If the equation x2−bxax−c=  m−1m+1  has roots equal numerically are opposite in sign then m=

Given equation is  x2−bxax−c=  m−1m+1  ⇒(m+1)x2  −bx(m+1)=  ax(m−1)  −c(m−1)  ⇒(m+1)x2−x{bm+b+am−a}+c  (m−1)=0 But, sum of the roots =0 ⇒    bm+b+am−am+1  =0 ⇒    m  (b+a)=a−b   ⇒  m=  a−ba+b