Question

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

• A. AP
• B. GP
• C. HP
• D. AGP

Answer 1

The correct option is A

AP

Explanation for the correct answer:

Let $m,n$ be the roots of the quadratic equation $a{x}^2+bx+c=0$

Hence the sum of the roots and product of the roots are given as

$m+n=\frac{-b}{a}$$...\left(i\right)$

$mn=\frac{c}{a}$$...\left(ii\right)$

It is given to us that sum of the roots is equal to sum of the square of its reciprocals

$\Rightarrow$$m+n=\frac{1}{m^2}+\frac{1}{n^2}$

$\Rightarrow$ $=\frac{m^2+n^2}{m^2{n}^2}$

$\Rightarrow$ $=\frac{m^2+n^2+2mn-2mn}{{\left(mn\right)}^2}$

$\Rightarrow$ $m+n=\frac{{\left(m+n\right)}^2-2\left(mn\right)}{{\left(mn\right)}^2}$ $...\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right]$

From $\left(i\right),\left(ii\right)$ we get

$\Rightarrow$$\frac{-b}{a}=\frac{{\left({\displaystyle \frac{-b}{a}}\right)}^2-2\left({\displaystyle \frac{c}{a}}\right)}{{\left({\displaystyle \frac{c}{a}}\right)}^2}$

$\Rightarrow$$\frac{-b}{a}=\frac{{\displaystyle \frac{b^2-2ac}{a^2}}}{{\displaystyle \frac{c^2}{a^2}}}$

$\Rightarrow$$\frac{-b}{a}=\frac{{\displaystyle b^2-2ac}}{{\displaystyle c^2}}$

$\Rightarrow$$-b{c}^2=a\left(b^2-2ac\right)$

$\Rightarrow$$2c{a}^2=a{b}^2+b{c}^2$

$\Rightarrow$$c{a}^2=\frac{a{b}^2+b{c}^2}{2}$

Hence $c{a}^2$ is the arithmetic mean of $a{b}^2$ and $b{c}^2$.

Hence $b{c}^2,c{a}^2,a{b}^2$ are in an arithmetic progression. So, option (A) is the correct answer.