Answer :

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

ax^{2} + bx + c = 0

a = b – c

b = c – a

c = a – b

Since the quadratic equations have real and equal roots,

b^{2} – 4ac = 0 for real and equal roots

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

⟹ c^{2} – 2ac + a^{2} – [4 × (ba – b^{2} – ca – bc)] = 0

⟹ c^{2} – 2ac + a^{2} – [4ba – 4b^{2} – 4ca + 4bc] = 0

⟹ c^{2} – 2ac + a^{2} – 4ba + 4b^{2} + 4ca - 4bc = 0

⟹ c^{2} + a^{2} – 4ba + 4b^{2} + 2ac - 4bc = 0

⟹ a^{2} + 4b^{2} + c^{2} – 4ab - 4bc + 2ac = 0

⟹ a^{2} + (-2b)^{2} + c^{2} + 2 × a(-2b) + 2 × (-2b)c + 2 × ac = 0

We have the following formula:

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 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.

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