Quadratic Irrational is Root of Quadratic Equation

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

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

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:

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


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 Rational Numbers form Field.

Hence the result.

$\blacksquare$