Inverse of fourth root function I have a function in the form
$$y=a\sqrt[^4]{b^2-(x-c)^2}, \qquad a, b, c\in \mathbb{R}$$
but I'm having trouble finding its inverse. That is, solving for $x$. The solution should seem pretty trivial with putting both sides to the $4^{\text{th}}$ power, rearranging, and applying the quadratic formula, but I can't seem to obtain the correct solution. Any suggestions? Thanks!

The equation is easier to work into the quadratic equation if we first get rid of radicals and avoid trying to "simplify" by using the $y^4/a^4$ rational. Expansion gives us the best view of a,b,c.
$$y=a\sqrt[^4]{b^2-(x-c)^2}\implies y^4=a^4\big(b^2-(c^2 - 2 c x + x^2)^2\big)\\
\implies 0=a^4 b^2 - a^4 c^2 + 2 a^4 c x - a^4 x^2-y^4\\
\implies      (a^4)x^2 - (2 a^4 c)x + (y^4 - a^4 b^2 + a^4 c^2) = 0\\
\implies x=  \frac{(2 a^4 c) \pm\sqrt{(2 a^4 c)^2-4(a^4)(y^4 - a^4 b^2 + a^4 c^2)}}{2a^4}\\
= \frac{(2 a^4 c) \pm\sqrt{4 a^8 b^2 - 4 a^4 y^4}}{2a^4} \\
\implies x = c \pm \frac{\sqrt{(a^2 b - y^2) (a^2 b + y^2)}}{a^2 }
\quad a,b,c,y\in\mathbb{R}\land a\ne 0\\
 x\in\mathbb{R}\iff y\le \sqrt{a^2b}$$
