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.
 A: Let me try to explain what this is saying. I am going to guess at some of the author's intentions, because what's written down is not quite precise.
First off, if you are doing problems at this level, what the author is really trying to communicate to you is: this problem looks too hard: it has a lot of terms (including a square root), with terms with pretty high degree, so from that you could guess that you probably don't have the mathematical tools to solve it algebraically, and thus will have to use a graphing calculator. This is not a precise, mathematical statement: any simple problem can me made to look hard just by adding a bunch of redundant terms that will cancel once you start simplifying it. But just based on how it looks, it's probably so complicated that you'll want to use a graphing calculator.
The author adds, as a side note/justification, that the problem is "impossible to solve".

Why is this equation considered imposible to solve?

That is a very hard question. The author is probably referring to the notion of "solvability in radicals": the ability to write down a root of a polynomial in terms of addition, subtraction, multiplication, division, and taking $n$-th roots. All polynomials of degree up to 4, that is, all equations of the form $ax^4 + bx^3 + cx^2 + dx + e = 0$, are solvable in radicals. There is a very famous theorem, the Abel-Ruffini theorem, which says that for polynomials of degree 5 and higher, there may not be solutions in radicals. For example, the equation $x^5 + x + 1 = 0$ has a solution, but you cannot write it down using just arithmetic and $n$-th roots.
For your problem, if you try to simplify it -- square both sides, then multiply by the denominator, then bring everything to one side -- you will see that you have a degree 10 polynomial. However, although some degree 10 polynomials do not have solutions in radicals, it is generally not at all easy to see which ones do and which ones do not. (In particular, 0 is a solution of your equation, and once you "eliminate" that one, you are left with a degree 6 polynomial -- almost down to degree 4!)
In fact it is true that you cannot write the solutions (other than 0) to that equation in radicals, but this is by no means easy to demonstrate or obvious. The author is just telling you, without expecting you to understand why at this stage.
A: 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.
