I am trying to show that the splitting field for an irreducible quadratic polynomial over $\ \mathbb{Q}$ is of the form $\ \mathbb{Q}(\sqrt{D})$ where $D$ is a square-free integer not equal to $0$ or $1$.
If we let $f(x) = ax^2+bx+c \in \mathbb{Q}[x]$ be irreducible in $\mathbb{Q}$ then the roots are
$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Then using field operations we can say that the splitting field of $f$ is $\mathbb{Q}(\sqrt{b^2-4ac})$. Since $f$ is irreducible over $\ \mathbb{Q} \implies$ no roots in $\mathbb{Q} \implies b^2-4ac \neq$ a square (including $0$ and $1$).
So I've shown that at least $D$ cannot be a square - but I am stuck on how to prove the stronger condition that $D$ must be square-free.
Any help would be appreciated!