If $K$ is a monoid (which we write multiplicatively) and $x$ is an element of $K$, then the element $x^2$ is the square of $K$. Conversely, if $x^2 = y$, then $x$ is a square root of $y$.
If $K$ is an integral domain, then (in classical mathematics) $x$ and $-x$ are the only square roots of $x^2$. If $y$ has a square root, then we often denote its square roots together as $\pm\sqrt{y}$, although there is no meaning of $\sqrt{y}$ itself.
If $K$ is a linearly ordered field, then every element $y$ has a unique nonnegative square root if it has a square root at all; this is the principal square root of $y$ and denoted $\sqrt{y}$.
In constructive mathematics, we cannot prove that $x$ and $-x$ are the only square roots of $x^2$. In the ordered field of real numbers, for example, the absolute value ${|x|}$ (like $-{|x|}$, for that matter) is also a square root of $x^2$, yet we cannot constructively prove that ${|x|} = x$ or ${|x|} = -x$. Without using the lesser limited principle of omniscience, if $x$ is close to zero (and we do not yet know whether it is exactly zero), we cannot decide whether $x$ is nonnegative (so that ${|x|} = x$) or nonpositive (so that ${|x|} = -x$).
However, in any linearly ordered field with an absolute value (including any real-closed field), we still have a unique nonnegative square root of $x^2$, which is in fact ${|x|}$. Thus, we can still use the notation $\sqrt{y}$, but we cannot prove that every square root of $y$ is one of $\pm\sqrt{y}$. However, we can prove, in any integral domain even, that if $x \neq \sqrt{y}$ and $x \neq -\sqrt{y}$, then $x^2 \neq y$. (We are using the weak notions of field and integral domain so that $\mathbb{R}$ will be an example.)
If we accept results in the complex numbers (or in $K[\mathrm{i}]$ for $K$ any real-closed field), then every element of $K$ has a square root. Even if $y$ is close to zero, we can still use
to construct a square root of $y$. However, we cannot prove that every complex number has a square root in the internal language of an aribtrary W-topos, although weak countable choice is enough to prove this. (See Richman (1998).) A counterexample is the topos of sheaves on the real line, since there is no continuous function $\sqrt{}\colon U \to \mathbb{C}$ for $U$ any neighbourhood of $0$ in $\mathbb{C}$. This is important for interpreting the quadratic formula.
How much choice is needed to prove that every element of $K[\mathrm{i}]$ has a square root, where $K$ is any real-closed field? Maybe COSHEP will do?
Note that there is never any trouble finding a square root of $y$ if we assume that $y \neq 0$, nor (obviously) is there any trouble if we assume that $y = 0$. Accordingly, the classical results hold for discrete fields, but this doesn't apply constructively to analysis.
square root