Argthtjtr

I was wondering the general method to solve

What is the value of $sqrt{a-bsqrt{c}}?$

The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$, but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$ which equals $sqrt{3}+sqrt{2}$. What is the general method to simplify these problems?(i.e. $sqrt{a-bsqrt{c}}=?$)

One way of approaching this problem is by viewing it as a zero of an equation. Let me explain. Let's say you want to compute $sqrt{x_0}$ where $x_0$ is a zero of some quadratic polynomial of the form $x^2-bx+1$. Now, one way to go is to note that if you have a zero of $x^2+ax+1$, then it will still be a zero if you multiply it with $x^2-ax+1$ which equals
$$x^4 + (2-a^2) x^2 + 1$$
Now the idea is to work backwards. So, in particular, if you can find you can find an $a$ such that $b=a^2-2$, then you can conclude that the square root of you polynomial is equal to one of the zeros of the polynomials $x^2-ax+1$ or $x^2+ax+1$. It is usually not too hard to find out which. If you found out which, you can rewrite your square root accordingly to the desired form :)

To conclude, one of the tricks is to find the right form of your polynomials such that you end up with something useful. This method will however require some puzzling.

Edit applying this method to your example, you will find that the polynomial you need (thus the one for which you want to calculate the square root of a zero) is $x^2-10x+1$. Then according to the above method (which you derive on the go), your $a=sqrt{12}$ and then you just need to solve $x^2-ax+1=0$ which is the only possibility since for the other one, filling in a positive number will yield a positive number. Solving this equation by completing the square is not too difficult. It turns out that the zeros lie at around 0.5 and 3. Hence, it is not difficult to note you need the larger zero which turns out to be exactly gicen by $sqrt{2}+sqrt{3}$. Does that make sense?

There are the following identities.
$$sqrt{a+sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}+sqrt{frac{a-sqrt{a^2-b}}{2}}$$ and
$$sqrt{a-sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}-sqrt{frac{a-sqrt{a^2-b}}{2}},$$
where all numbers under radicals they are non-negatives.

For example:
$$sqrt{5+2sqrt6}=sqrt{5+sqrt{24}}=sqrt{frac{5+sqrt{5^2-24}}{2}}+sqrt{frac{5-sqrt{5^2-24}}{2}}=sqrt3+sqrt2.$$
This is interesting, when $a$ and $b$ are rationals and $a^2-b$ is a square of a rational number.

The first identity is true because
$$left(sqrt{frac{a+sqrt{a^2-b}}{2}}+sqrt{frac{a-sqrt{a^2-b}}{2}}right)^2=$$
$$=frac{a+sqrt{a^2-b}}{2}+frac{a-sqrt{a^2-b}}{2}+2sqrt{frac{a+sqrt{a^2-b}}{2}}cdotsqrt{frac{a-sqrt{a^2-b}}{2}}=a+sqrt{b}.$$