Question related to a proof in Quadratic equations

Question

If the roots of the equation $a\left(b - c\right)x^{2} + b\left(c - a\right)x + c\left(a - b\right) = 0$ are equal then prove that $2/b = 1/a + 1/c$, i.e. $a,b,c$ are on H.P. I have tried to prove this using the quadratic formula and also with the zero-coefficient relationship but at the end the expression becomes too cumbersome to solve.

Answer 1

So it discriminat must be $0$:

$$ b^2(c-a)^2 = 4ac(a-b)(b-c)$$

Let $x=b/a$ and $z=b/c$. Then $$ (z-x)^2 = 4(x-1)(1-z)$$ so $$ x^2+z^2-2zx = 4z+4x-4-4zx$$ so $$ (x+z)^2 = 4(x+z)-4$$ thus $$(x+z-2)^2 =0$$ and we are done.

Answer 2

Set $1/a=A,1/b=B,1/c=C$ to find

$$(C-B)x^2+(A-C)x+(B-A)=0$$

So, we need to establish $A,B,C$ are in A.P.

As the discriminant has to be $=0,$

$$0=(A-C)^2-4(C-B)(B-A)=(C-B+B-A)^2-4(C-B)(B-A)=?$$