Question

Let a, b, c be positive real numbers. If $ \frac{{{x}^{2}}-bx}{ax-c}=\frac{m-1}{m+1} $ has two roots which are numerically equal but opposite in sign, then the value of m is

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

Answer 1

Answer: (E)

Given, $ \frac{{{x}^{2}}-bx}{ax-c}=\frac{m-1}{m+1} $
$ \Rightarrow $ $ (m+1){{x}^{2}}-b(m+1)x $
$=a(m-1)x-c(m-1) $
$ \Rightarrow $ $ (m+1){{x}^{2}}-(bm+b+am-a)x $ $ +c(m-1)=0 $
$ \Rightarrow $ $ (m+1){{x}^{2}}-\{(a+b)m+(b-a)\}x $ $ +c(m-1)=0 $
Let the roots of the equation is $ (a,-a) $ .
Then sum of the roots
$=\frac{(a+b)m+(b-a)}{(m+1)} $ $ \alpha -\alpha =\frac{(a+b)m+(b-a)}{(m+1)}=0 $
$ \Rightarrow $ $ (a+b)m-(a-b)=0 $
$ \Rightarrow $ $ m=\frac{a-b}{a+b} $