Why is $\sqrt{x}$ only positive when $x >0$? (i.e., why is $\sqrt{4}$ not both $2$ and $-2$?) I have always asked myself why this happens. 
If $x = 4$, then $\sqrt{x} = 2$, but if I search for the $\sqrt{4}$,  I get $2$ & $-2$.
 A: $\sqrt{\cdot}$ is a function. As a function it cannot return $2$ values. Actually mathematicians defined that function to show the positive root. What does it mean?
Suppose $x^2=b$. Then $x$ could take two vales: $\sqrt b$ and $-\sqrt b$. Both of them satisfy the equation. Meanwhile, if we have $x=\sqrt c$, then there is only $1$ value of $x$ that satisfy the equation, namely $\sqrt c$, which is a positive number.
If we put this more formally: Let $f:[0,\infty) \longrightarrow \mathbb R$ be a function defined by $x\longmapsto f(x)=\sqrt{x}$. We know that $f\subseteq \mathbb R\times \mathbb R:[(x,z),(x,y)\in f\Longrightarrow z=y]\wedge [(x,f(x))\in f\,\forall x \in [0,\infty)]$. Also $(x,\sqrt x)\in f$ iff $\sqrt x>0\wedge (\sqrt x)^2=x$. If we accept $(x,-\sqrt x)\in f$, then $-\sqrt x = \sqrt x$, and that's nonsense, the square root wouldn't be a function. However IT IS  a function, and basically, it's just a definition that you take the positive root always.
A: The problem you are having is in how to interpret square roots.  You have probably been taught that when we say $\sqrt{4}$, we are thinking about the "number that when you square it, you get $4$".  If you are thinking about it this way, then of course $\sqrt{4}$ could be $2$ or $-2$ since when you square each of those, you get $4$.
But you should not think about square root this way.  The square root is a positive number.  So $\sqrt{4}$ is the positive number that you square to get $4$, which means $\sqrt{4} = 2$.
Now, when you have the equation $x^{2} = 4$, and you want to solve this equation for $x$, this is the equation where you ask yourself: "what number squared gives me $4$?"  And in this case, it's both $2$ and $-2$, i.e., $+ \sqrt{4}$ and $- \sqrt{4}$.  Notice that here, $\sqrt{4}$ is a positive number.  When we want to express the $-2$ answer, we write $- \sqrt{4}$.  This is because $\sqrt{4} = 2$.
So don't forget that the square root is a positive number.  For example, $\sqrt{9} = 3$.  But the solutions to $x^{2} = 9$ are $+ \sqrt{9}$ and $- \sqrt{9}$.
A: There is a difference.
$$\sqrt {x^2}=|x|$$
And,If
$$x^2=k \implies x=\pm\sqrt{k}$$
A: The map z->z^2 is taken to be ramification of order 2. Setting w=z^2 we have that the inverse function z=sqrt(w) has a branch point at zero, so that there is nontrivial monodromy about the origin in the sense of riemann surface theory. In this sense and modern algebraic senses, the square root fuction is a multivalued function. The convention is to choose the positive solution as the principle branch.
when talking about real numbers this means that sqrt(x^2)=±x, but we have made it a convention to select +x as the principal solution because of the most important applications, where physical measurements are positive.
In general, we have z^n=|w|*e^(2*i*pi) -> z=|w|^(1/n)*e^(2*i*(k/n)*pi) for k=0,1,...,n-1.
So, w=z^2 -> z=sqrt(|w|)e^(ik*pi) for k=0,1, which simplifies to z=±sqrt(|w|)
Also important, in the theory of generalized functions,
sqrt(z^2)=z*csgn(z)
