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

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lightjazz | Student, Undergraduate | eNoter

Posted June 10, 2012 at 3:47 PM via web

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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 10, 2012 at 4:14 PM (Answer #1)

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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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