Question

How do you simplify `sqrt(3-2*sqrt(2))`?

Answer 1

We have that

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

Answer 2

Take the needed answer form with variables and solve for them

Explanation:

The answer above is completely correct, but it may help the student to see how to reach it:

We seek a square root of `3-2sqrt(2)`. This must have the form `a+bsqrt(2)`.

`(a+bsqrt(2))^2=a^2+2ab sqrt(2)+2b^2`

So
`a^2+2b^2=3`
and
`2ab=-2rArrab=-1`
These are two simultaneous equations for `a` and `b`.

Rearrange the second:
`b=-1/a`
Substitute into the first:
`a^2+2/a^2=3`
`a^4+2=3a^2`
`a^4-3a^2+2=0`

Note that this is a quadratic in `a^2`:
`(a^2)^2-3(a^2)+2=0`
Factorise:
`(a^2-2)(a^2-1)=0`

This gives us two possible solutions for `a^2`: `2` and `1`, and so the four solutions for `a`: `+-sqrt(2)` and `+-1`.

We are looking for integer solutions for `a`, and so `+-1` are possible solutions. But the other two are possible too - they can simply be folded in to the `sqrt(2)` term. This wouldn't have been possible if we'd had the root of some other number in the solution for `a`, but this solution is a special case.

Now use the second equation to deduce the four equivalent solutions for `b`:
`b=-1/a`
`b=bar(+)1/sqrt(2)=bar(+)1/2sqrt(2)` and `bar(+)1`.

So we have the four solution pairs `(a,b)`:
`(sqrt(2),-1/2sqrt(2))`
`(-sqrt(2),1/2sqrt(2))`
`(1,-1)`
`(-1,1)`

This is a bit suspicious - we expect only two solutions, positive and negative square roots, so we wonder if some of these are identical to each other: When we substitute them in to our desired expression `a+bsqrt(2)`, we get:

`sqrt(2)-1/2sqrt(2)sqrt(2)=-1+sqrt(2)`
`-sqrt(2)+1/2sqrt(2)sqrt(2)=1-sqrt(2)`
`1-sqrt(2)`
`-1+sqrt(2)`

So the two solutions with `sqrt(2)` are identical to the two simpler solutions, so we can get rid of them. We now have two solutions, positive and negative square roots:
`1-sqrt(2)`
`-1+sqrt(2)`

When we take the written square root of a quantity, it is implied that the desired root is the positive root, the "principal value" of the square root function. So we take the single solution that comes from `a=+1`:
`1-sqrt(2)`

Double check: Make sure that this produces the desired answer:
`(1-sqrt(2))^2=1-2sqrt(2)+2=3-2sqrt(2)`