Easy question about the root (of a real number)

  • #1
[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:
[tex]\sqrt{3 - 2\sqrt{2}} = \sqrt{2} - 1[/tex]

?
 
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Answers and Replies

  • #2
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61
Hi, i'm a bit embarresed to ask this but does anybody know how to get this:
[tex]\sqrt{3 - 2\sqrt{2}} = \sqrt{2} - 1[/tex]

?
[tex]\sqrt{3 - 2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}[/tex]

what is the relationship between [tex]\left(2-2\sqrt{2}+1\right)[/tex] and [tex]\left(\sqrt{2} - 1\right)[/tex]??
 
  • #3
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0
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

[tex]\sqrt{3-2\sqrt{2}}=\sqrt{2}-1[/tex]

and solve it as is (this will ultimately lead you to what S_David is pointing out).
 
  • #4
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

[tex]\sqrt{3-2\sqrt{2}}=\sqrt{2}-1[/tex]

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!
 
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  • #5
299
0
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:
 

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