This was posted in Beginning Algebra. I am sure that Dr. Peterson is correct that use of the quartic formula could be used to get an answer by purely algebraic means. But the quartic formula is not beginning algebra. And Dr. Peterson‘s graphic method is very nice. But the OP wanted an analytic answer. A very simple one is available through calculus.
Consider the family of real functions specified by
[math]x \ne 1 \text { and } f(x) = y = x^2 + \left ( \dfrac{x}{x - 1} \right )^2 - a.[/math]
How can we describe the real roots of such a function, realizing that the roots will be the solutions to equations of the form
[math]x^2 + \left ( \dfrac{x}{x - 1} \right )^2 = a.[/math]
It should be obvious that if a is negative, the function will have no real roots.
What is the derivative of f(x)?
[math]\dfrac{dy}{dx} = 2x + 2x(x - 1)^{-2} + x^2(-2)(x - 1)^{-3} = 2x * \left ( 1 + \dfrac{x - 1}{(x - 1)^3} - \dfrac{x}{(x - 1)^3}\right ) =[/math]
[math]\dfrac{2x}{(x - 1)^3} * \left ( x^3 - 3x^2 + 3x - 2 \right ) = \dfrac{2x(x - 2)(x^2 - x + 1)}{(x - 1)^3.}[/math]
The term [math]x^2 - x + 1[/math] has no real roots so the derivative is equal to 0 only at x = 0 and x = 2.
If x < 0, the derivative is negative. If 0 < x < 1, the derivative is positive. Thus the function has a minimum at x = 0.
If 1 < x < 2, the derivative is negative. If x > 0, the derivative is positive. Thus the function has another minimum at x = 2.
At x = 0, the function equals - a. Thus, if a = 0, the function has a duplicated real root at x = 0. If a = 0, the function equals 8 at x = 2.
If a = 8, we have three real roots. If a > 8, there are four real roots.
Going back to the equation, we can summarize
If a is negative, there are no real solutions.
If a is zero, there is one real solution, namely zero.
0 < a < 8, there are two real solutions.
If a = 8, there are three real solutions, one at x = 2.
If a > 8, there are four real solutions.
If there is more than one real solution, one is negative and another is between 0 and 1.
If there are four solutions, one is between 1 and 2, and another exceeds 2.
Relatively easy calculus problem. Horrific algebra problem.