Quadratic Irrational is Root of Quadratic Equation

Theorem

Let $x$ be a quadratic irrational.

Then $x$ is a solution to a Quadratic Equation with rational coefficients.

Proof

Let $x = r + s \sqrt n$.

From Quadratic Equation, the solutions of $a x^2 + b x + c$ are:

$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c}} {2a}$

given the appropriate condition on the discriminant.

So if $x = r + s \sqrt n$ is a solution, then so is $x = r - s \sqrt n$.

Hence we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({x - r + s \sqrt n}\right) \left({x - r - s \sqrt n}\right)\)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({x-r}\right)^2 - \left({s \sqrt n}\right)^2\)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Difference of Two Squares
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x^2 - 2 r x + r^2 - s^2 n\)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

As $r$ and $s$ are rational and $n$ is an integer, it follows that $-2 r$ and $r^2 - s^2 n$ are also rational from the fact that rational numbers form a field.

Hence the result.

$\blacksquare$