Question

If the roots if the equation $(a-b)x^2+(b-c)x+c-a=0$, are equal. Prove that a,b,c are in AP.

Answer 1

Given quadratic equation is
$(a-b)x^2+(b-c)x+(c-a)=0$
Since the root are equal, discriminent of the quadritic equation = 0
$Hence(b-c{)}^2=4(a-b\left)\right(c-a)\Rightarrow b^2+c^2-2bc=4(ac-a^2-bc+ab)\Rightarrow b^2+c^2-2bc=4ac-4a^2-4bc+4ab\Rightarrow b^2+c^2-2bc-4ac+4a^2+4bc-4ab\Rightarrow b^2+c^2+4a^2+2bc-4ac-4ab=0\Rightarrow b^2+c^2+(-2a{)}^2+2\left(b\right)\left(c\right)+2(-2a)c+2(-2a)b=0\Rightarrow (b+c-2a{)}^2=0\Rightarrow b+c-2a=0\therefore b+c=2a$
Hece a,b and c are in AP.