Question

if x^2-bx/ax-c =m-1/m+1
has roots which are numerically equal but of opposite sign, then the value of'm'must be
(A) a-b/a+b
(B) a+b/a-b
(C) O
(D) 1
soneone please answer this​

Answer 1

$\textbf{Given:}$

$\dfrac{x^2-bx}{ax-c}=\dfrac{m-1}{m+1}$

$\textbf{To find:}$

$\text{The value of m}$

$\textbf{Solution:}$

$\text{Let the roots of the given quadratic equation be \alpha and -\alpha }$

$\text{Consider,}$

$\dfrac{x^2-bx}{ax-c}=\dfrac{m-1}{m+1}$

$(m+1)(x^2-bx)=(ax-c)(m-1)$

$(m+1)x^2-b(m+1)x-a(m-1)x+c(m-1)=0$

$(m+1)x^2-[b(m+1)+a(m-1)]x+c(m-1)=0$

$\text{Now,}$

$\text{Sum of the roots= \dfrac{-b}{a} }$

$\implies\alpha+(-\alpha)=\dfrac{b(m+1)+a(m-1)}{m+1}$

$\implies\,0=\dfrac{bm+b+am-a}{m+1}$

$\implies\,bm+b+am-a=0$

$\implies\,bm+am=a-b$

$\implies\,m(a+b)=a-b$

$\implies\boxed{\bf\,m=\dfrac{a-b}{a+b}}$

$\therefore\textbf{Option (A) is correct}$

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