 Solution

Because $L/K$ is a finite Galois extension,

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, $\mathrm{Gal}(L/K(\alpha ))$ 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 $L[x]$.