Let $ax^2+bx+c$ be in $\mathbb{Q}[x]$, where $a\ne 0$. I want to find a primitive element for a splitting field of $f$ over $\mathbb{Q}$.
I'm not sure if I get the idea completely. To find the splitting field, I need to do the following:
(1) Find the zeroes of $f$, which there are two, $\alpha_1, \alpha_2$.
(2) Adjoin the two roots to make the extension $\mathbb{Q}(\alpha_1, \alpha_2)$ and show that $\mathbb{Q}(\alpha_1, \alpha_2)=\mathbb{Q}(\alpha_1+ \alpha_2)$, correct?
(3) Thus $\alpha_1+\alpha_2$ is a primitive element of the field?
Would appreciate if someone could please clarify this for me.
You can factor your polynomial as
$$ax^2 + bx + c = a(x-\alpha_1)(x-\alpha_2) = a(x^2 - (\alpha_1 + \alpha_2)x + \alpha_1 \alpha_2)$$
From the last expression, you can see that the coefficient of $x$ is $b = -a(\alpha_1 + \alpha_2)$, i.e., the sum of the roots is $-b/a \in \mathbb{Q}$.
And so when you adjoin just one of the roots, say, $\alpha_1$, you are able to generate the other one as well: $-b/a = \alpha_1 + \alpha_2 \implies -b/a - \alpha_1 = \alpha_2$. Since just one of the roots suffices to get you the other, you may conclude that $\mathbb{Q}(\alpha_1) = \mathbb{Q}(\alpha_1, \alpha_2)$.
Note that the above shows $\mathbb{Q}(\alpha_1 + \alpha_2) = \mathbb{Q}$, so this will not be the same as the splitting field $\mathbb{Q}(\alpha_1)$ if $\alpha_1 \not\in \mathbb{Q}$. You can adjoin either root to get the other, but adjoining their sum gets you nothing new!