Question

If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f\left(x\right)=3x^2-4x+1$, then find a quadratic polynomial whose zeros are $\frac{{\alpha }^2}{\beta }$ and $\frac{{\beta }^2}{\alpha }$

Answer 1

We have,

$\begin{array}{c}3x^2-4x+1=0\\ \alpha +\beta =\frac{4}{3}\alpha .\beta =\frac{1}{3}\\ \frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{\left(\alpha +\beta \right)\left({\alpha }^2+{\beta }^2-\alpha .\beta \right)}{\alpha \beta }\\ =\frac{\left(\alpha +\beta \right)\left({\left(\alpha +\beta \right)}^2-3\alpha \beta \right)}{\frac{1}{3}}\\ =4\left(\frac{16}{9}-1\right)=\frac{4\times 7}{9}=\frac{28}{9}\\ \frac{{\alpha }^2}{\beta }\times \frac{{\beta }^2}{\alpha }=\alpha \beta =\frac{1}{3}\end{array}$

Hence,

quadratic equation whose roots are $\frac{{\alpha }^2}{\beta }$ and $\frac{{\beta }^2}{\alpha }$:

$\begin{array}{c}\Rightarrow x^2-\frac{28}{9}x+\frac{1}{3}=0\\ 9x^2-28x+3=0\end{array}$