If the roots of the equation (a – b) x2 + (b – c) + (c – a) = 0 are equal, prove that b + c = 2a.

As the equation is in the form Ax2 + Bx + C = 0 with non-zero A.


In which,


A = a - b


B = b - c


C = c - a


And we know that if the roots of a equation are equal then we have


Discriminant, D = 0


Where, D = b2 - 4ac


b2 - 4ac = 0


(b - c)2 - 4(a - b)(c - a) = 0


(b - c)2 + 4(a - b)(a - c) = 0


b2 + c2 - 2bc + 4(a2 - ac - ab + bc) = 0


b2 + c2 - 2bc + 4a2 - 4ac - 4ab + 4bc = 0


4a2+ b2 + c2 - 4ac + 2bc - 4ac = 0


(-2a)2+ b2 + c2 + 2(-2a)c + 2bc + 2(-2a)c = 0


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


[using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2za ]


-2a + b + c = 0


b + c = 2a


Hence Proved.


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