2bc – 4ac – 4ab = 0 ⇒- (-2a + b + c)² = 0 [∵ (a + b + c) = a²+ b² + c²+ 2ab + 2bc + 2ca)] ⇒ 2a + b + c = 0 ⇒ 2 a = b + c ∴ a, b, c are in A.P.

If the roots of (a – b)x2 + (b – c)x + (c – a) = 0 are real and equal, then prove that b, a, c are in arithmetic progression.

Answer»

(a – b)x² + (b – c)x + (c – a) = 0

A = (a – b), B = (b – c), C = (c – a)

Δ = b² – 4ac = 0

⇒ (b – c)² – 4(a – b)(c – a)

⇒ b² – 2bc + c² - 4 (ac – bc – a² + ab)

⇒ b² – 2bc + c² – 4ac + 4bc + 4a² – 4ab = 0

⇒ 4a² + b² + c² + 2bc – 4ac – 4ab = 0

⇒- (-2a + b + c)² = 0 [∵ (a + b + c) = a²+ b² + c²+ 2ab + 2bc + 2ca)]

⇒ 2a + b + c = 0

⇒ 2 a = b + c

∴ a, b, c are in A.P.