If extension $E/F$ has degree $2$, how can we show that $E=F(\alpha)$ where $\alpha$ is a root of $t^2 - b$, $b \in F$? ($F$ in characteristic $0$) Note: Assume we are in characteristic 0 (which I forgot to include in the original post, hence the comments claiming the statement to be false and providing counterexamples). I've already shown that $E=F(\alpha)$ for some $\alpha \notin F$. But showing that $\alpha$ is a root of a polynomial $t^2 - b$ with $b \in F$ is giving me trouble. Also, there's a hint to use the quadratic formula (from high school algebra) somewhere in the problem -- not sure how this helps, though.
 A: 
Theorem(1): Let be $E/F$ finite extension and $|Aut(E,F)|=[E:F]$ then the following are true:
$1)$ $E_{Aut(E,F)}=F$
$2)$ $E/F$ separable extension
$3)$ $E/F$ normal extension
$4)$ $\alpha , \sigma (\alpha) $ roots of same polynomial for $\sigma \in Aut(E,F)$
$5)$ $E$ is spilitting extension of $F$

Theorem(2): if $E$ is an extension field of $F$ with $[E:F]=2$ then $E$ is a normal extension of $F$
We have $[E:F]=2$, then from theorem (2) $E$ is a normal exension of $F$ then $E$ is a spilitting field of $f(x)=x^2+ax+b \in F[x]$ where $\{\alpha,-a-\alpha \}$ are roots of $f(x)$ in $E$.
if $f(x)$ isnot separable polynomial then $\alpha=-a-\alpha\Rightarrow a=-2\alpha \in E$ 'contradition'. Therefore $E$ is a normal and separable extension of $F$, then $Aut(E,F)=[E:F]=2\Rightarrow Aut(E,F)=\{I, \sigma \}; I(\alpha)=\alpha , \sigma(\alpha)=-\alpha $, then from theorem (1)- number (3)-  we have $\{\alpha, -\alpha \}$ roots the same polynomial, i.e. $f(x)=(x-\alpha)(x+\alpha)=x^2-\alpha^2=x^2-b; b=\alpha^2 \in F$ and $E=F[\alpha]$
