Show $r-s\sqrt{2}$ is a root if $r+s\sqrt{2}$ is

Question

Given that $x = r+s\sqrt{2}$ is a solution to $x^2+ax+b=0$ for some $a,b,r,s\in \mathbb{Q}$ ($s\ne 0$), show that $x = r - s\sqrt{2}$ is also a solution.

If we let $f(x) = x^2+ax+b$, then: $$f(r+s\sqrt{2})=r^2+(2s\sqrt{2}+a)r+(2s^2+as\sqrt{2}+b)=0 $$ We know the above must be true.

Then, substituting $x = r-s\sqrt{2}$ (or setting $s = -s$ in the above equation) gives: $$f(r-s\sqrt{2})=r^2+(a-2s\sqrt{2})r+(2s^2-as\sqrt{2}+b) $$ $$ = f(r+s\sqrt{2})-2s\sqrt{2}(2r+a)$$ $$ =-2s\sqrt{2}(2r+a)$$ How can I show this is equal to $0$? I feel like I'm missing something very obvious, or perhaps have made a mistake in my substitutions somewhere.

Additionally, the result seems similar to the theorem in complex numbers (which I've been taught without proof) that $f(z) = 0\iff f(\bar{z}) = 0$. Here we're dealing with $\sqrt{2}\notin \mathbb{Q}$ rather than $i \notin \mathbb{R}$, so are there any simple proofs of the complex conjugate theorem that could be repurposed here?

Answer 1

Putting $x = r+s\sqrt{n}$ where $\sqrt{n}$ is irrational (using $n$ for $2$ in the hope of generalizing) into $x^2+ax+b = 0$ gives

$\begin{array}\\ 0 &=(r+s\sqrt{n})^2+a(r+s\sqrt{n})+b\\ &=r^2+2rs\sqrt{n}+s^2n+ar+as\sqrt{n}+b\\ &=r^2+s^2n+ar+b+s(2r+a)\sqrt{n}\\ \end{array} $

Since $\sqrt{n}$ is irrational, we must have both $r^2+s^2n+ar+b = 0$ and $s(2r+a) = 0$.

Replicating the calculation above but with $-s$ for $s$,

$(r-s\sqrt{n})^2+a(r-s\sqrt{n})+b =r^2+s^2n+ar+b-s(2r+a)\sqrt{n} =0 $ so $r-s\sqrt{n}$ is also a root.

Answer 2

Sum of roots $\in \mathbb Q$, so second root must be $k - s\sqrt 2$ ; $k \in \mathbb Q$
Also product of roots $\in \mathbb Q$
So, $rk + (k-r)s\sqrt 2 - 2s^2 \in \mathbb Q$ which is true iff $k=r$
Therefore, second root is $r-s\sqrt 2$.

Answer 3

Here is a proof for a general polynomial.

Suppose $r+s\sqrt{n}$ is a root of $\sum_{k=0}^d a_kx^k =0 $ where $r, s$, and the $a_k$ are rational and $\sqrt{n}$ is irrational.

Then

$\begin{array}\\ 0 &=\sum_{k=0}^d a_k(r+s\sqrt{n})^k\\ &=\sum_{k=0}^d a_k\sum_{j=0}^k\binom{k}{j}r^{k-j}(s\sqrt{n})^{j}\\ &=\sum_{k=0}^d a_k\left(\sum_{j=0,even}^k\binom{k}{j}r^{k-j}(s\sqrt{n})^{j}+\sum_{j=0,odd}^k\binom{k}{j}r^{k-j}(s\sqrt{n})^{j}\right)\\ &=\sum_{k=0}^d a_k\left(\sum_{j=0,even}^k\binom{k}{j}r^{k-j}s^jn^{j/2}+\sum_{j=0,odd}^k\binom{k}{j}r^{k-j}s^jn^{j/2}\right)\\ &=\sum_{k=0}^d a_k\left(\sum_{j=0,even}^k\binom{k}{j}r^{k-j}s^jn^{j/2} +sn^{1/2}\sum_{j=0,odd}^k\binom{k}{j}r^{k-j}s^{j-1}n^{(j-1)/2}\right)\\ &=\sum_{k=0}^d a_k\sum_{j=0,even}^k\binom{k}{j}r^{k-j}s^jn^{j/2} +sn^{1/2}\sum_{k=0}^d a_k\sum_{j=0,odd}^k\binom{k}{j}r^{k-j}s^{j-1}n^{(j-1)/2}\\ \end{array} $

Since $n^{1/2}$ is irrational, both double sums must be zero.

Both double sums have integral exponents for $n$ and even exponents for $s$. Therefore their value is unchanged when $s$ is replaced by $-s$.

Therefore, if $r+s\sqrt{n}$ is a root, so is $r-s\sqrt{n}$.