How do you simplify #sqrt(21+12sqrt3)#?

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Answer 1 · 2015-10-27T19:54:40

Find rational solution of #(a+b sqrt(3))^2 = 21+12sqrt(3)#, hence

#sqrt(21+12sqrt(3)) = 3 + 2sqrt(3)#

Is #(21+12sqrt(3))# a perfect square of something?

#(a+b sqrt(3))^2 = (a^2+3b^2)+2ab sqrt(3)#

So are there rational numbers #a# and #b# such that:

#a^2+3b^2 = 21# and #2ab = 12# ?

From #2ab = 12#, we get #b = 6/a#

Substitute that into #a^2+3b^2 = 21# to find:

#a^2+3(6/a)^2 = 21#

Subtract #21# from both sides and multiply through by #a^2# to get:

#0 = (a^2)^2 - 21(a^2)+108 = (a^2-9)(a^2-12)#

This has rational roots #a = +-3#.

We might as well choose #a=3# hence #b = 2# and we have found that:

#(21+12sqrt(3)) = (3+2sqrt(3))^2#

So:

#sqrt(21+12sqrt(3)) = 3 + 2sqrt(3)#