Proof/Show with polynomials under the radical

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Homework Help Overview

The problem involves showing that the expression \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=2\) holds true. The subject area relates to properties of square roots and polynomials under radicals.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the concept of nested square roots and the potential use of modulus. There are suggestions to express the nested square roots in a different form, such as letting \(\sqrt{3+2\sqrt{2}}=a+b\sqrt{2}\), and to square the left-hand side to simplify the expression.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants are questioning the necessity of manipulating the right-hand side of the equation, while others are providing guidance on how to handle the nested square roots.

Contextual Notes

There is mention of constraints regarding the assumptions that can be made about the equality and the forms of the expressions involved. Participants are also expressing uncertainty about the rules that apply to the manipulation of square roots.

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.
 

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