Question

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

Answer 1

( a - b)x ^2 +( b - c )x +(c - a) =0
because roots are equal
so,
D = b^2 -4ac =0

now,
(b - c )^2 -4( c - a)( a - b) =0

(b - c )^2 = 4( c -a )( a - b)

b^2 + c^2 -2bc =4{ a(b+c ) -a^2 -bc }

b^2 +c^2 +2bc = 4a(b+c) -4a^2

b^2 +4a^2 +c^2 +2bc -4ab-4ac =0

(b)^2 +(-2a)^2 +(c)^2 +2(bc) +2(-2a)(b) +2(-2a)(c) =0

use (a + b +c )^2= a^2+b^2+c^2+2ab+2bc+2ca

formula in here

{ b + c - 2a}^2 =0

so,
b + c = 2a
hence proved

Answer 2

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
please mark as brainliest answer