Question

If the roots $ \alpha $ , $ \beta $ of the equation $ \frac{{{x}^{2}}-bx}{ax-c}=\frac{\lambda -1}{\lambda +1} $ are such that $ \alpha +\beta =0, $ then the value of $ \lambda $ is:

• A. $ \frac{a-b}{a+b} $
• B. $ c $
• C. $ \frac{1}{c} $
• D. $ \frac{a+b}{a-b} $
• E. $ \frac{1}{ab} $

Answer 1

Answer: (A)

The given equation can be rewritten as $ {{x}^{2}}(\lambda +1)-(b\lambda +b+a\lambda -a)x+x(\lambda -1)=0 $
$ \because $ $ \alpha +\beta =\frac{(b\lambda +b+a\lambda +a)}{\lambda +1} $
Also $ \alpha +\beta =0 $
$ \therefore $ $ b\lambda +b+a\lambda -a=0 $
$ \Rightarrow $ $ \lambda =\frac{a+b}{a+b} $