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.
 A: 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 $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$. 
A: If you label:
$$\begin{cases}a(b-c)=p\\
b(c-a)=q\\
c(a-b)=r\end{cases} \Rightarrow p+q+r=0$$
then the equation becomes:
$$px^2+qx+r=0,\\
D=q^2-4pr=(-p-r)^2-4pr=(p-r)^2=0 \Rightarrow p=r\Rightarrow \\
a(b-c)=c(a-b) \Rightarrow 2ac=ab+bc \stackrel{\cdot \frac1{abc}}{\Rightarrow} \frac2b=\frac1a+\frac1c.$$
A: Since $a(b-c)+b(c-a)+c(a-b)=0$, you know that $x=1$ is a root, as you observed.
With synthetic division by $x-1$, we get, assuming $a\ne0$ and $b\ne c$,
$$
\begin{array}{c|cc|c}
  & ab-ac      & bc-ab & ac-bc \\
1 & \downarrow & ab-ac & bc-ac \\
\hline
  & ab-ac      & bc-ac & 0
\end{array}
$$
The other root is
$$
-\frac{bc-ac}{ab-ac}
$$
Therefore $bc-ac=-ab+ac$ is the condition that the equation has two equal roots. This becomes
$$
2ac=bc+ab
$$
and, dividing by $abc$, we get
$$
\frac{2}{b}=\frac{1}{a}+\frac{1}{c}
$$
If $a=0$ or $b=c$ the equation is not a quadratic. Can it be $b=0$? In this case we get $-acx^2+ac=0$ and we need $c\ne0$; the roots are $1$ and $-1$. With $c=0$ it would be the same.
