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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.

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    The basis $ 1 , \alpha, \alpha^2..\alpha^n$ satisfies a polynomial of degree n . As the extension is 2, the degree is 2. So now $ \alpha$ is a root of a quadratic polynomial. Now you can change the variable to omit the linear term and get the required equation.. – 2017-02-12
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    It should give you trouble! Because it is false in general. In characteristic two we have counterexamples. If the characteristic is not two, then you can "complete the square" in the minimal polynomial of $\alpha$ and replace $\alpha$ with another generator whose minimal polynomial has this form. – 2017-02-12
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    @ChirantanChowdhury: That is the right idea! Unfortunately if $1+1=0$ in $F$ you cannot divide by two, and for the change of variable to work the way you described, it is necessary to be able to divide by two. That is the only obstruction to your plan, so assuming $2\neq0$ in $F$ it works (and the claim is false in general). – 2017-02-12
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    A counterexample: $F=\Bbb{F}_2=\{0,1\}$, $E=\Bbb{F}_4=\{0,1,t,t+1\}$, where $t^2=t+1$ and $(t+1)^2=t$. The elements $t$ and $t+1$ are the only elements of $E$ not in $F$, but neither has a square in $F$. – 2017-02-12
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    So I know that $\{ 1, \alpha \}$ is a basis of $E$ over $F$ and I know that the irreducible polynomial of $\alpha$ is of the form $t^2 + b_1t+b_2$. Now, it seems I need to show that $b_1=0$ to complete the proof, right? Otherwise the last term wouldn't be in $F$. – 2017-02-12
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    Also, thanks for catching the error in my post. We are supposed to be in characteristic $1$. – 2017-02-12
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    The characteristic of a field is **always** zero or else a *prime number*. There is no "characteristic" $\;1\;$ , so what did you really mean? – 2017-02-12
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    Sorry! I meant characteristic $0$. (*face palm*) – 2017-02-12

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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 $E=F[\alpha]$