Only for you

It is obvious that $x=m$ works as a solution. Thus, we only have to find the condition that $x=-m$ works as a solution.

$-\frac{1}{m}+\frac{1}{b-m}=\frac{1}{m}+\frac{1}{m+b}$

$\frac{2}{m}=\frac{1}{b-m}-\frac{1}{b+m}$

$\frac{2}{m}=\frac{2m}{b^2-m^2}$

$b^2-m^2=m^2$

$\boxed{b^2=2m^2}$

We shall first rewrite both sides of the equation as single fractions:

(2x+b)/(x(x+b))= (2m+b)/(m(m+b))

Now, we shall rewrite the equation in the standard quadratic form, Ax²+Bx+C= 0

(2m+b)x²-(2m²-b²)x-(bm²+b²m)= 0

A= (2m+b) B= -(2m²-b²) C= -(bm²+b²m)

Now we are asked for the condition necessary for the equation to have real roots that are equal in magnitude but opposite in sign, which is synonymous with asking for the condition necessary for the equation to have real, distinct roots which are equal in magnitude.

Such a quadratic equation is of the form (x+α)(x-α)= 0. Notice that this is a difference of two squares and can be written in the form x²-α²= 0. Another thing to note is that B= 0 for such an equation.

Looking at the equation in the problem, [B=0]⇒ [-(2m²-b²)= 0].

Since [-(2m²-b²)= 0]⇒[(2m²-b²)= 0], [B=0]⇒ [(2m²-b²)= 0].

Since [(2m²-b²)= 0]⇒ [2m²=b²], [B=0]⇒[2m²=b²].

Thus, the condition necessary for the equation to have real roots that are equal in magnitude but opposite in sign is that 2m²=b².