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

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Answer 1 · 2017-02-26T00:41:36

#1+sqrt(7)#

#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)#