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

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

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

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

The equation whose roots are numerically equal but opposite in sign of the roots of 3x^2-5x-7=0 is

The equation whose roots are numerically equal but opposite in sign to the roots of 3x^(2) - 5x - 7 = 0 is