Question

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

Answer 1

Given, $(a-b)x^2+(b-c)x+(c-a)=0$ are equal.

Then the discriminant $=0$.

Then

$(b-c{)}^2-4(c-a\left)\right(a-b)=0$

or, $(b^2-2bc+c^2)-4(ac-bc-a^2+ab)=0$

or, $(b^2+2bc+c^2)-2a(b+c)+4a^2=0$

or, $(b+c{)}^2-2a(b+c)+4a^2=0$

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

or, $2a=b+c$.