Question

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

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

Answer 1

The correct option is A

A.P

Explanation for the correct option.

Let the roots of the given quadratic equation be $\alpha \mathrm{and}\beta$.

$\begin{array}{rcl}& \Rightarrow & \alpha +\beta =\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}\\ & =& \frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}\\ & =& \frac{{\left(\alpha +\beta \right)}^2-2\alpha \beta }{{\displaystyle {\alpha }^2{\beta }^2}}\end{array}$

The sum of roots of quadratic equation is $-\frac{b}{a}$ and the product of roots is $\frac{c}{a}$.

$\begin{array}{rcl}\Rightarrow -\frac{b}{a}& =& \frac{{\left(-{\displaystyle \frac{b}{a}}\right)}^2-{\displaystyle \frac{2c}{a}}}{{\left({\displaystyle \frac{c}{a}}\right)}^2}\\ \Rightarrow \frac{-b{c}^2}{a^3}& =& \frac{b^2-2ca}{a^2}\\ \Rightarrow -b{c}^2& =& a{b}^2-2c{a}^2\\ \Rightarrow 2c{a}^2& =& a{b}^2+b{c}^2\\ & \Rightarrow & c{a}^2-b{c}^2=a{b}^2-c{a}^2\end{array}$

Therefore, $b{c}^2,c{a}^2,a{b}^2$are in A.P

Hence, option A is correct.