If , , where

Question

If $\alpha ,\beta $, where $\alpha <\beta $, are the roots of the equation $\lambda {x}^{2}-(\lambda +3)x+3=0$ such that $\frac{1}{\alpha }-\frac{1}{\beta }=\frac{1}{3}$, then the sum of all possible values of $\lambda $ is

Accepted Answer

The given equation is $\lambda {x}^{2}-(\lambda +3)x+3=0$. We can find the roots by factoring:

$\lambda {x}^{2}-\lambda x-3x+3=0$

$\Rightarrow$ $\lambda x(x-1)-3(x-1)=0$

$\Rightarrow$ $(x-1)(\lambda x-3)=0$

The two roots are : $x=1,\, x=\frac{3}{1}$

We are given that $\alpha $ and $\beta $ are the roots such that $\alpha $\frac{1}{\alpha }-\frac{1}{\beta }=\frac{1}{3}$

Because we don't know if 1 is smaller or larger than $\frac{3}{\lambda }$, we must consider two distinct cases.

Case A : $\alpha =1$ and $\beta =\frac{3}{\lambda }$

Substitute these into the condition $\frac{1}{\alpha }-\frac{1}{\beta }=\frac{1}{3}$ :

$$$\begin{gathered} \frac{1}{1}-\frac{1}{3/\lambda }=\frac{1}{3} \ 1-\frac{\lambda }{3}=\frac{1}{3} \ 1-\frac{1}{3}=\frac{\lambda }{3} \ 1-\frac{1}{3}=\frac{\lambda }{3} \ \frac{2}{3}=\frac{\lambda }{3}\implies \lambda =2 \end{gathered}$$$

Check the constraint $\alpha Case B : $\alpha =\frac{3}{\lambda }$ and $\beta =1$

Substitute these into the condition $\frac{1}{\alpha }-\frac{1}{\beta }=\frac{1}{3}$ :

$\begin{gathered} \frac{1}{3/\lambda }-\frac{1}{1}=\frac{1}{3} \ \frac{\lambda }{3}-1=\frac{1}{3} \ \frac{\lambda }{3}=1+\frac{1}{3} \ \frac{\lambda }{3}=\frac{4}{3}\implies \lambda =4 \end{gathered}$

Check the constraint $\alpha The possible values for $\lambda $ are 2 and 4 .

$=2+4=6$

The sum of all possible values of $

Solutions · 1