Question

If $\alpha$ and $\beta$ are the roots of the equation $4x^2-5x+2=0$, find the equation whose roots are
$\alpha +\frac{1}{\alpha }\mathrm{and}\beta +\frac{1}{\beta }$

Answer 1

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

$\mathrm{\alpha \beta }=\frac{2}{4}=\frac{1}{2}$
$$\begin{gathered} \mathrm{Let}{\mathrm{\alpha }}_1=\mathrm{\alpha }+\frac{1}{\mathrm{\alpha }}\mathrm{and}{\mathrm{\beta }}_1=\mathrm{\beta }+\frac{1}{\mathrm{\beta }} \\ \mathrm{Then},\mathrm{we}\mathrm{get}: \\ {\mathrm{\alpha }}_1+{\mathrm{\beta }}_1=\mathrm{\alpha }+\frac{1}{\mathrm{\alpha }}+\mathrm{\beta }+\frac{1}{\mathrm{\beta }} \\ =\left(\mathrm{\alpha }+\mathrm{\beta }\right)+\left(\frac{1}{\mathrm{\alpha }}+\frac{1}{\mathrm{\beta }}\right) \\ =\frac{5}{4}+\left(\frac{{\displaystyle \frac{5}{4}}}{{\displaystyle \frac{1}{2}}}\right)[\mathrm{From}(1)\mathrm{and}(2)] \\ =\frac{5}{4}+\frac{5}{2}=\frac{15}{4} \end{gathered}$$
_{$$\begin{gathered} {\alpha }_1{\beta }_1=\left(\alpha +\frac{1}{\alpha }\right)\left(\beta +\frac{1}{\beta }\right) \\ =\left(\frac{{\alpha }^2+1}{\alpha }\right)\left(\frac{{\beta }^2+1}{\beta }\right) \\ =\frac{{\alpha }^2{\beta }^2+{\alpha }^2+{\beta }^2+1}{\alpha \beta } \\ =\frac{{\left({\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{5}{4}}\right)}^2-\left(2\times {\displaystyle \frac{1}{2}}\right)+1}{{\displaystyle \frac{1}{2}}} \\ =\frac{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{25}{16}}}{{\displaystyle \frac{1}{2}}} \\ =\frac{29}{8} \end{gathered}$$}

$$\begin{gathered} \text{W}e\mathrm{know}\mathrm{that}\mathrm{if}{\mathrm{\alpha }}_1\text{and}{\mathrm{\beta }}_1\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{a}\mathrm{quadratic}\mathrm{equation},\mathrm{then}\mathrm{the}\mathrm{quadratic} \\ \mathrm{equation}\mathrm{is} \\ {x}^2-\left({\alpha }_1+{\beta }_1\right)x+{\alpha }_1{\beta }_1=0 \\ \mathrm{On}\mathrm{substituing}{\alpha }_1+{\beta }_1=\frac{15}{4}\mathrm{and}{\alpha }_1{\beta }_1=\frac{29}{8},\mathrm{we}\mathrm{get}: \\ {x}^2-\frac{15}{4}x+\frac{29}{8}=0 \\ \Rightarrow 8x^2-30x+29=0 \end{gathered}$$