Question

If $\alpha ,\beta$ are roots of the equation $p{x}^2+qx-r=0$, then the value of $\frac{\alpha }{{\beta }^2}+\frac{\beta }{{\alpha }^2}$ is equal to?

Answer 1

We are given,

the roots of the equation

$p{x}^2+qx-r=0$ are $\alpha$ & $\beta$

So, some attributes of quadratic equation can help here

as we know, sum of two roots are

given by, $\alpha +\beta =\frac{-b}{a}=\frac{-a}{p}---\left(1\right)$

& Multiplication of two roots are given by, $\alpha \beta =\frac{c}{a}=\frac{-r}{p}----\left(2\right)$

So,

$\frac{\alpha }{{\beta }^2}+\frac{\beta }{{\alpha }^2}=\frac{{\alpha }^3+{\beta }^3}{(\alpha \beta {)}^2}=\frac{(\alpha +\beta )({\alpha }^2-\alpha \beta +{\beta }^2)}{(\alpha \beta {)}^2}----\left(3\right)$

here ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$ (From $\left(1\right)$ & $\left(2\right)$)

$=\left({\frac{-9}{p}}^2\right)-2\left(\frac{-r}{p}\right)$

${\alpha }^2+{\beta }^2=\frac{q^2+2r}{p}$

$\frac{\alpha }{{\beta }^2}+\frac{\beta }{{\alpha }^2}=\frac{\left(\frac{-q}{p}\right)\times (\frac{q^2+2r}{p}-\left(\frac{-r}{p}\right))}{{\left(\frac{-r}{p}\right)}^2}$

$=\frac{\left(\frac{-q}{p}\right)\left(\frac{q^2+3r}{p}\right)}{\frac{r^2}{p^2}}$

$=\frac{-q^3-3qr}{r^2}$

$\frac{\alpha }{{\beta }^2}+\frac{\beta }{{\alpha }^2}=\frac{-q^3}{r^2}-\frac{3q}{r}$