Proving a result by making discriminant zero 
If the roots of given Quadratic equation 
  $$a(b-c)x^2 +b(c-a)x + c(a-b)=0$$ 
  are equal, prove the following:
  $$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$$.  

MY approach: 
Method 1: put Discriminant=0 and get stuck.
Method2:add $acx$ on both sides and get $(x-1)$ as factor ,so other root is also 1 and hence the result.
But if someone can prove the result by making discriminant zero (method 1), that would be more rigorous.
Thank you.

Following your first approach, we have that the discriminant is
$$\begin{align}
\Delta&=b^2(c−a)^2-4a(b−c)c(a−b)\\
&=b^2(c−a)^2+4ac(b^2-b(a+c)+ac)\\
&=b^2(c+a)^2+(2ac)^2-2b(a+c)(2ac)\\
&=(b(a+c)-2ac)^2
\end{align}.$$
and by letting $\Delta=0$, it is easy to show that $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$.
As regards your second approach, since you already noted that $x=1$ is a root, we have a double root if and only if the following polynomial identity holds (polynomial factorization is unique):
$$a(b-c)x^2 +b(c-a)x + c(a-b)=0=a(b-c)(x-1)^2.$$
Now by letting $x=0$ we find $$c(a-b)=a(b-c)$$ which implies $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$. 
