The equation $(\dfrac{x}{x+1})^{2}+(\dfrac{x}{x-1})^{2}=a(a-1)$ has
- $four \ real \ roots \ if \ a>2 $
- $four \ real \ roots \ if \ a<-1$
- $two \ real \ roots \ if \ a= 1$
- $no \ real \ root \ if a<-1$
Let f(x) = (x/(x+1))^2 + (x/(x-1))^2. This simplifies to 2(x^4 + x^2)/(x^2-1)^2. Analysis of the range of this function shows it has four real roots for specific values of a.
Take the square root of the entire equation to get the absolute value of x over (x + 1) plus the absolute value of x over (x minus 1) equal to the square root of (a squared minus a). When we clear the denominators, this becomes the absolute value of x squared minus the absolute value of x times the square root of (a squared minus a) equals 1, which requires the right side to be greater than 2 to yield four distinct roots. This condition is met when the square root of (a squared minus a) is strictly greater than 2, meaning a squared minus a is greater than 4. Solving the inequality a squared minus a minus 4 is greater than 0 shows this holds true when a is greater than approximately 2.56 or less than approximately negative 1.56, making the condition a > 2 a correct representation of a range yielding four real roots.