" roots of "(x^(2)-bx)/(ax-c)=(K-1)/(K+1)" are numerically equal but opposite in sign,then "K=

If (x^(2)-bx)/(ax-b)=(m-1)/(m+1) has roots which are numerically equal but of opposite signs,the value of m must be:

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

If the roots of the equation (x^(2)-bx)/(ax-c)=(m-1)/(m+1) are equal to opposite sign,then the value of m will be (a-b)/(a+b) b.(b-a)/(a+b) c.(a+b)/(a-b) d.(b+a)/(b-a)

. For what value of m will the equation (x^(2)-bx)/(ax-c)=(m-1)/(m+1) have roots equal in magnitude but opposite in sign?

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

For what value of m will the equation (x^2-bx)/(ax-c)=(m-1)/(m+1) have roots equal in magnitude but opposite in sign?

For what value of m will the equation (x^2-bx)/(ax-c)=(m-1)/(m+1) have roots equal in magnitude but opposite in sign?