Question

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

Answer 1

Since the roots are equal

Discriminent = 0
=> (b-c) ^2 - 4(a-b)(c-a) = 0

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

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

=> - 2a +b +c = 0

=> b+c = 2a