Question

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

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

If the roots of quadratic equation are equal, discriminant is 0 .

b² - 4ac = 0

(b-c) ² - 4 ( a-b)(c-a) = 0

b²+c²-2bc -4 { ac-a²-bc+ba} = 0

b²+c²-2bc-4ac+4a²+4bc-4ab= 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 .

Hope helped!