How can a square root be defined since it has two answers? I am aware that 1/0 is undefined for two reasons:


*

*If you would have to give an answer to this it is infinity which is not a number but a concept;

*The limit of 1/x for $x \to 0$ is either positive or negative infinity and since it has two limits we cannot state what its exact value is


How can we use this for square roots? It has multiple answers so why do we pick the positive one? if $x^2 = 16 \implies x = \sqrt{16} $ or $x = -\sqrt{16}$ for respectively the positive and negative solution. This implies that the square root function has a single answer and we must negate its answer to obtain the second solution. I understand this is extremely practical but it feels like 'cheating' since we know that there are 2 solutions, but in the function we simply ignore one of them. Is this correct?
 A: Take it this way: $a = \sqrt{16}$ is just a symbol denoting something.
It denotes (by definition) the non-negative root of $x^2 = 16$.
Since $16 = a^2 = (-a)^2$, it means the other solution of this equation is $-a = -\sqrt{16}$ 
A: In general, if $x$ is a positive number, there are $n$ complex numbers whose $n$th power is $x$. But there is only one real, positive number. That is defined to be $x^{1/n}$.  
A: Deciding whether $1/0$ is positive or negative is not really a problem. The real problem is there is no number $x$ such that $0 \times x = 1$. In contrast, there are exactly two distinct numbers $x_+$ and $x_-{}$ such that 
$(x_+)^2 = (x_-)^2 = 16$.
It's useful to be able to say sometimes which of the possible square
roots we mean in a certain expression. If you want the negative root,
write $-\sqrt{16}$. If you want to leave open the possibility of using
either square root, write $\pm\sqrt{16}$ as is done in most
presentations of the formula for the roots of a quadratic equation.
A: This has to be taken into account when solving equations.  
If you have (amongst other constraints) :
 $$x^2 = 16$$
then at first take you must consider: 
Not:
$$x = \sqrt{16} $$
But:
$$x = \pm \sqrt{16}$$
A: If you want to solve the equation $x^2=16$ you can rewrite it $$0=x^2-16=(x+4)(x-4)$$One of the factors must be zero and you get $x=-4$ or $x=4$. A quadratic equation typically has two roots (occasionally a double root, and sometimes the roots are complex numbers).
The interpretation of the expression $\sqrt {16}$, as a matter of convention, is to set the value equal to the positive square root.
Quite often with expressions which can take multiple values, convention (or choice) will assign a principal value, so that the expression can be used with confidence. 
When working in more general contexts it is not always true that principal values so chosen will always respect the normal laws of arithmetic, so care needs to be taken with expressions which involve multiple choices of principal value. Various common "paradoxes" in the manipulation of complex numbers can be attributed to this.
For example define $i=\sqrt {-1}$ then a "paradox" arises from $$i=\sqrt {-1}=\sqrt {\frac 1{-1}}=\frac {\sqrt 1}{\sqrt {-1}}=\frac 1i=\frac i{i^2}=-i$$
