Question

If $\alpha$ and $\beta$ are the roots of the equation $4x^2-5x+2=0$, find the equation whose roots are
$\frac{{\alpha }^2}{\beta }\mathrm{and}\frac{{\beta }^2}{\alpha }$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}4x^2-5x+2=0 \\ \mathrm{On}\mathrm{comparing}\mathrm{this}\mathrm{equation}\mathrm{with}a{x}^2+bx+c=0,\mathrm{we}\mathrm{get}: \\ \mathit{}a=4,b=-5,c=2 \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{\alpha }+\mathrm{\beta }=\frac{-b}{a}\mathrm{and}\mathrm{\alpha \beta }=\frac{c}{a} \\ \mathrm{Thus},\mathrm{\alpha }+\mathrm{\beta }=-\left({\displaystyle \frac{-5}{4}}\right)=\frac{5}{4}...(1) \end{gathered}$$

$\alpha \beta =\frac{2}{4}=\frac{1}{2}...\left(2\right)$
$$\begin{gathered} \mathrm{Let}{\mathrm{\alpha }}_1=\frac{{\mathrm{\alpha }}^2}{\mathrm{\beta }}\mathrm{and}{\mathrm{\beta }}_1=\frac{{\mathrm{\beta }}^2}{\mathrm{\alpha }} \\ \mathrm{Then}, \\ {\mathrm{\alpha }}_1+{\mathrm{\beta }}_1=\frac{{\mathrm{\alpha }}^2}{\mathrm{\beta }}+\frac{{\mathrm{\beta }}^2}{\mathrm{\alpha }} \\ =\frac{{\mathrm{\alpha }}^3+{\mathrm{\beta }}^3}{\mathrm{\alpha \beta }} \\ =\frac{{\left(\mathrm{\alpha }+\mathrm{\beta }\right)}^3-3\mathrm{\alpha \beta }\left(\mathrm{\alpha }+\mathrm{\beta }\right)}{\mathrm{\alpha \beta }} \\ =\frac{{\left({\displaystyle \frac{5}{4}}\right)}^3-3\times {\displaystyle \frac{1}{2}}\times \left({\displaystyle \frac{5}{4}}\right)}{{\displaystyle \frac{1}{2}}}[\mathrm{From}(1)\mathrm{and}(2\left)\right] \\ =2\left(\frac{125}{64}-\frac{15}{8}\right) \\ =2\left(\frac{5}{64}\right)=\frac{5}{32} \\ \mathrm{And}{\mathrm{\alpha }}_1{\mathrm{\beta }}_1=\frac{{\mathrm{\alpha }}^2}{\mathrm{\beta }}\times \frac{{\mathrm{\beta }}^2}{\mathrm{\alpha }}=\mathrm{\alpha \beta }=\frac{1}{2}[\mathrm{From}(2\left)\right] \\ \text{We}\mathrm{know}\mathrm{that}\mathrm{if}{\mathrm{\alpha }}_1\mathrm{and}{\mathrm{\beta }}_1\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{quadratic}\mathrm{equation}. \\ \mathrm{The}\text{refore},\mathrm{the}\mathrm{quadratic}\mathrm{equation}\mathrm{is} \\ {x}^2-\left({\mathrm{\alpha }}_1+{\mathrm{\beta }}_1\right)x+{\mathrm{\alpha }}_1{\mathrm{\beta }}_1=0 \\ \mathrm{On}\mathrm{substituting}{\mathrm{\alpha }}_1+{\mathrm{\beta }}_1=\frac{5}{32}\mathrm{and}{\mathrm{\alpha }}_1{\mathrm{\beta }}_1=\frac{1}{2},\mathrm{we}\mathrm{get}: \\ {x}^2-\left(\frac{5}{32}\right)x+\frac{1}{2}=0 \\ \Rightarrow 32x^2-5x+16=0 \end{gathered}$$