When is irreducibility of a polynomial over a field equivalent to not having any roots in it? 
When is irreducibility of a polynomial over a field equivalent to not having any roots in it?

Apart from of course, the simple cases when a polynomial $f \in K[x]$ is of degree less than or equal to three. 
One direction is clear: If a polynomial is irreducible in $K$, it can have no roots in it. But the converse is much more bizarre. So I pose:


*

*What conditions must be put on $f$ so that this happens? 

*How does information about $K$ alter this?


I do not even know where to start. Links to any research done in this area is also appreciated :) Thanks, guys!
 A: Here are two results (to be found in  Lang's Algebra, for example): 


*

*Let $k$ be a field of any characteristic $\geq 0$ and $p$ an odd
prime  number (not supposed to equal $\operatorname{char} k$) If
$x^p-a\in k(x)$ has no root in $k$, then it is irreducible  and,
more strongly, also $x^{p^n}-a$ is  irreducible over $k$ for all 
integers $n\geq 0$  (Capelli).

*Over a field $k$ of characteristic $p\gt0$ the polynomial
    $f(x)=x^p-x-a\in k[x]$ is irreducible if and only if it has no root
    in $k$ (Artin-Schreyer).

A: This is not an answer in the strict sense but  a result strongly related to your question.    
Let $k$ be a field and let $f(x)\in k[x]\setminus k$ be a polynomial of degree $n\gt0$. We have the equivalence: 

$f(x)$  irreducible over $k\\$ $\iff$ For any extensions $k\subset K$ of degree $\leq n/2$, $f(x)$  has no root in $K$.   

Proof of $\Longrightarrow:$
Any field $K$ in which $f(x)$ has a root contains a copy of $k[x]/(f(x))$ and has thus degree $\geq n$
Proof of $\Longleftarrow:$
If $f(x)$ were reducible one of its irreducible factors $g(x)$ would have degree $deg(g(x))\leq n/2$ and thus the extension $K:=k[x]/(g(x))$ would contain a root of $g(x)$ and a fortiori a root of $f(x)$, contradicting the hypothesis.   
