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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):

enter image description here

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:

enter image description here

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

1 Answer 1

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

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  • $\begingroup$ Haha of course, thank you!! Physically, I should have been using the multi-valued function g(x) = [x(±(A^2 - x^2)^0.5 + 1)]/[x^2 - (±(A^2 - x^2)^0.5)] $\endgroup$ Commented Nov 30, 2023 at 5:00
  • $\begingroup$ Indeed when substituting +(A^2 - x^2)^0.5 for -(A^2 - x^2)^0.5 in desmos the root shows up: desmos.com/calculator/0uelidix7o $\endgroup$ Commented Nov 30, 2023 at 5:02

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