Question

If the roots of $a(b-c)x^2+b(c-a)x\hspace{0.17em}+c(a-b)=0$ be equal , then $a,b,c$ are in

• A. A.P
• B. G.P
• C. H.P
• D. None of these

Answer 1

The correct option is C

H.P

Step 1: Apply condition for equal roots of a quadratic equation

For a standard quadratic equation $A{x}^2+Bx+C=0$, The Discriminant, $D=B^2-4AC=0$

Comparing the given equation with the above we get

$$\begin{gathered} A=a(b-c) \\ B=b(c-a) \\ C=c(a-b) \end{gathered}$$

Putting values of A,B, and C in $D=B^2-4AC=0$

$$\begin{gathered} \Rightarrow b^2{(c-a)}^2-4a(b-c)c(a-b)=0 \\ \Rightarrow b^2(c^2+a^2-2ca)-4ac(b-c)(a-b)=0 \\ \Rightarrow b^2{c}^2+a^2{b}^2-2b^{2}ac-4ac(ab-b^2-ac+bc)=0 \\ \Rightarrow b^2{c}^2+a^2{b}^2-2a{b}^{2}c-4a^{2}bc+4ac{b}^2+4a^2{c}^2-4ab{c}^2=0 \\ Rearranging \\ \Rightarrow a^2{b}^2+b^2{c}^2+4a^2{c}^2+2a{b}^{2}c-4a^{2}bc-4ab{c}^2=0 \end{gathered}$$

Step 2: Using identity ${(x+y+z)}^2=x^2+y^2+z^2+2xy+2yz+2zx$

$$\begin{gathered} \Rightarrow {(ab+bc-2ac)}^2=0 \\ \Rightarrow ab+bc-2ac=0 \\ \Rightarrow ab+bc=2ac \\ Dividebothsidebyabc \\ \Rightarrow \frac{ab}{abc}+\frac{bc}{abc}=\frac{2ac}{abc} \\ \Rightarrow \frac{1}{c}+\hspace{0.17em}\frac{1}{a}=\frac{2}{b} \end{gathered}$$

As we know the above condition only holds when $a,b$ and $c$ are in H.P..

Therefore, option (c) is correct as $\frac{1}{c}+\frac{1}{a}=\frac{2}{b}$ forms H.P.