# Proof/Show with polynomials under the radical

1. Mar 10, 2013

1. The problem statement, all variables and given/known data
show that:

$\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=2$

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

3. The attempt at a solution
I'm still at the conceptual stage

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

2. Mar 10, 2013

### Mentallic

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

$$\sqrt{3+2\sqrt{2}}=a+b\sqrt{2}$$

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.

3. Mar 10, 2013

With something where I have to show, am I allowed to touch the RHS?

4. Mar 10, 2013

### Mentallic

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

As for $\sqrt{3+2\sqrt{2}}=a+b\sqrt{2}$, 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.

5. Mar 10, 2013

### Curious3141

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.