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 $\alpha +\beta =0$
$\therefore$ $b\lambda +b+a\lambda -a=0$
$\Rightarrow$ $\lambda =\frac{a+b}{a+b}$