Question

If $\alpha$ and $\beta$ are the roots of the equation $3x^2-4x+1=0$ form a quadratic equation whose roots are $\frac{{\alpha }^2}{\beta }$ and $\frac{{\beta }^2}{\alpha }$

Answer 1

Let $\alpha ,\beta$ be the roots of $3x^2-4x+1=1$
Thus, $\alpha +\beta =\frac{-b}{a}$
$=\frac{-(-4)}{3}=\frac{4}{3}$
and $\alpha \beta =\frac{c}{a}=\frac{1}{3}$
The roots are $\frac{{\alpha }^2}{\beta }$ and $\frac{{\beta }^2}{\alpha }$.
$\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }=\frac{{\alpha }^3+{\beta }^3}{\alpha \beta }$
$=\frac{{(\alpha +\beta )}^3-3\alpha \beta (\alpha +\beta )}{\alpha \beta }$
$=\frac{{\left(\frac{4}{3}\right)}^3-3\left(\frac{1}{3}\right)\left(\frac{4}{3}\right)}{\frac{1}{3}}$
$=\frac{\frac{64}{27}-\frac{4}{3}}{\frac{1}{3}}=\frac{\frac{64-36}{27}}{\frac{1}{3}}$
$=\frac{\frac{28}{27}}{\frac{1}{3}}=\frac{28}{27}\times \frac{3}{1}=\frac{28}{9}$
Therefore, $\frac{{\alpha }^2}{\beta }\frac{{\beta }^2}{\alpha }=\frac{{\left(\alpha \beta \right)}^2}{\alpha \beta }=\alpha \beta =\frac{1}{3}$
The required equation is
$x^2-$ (Sum of the roots) $x+$ Product of roots $=0$
$\Rightarrow x^2-\frac{28}{9}x+\frac{1}{3}=0$
$\Rightarrow 9x^2-28x+3=0$