Roots of Quadratic with Rational Coefficients of form r plus s Root 2

Theorem

$(1): \quad a^2 x + b x + c = 0$

where $a, b, c$ are rational.

Let $\alpha = r + s \sqrt 2$ be one of the roots of $(1)$.

Then $\beta = r - s \sqrt 2$ is the other root of $(1)$.

We have that:

$\ds a \paren {r + s \sqrt 2}^2 + b \paren {r + s \sqrt 2} + c$

$=$

$\ds 0$

$\ds \leadsto \ \ $

$\ds \paren {a r^2 + 2 a s + br + c} + \paren {2 a + b} s \sqrt 2$

$=$

$\ds 0$

Because $a$, $b$, $c$, $r$ and $s$ are rational, it must be that $\paren {2 a + b} s = 0$.

Hence:

$\ds a \paren {r - s \sqrt 2}^2 + b \paren {r - s \sqrt 2} + c$

$=$

$\ds \paren {a r^2 + 2 a s + br + c} - \paren {2 a + b} s \sqrt 2$

$\ds $

$=$

$\ds 0$

and so $\beta$ is also a root of $(1)$.

$\blacksquare$