If the roots of the quadratic equation (b – c)x² + (c – a)x + (a – b) = 0 are real and equal then prove that 2b = a + c.

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = b – c

b = c – a

c = a – b

Since the quadratic equations have real and equal roots,

b² – 4ac = 0 for real and equal roots

⟹ (c – a) ² – [4 × (b – c) × (a – b)] = 0

⟹ c² – 2ac + a² – [4 × (ba – b² – ca – bc)] = 0

⟹ c² – 2ac + a² – [4ba – 4b² – 4ca + 4bc] = 0

⟹ c² – 2ac + a² – 4ba + 4b² + 4ca - 4bc = 0

⟹ c² + a² – 4ba + 4b² + 2ac - 4bc = 0

⟹ a² + 4b² + c² – 4ab - 4bc + 2ac = 0

⟹ a² + (-2b)² + c² + 2 × a(-2b) + 2 × (-2b)c + 2 × ac = 0

We have the following formula:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

So according to the formula

⟹ (a + (-2b) + c) ²= 0

Taking square root of both sides

⟹ a + (-2b) + c = 0

∴ 2b = a + c

Hence Proved.