Easy question about the root (of a real number)

[joris_pixie]

[SOLVED] Easy question about the root (of a real number)

Hi, I'm a bit embarresed to ask this but does anybody know how to get this:
\sqrt{3 - 2\sqrt{2}} = \sqrt{2} - 1

?

 

[EngWiPy]

Hi, I'm a bit embarresed to ask this but does anybody know how to get this:
\sqrt{3 - 2\sqrt{2}} = \sqrt{2} - 1

?

\sqrt{3 - 2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}

what is the relationship between \left(2-2\sqrt{2}+1\right) and \left(\sqrt{2} - 1\right)??

 

[phyzmatix]

I must say that this one does require a bit of clever thinking. I can add that the best way to "see" the answer is to take your equation

\sqrt{3-2\sqrt{2}}=\sqrt{2}-1

and solve it as is (this will ultimately lead you to what S_David is pointing out).

 

[joris_pixie]

OK! Got it !
Sorry for wasting your time and thank you ! :)

This one does require a bit of clever thinking I must say. I can add that the best way to "see" the answer is to take your equation

\sqrt{3-2\sqrt{2}}=\sqrt{2}-1

and solve it as is (this will ultimately lead you to what S_David is pointing out).

It's true that it is one you have to 'see' !
And if you 'see it' it's easy, but if you don't ...

But thanks a lot you guys, got it now!

 

[phyzmatix]

OK! Got it !
Sorry for wasting your time and thank you ! :)

Definitely didn't waste my time. You forced me to think, which is always good!