Consider the irreducible cubic equation $x^3 - x - 1 = 0$ and suppose we one of the roots $x$. The other two are $a,b$ such that $x + a + b = 0$ and $xab = 1$. Then $a$ and $b$ satisfy a quadratic
$$ z^2 + xz + \frac{1}{x} = z^2 + xz + (x^2 - 1) =0 $$
which is reducible in $\mathbb{Q}(x)$. Using the quadratic formula we can write down the roots explicitly in terms of $x$:
$$ a,b = \frac{-x \pm \sqrt{x^2 - 4(x^2 - 1)}}{2}$$
Since this polynomial has leading coefficient $1$ (is monic) the discriminant of this quadratic $d = 4 - 3x^2 $ should be a perfect square in the ring of integers $\mathbb{Z}[x]$.
What is the square root? $\sqrt{d}$
In Wikipedia, the solution to $x^3 - x - 1$ is called the plastic number. In fact, it has explicit formula:
$$ x = \frac{\sqrt[3]{108 + 12\sqrt{69}}+ \sqrt[3]{108 - 12\sqrt{69}}}{6}$$
In fact, I have written a pisot number, but this question could be redone for any integer cubic in $\mathbb{Z}[x]$.
EDIT The response below incidate I have not written a Galois extension.
The question could make more sense if I ask for $\sqrt{4 - 3x^2}$ in the $\mathbb{Q}(x, \sqrt{-23})$ ring of integers, which I believe is $\mathbb{Z}[x, \frac{1 + \sqrt{-23}}{2}]$.
$((x-a)(x-b)(a-b))^2 = -23$ which is not a square in $\Bbb Q$.
And so $\sqrt{-23} \in \Bbb Q(x,a)$.
This extension must have even degree over $\Bbb Q$. If $\Bbb Q(x,a)$ were equal to $\Bbb Q(x)$ it would have degree $3$, which is impossible.
Hence those fields are different, and so $a \notin \Bbb Q(x)$. $d$ cannot be a square in $\Bbb Q(x)$.