Solving a radical equation for real $x$

Solve for $x \in \mathbb{R}$

$$\dfrac{\sqrt{x^2-x+2}}{1+\sqrt{-x^2+x+2}} - \dfrac{\sqrt{x^2+x}}{1+\sqrt{-x^2-x+4}} = x^2-1$$

I tried squaring the equation but it became a sixteen degree equation. I also tried substitutions, but that didn't help. There must be some elegant solution to it in its current form.

Any help will be appreciated.
Thank you.

Solution 1:

Firstly, we need to find what values of $x$ are acceptable. Solving the inequalities: $$\begin{array}[t]{rl} x^2-x+2 &\ge 0\\ x^2+x &\ge 0 \\ -x^2+x+2 &\ge 0\\ -x^2-x+4 &\ge 0 \end{array} $$ yields $$\left.\begin{array}{l} x \in \mathbb R\\ x \in (-\infty,-1] \cup [0,+\infty)\\ x \in [-1,2] \\ x\in \left[ \frac{-1-\sqrt{17}}{2},\frac{-1+\sqrt{17}}{2}\right] \end{array}\right\}\implies x\in \{-1\} \cup \left[0,\frac{-1+\sqrt{17}}{2}\right]. $$

We can see that $x = -1$ is not a solution, so we are looking for non - negative solutions.

For $ x \in [0,1)$ the RHS is negative. But in that case we have that the first numerator is greater than the second one (this can be derived easily by solving the inequality $\sqrt{x^2-x+2} \gt \sqrt{x^2+x}$) and the first denominator is less than the second denominator, thus the first fraction is always greater than the second one, so their difference is positive. Thus there are no solutions in $[0,1)$.

Similarly, for $x\in \left(1,\frac{-1+\sqrt{17}}{2}\right]$ we have that the RHS is positive and the LHS is negative.

This implies that the only real solution is $x = 1$.