Question

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

Answer 1

Find the form of reciprocal.

Let us consider that $\mathrm{\alpha }$ and $\mathrm{\beta }$ are the roots of the given equation ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$.

So,

$\Rightarrow$ $\mathrm{\alpha }=-\frac{\mathrm{b}}{\mathrm{a}}$

and

$\Rightarrow$ $\mathrm{\beta }=\frac{\mathrm{c}}{\mathrm{a}}$

From the given condition, The squares of their reciprocals are equal to the sum of the roots of the quadratic equation.

So,

$\Rightarrow$ $\mathrm{\alpha }+\mathrm{\beta }=\frac{1}{{\mathrm{\alpha }}^2}+\frac{1}{{\mathrm{\beta }}^2}$

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

$\Rightarrow$ $2{\mathrm{ca}}^2={\mathrm{ab}}^2+{\mathrm{bc}}^2$

As we know that, Harmonic progression is a progression formed by taking the reciprocals of an Arithmetic progression.

The first term of a HP is $\frac{1}{\mathrm{a}}$.

If the H.P. be as $\frac{1}{\mathrm{a}}$, $\frac{1}{\mathrm{a}+\mathrm{d}}$, $\frac{1}{\mathrm{a}+2\mathrm{d}},...$ then corresponding A. P. is $\mathrm{a},\mathrm{a}+\mathrm{d},\mathrm{a}+2\mathrm{d},...$.

Hence, the reciprocal $\frac{\mathrm{a}}{\mathrm{c}},\frac{\mathrm{b}}{\mathrm{a}},\frac{\mathrm{c}}{\mathrm{b}}$ are in Harmonic progression (H. P.).