Question

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

Answer 1

Given equation is (a –b)$x^2$ + (b – c)x + (c – a) = 0

On comparing with A$x^2$ + Bx +C = 0, we get A =(a- b), B = (b – c) and C = (c – a)

Since it has equal roots, Discriminant D = $B^2$ – 4AC = 0

(b- c)${}^2$ – 4(a-b)(c-a)= 0

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

$4a^2+b^2+c^2$ – 4ab + 2bc - 4ac = 0

$(2a{)}^2+b^2+c^2$ + 2(2a) (-b) + 2(-b)(-c) + 2(2a)(-c) = 0

$(2a–b–c{)}^2$ = 0

2a – b – c = 0

2a = b + c