# 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)?

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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, $y^2 - 2y - 7$.)

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Your number will satisfy $x^4-18x^2+49=0$, and that will be minimal unless your number is already in ${\bf Q}(\sqrt2)$, which it is, since $(1-2\sqrt2)^2=9-4\sqrt2$.

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$\rm x^4-18\,x^2+49\:$ will be the minimal polynomial, unless $\rm\:\sqrt{9-4\sqrt{2}}\:$ denests to $\rm\:a + b\sqrt{2}.\:$
This can be tested by a radical denesting formula that I discovered as a teenager.

Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace}$

Recall $\rm\: w = a + b\sqrt{n}\:$ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2$

and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$

Here $\:9-4\sqrt{2}\:$ has norm $= 49.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 7\$ yields $\ 2-4\sqrt{2}\:$

and this has $\rm\ \sqrt{trace}\: =\: 2,\ \ so,\ \ \ \color{brown}{dividing\ it\ out}\$ of this yields the sqrt: $\:1-2\sqrt{2}.$

Checking we have $\ \smash[t]{\displaystyle \left(1-2\sqrt{2}\right)^2 =\ 1+8 -4\sqrt{-3}\ =\ 9-4 \sqrt{2}}.$

$-7+2z+z^2$