Multiplicative inverse of a quadratic algebraic number $\,a+b\sqrt 2$ Find the multiplicative inverse of $1+ 3\sqrt{2}$ in the ring $\mathbb{Q}(\sqrt{2})$ and use it to solve the equation $(1+3\sqrt{2})x=1-5\sqrt{2}$.
I think that the inverse is the conjugate, so it would be $1-3\sqrt{2}$, but then I don't know where to use in the equation that needs to be solved.

Instead of finding inverse, you can directly find(less computation) $x$
Let $x=a+b\sqrt 2$
Then, $(1+3\sqrt 2)(a+b\sqrt 2)=1-5\sqrt 2\implies (a+6b)+(3a+b)\sqrt 2=1-5\sqrt 2$
$\implies a+6b=1$ and $3a+b=-5$.
Solving these equations, we get
$a=-\frac{31}{17}$ and $b=\frac{8}{17}\implies x=-\frac{31}{17}+\frac{8}{17}\sqrt 2$
