Denesting radicals

Problem:

Solution:

This is not a particular hard question, but I learn an interesting trick from it called the denesting of radicals, so I wanted to write this to remember.

The interesting thing to tackle is the $ \sqrt{5 - 2\sqrt{6}} $ and the $ \sqrt{8 - 2 \sqrt{15}} $ thing. They seems complicated, we would like to get rid of the nesting, so how?

Consider the general form and take a square of it and see what they look like:

$ (\sqrt{a} - \sqrt{b})^2 = a + 2\sqrt{ab} + b $

So if we wanted to make $ \sqrt{5 - 2\sqrt{6}}= \sqrt{a} - \sqrt{b} $, we need to find $ a + b = 5 $ and $ ab = 6 $, which is trivial in this case $ a = 3 $, $ b = 2 $. Obviously, we have to do big minus small to make sure it is positive.

Similarly, we have $ \sqrt{8 - 2 \sqrt{15}} = \sqrt{5} - \sqrt{3} $, the rest follows, in a very pleasing way, so I will do it here.

$ \begin{eqnarray*} \frac{x}{\sqrt{3}} + y &=& \sqrt{3} - 1 \\ x + \sqrt{3}y &=& 3 - \sqrt{3} \\ x &=& 3 - \sqrt{3} - \sqrt{3}y \end{eqnarray*} $
From the first equation, we can greatly simplify it as:

$ \begin{eqnarray*} x(\sqrt{3} - \sqrt{2}) + y(\sqrt{5} - \sqrt{3}) &=& \sqrt{3}(x + 1) + \sqrt{5}(y - 3) + 3(\sqrt{5} - \sqrt{2}) \\ \sqrt{3}x - \sqrt{2}x + \sqrt{5}y - \sqrt{3}y &=& \sqrt{3}x + \sqrt{3} + \sqrt{5}y - 3\sqrt{5} + 3\sqrt{5} - 3\sqrt{2} \\ \sqrt{3}x - \sqrt{2}x + \sqrt{5}y - \sqrt{3}y - \sqrt{3}x - \sqrt{5}y &=& \sqrt{3} - 3\sqrt{5} + 3\sqrt{5} - 3\sqrt{2} \\ \sqrt{3}x - \sqrt{3}x - \sqrt{2}x + \sqrt{5}y - \sqrt{5}y - \sqrt{3}y &=& \sqrt{3} - 3\sqrt{5} + 3\sqrt{5} - 3\sqrt{2} \\ - \sqrt{2}x - \sqrt{3}y &=& \sqrt{3} - 3\sqrt{2} \\ \sqrt{2}x + \sqrt{3}y &=& 3\sqrt{2} - \sqrt{3} \\ \end{eqnarray*} $
So at this point we simply substitute

$ \begin{eqnarray*} \sqrt{2}(3 - \sqrt{3} - \sqrt{3}y) + \sqrt{3}y &=& 3\sqrt{2} - \sqrt{3} \\ 3\sqrt{2} - \sqrt{2}\sqrt{3} - \sqrt{2}\sqrt{3}y + \sqrt{3}y &=& 3\sqrt{2} - \sqrt{3} \\ -\sqrt{2}\sqrt{3}y + \sqrt{3}y &=& 3\sqrt{2} - \sqrt{3} - 3\sqrt{2} + \sqrt{2}\sqrt{3} \\ (-\sqrt{2}\sqrt{3}+\sqrt{3})y &=& 3\sqrt{2} - \sqrt{3} - 3\sqrt{2} + \sqrt{2}\sqrt{3} \\ (-\sqrt{2}\sqrt{3}+\sqrt{3})y &=& - \sqrt{3} + \sqrt{2}\sqrt{3} \\ y &=& -1 \end{eqnarray*} $
$ \begin{eqnarray*} x &=& 3 - \sqrt{3} - \sqrt{3}y \\ x &=& 3 - \sqrt{3} + \sqrt{3} \\ x &=& 3 \end{eqnarray*} $

Does it looks very complicated, yes, it is intended to make it look like so :p