TITLE: Proving a result by making discriminant zero
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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:
$$px^2+qx+r=0,\\
D=q^2-4pr=(-p-r)^2-4pr=(p-r)^2=0 \Rightarrow p=r\Rightarrow \\
a(b-c)=c(a-b) \Rightarrow 2ac=ab+bc \stackrel{\cdot \frac1{abc}}{\Rightarrow} \frac2b=\frac1a+\frac1c.$$