Proof/Show with polynomials under the radical

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miniradman
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Homework Statement


show that:

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


Homework Equations


I remember over-hearing someone talking about the modulus? I don't know how that's suppose to help me


The Attempt at a Solution


I'm still at the conceptual stage :cry:

I don't know which rules apply to polynomials under the square roots. Can someone please give me some advice please?
 
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miniradman said:

Homework Statement


show that:

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


Homework Equations


I remember over-hearing someone talking about the modulus? I don't know how that's suppose to help me


The Attempt at a Solution


I'm still at the conceptual stage :cry:

I don't know which rules apply to polynomials under the square roots. Can someone please give me some advice please?

These are called nested square roots, and you'll want to remove the "nest". In other words, try letting

[tex]\sqrt{3+2\sqrt{2}}=a+b\sqrt{2}[/tex]

and see if you can find the integer values of a and b that satisfy this equality (note: the unknowns will only be integers in some special cases such as this one).
To begin, square both sides and equate the rational parts and the irrational parts separately. You'll have two equations in two unknowns.
 
With something where I have to show, am I allowed to touch the RHS?
 
miniradman said:
With something where I have to show, am I allowed to touch the RHS?

You don't need to touch the RHS in [itex]\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=2[/itex]. Once you convert the nested square roots into the form I showed you earlier, the LHS will fall into place.

As for [itex]\sqrt{3+2\sqrt{2}}=a+b\sqrt{2}[/itex], there's no problem beginning with the assumption that the LHS can be converted into the form on the RHS, and if they couldn't, there would be a contradiction somewhere in the maths.
 
miniradman said:

Homework Statement


show that:

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

Homework Equations


I remember over-hearing someone talking about the modulus? I don't know how that's suppose to help me

The Attempt at a Solution


I'm still at the conceptual stage :cry:

I don't know which rules apply to polynomials under the square roots. Can someone please give me some advice please?

Apart from Mentallic's suggestion, a more direct approach is to simply square the LHS (left hand side).

Use ##(x - y)^2 = x^2 - 2xy + y^2## and ##(x+y)(x-y) = x^2 - y^2## to simplify.

You'll find all the square roots disappear very rapidly.