# Confused about the extraneous root of $\frac{x \sqrt{A^2 - x^2} + x}{x^2 - \sqrt{A^2 - x^2}}$

I have:

$$f(x) = \frac{x \sqrt{A^2 - x^2} + x}{x^2 - \sqrt{A^2 - x^2}}$$

to find the roots:

$$x \sqrt{A^2 - x^2} + x = 0$$ $$\sqrt{A^2 - x^2} + 1 = 0$$ $$\sqrt{A^2 - x^2} = -1$$ $$A^2 - x^2 = 1$$ $$A^2 - 1 = x^2$$

which gives these roots (real for A > 1):

$$x_{root} = 0, \pm \sqrt{A^2 - 1}$$

but plugging the non-zero roots back in gets:

$$f(x_{root}) = \frac{\pm 2 \sqrt{A^2 - 1}}{A^2 - 2} \neq 0$$

which can be verified graphically in desmos: https://www.desmos.com/calculator/zjsscvsiun and even in the complex plane the roots don't show up off the real axis (for A = 1.5, fake root = 1.118):

Why doesn't this root show up? Is there an invalid step in finding the root or additional condition that's being missed?

Edit/Recap:

I should have been considering the multi-valued function

$$f(x) = \frac{x \pm \sqrt{A^2 - x^2} + x}{x^2 \mp \sqrt{A^2 - x^2}}$$

where the roots show up for the negative branch. For completeness, here's the complex plot for the negative branch:

• I think there is an error when you found the roots. How did you find them? Commented Nov 30, 2023 at 3:12
• By setting the numerator equal to zero and solving, I've updated with the details. Sympy.solve gives the same result. Commented Nov 30, 2023 at 4:12

$$\sqrt{A^2 - x^2} = -1$$ has no real solution.
LHS is defined only when $$A^2 - x^2 \ge 0$$ in which case, LHS is non-negative and so, the equation has no real solution.
Thus, $$x=0$$ is the only real solution provided the denominator of LHS in the original equation $$x^2 -\sqrt{A^2 - x^2}$$ is non-zero at $$x=0$$ (which holds when $$A\ne 0$$).
Rewriting this, we get no real solutions for $$x$$ if $$A=0$$ and only real solution $$x=0$$ if $$A\ne 0$$.