Minimal polynomial of the root of algebraic number I have obtained the minimal polynomial of $9-4\sqrt{2}$ over $\mathbb{Q}$ by algebraic operations:
$$ (x-9)^2-32 = x^2-18x+49.$$
I wonder how to calculate the minimal polynomial of $\sqrt{9-4\sqrt{2}}$ with the help of this sub-result? Or is there a smarter way to do this (not necessarily algorithmic)?

Since $x = 9 - 4\sqrt{2}$ satisfies $x^2 - 18x + 49 = 0$, your number $y = \sqrt{x} = \sqrt{9 - 4\sqrt{2}}$ satisfies $y^4 - 18y^2 + 49 = 0$. This could be your minimal polynomial, but the polynomial factorizes as $$y^4 - 18y^2 + 49 = (y^2 + 2y - 7)(y^2 - 2y - 7).$$ Since the product is zero if and only if at least one of them is zero, we get that either $y^2 + 2y - 7 = 0$ or $y^2 - 2y - 7 = 0$. Since the minimal polynomial must have degree at least $2$, one of those must be your minimal polynomial. (In this case it is the latter, $y^2 - 2y - 7$.)
