View Full Version : Easy question about the root (of a real number)
joris_pixie
Oct23-09, 10:18 AM
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
?
S_David
Oct23-09, 11:04 AM
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
Oct23-09, 11:11 AM
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
Oct23-09, 11:14 AM
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
Oct23-09, 11:16 AM
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! :biggrin:
vBulletin® v3.7.6, Copyright ©2000-2009, Jelsoft Enterprises Ltd.