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

##### 1 Answer

#### 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))^2 = (a^2+7b^2) + 2ab sqrt(7)#

Equating coefficients, we want to find

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

In particular, we would like

We quickly find that

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

So

Then from

#(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

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