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
$=\frac{(a+b)m+(b-a)}{(m+1)}$ $\alpha -\alpha =\frac{(a+b)m+(b-a)}{(m+1)}=0$
$\Rightarrow$ $(a+b)m-(a-b)=0$
$\Rightarrow$ $m=\frac{a-b}{a+b}$