1. ## Square root manipulation

Hello. From my calculator, I know that this is a true equality:

$\sqrt{5+2\sqrt{6}} = \sqrt{3} + \sqrt{2}$

However, I simply cannot come up with the method used to "simplify" the first expression to arrive at the second expression (or vice versa). Could someone demonstrate or explain what I am missing?

2. Originally Posted by drumist
Hello. From my calculator, I know that this is a true equality:

$\sqrt{5+2\sqrt{6}} = \sqrt{3} + \sqrt{2}$

However, I simply cannot come up with the method used to "simplify" the first expression to arrive at the second expression (or vice versa). Could someone demonstrate or explain what I am missing?
Ok, let's start from

$C = \sqrt{3} + \sqrt{2}$

Square both sides

$C^2 = (\sqrt{3} + \sqrt{2})^2 = (\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}$

Now take the positive root to find $C = \sqrt{5 + 2\sqrt{6}}$

3. Well, to be clear, seeing the equality, I can definitely prove that it is true, just like you showed.

I see that you can generalize it into an identity like this

$\sqrt{a+b\pm2\sqrt{ab}}=\sqrt{a}\pm\sqrt{b}$

I guess my point is: are there any other similar identities like this that would be useful in simplifying stacked square root expressions? I don't remember ever being taught this particular identity, although I'm not sure if it really occurs very often.

4. $5+2\sqrt{6}=3+2+2\cdot\sqrt{2}\cdot\sqrt{3}=(\sqrt {2}+\sqrt{3})^2$

you've got to recognize the terms of the identity/ies $(a\pm b)^2=a^2+b^2\pm 2ab$