<s>[INST] Solving $\sqrt{9-x^2} > x^2 + 1$ without graphic calculator for the exact form 
Is there any way to solve this inequality without using a graphic calculator to get the exact form?
$$\sqrt{9-x^2} > x^2 + 1$$

I've tried completing the square but I end up with
$$\frac{3 - \sqrt{41}}{2} < x^2 < \frac{3 + \sqrt{41}}{2}$$
which does not match with the answer on Desmos.
 [/INST] Hint:

*

*If $A>B>0$ then $A^2>B^2$. Apply this to $\sqrt{9-x^2}>x^2+1$.

*As $x^2+1$ is always positive then $9-x^2>0$.

Both must be satisfied. So you need $\cap$ to intersect the solution sets from both cases.
The hint is exapanded.
\begin{gather}
9-x^2>x^4+2x^2+1\\
x^4+3x^2-8<0\\
(x^2-\frac{-3+\sqrt{41}}{2})(x^2-\frac{-3-\sqrt{41}}{2})<0
\end{gather}
that must be intersected with
\begin{gather}
9-x^2>0\\
x^2<9
\end{gather}
The solution is $x^2< \frac{-3+\sqrt{41}}{2}$ or $$-\sqrt{\frac{-3+\sqrt{41}}{2}}<x< \sqrt{\frac{-3+\sqrt{41}}{2}}$$
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