Let $\mathbb{Q}\leq K$ be a field extension of degree 2. Show that there exists $\psi\in K$ such that $K = \mathbb{Q}[\psi]$ and $\psi^{2}$ is a square free integer.
Since $\mathbb{Q}\leq K$ is a finite field extension, then we know that $\mathbb{Q}\leq K$ is indeed algebraic.
Now, I know that $K$ is generated by $\{1,\psi\}$ over $\mathbb{Q}$ for $1,\psi\in K$.
Since the extension is finite, we also have that $\psi$ is algebraic over $\mathbb{Q}$, then we have a minimum polynomial $f\in\mathbb{Q}[]$ such that $f(\psi)=0\in K$.
Then, $\deg(f)=2$, but now I'm unsure how to proceed further.
Multiplying by denominators, if necessary, we get $f \in \Bbb Z[X]: f(X)=aX^2+bX+c$. Then the quadratic formula tells us the roots of $f$ are:
$$\frac{-b \pm \sqrt{b^2-4ac}}{2a}.$$
Thus, $K= \Bbb Q\left [\sqrt{b^2-4ac} \right ]$. Moreover, $b^2-4ac$ cannot itself be a square, or $\Bbb Q = K$, which we know is not the case.
Therefore, $b^2-4ac=m^2n$ where $n$ is square-free, and $K=\Bbb Q [\sqrt n ]$.