Question

If in the equation $a{x}^2+bx+c=0$, the sum of the roots is equal to the sum of the squares of their reciprocals, then $\frac{c}{a},\frac{a}{b},\frac{b}{c}$ are in

• A. AP
• B. GP
• C. HP
• D. None of these

Answer 1

The correct option is A

AP

Explanation for the correct option.

Find the relation between the terms.

Let the roots of the equation $a{x}^2+bx+c=0$ be $p$ and $q$.

Now, the sum of roots is given as $p+q=-\frac{b}{a}$ and the product of roots is given as: $pq=\frac{c}{a}$.

It is given that the sum of the roots is equal to the sum of the squares of their reciprocals, so $p+q=\frac{1}{p^2}+\frac{1}{q^2}$.

Now the equation can be simplified as:

$$\begin{gathered} p+q=\frac{q^2+p^2}{p^2{q}^2} \\ \Rightarrow p+q=\frac{p^2+q^2+2pq-2pq}{{\left(pq\right)}^2} \\ \Rightarrow p+q=\frac{{\left(p+q\right)}^2-2pq}{{\left(pq\right)}^2} \\ \Rightarrow -\frac{b}{a}=\frac{{\left(-{\displaystyle \frac{b}{a}}\right)}^2-2{\displaystyle \frac{c}{a}}}{{\left({\displaystyle \frac{c}{a}}\right)}^2}\mathbf{[}\mathbf{p}\mathbf{+}\mathbf{q}\mathbf{=}\mathbf{-}\frac{\mathbf{b}}{\mathbf{a}}\mathbf{,}\mathbf{}\mathbf{p}\mathbf{q}\mathbf{=}\frac{\mathbf{c}}{\mathbf{a}}\mathbf{]} \\ \Rightarrow -\frac{b}{a}\frac{c^2}{a^2}=\frac{b^2}{a^2}-2\frac{c}{a} \\ \Rightarrow b{c}^2=-a{b}^2+2a^{2}c \end{gathered}$$

Now divide both sides by $abc$.

$$\begin{gathered} \frac{b{c}^2}{abc}=\frac{-a{b}^2+2a^{2}c}{abc} \\ \Rightarrow \frac{c}{a}=-\frac{b}{c}+2\frac{a}{b} \\ \Rightarrow \frac{c}{a}+\frac{b}{c}=2\frac{a}{b} \end{gathered}$$

As, $2\frac{a}{b}=\frac{c}{a}+\frac{b}{c}$ so the terms $\frac{c}{a},\frac{a}{b},\frac{b}{c}$ are in AP.

Hence, the correct option is A.