$x^4-3x^2+4$ irreducible over over $\mathbb{Q}$

Question

I need to prove irreducibility of $x^4-3x^2+4$ over $\mathbb{Q}$. It can't have any linear factor since it doesn't have any root in $\mathbb{Q}$ because any $\alpha \in \mathbb{Q}$ is a root only if $\alpha|4$ i.e. only possibility for $\alpha=\pm1,\pm2,\pm4$ but none of them is a root. But how shall I prove that it can't have quadratic factor over $\mathbb{Q}$?

Answer 1

Hint: $x^2-3x^2+4=(x^2+2)^2-7x^2=(x^2-\sqrt{7}\,x+2)(x^2+\sqrt{7}\,x+2)$ is the decomposition into irreducible factors over $\mathbb{R}$.