Charles Pinter A Book of Abstract Algebra chapter 31 problem C5:

Prove:

If $$p(x) = x^4 + ax^2 + b$$ is irreducible in F[x], then F[x]/$$$$ is the root field of p(x) over F.

Discussion:

As the hint in the back of the book states, F[x]/$$$$ = F(c) where c is a root of p(x). The hint then goes on to rewrite p(x) as $$(x^2 +a/2)^2 – (a^2/4 – b)$$ (actually it has $$(x^2 - a/2)^2 – (a^2/4 – b)$$, but that’s a typo). The hint then says to let $$d_1$$ be a root of $$x^2 – [(a^2/4) – b]$$ and $$d_2$$ be a root of $$x^2 – (a/2 +d_1)$$. It says then that F($$d_1$$, $$d_2$$) = F(c) and that F($$d_1$$, $$d_2$$) contains all the roots of p(x).

$$d_1 = {±\sqrt{a^{2}/4-b}}$$

$$d_2 = {±\sqrt{a/2±\sqrt{a^{2}/4-b}}}$$

Note that $$d_2$$ isn’t necessarily a member of F[x]. I had considered that p(x) may have all its roots in F(c) because it has at least one root in the root field of $$x^2 – (a/2 +d_1)$$, but because $$x^2 – (a/2 +d_1)$$ isn't in F[x], that isn't applicable.

The roots of p(x) are $${±\sqrt{a/2±\sqrt{a^{2}/4-b}}}$$, the same as $$d_2$$.

So if I select $${\sqrt{a/2+\sqrt{a^{2}/4-b}}}$$ as c, then it’s clear that F($$d_1$$, $$d_2$$) = F(c). Since $$d_1 = d_2^2 – a/2$$, the $$d_1$$ is extraneous, and F($$d_1$$, $$d_2$$) = F($$d_2$$) = F(c).

But for F(c) to be the root field, it would also have to contain $${\sqrt{a/2-\sqrt{a^{2}/4-b}}}$$.

Does there exist some set of $$k_0$$ to $$k_n$$ member of F such that $$k_0 + k_1c + k_2c^2 + … + k_nc^n = {\sqrt{a/2-\sqrt{a^{2}/4-b}}}$$?

The only members of F specified are 0, 1, a and b. The roots of p(x) are specifically excluded. There could be other members, but nothing I can assume.

Letting c = $${\sqrt{a/2+\sqrt{a^{2}/4-b}}}$$ and d = $${\sqrt{a/2-\sqrt{a^{2}/4-b}}}$$, then F(c, d) would contain all the roots of p(x). In addition, F[x]/$$$$ is isomorphic to F(c) is isomorphic to F(d), but that’s not the same as F(c, d).

What am I missing here?

I don't have the book with me; I assume that by root field the author means what is more commonly called a splitting field, i.e. the field extension obtained by adjoining all the roots of $$p(x)$$.
With this interpretation, the statement is false; $$F(c)$$ need not be the splitting field. As indicated in this answer, the statement will be false precisely when both $$b$$ and $$b(a^2-4b)$$ are not squares in $$F$$ (e.g. take $$F = \mathbb{Q}$$, $$a=1, b=2$$).
To see that this is the case, note that the field $$F(c, d)$$ contains the element $$\sqrt{b} = cd$$ and the element $$\sqrt{a^2 - 4b} = c^2 - d^2$$. Hence it also contains $$\sqrt{b(a^2 - 4b)} = \sqrt{b}\sqrt{a^2 - 4b}$$, whereby $$F' = F(\sqrt{b(a^2-4b)})$$ is a subfield of $$F(c, d)$$. But the polynomial $$p(x)$$ is still irreducible over $$F'$$, as $$F'$$ does not contain the square root of the discriminant $$a^2 - 4b$$. It follows that $$F(c, d)$$ has degree at least $$4$$ over $$F'$$, whereby $$F(c, d)$$ has degree at least $$8$$ over $$F$$. Thus we cannot have $$F(c, d) = F(c)$$.