Radicals Within Radical

Sep 2014
995
42
NYC
I set out to prove, without a calculator, that
sqrt{sqrt{6 + 4sqrt{2}} = sqrt{2 + sqrt{2}}.

I let d = sqrt{2 + sqrt{2}}.

I was able to complete the prove by hand but NOT without using Wolfram to simplify a portion of the problem.

I factored the radicand of sqrt{6 + 4sqrt{2}} to become sqrt{2(3 + 2sqrt{2})}.

According to Wolfram:

sqrt{6 + 4sqrt{2}} = sqrt{2(3 + 2sqrt{2})} = 2 + sqrt{2}.

QUESTION:

How do I manipulate the radical sqrt{2(3 + 2sqrt{2})} to become 2 + sqrt{2}?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
IF it is true that \(\displaystyle \sqrt{\sqrt{6+ 4\sqrt{2}}}= \sqrt{2+ \sqrt{2}}\) then it must be true (squaring both sides) that \(\displaystyle \sqrt{6+ 4\sqrt{2}}= 2+ \sqrt{2}\). Squaring again, \(\displaystyle 6+ 4\sqrt{2}= 4+ 4\sqrt{2}+ 2= 6+ 4\sqrt{2}\). Note that every step is invertible (as long as we take the positive square root).
 
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Dec 2014
129
101
USA
$\sqrt{2(3 + 2\sqrt{2})} = \sqrt{2(1 + 2\sqrt{2} + 2)} = \sqrt{2(1+\sqrt{2})^2} = \sqrt{2}(1+\sqrt{2}) = \sqrt{2} + 2$
 
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Feb 2015
2,255
510
Ottawa Ontario
I set out to prove, without a calculator, that
sqrt{sqrt{6 + 4sqrt{2}} = sqrt{2 + sqrt{2}}.

I let d = sqrt{2 + sqrt{2}}.
I see nothing wrong with substituting as you did, then solving.
Making it a simpler substitution:
Let d = √2 *****

√(√(6 + 4d)) = √(2 + d)
square both sides:
√(6 + 4d) = 2 + d
square both sides:
6 + 4d = 4 + 4d + d^2
d^2 = 2
d = √2 *****
 
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