Pythagorean Closure of $\mathbb{Q}$ is extremely obscure. Can someone give a clear answer? My previous question didn't get me a definite answer. I've refined it so with a specific example so it might help someone clear my confusion.
In Stewart's book, The Pythagorean closure is defined as
$\textbf{Definition}$ The Pythagorean Closure $\mathbb{Q}^{\operatorname{Py}}$ of $\mathbb{Q}$ is the smallest subfield $K \subseteq \mathbb{C}$ with the property
$$z \in K \Rightarrow  \pm \sqrt{z} \in K   \tag{Eq1}$$
I am not clear about what this definition is saying. Specifically, in the proof of the theorem (7.11 in Stewart Galois theory 4th Ed)
$\textbf{Theorem}$
A complex number $\alpha$ is an element of $\mathbb{Q}^{\operatorname{Py}}$ if and only if there is a tower of field extensions
$$\mathbb{Q}=K_0 \subseteq K_1 \subseteq \cdots \subseteq =K_n \mathbb{Q}(\alpha)$$
such that
$$   [K_{j+1}:K_j]=2$$
for $0 \leq j \leq n-1$
$\textbf{proof}$
($\Leftarrow$) Suppose that such a tower exists. We prove by induction on $j$ that $K_j \subseteq \mathbb{Q}^{\operatorname{Py}}$. This is clear for $j=0$.
$\textbf{Note}$:Since I am not clear of the definition of $\mathbb{Q}^{\operatorname{Py}}$, I do not truly understand why $K_0=\mathbb{Q} \in \mathbb{Q}^{\operatorname{Py}}$. But since Stewart says that $\mathbb{Q}^{\operatorname{Py}}$ is the intersection of all subfields of $\mathbb{C}$ satisfying Eq1, for now I am taking it for granted and acquiesced.
Now $K_{j+1}$ is an extension of $K_j$ with degree $2$ so $K_j+1=K_j(\beta)$ where the minimum polynomial of $\beta$ over $K_j$ is a quadratic. Since quadratics can be solved by extracting square roots, $\beta \in \mathbb{Q}^{\operatorname{Py}}$, so $K_{j+1} \in \subseteq \mathbb{Q}^{\operatorname{Py}}$. (Pause proof)
The statement $\beta \in \mathbb{Q}^{\operatorname{Py}}$ is what I am not convinced with. So, if $\beta \in \mathbb{Q}^{\operatorname{Py}}$, then $\pm \sqrt{\beta} \in \mathbb{Q}^{\operatorname{Py}}$. I do not see $\textit{what}$ fact assures this. I'll be more specific with a break down of how my train of thought is running now and where my confusion is coming from.
So I know $m$ is a quadratic $ax^2+bx+c=m(x)$. Where, $m(\beta)=0$.
Thus $a \beta^2+b \beta+c=0$. I believe completing the square gives,
$$\Big(\beta + \frac{b}{2a}\Big)^2=\frac{b^2-4ac}{4a^2}$$
So,
$$\beta =\pm \sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}$$
Coinciding with the result using the quadratic formula(I did this since the proof says "extract square roots"). Hence,
$$\pm \sqrt{\beta} = \pm \sqrt{\pm \sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}} \tag{Eq2}$$
Which looks ugly. What assures Eq2 $\in \mathbb{Q}^{\operatorname{Py}}$? I get stuck answering this; simply because, there is no further information given in the definition to determine whether or not a particular number is in or not in $\mathbb{Q}^{\operatorname{Py}}$. 
$\textbf{Question }$:Like in this case, I have $\beta$. And $\pm \sqrt{\beta}$ is Eq2. Is Eq2 in $\mathbb{Q}^{\operatorname{Py}}$? It is IF $\beta \in \mathbb{Q}^{\operatorname{Py}}$ but then I must turn to and how can I answer the question "is $\beta$ in $\mathbb{Q}^{\operatorname{Py}}$"? Well, if $\beta^2 \in \mathbb{Q}^{\operatorname{Py}}$ then it is, but then what assures me that $\beta^2 \in \mathbb{Q}^{\operatorname{Py}}$? What is the determining factor? I do not know the particular form of the elements of $\mathbb{Q}^{\operatorname{Py}}$ or anything. And all I know about $\beta$ in this particular case is that it's implied that it's  in $\mathbb{C}$.
Essentially, I do not understand the statement "Since quadratics can be solved by extracting square roots, $\beta \in \mathbb{Q}^{\operatorname{Py}}$"
I suspect my question is caused by my lack of comfortability and understanding of the definition given by Stewart. Further, "Pythagorean Closure of $\mathbb{Q}$" doesn't seem like a common notion on the internet; Pythagorean field is the most hit I have which, Wikipedia claims 
"In algebra, a Pythagorean field is a field in which every sum of two squares is a square"
And I cannot link THIS definition to the one Stewart is giving. So I have assumed they may be related, but different. 
 A: The statement 


*

*$\beta \in \mathbb{Q}^{Py}$


is equivalent to the statement 


*

*for every subfield $K \subset \mathbb{C}$ having the property that $z \in K \implies \pm \sqrt{z} \in K$ we have $\beta \in K$.


Let's prove that $\beta \in \mathbb{Q}^{Py} \implies \pm\sqrt{\beta} \in \mathbb{Q}^{Py}$. 
Assume $\beta \in \mathbb{Q}^{Py}$. Assume $K \subset \mathbb{C}$ is any subfield having the property that $z \in K \implies \pm \sqrt{z} \in K$. Since $\mathbb{Q}^{Py}$ is the intersection of all such subfields, it follows that $\beta \in K$. By assumption on $K$ it follows that $\pm \sqrt{\beta} \in K$. Since $K$ was arbitrary (amongst all subfields having the property that $z \in K \implies \pm \sqrt{z} \in K$), it follows that $\pm\sqrt{\beta}$ is in the intersection of all those $K$. That intersection equals $\mathbb{Q}^{Py}$. We have proved that $\pm\sqrt{\beta} \in \mathbb{Q}^{Py}$. 
In your example, you are starting with the assumption that the coefficients $a,b,c$ are all in $\mathbb{Q}^{Py}$. And in (Eq2) you have written out an explicit expression for $\pm\sqrt{\beta}$ formed from $a,b,c$ using the operations of arithmetic and square root. But  $\mathbb{Q}^{Py}$ is closed under the operations of arithmetic because it is a field, and I've just shown it is also closed under the operation of square root. It follows that $\pm \sqrt{\beta} \in \mathbb{Q}^{Py}$.
