Find the square root of the following binomial surd: $\sqrt{27}+2\sqrt{6}$

Question

Find the square root of the following binomial surd: $\sqrt{27}+2\sqrt{6}$

$\sqrt{\sqrt{27}+2\sqrt{6}}=\sqrt{x}+\sqrt{y}$

$(\sqrt{\sqrt{27}+2\sqrt{6}})^2=(\sqrt{x}+\sqrt{y})^2$

$3\sqrt{3}+2\sqrt{6}=x+y+2\sqrt{xy}$

$x+y=0$ because there is no rational component, and $2\sqrt{xy} = 3\sqrt{3}+2\sqrt{6}$

Now $(x-y)^2=(x+y)^2-4xy \Rightarrow (x-y)^2 = (0)^2-(3\sqrt{3}+2\sqrt{6})^2$.

$(x-y)^2= -(27+24+12\sqrt{18}) = -(51+2\sqrt{648})$

$x-y=\sqrt{-(51+2\sqrt{648})}$

I'm stuck here. First I don't know what to do with the minus sign. Second, if the minus sign didn't exist, I suppose I would repeat the process, namely $x-y=\sqrt{51+2\sqrt{648}}=\sqrt{x}+\sqrt{y}$. But then I feel it's wrong for me to set $x-y = \sqrt{x}+\sqrt{y}$. Thanks for the help.

Answer 1

Observe that $$\sqrt{27} + 2\sqrt{6} = 3\sqrt{3} + 2\sqrt{6} = \sqrt{3}(3 + 2\sqrt{2}) = \sqrt{3}(1 + 2\sqrt{2} + 2) = \sqrt{3}(1 + \sqrt{2})^2$$ Hence, $$\sqrt{\sqrt{27} + 2\sqrt{6}} = \sqrt[4]{3}(1 + \sqrt{2})$$

Answer 2

Let$$t＝\sqrt {\sqrt {27}＋2\sqrt {6}}\\~\\t^2＝\sqrt {27}＋2\sqrt {6}\implies\;t^2＝\sqrt {9×3}＋2\sqrt {2×3}＝3\sqrt {3}＋2\sqrt {2}\sqrt {3}＝\sqrt {3}\left({3＋2\sqrt {2}}\right)\\~\\\implies\;t^2＝\sqrt {3}\left({3＋2\sqrt {2}}\right)＝\sqrt {3}\left({1＋}\left({\sqrt {2}}\right)^2＋2\sqrt {2}\right)＝\sqrt {3}\left({1＋\sqrt {2}}\right)^2＝\left({\sqrt[4]{3}\left({1＋\sqrt {2}}\right)}\right)^2$$ Finally, $$t＝\sqrt {\sqrt {27}＋2\sqrt {6}}={\sqrt[4]{3}\left({1＋\sqrt {2}}\right)}$$

Answer 3

Let $m>n$

$(\sqrt m + \sqrt n)^2=m+n+2\sqrt{mn}=\sqrt27+2\sqrt6=3\sqrt3+2\sqrt6$

$2\sqrt{mn}=2\sqrt6$

$\sqrt{mn}=\sqrt6$

$m+n=3\sqrt3$

then

$m=2\sqrt3$ and $n=\sqrt3$

$\sqrt{\sqrt27+2\sqrt6}=\sqrt{2\sqrt3}+\sqrt{\sqrt3}=\sqrt[4]{12}+\sqrt[4]{3}$

Answer 4

Alt. hint: $\;$ let $\,a = \sqrt{\sqrt{27} + 2\sqrt{6}}\,$, $\,b = \sqrt{\sqrt{27} - 2\sqrt{6}}\,$, then $\require{cancel}\,a^2 + b^2 = 6 \sqrt{3}\,$ and $\,p = ab = \sqrt{3}\,$, so $\,s = a + b = \sqrt{a^2+b^2+2ab} = 2 \sqrt{2\sqrt{3}}\,$.

It follows that $\,a,b\,$ are the roots of the quadratic $\,x^2 - s x + p = 0\,$, or $\,x^2 - 2 \sqrt{2\sqrt{3}}\,x + \sqrt{3} = 0\,$, so $\,a, b = \sqrt{2\sqrt{3}} \pm \sqrt{2 \sqrt{3} - \sqrt{3}} = \sqrt[4]{3}\left(\sqrt{2} \pm 1\right)\,$.