<s>[INST] 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.
 [/INST] 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 $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$. 
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