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!

\begin{align}
y=a\sqrt[^4]{b^2-(x-c)^2} &\implies y^{4} = a^{4}(b^{2} - (x-c)^2)\\
&\implies \frac{y^{4}}{a^{4}} = b^{2} - x^{2}+2cx-c^{2}\\
&\implies x^{2} - 2cx + \left(\frac{y^{4}}{a^{4}}-b^{2}+c^{2}\right) = 0\\
&\implies x = \frac{2c \pm \sqrt{4c^2-4\left(\frac{y^{4}}{a^{4}}-b^{2}+c^{2}\right)}}{2}\\
&\implies\boxed{x = c\pm\sqrt{b^2-\frac{y^{4}}{a^{4}}}}
\end{align}
