Question

If lf $\alpha ,\beta$ are the roots of the equation :

$(\pi -1){\lambda }^2+2\lambda +1=0$

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=?$

Answer 1

$\Rightarrow$ $(\pi -1){\lambda }^2+2\lambda +1=0$

$\Rightarrow$ Here, $a=\pi -1,b=2,c=1$

$\Rightarrow$ $\alpha \beta =\frac{c}{a}=\frac{1}{\pi -1}$ ----- ( 1 )

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}=\frac{-2}{\pi -1}$ ------- ( 2 )

$\Rightarrow$ $(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

$\Rightarrow$ ${\left(\frac{-2}{\pi -1}\right)}^2={\alpha }^2+{\beta }^2+2\times \frac{1}{\pi -1}$ [ Using ( 1 ) and ( 2 ) ]

$\Rightarrow$ $\frac{4}{(\pi -1{)}^2}-\frac{2}{\pi -1}={\alpha }^2+{\beta }^2$

$\Rightarrow$ $\frac{4-2\pi +2}{(\pi -1{)}^2}={\alpha }^2+{\beta }^2$

$\Rightarrow$ ${\alpha }^2+{\beta }^2=\frac{-2\pi +6}{(\pi -1{)}^2}$ ----- ( 3 )

Now,

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

$=\frac{\frac{-2\pi +6}{(\pi -1{)}^2}}{\frac{1}{\pi -1}}$

$=\frac{-2\pi +6}{(\pi -1{)}^2}\times \frac{(\pi -1)}{1}$

$=\frac{-2\pi +6}{\pi -1}$

$=\frac{-2\times \frac{22}{7}+6}{\frac{22}{7}-1}$

$=\frac{-44+42}{22-7}$

$=\frac{-2}{15}$