If both roots of the equation $(a-b)x^2+(b-c)x+(c-a)=0$ are equal, prove that $2a=b+c$.
Things should be known:
Roots of a Quadratic Equations can be identified by:
The roots can be figured out by:
$$\frac{-b \pm \sqrt{d}}{2a},$$
where
$$d=b^2-4ac.$$When the equation has equal roots, then $d=b^2-4ac=0$.
That means $d=(b-c)^2-4(a-b)(c-a)=0$
Best Answer
The expression for $d$ factors as $(2a-b-c)^2$, so we must have $2a-b-c=0$.
The easiest way to see this is to let $x=a-b,y=c-a$ and note that $b-c=-x-y$: $d=(-x-y)^2-4xy=x^2+y^2+2xy-4xy=(x-y)^2$.
Alternative solution: Note that $(a-b)+(b-c)+(c-a)=0$, i.e. $x=1$ is a root of the equation. If both roots are equal, it means that the other root is also $1$, showing that $(a-b)x^2+(b-c)x+(c-a)=(a-b)(x-1)^2$.
Equating the constant term, we find: $c-a=a-b$, which means $2a=b+c$. (This can be done with the coefficient of $x$ too).