Question

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

Answer 1

$4x^2-5x+2=0$
Here, $a=4,b=-5,c=2$
$\alpha +\beta =\frac{-b}{a}=-\frac{-5}{4}=\frac{5}{4}$
and $\alpha \beta =\frac{c}{a}=\frac{2}{4}=\frac{1}{2}$
Roots are $\alpha +\frac{1}{a}$ and $\beta +\frac{1}{\beta }$
$\alpha +\frac{1}{\alpha }+\beta +\frac{1}{\beta }=\alpha +\beta +\frac{1}{\alpha }+\frac{1}{\beta }=\alpha +\beta +\frac{\beta +\alpha }{\alpha \beta }$
$=\frac{5}{4}+[\frac{5}{4}÷\frac{1}{2}]$
$=\frac{5}{4}+\frac{5}{2}=\frac{15}{4}....\left(1\right)$
$(\alpha +\frac{1}{\alpha })(\beta +\frac{1}{\alpha })=\alpha \beta +\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+\frac{1}{\alpha \beta }$
$=\alpha \beta +\frac{1}{\alpha \beta }+\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$
$={\left(\frac{5}{4}\right)}^2-2\left(\frac{1}{2}\right)=\frac{25}{16}-1=\frac{9}{16}$
$\therefore \alpha \beta +\frac{1}{\alpha \beta }+\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{1}{2}+[1÷\frac{1}{2}]+\frac{9}{16}÷\frac{1}{2}$
$\therefore x^2-\frac{15}{4}x+\frac{29}{8}=0$ ... from $\left(1\right)$ and $\left(2\right)$
$\therefore 8x^2-30x+29=0$ (multiplying by $8$)
Hence the required equation is $8x^2-30x+29=0$