- #1

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**[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]

?

Last edited:

- Thread starter joris_pixie
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- #1

- 25

- 0

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]

?

Last edited:

- #2

- 1,367

- 61

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

?

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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[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

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OK! Got it !

Sorry for wasting your time and thank you !! :)

And if you 'see it' it's easy, but if you don't ...

But thanks a lot you guys, got it now!

Sorry for wasting your time and thank you !! :)

It's true that it is one you have to 'see' !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).

And if you 'see it' it's easy, but if you don't ...

But thanks a lot you guys, got it now!

Last edited:

- #5

- 299

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

Sorry for wasting your time and thank you !! :)

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