Surds

The best solution for finding the exact values of $x$, $y$ and $a$ satisfying the following system of equations came from Bithian.

The first equation gives: From the second and third equations: Now we use a neat trick that is well worth knowing, for getting rid of surds in the denomenator of a fraction. This trick uses the difference of two squares. In this case you use $(\sqrt{2} + 1)(\sqrt{2} - 1) = 2 - 1 = 1$.

Let's treat the two possible values for $a$ separately:

1. For $a = \sqrt{2}$ we have to multiply the top and bottom by $\sqrt 2 - 1$ to give $$y=4-2\sqrt{2}$$ Then use $x = ay = \sqrt 2 (4-2\sqrt 2) = 4\sqrt{2} - 4$ to get $$\begin{array}{rl}a& =\sqrt{2}\\ x& =4\sqrt{2}-4\\ y& =4-2\sqrt{2}\end{array}$$
2. For $a = -\sqrt{2}$ we have to multiply the top and bottom by $\sqrt 2 + 1$ to give $$y=4+2\sqrt{2}$$ Then use $x = ay = -\sqrt 2 (4+2\sqrt 2) = -4\sqrt{2} - 4$ to get $$\begin{array}{rl}a& =-\sqrt{2}\\ x& =-4\sqrt{2}-4\\ y& =4+2\sqrt{2}\end{array}$$

[NOTE: When $(a\sqrt{2} + b)$ occurs as a factor in the denominator (where $a$ and $b$ are whole numbers) you multiply top and bottom of the fraction by exactly the same thing, by $(a\sqrt{2} - b)$. In effect you just multiply the whole fraction by one and it will always give the whole number $2a^2 - b^2$ in the denominator.]