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