In A Book of Abstract Algebra by Charles C. Pinter, part 5 of exercise C page 318 is the following:
If $p(x)=x^4+ax^2+b\in F[x]$, then $F[x]/\langle p(x)\rangle$ is the root field of $p(x)$ over $F$.
(The root field is the splitting field)
We should of course assume that $p(x)$ is irreducible over $F$ lest $F[x]/\langle p(x)\rangle$ wouldn't be an integral domain. Assuming this, we have $F[x]/\langle p(x)\rangle\cong F(m)$ for any root $m$ of $p(x)$ in an extension of $K$. As $\deg p(x)=4$, we have $[F(m):F]=4$ for any root $m$ of $p(x)$.
Let $\Delta=a^2-4b$, $m=\sqrt{\frac{-a-\sqrt{\Delta}}{2}}$ and $n=\sqrt{\frac{-a+\sqrt{\Delta}}{2}}$. Then $m$, $-m$, $n$ and $-n$ are the distinct roots of $p(x)$. Thus, to show that $F(m)$ is the splitting field of $p(x)$, it is enough to show that $n\in F(m)$. That's what I'm unable to show.
$F(m)$ is a $4$-dimensional vector space over $F$ and $\{1,m,m^2,m^3\}$ is a basis of it. We have $\sqrt{\Delta}=-a-2m^2\in F(m)$ and $n^2=-a-m^2\in F(m)$. If $n\in F(m)$ then $\exists !\,x,y,z,t\in F,\, n=x+ym+zm^2+tm^3$. Using the equality $p(m)=0$, we find, unless there's an error, that $$n^2=x^2+abt^2-(2ty+z^2)b+2(xy-btz)m+((a^2-b)t^2+y^2+2xz-a(tz+z^2))m^2+2(tx-atz+yz)m^3$$ leading to a pretty complicated nonlinear system: $$\begin{cases} x^2+abt^2-(2ty+z^2)b &= &-a \\ 2(xy-btz) &= &0 \\ (a^2-b)t^2+y^2+2xz-a(tz+z^2) &= &-1 \\ 2(tx-atz+yz) &= &0\end{cases} $$ (we don't even know what is the caracteristic of this field).
This suggests that there might be another method, but I couldn't find anything. Would you please help me? Thank you in advance!
This is false in general. As a counterexample I proffer the case $F=\Bbb{Q}$ and $$p(x)=x^4-2.$$
It is irreducible by Eisenstein. But its splitting field cannot be gotten by adjoining one of the zeros. The zero $\root4\of2$ is real, and obviously won't generate the non-real roots.
More precisely, the splitting field is $K=\Bbb{Q}(\root4\of2,i)$. We easily see that $[K:\Bbb{Q}]=8$. It is not hard to show that the Galois group is isomorphic to the dihedral group $D_4$ of order eight. (I'm fairly sure this has been done on our site)
OTOH, if $F$ is finite, then the claim is true. This is because any finite extension of a finite field is Galois (actually cyclic).