Question

If $a(b-c)x^2+b(c-a)x+c(a-b)=0$ has equal roots, then P.T.o $\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$

Answer 1

$\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$

Answer 2

The sum of coefficients is $a(b-c)+b(c-a)+c(a-b) = ab-ac+bc-ab+ac-bc=0$. This implies $x=1$ is a root. For equal roots, $x=1$ must be a double root. For a quadratic $Ax^2+Bx+C=0$ to have a double root at $x=1$, we need $C=A$. Thus, $c(a-b) = a(b-c)$. Expanding gives $ac-bc = ab-ac$, so $2ac = ab+bc$. Dividing by $abc$ (assuming $a,b,c \neq 0$) yields $\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$.