= Solution
Because $L/K$ is a finite <Galois extension>,
$$
[L:K]=|\operatorname{Gal}(L/K)|=|A_{10}|=\frac{10!}{2}=1\,814\,400.
$$
Suppose the irreducible quadratic $f$ became reducible over $L$. It would then have a root $\alpha\in L$, and $K(\alpha)/K$ would be a quadratic intermediate extension. By the <Fundamental theorem of Galois theory>, $\operatorname{Gal}(L/K(\alpha))$ would be a subgroup of index two in $A_{10}$ and hence a <normal subgroup> by the <index-two subgroup is normal> result. But the <Simplicity of alternating groups> says that $A_{10}$ is simple, while this subgroup would be proper and nontrivial. This contradiction proves that $f$ remains irreducible in $L[x]$.
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