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

2
by 426ri543

Log in to add a comment

by 426ri543

Log in to add a comment

The Brainliest Answer!

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

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

Hence Proved