Definition of " the principal n-th root of " using a sign condition.[modified title] [ Edited] 
Is it ok to say that, the principal n-th root of a is simply the number x such that : 
(1) x to the n-th power is equal to a
and 
(2)  x has the same sign as a ? 
My question deals specially with the second alleged condition. 
If I am correct, this is not the usual way to define "the n-th root of" a quantity. Why?  Is it because the presence of a sign condition would make the definition useless in algebra in cases where we do not know the sign of a litteral expression such as "  x cubed"?  
Would this definition be ok at least for integers, rational and real numbers? 
 A: "$x$ to the $n^{th}$ power is $a$" is the only rule.
For positive $a$,


*

*if $n$ is odd, then $x$ must be positive;

*if $n$ is even, then $x$ can have any sign (if you want a single-valued function, take positive).
For negative $a$,


*

*if $n$ is odd, then $x$ must be negative;

*if $n$ is even, then the root is not defined.
Other conventions would be impractical.
In particular, because of
$$\sqrt{-1}:=-1$$ you wouldn't be allowed to write
$$(\sqrt[n]a)^n=a.$$
A: This answer addresses the original question, before the title was changed.
It is useful to distinguish the concept of "an $n$-th root of a number" from the concept of "the $n$-th root function". Notice the choice of articles: the indefinite article "an" in the first case; the definite article "the" in the second case.
Notice that "an $n$-root of a number" need not be unique. For example, both $1$ and $-1$ are square roots of $1$. Also, each of $1,-1,i,-i$ are fourth roots of $1$. This is a very useful mathematical concept, and it is not worthwhile to try to resolve this non-uniqueness. In fact, the non-uniqueness itself is very important in the general study of roots of polynomials. In general the equation $x^n-a=0$ has exactly $n$ roots in the complex numbers, for any complex number $a=x+iy$; the case $a=0$ is special in that the root $x=0$ is counted with multiplicity $n$. More generaly there is a beautiful theorem, the fundamental theorem of algebra, saying that for every polynomial (with complex coefficients) of degree $n$, the number of roots of that polynomial (in the complex numbers), when counted with multiplicity, is exactly equal to $n$.
On the other hand, it is also useful to have "the $n$-th root function", often denoted $f(x) = \sqrt[n]{x}$. This is very important in calculus, for instance. What you write almost works to express the definition of the $n$-th root function, but it does not quite work. In order to define that function, as with any function you should also be very explicit about the domain and range (and here I shall assume real number values of the input $x$):

If $n$ is even then $f(x) = \sqrt[n]{x}$ is the function with domain $[0,+\infty)$ and range $[0,+\infty)$, which for each $x \in [0,\infty)$ is  defined so that $\sqrt[n]{x}$ is the unique $n$-th root of $x$ that is contained in $[0,\infty)$.
If $n$ is odd then $f(x) = \sqrt[n]{x}$ is the function with domain $(-\infty,+\infty)$ and range $(-\infty,+\infty)$, which for each $x \in (-\infty,+\infty)$ is defined so that $\sqrt[n]{x}$ is the unique $n$-th root of $x$ that is contained in $(-\infty,+\infty)$.

Notice what I have done here: instead of expressing the signs of the inputs and outputs, I have simply expressed the domain and range of the function appropriately.
But there is still one more feature of this definition to take careful note of: in the phrase "the unique $n$-th root of $x$ that is contained in ...", the uniqueness must be proved before the definition is stated, otherwise one does not know that the function $f(x)=\sqrt[n]{x}$ is well-defined.
