Question

If $\alpha$ and $\beta$ are roots of the quadratic equation $5x^2-3x-1=0$, find a quadratic equation with integral coefficients which have the roots.
(a) $\frac{1}{{\alpha }^2}$ and $\frac{1}{{\beta }^2}$
(b) $\frac{{\alpha }^2}{\beta }$ and $\frac{{\beta }^2}{\alpha }$

Answer 1

From the given quadratic equation, we obtain $\alpha +\beta =\frac{3}{5}$ and $\alpha \beta =\frac{-1}{5}$

a. In order to obtain quadratic equation having roots $\frac{1}{{\alpha }^2}$ and $\frac{1}{{\beta }^2},$ we need to find out $\frac{-b}{a}$ and $\frac{c}{a}$.

$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{-b}{a}$

=$\frac{{\beta }^2+{\alpha }^2}{{\alpha }^2{\beta }^2}$

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

Substituting values of $\alpha +\beta$ and $\alpha \beta$ in above equation, we obtain $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=19$

and $\frac{1}{(\alpha \beta {)}^2}=\frac{c}{a}=25$

Therefore quadratic equation is written as $x^2-19x+25=0$.

b. sum of roots= $\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }=\frac{{\alpha }^3+{\beta }^3}{\alpha \beta }$

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

Substituting values of $\alpha +\beta$ and $\alpha \beta$, we obtain

$\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }=\frac{72}{25}$

product of roots=$\alpha \beta =\frac{-1}{5}$

The quadratic equation is written as $x^2-\frac{72}{25}x-\frac{1}{5}=0$