<s>[INST] Why is $ \frac{x}{\sqrt{x^{2}+1}}=x^{4}-x$ considered impossible to solve? This is from Stewart's Transcendentals.

If we were to try to find the exact intersection points, we
would have to solve the equation $$ \frac{x}{\sqrt{x^{2}+1}}=x^{4}-x$$
This looks like a very difficult equation to solve exactly (in
fact, it's impossible), so instead we use a graphing device to draw
the graphs of the two curves in Figure 7.

Why is this equation considered imposible to solve?
What is the factor that indicates you to not even try and just go directly with a graphing program.
 [/INST] Of course it is possible to solve this equation, but answer cannot be written in closed form.
$$\frac{x}{\sqrt{x^{2}+1}}=x^{4}-x,x\in\mathbb{R}$$
$$\frac{x}{\sqrt{x^2+1}}\left((x^3-1)\sqrt{x^2+1}-1\right)=0$$
$$x=0\lor (x^3-1)\sqrt{x^2+1}=1$$
Let find real root other than $x=0$.
$$(x^3-1)\sqrt{x^2+1}=1,x\in\mathbb{R}$$
$$(x^6-2x^3+1)(x^2+1)=1 \land x^3-1>0$$
$$x^8-2x^5+x^2+x^6-2x^3+1-1=0 \land x>1$$
$$x^2(x^6+x^4-2x^3-2x+1)=0 \land x>1$$
$$x^6+x^4-2x^3-2x+1=0 \land x>1$$
Real root of original equation other than $x=0$ is real root of $x^6+x^4-2x^3-2x+1=0$ which is greater than $1$. This root can be written in Wolfram Root notation (which is not part of commonly used closed form arrangement) as $Root(x^6+x^4-2x^3-2x+1,i)$, where $i$ is some integer number from $1$ to $6$. Numerical value of this root can be found using numerical methods or plotting.
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