Let $[L:K]=2$ and char $K\neq 2$ show that $L/K$ is a simple 2-radical extension. I want to show that
$[L:K]=2$ and char $K\neq 2$ $\Rightarrow$ $L/K$ is a simple 2-radical extension.
I know that, since $[L:K]$ is prime, the extension is simple. I also concluded that, since $[L:K]=[K(a):K]=[a:K]$ for all $a\in L/K$, the minimal polynomial of $a$ has to have degree 2. But how do I proceed from here ? And how does char$K\neq 2$ help me ?

Choose any
$a \in L \setminus K; \tag 1$
then since
$[L:K] = 2, \tag 2$
it follows that
$\{1, a \} \subset L \tag 3$
forms a basis for $L$ over $K$; thus the set
$\{1, a, a^2 \} \tag 4$
is linearly dependent over $K$; hence we have
$\sigma, \tau, \rho \in K, \tag 5$
not all zero, such that
$\sigma a^2 + \tau a + \rho = \sigma a^2 + \tau a + \rho 1 = 0; \tag 6$
we note that
$\sigma \ne 0, \tag 7$
lest
$\tau a + \rho = 0; \tag 8$
but then 
$\tau \ne 0 \Longrightarrow a = -\dfrac{\rho}{\tau} \in K \Rightarrow \Leftarrow a \in L \setminus K, \tag 9$
whereas
$\tau = 0 \Longrightarrow \rho = 0 \Longrightarrow \sigma, \tau, \rho = 0, \tag{10}$
contradicting our hypothesis that there is a non-zero element amongst $\sigma$, $\tau$, $\rho$; thus (7) binds and setting
$\alpha = \dfrac{\tau}{\sigma}, \; \beta = \dfrac{\rho}{\sigma}, \tag{11}$
we write (6) in the form
$a^2 + \alpha a + \beta = 0, \tag{12}$
that is, $a$ satisfies the monic quadratic polynomial
$x^2 + \alpha x+ \beta \in K[x]; \tag{13}$
(12) yields
$a^2 + \alpha a = -\beta, \tag{14}$
whence we have, since $\text{char}(K) \ne 2$,
$\left (a + \dfrac{\alpha}{2} \right )^2 = a^2 + \alpha a + \dfrac{\alpha}{4} = \dfrac{\alpha}{4} - \beta \in K; \tag{15}$
we thus see that
$b = a + \dfrac{\alpha}{2} \in L \setminus K \tag{16}$
with
$b^2 \in K. \tag{17}$
It is now manifestly evident that
$L = K(a) = K \left (a +  \dfrac{\alpha}{2} \right ) = K(b) \tag{18}$
is a simple extension of $K$ (since it is generated by adjoining either of the single elements $a$ or $b$ to $K$), and by virtue of (17), is what our OP Christian Singer refers to a a "2-radical extension", since we may write
$b = \sqrt  { \dfrac{\alpha}{4} - \beta  }. \tag{19}$
