Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}4x^2-5x+2=0 \\ \mathrm{On}\mathrm{comparing}\mathrm{this}\mathrm{equation}\mathrm{with}a{x}^2+bx+c=0,\mathrm{we}\mathrm{get}: \\ \mathit{}a=4,b=-5,c=2 \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{\alpha }+\mathrm{\beta }=\frac{-b}{a}\mathrm{and}\mathrm{\alpha \beta }=\frac{c}{a}. \\ \mathrm{Thus},\mathrm{\alpha }+\mathrm{\beta }=-\left({\displaystyle \frac{-5}{4}}\right)=\frac{5}{4}...(1) \end{gathered}$$
$\mathrm{\alpha \beta }=\frac{2}{4}=\frac{1}{2}...\left(2\right)$
$$\begin{gathered} {\mathrm{\alpha }}_1=\frac{\mathrm{\alpha }}{\mathrm{\beta }}\mathrm{and}{\mathrm{\beta }}_1=\frac{\mathrm{\beta }}{\mathrm{\alpha }} \\ \mathrm{Then}, \\ {\mathrm{\alpha }}_1+{\mathrm{\beta }}_1=\frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }} \\ =\frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2}{\mathrm{\alpha \beta }} \\ =\frac{{\left(\mathrm{\alpha }+\mathrm{\beta }\right)}^2-2\mathrm{\alpha \beta }}{\mathrm{\alpha \beta }} \\ =\frac{{\left({\displaystyle \frac{5}{4}}\right)}^2-2\times {\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}}[\mathrm{From}(1)\mathrm{and}(2)] \\ =2\left(\frac{25}{16}-1\right)=2\times \frac{9}{16}=\frac{9}{8} \\ \mathrm{And}{\mathrm{\alpha }}_1{\mathrm{\beta }}_1=\frac{\mathrm{\alpha }}{\mathrm{\beta }}\times \frac{\mathrm{\beta }}{\mathrm{\alpha }}=1 \\ \mathrm{Now},\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{if}{\mathrm{\alpha }}_1\mathrm{and}{\mathrm{\beta }}_1\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{quadratic}\mathrm{equation},\mathrm{then}\mathrm{the}\mathrm{quadratic}\mathrm{equation}\mathrm{is} \\ {x}^2-\left({\mathrm{\alpha }}_1+{\mathrm{\beta }}_1\right)x+{\mathrm{\alpha }}_1{\mathrm{\beta }}_1=0 \\ \mathrm{On}\mathrm{substituting}{\mathrm{\alpha }}_1+{\mathrm{\beta }}_1=\frac{9}{8}\mathrm{and}{\mathrm{\alpha }}_1\mathrm{\beta }=1,\mathrm{we}\mathrm{get}: \\ {x}^{\mathit{2}}-\left(\frac{9}{8}\right)x+1=0 \\ \Rightarrow 8x^2-9x+8=0 \end{gathered}$$