Question

If $\alpha$ and $\beta$ are the zeros of quadratic polynomial: $Px=-3x^2-4x+1$. Find the quadratic polynomial whos zeros are $\frac{{\alpha }^2}{\beta }$ and $\frac{{\beta }^2}{\alpha }$

Answer 1

Let $\alpha$ and $\beta$ be the roots of the quadratic polynomial.

Sum$=\alpha +\beta =\frac{-(-4)}{-3}=\frac{-4}{3}$

Product$=\alpha \beta =\frac{-\left(1\right)}{3}=\frac{-1}{3}$

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

$=\frac{{\alpha }^3+{\beta }^3}{\alpha \beta }$

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

$=\frac{{\left(\frac{-4}{3}\right)}^3-3\times \frac{-1}{3}\left(\frac{-4}{3}\right)}{\frac{-1}{3}}$

$=\frac{\frac{-64}{27}-\frac{4}{3}}{\frac{-1}{3}}$

$=\frac{\frac{-64-36}{27}}{\frac{-1}{3}}$

$=\frac{\frac{-100}{27}}{\frac{-1}{3}}$

$=\frac{100}{9}$

Prioduct $=\frac{{\alpha }^2}{\beta }\times \frac{{\beta }^2}{\alpha }=\frac{{\alpha }^2{\beta }^2}{\alpha \beta }=\alpha \beta =\frac{-1}{3}$

$\therefore$ Required polynomial is $x^2-\frac{100x}{9}-\frac{1}{3}$