# How should prove identity `sqrt(3+2sqrt(2))-sqrt(3-2sqrt(2))=2` without raise to 2?

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### 1 Answer

You should notice that `3 + 2sqrt2 ` is expansion of squared binomial `(1+sqrt2)^2` such that:

`(1+sqrt2)^2 = 1 + 2sqrt2 + 2`

`(1+sqrt2)^2 = 3 + 2sqrt2`

Hence, `sqrt(3 + 2sqrt2) = sqrt((1+sqrt2)^2 ) = |(1+sqrt2)|`

`|(1+sqrt2)| =(1+sqrt2)`

Notice that 3 - 2sqrt2 is also the expansion of the squared binomial (1-sqrt2)^2, hence you may evaluate `sqrt(3- 2sqrt2)` such that:

`sqrt(3 - 2sqrt2)= sqrt((1-sqrt2)^2 ) = |(1-sqrt2)|`

`|(1-sqrt2)| = sqrt2 - 1`

You should substitute `(1+sqrt2)` for `sqrt(3 + 2sqrt2)` and `sqrt2 - 1` for `sqrt(3 - 2sqrt2)` such that:

`sqrt(3 + 2sqrt2)- sqrt(3- 2sqrt2) = 1 + sqrt2 - (sqrt2 - 1)`

`sqrt(3 + 2sqrt2)- sqrt(3 - 2sqrt2) = 1 + sqrt2 - sqrt2+ 1`

Reducing like terms yields:

`sqrt(3 + 2sqrt2)- sqrt(3 - 2sqrt2) = 2`

**Hence, evaluating the expression `sqrt(3 + 2sqrt2)- sqrt(3 - 2sqrt2)` using the properties of absolute value yields that `sqrt(3 + 2sqrt2)- sqrt(3 - 2sqrt2) = 2.` **

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