Question

lf $\alpha ,\beta$ are the roots of the equation $\lambda (x^2-x)+x+5=0,{\lambda }_1$ and ${\lambda }_2$ are the two values of $\lambda$ for which the roots, $\alpha ,\beta$ are connected by the relation $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{5}$, then the value of $\frac{{\lambda }_1}{{\lambda }_2}+\frac{{\lambda }_2}{{\lambda }_1}=$

• A. $150$
• B. $254$
• C. $180$
• D. $1022$

Answer 1

Answer: (B)

$254$

The given quadratic equation is $\lambda x^2+(1-\lambda )x+5=0$

Given $\frac{({\alpha }^2+{\beta }^2)}{\alpha \beta }=\frac{4}{5}$

$\therefore \frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }=\frac{4}{5}$

$\therefore \frac{{\left(\frac{\lambda -1}{\lambda }\right)}^2-2\frac{5}{\lambda }}{\frac{5}{\lambda }}=\frac{4}{5}$

$\therefore {\lambda }^2-16\lambda +1=0$

We require,

$\frac{({\lambda }_1^2+{\lambda }_2^2)}{{\lambda }_1{\lambda }_2}$

$=\frac{({\lambda }_1+{\lambda }_2{)}^2-2{\lambda }_1{\lambda }_2}{{\lambda }_1{\lambda }_2}=\frac{{16}^2-2}{1}=254$