Let $a, b$ and $c$ be positive real numbers. If $\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}$ has two roots which…
Let $a, b$ and $c$ be positive real numbers. If $\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}$ has two roots which are numerically equal but opposite in sign, then the value of $m$ is
$c$
$\frac{1}{c}$
$\frac{a+b}{a-b}$
$\frac{a-b}{a+b}$
Solution
$
\begin{aligned}
& \text { Given, } \frac{x^2-b x}{a x-c}=\frac{m-1}{m+1} \\
& \Rightarrow\left(x^2-b x\right)(m+1)=(m-1)(a x-c) \\
& \Rightarrow x^2(m+1)-x(b(m+1)+a(m-1)+c(m-1)=0
\end{aligned}
$
Let $p$ and $-p$ are roots of Eq. (i), then
$
\begin{aligned}
& \text { Sum of roots } \Rightarrow p+(-p)=\frac{b(m+1)+a(m-1)}{m+1} \\
& 0=\frac{(b+a) m+(b-a)}{m+1} \\
& \therefore m(b+a)+(b-a)=0 \\
& \Rightarrow \quad m=\frac{a-b}{a+b}
\end{aligned}
$