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.
$\blacksquare$