If the roots of the equation

Question

If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal then $a,b,c$ are in

1. arithmetic progression
2. geometric progression
3. harmonic progression
4. none of the above

My Attempt: Two roots are equal if $$b^2-4ac=0$$ $$[b(c-a)]^2-4a(b-c).c(a-b)=0$$ $$b^2(c^2-2ac+a^2)-4ac(ab-b^2-ac+bc)=0$$ $$b^2c^2-2ab^2c+a^2b^2-4a^2bc+4ab^2c+4a^2c^2-4abc^2=0$$

Answer 1

Whenever cyclic coefficients of $x^2, x, 1 $ are observed, Like $ (a-b),(b-c),(c-a) $, then $ x= 1$ must be a solution.

Now $ x=1 $, since roots of this quadratic are equal, so Roots are $ z_1=z_2=1 $

Product of roots $ = z_1z_2 = \frac{c(a-b)}{a(b-c)} $

$ \implies 1 = \frac{c(a-b)}{a(b-c)} $

$ \implies ab - ac = ac - bc $

Dividing throughout by $ abc $ we get

$ \frac{1}{c} + \frac{1}{a} = \frac{2}{b} $

Which are clearly in H.P.

Answer 2

prove that $$x_1=1$$ or $$x_2=\frac{c(a-b)}{a(b-c)}$$ are Solutions of your equation an $AP$ is not possible, if we assume that $$x_1=1,x_2=qx_1=q$$ then we get $$q={\frac{c(a-b)}{a(b-c)}}$$

Answer 3

Your start is correct. We thus have, $$\text{ Discriminant } =0 \implies (b(c-a))^2 = 4ac(b-c)(a-b)$$ $$\implies b^2c^2+2ab^2c+a^2b^2-4a^2bc+4a^2c^2-4abc^2=0$$ Dividing by $a^2b^2c^2$, we get, $$\frac{1}{a^2}+\frac{2}{ac}+\frac{1}{c^2}+\frac{4}{b^2}=\frac{4}{bc}+\frac{4}{ab}$$ $$\implies (\frac{1}{a}+\frac{1}{c})^2+\frac{4}{b^2}=\frac{4}{b}(\frac{1}{a}+\frac{1}{c})$$ $$(\frac{1}{a}+\frac{1}{c}-\frac{2}{b})^2=0$$ $$\boxed{\frac{1}{a}+\frac{1}{c}=\frac{2}{b}}$$

Answer 4

After the last equation in the question, this will require some rearrangements… $$b^2c^2-2ab^2c+a^2b^2 -4a^2bc+4ab^2c+4a^2c^2-4abc^2=0$$ $$(ab)^2+4(ac)^2+(bc)^2-4(ac)(bc)+2(ab)(bc)-4(ab)(bc)=0$$ $$(ab-2ac+bc)^2=0$$ $$ab-2ac+bc=0$$ $$ab+bc=2ac$$ $$\frac1c+\frac1a=\frac2b$$ and thus $a,b,c$ are in harmonic progression.