Question

If ${\lambda }_1$ and ${\lambda }_2$ are the two values of $\lambda$ such that the roots $\alpha$ and $\beta$ of the quadratic equation, $\lambda (x^2-x)+x+5=0$ satisfy $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+\frac{4}{5}=0$, then $\frac{{\lambda }_1}{{\lambda }_2^2}+\frac{{\lambda }_2}{{\lambda }_1^2}=0$ is equal to:

• A. $536$
• B. $512$
• C. $504$
• D. $488$

Answer 1

Answer: (D)

$488$

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

$\Rightarrow \lambda x^2+(1-\lambda )x+5=0{<}_{\beta }^{\alpha }$

Since $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+\frac{4}{5}=0$

Therefore, $\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }+\frac{4}{5}=0$

$\Rightarrow \frac{{\left(\frac{\lambda -1}{\lambda }\right)}^2-\frac{10}{\lambda }}{\left(5/\lambda \right)}=-\frac{4}{5}$

$\Rightarrow \frac{{\lambda }^2-2\lambda +1}{{\lambda }^2}-\frac{10}{\lambda }=\frac{-4}{\lambda }$

$\Rightarrow {\lambda }^2-2\lambda +1=6\lambda$

$\Rightarrow {\lambda }^2-8\lambda +1=0$

${\lambda }_1$ and ${\lambda }_2$ are two values of $\lambda$.

Now, $\frac{{\lambda }_1}{{\lambda }_2^2}+\frac{{\lambda }_2}{{\lambda }_1^2}=\frac{{\lambda }_1^3+{\lambda }_2^3}{({\lambda }_1\cdot {\lambda }_2{)}^2}=\frac{({\lambda }_1+{\lambda }_2{)}^3-3{\lambda }_1{\lambda }_2({\lambda }_1+{\lambda }_2)}{({\lambda }_1\cdot {\lambda }_2{)}^2}=512-24=488$