Question

Simplify `sqrt(29-2sqrt(28))` ?

Answer 1

`1+sqrt(7)`

Explanation:

`sqrt(29-2sqrt(28)) = sqrt(29-2sqrt(2^2*7))`

`color(white)(sqrt(29-2sqrt(28))) = sqrt(29-4sqrt(7))`

Is there a number of the form `a+bsqrt(7)` with square `29-4sqrt(7)`?

`(a+bsqrt(7))^2 = (a^2+7b^2) + 2ab sqrt(7)`

Equating coefficients, we want to find `a, b` satisfying:

`{ (a^2+7b^2 = 29), (2ab = -4) :}`

In particular, we would like `29-7b^2` to be a perfect square.

We quickly find that `b = +-2` works, resulting in:

`a^2 = 29-7b^2 = 29-7(2^2) = 1`

So `a = +-1`

Then from `2ab = -4` we find two possible solution pairs:

`(a, b) = (1, -2)" "` or `" "(a, b) = (-1, 2)`

The second of these results in the positive square root:

`sqrt(29-4sqrt(7)) = -1+2sqrt(7)`

Next we have:

`9 + sqrt(29-4sqrt(7)) = 9+(-1+2sqrt(7)) = 8+2sqrt(7)`

and we would like to find the square root of this.

Attempt to solve:

`8+2sqrt(7) = (c+dsqrt(7))^2 = (c^2+7d^2)+2cdsqrt(7)`

Hence:

`{ (c^2+7d^2 = 8), (2cd = 2) :}`

We can fairly quickly spot that `(c, d) = (1, 1)` results in a positive square root:

`sqrt(8+2sqrt(7)) = 1+sqrt(7)`