<s>[INST] $h^2 = x^2 + (x+1)^4$ This is a question I had of finding an exact value of $h$ as I was interested in the process and techniques used. The question was one to do with Pythagoras and the math shown is the question in $a^2 + b^2 = c^2$ form. 
The question was what is the size of the hypotenuse of a right-angled triangle if one side is a size $x$ and the other is a size $x^2 + 2x + 1$ (which was rewritten as $(x + 1)^2$). I got an answer from Wolfram Alpha of a value for $h$ that does not include $x$, only numbers. 
Wolfram Input:

Wolfram Answer:

What would be the process of finding the exact value of h as to where there is no sign of x in the h = equation?
P.S. If the process is simply too lengthy or tedious, feel free to take this question down.
 [/INST] \begin{align}
c^2 &= a^2 + b^2 \\[1em]
h^2 &= x^2 + \left[(x+1)^2\right]^2 \\
h^2 &= x^2 + (x+1)^4 \\
h^2 &= x^2 + x^4 + 4x^3 + 6x^2 + 4x +1\\
h^2 &= x^4 + 4x^3 + 7x^2 + 4x +1\\
h   &= \sqrt{x^4 + 4x^3 + 7x^2 + 4x +1}
\end{align}
You probably put into Wolfram Alpha some other task - you wanted the square root, and you obtained the root (= solution) of the system of equations with unknowns $h$ and $x$.
Addendum:
If you wanted from Wolfram Alpha the result of my computing, enter into it 
solve(h^2 = x^2 + (x+1)^4, h)

(h at the end determines which you want to compute), and you will obtain
$$h   = \pm \sqrt{x^4 + 4x^3 + 7x^2 + 4x +1}$$
If you want the result for a particular $x$, enter its value, e. g.
h^2 = x^2 + (x+1)^4, x=1

to obtain plot

and solutions
$$\begin{array}{l}{h=-\sqrt{17}, \quad x=1} \\ {h=\sqrt{17}, \quad x=1}\end{array}$$
</s>