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

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 (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:

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)

Find m so that the roots of the equation (x^(2)-bx)/(ax-c)=(m-1)/(m+1) may be equal in magnitude and opposite in sign.

If b=0,c<0, is it true that the roots of x^(2)+bx+c=0 are numerically equal and opposite in sign? Justify your answer.

Find k so that the roots of the equation (x^(2)-qx)/(px-r) = (k-1)/(k+1) may be equal in magnitude but opposite in sign.