Showing isomorphism of field extensions I need to show that if $\alpha$ has minimal polynomial $t^2-2$ over $\mathbb{Q}$ and $\beta$ has minimal polynomial $t^2-4t+2$ over $\mathbb{Q}$, then the extensions $\mathbb{Q}(\alpha):\mathbb{Q}$ and $\mathbb{Q}(\beta):\mathbb{Q}$ are isomorphic. 
I want to say that somehow $t^2-2 \equiv t^2-4t+2$ modulo something and this will be of help in the proof? I see that $t^2-4t+2-(t^2-2)=-4t+4$, so maybe I should work modulo $1-t$? Does this make any sense and am I on the right track? 
Thanks.

Working in $\mathbb{C}$ (which includes the algebraic closure of $\mathbb{Q}$), you can see that
$$\alpha=\pm\sqrt{2}$$
$$\beta=\frac{4\pm\sqrt{16-8}}{2}=2\pm\sqrt{2}$$
hence $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$.
If you want to reason purely algebraically, then you need to show that $\beta\in\mathbb{Q}(\alpha)$ (and viceversa for $\alpha$). You know that every element of $\mathbb{Q}(\alpha)$ is of the form $x\alpha+y$  for $x,y\in\mathbb{Q}$. Can you find suitable $x,y$ such that $\beta=x\alpha+y$?
Use the fact that $$0=\beta^2-4\beta+2=(x^2\alpha^2+y^2+2xy\alpha)-4(x\alpha+y)+2=(2x^2+y^2-4y+2)+\alpha(2xy-4x)$$
hence $x=1$ and $y=2$. Something similar shows that $\alpha\in\mathbb{Q}(\beta)$.
Edit: note that writing elements of $\mathbb{Q}(\beta)$ as elements of $\mathbb{Q}(\alpha)$ makes sense since we are viewing both fields as subfields of the same algebraic closure (which exists, assuming the axiom of choice).
