What are some sufficient conditions for $\mathbb{Q} (\alpha , \beta )=\mathbb{Q} (\alpha +c\beta ) \forall c\in \mathbb{Q}^*$ with $\alpha , \beta $ algebraic over $\mathbb{Q}$?
We know that, for example,
$[\mathbb{Q} (\alpha ):\mathbb{Q}]=[\mathbb{Q} (\beta ):\mathbb{Q}]=2 \wedge\mathbb{Q}(\alpha)\neq\mathbb{Q}(\beta)$ is a sufficient condition.
We can prove it by adapting the proof of the existence of the primitive element.
Proof. Let $\sigma _1,...,\sigma _4$ the be the elements of $Gal(\mathbb{Q}(\alpha, \beta)/\mathbb{Q})\cong C_2\times C_2 $. Then the polynomial $$p(x)=\Pi_{i<j}(\sigma_i(\alpha)+x\sigma _j(\beta)-\sigma_j(\alpha)-x\sigma_j(\beta))$$ is not $0$.
If we prove that $p(c)\neq0 \forall c\in \mathbb{Q}^* $ we are done, since it means that $Gal(\mathbb{Q}(\alpha, \beta)/\mathbb{Q})\subseteq Gal(\mathbb{Q}(\alpha+c \beta)/\mathbb{Q})$ which implies $[\mathbb{Q} (\alpha+c\beta ):\mathbb{Q}]\geq4$.
But, expanding $p(x)$ we get that $$p(c)=0 \iff (\alpha+c\beta=0 \lor\alpha-c\beta=0)$$ which is impossible, since $\mathbb{Q}(\alpha)\neq\mathbb{Q}(\beta)$.$\square$
Moreover I think that
$Gal(\mathbb{Q}(\alpha,\beta)/\mathbb{Q})\cong Gal(\mathbb{Q}(\alpha)/\mathbb{Q})\times Gal(\mathbb{Q}(\beta)/\mathbb{Q})$ is a sufficient condition.
Proof. by generalizing the above reasoning we get that $$p(c)=0\iff \exists i,j(i\neq j) s.t. (\sigma_j(\beta)-\sigma_i(\beta))c=\sigma_i(\alpha)-\sigma_j(\alpha)$$ which implies that $\sigma_j(\beta)-\sigma_i(\beta)$ and $\sigma_i(\alpha)-\sigma_j(\alpha)$ are both rational, and one of them is not $0$. Say $\sigma_j(\beta)-\sigma_i(\beta)=q\in\mathbb{Q}^*$ then $\sigma_i^{-1}\sigma_j(\beta)=\beta+q$.
Then $$(\sigma_i^{-1}\sigma_j)^n, n\in\mathbb{N}$$ are all distinct elements of $Gal(\mathbb{Q}(\beta)/\mathbb{Q})$ which is finite, absurd.$\square$