solve

If the two roots are equal then,

B²-4AC = 0

(b-c)² - 4(a-b)(c-a) =

b²-2bc+c²-4ac+4a²+4bc-4ab = 0

b²+2bc+c²-4ac+4a²-4ab = 0

b²+2bc+c² = 4a(c-a+b)

(b+c)² = 4a(c-a+b)

(b+c)² = 4a(c+b - a)

Let's put b+c = 2a

(2a)² = 4a(2a - a)

4a² = 4a²

LHS = RHS

Hence 2a = b+c.

regards,

Team,

TopperLearning.