Find the fourth roots of the following binomial surd: $X=14+8\sqrt{3}$
I attempt to find the square root first: $\sqrt{X}=\sqrt{14+8\sqrt{3}}$
$\sqrt{14+8\sqrt{3}}=\sqrt{x_1}+\sqrt{y_1}$
$(\sqrt{14+8\sqrt{3}})^2=(\sqrt{x_1}+\sqrt{y_1})^2$
$14+8\sqrt{3}=x_1+y_1+2\sqrt{x_1y_1}$
$x_1+y_1=14$; $2\sqrt{x_1y_1}=8\sqrt{3} \Rightarrow x_1y_1=48$
$(x_1-y_1)^2=(x_1+y_1)^2-4x_1y_1 \Rightarrow (x_1-y_1)^2=(14)^2-4(48)=4$
$x_1-y_1=2$ (Ignore the other solution $-2$ by making the assumption $x_1>y_1$)
$x_1+y_1=14$; $x_1-y_1=2$; $x_1=8, y_1=6$
$\sqrt{14+8\sqrt{3}}=\sqrt{8}+\sqrt{6}=2\sqrt{2}+\sqrt{6}$
Now to find the square root again to get the fourth root:$\sqrt[4]{X}=\sqrt{2\sqrt{2}+\sqrt{6}} $
$\sqrt{2\sqrt{2}+\sqrt{6}}=\sqrt{\sqrt{2}(2+\sqrt{3})}=\sqrt[4]{2}\sqrt{2+\sqrt{3}}$
Let $\sqrt{2+\sqrt{3}}=\sqrt{x_2}+\sqrt{y_2}$
$2+\sqrt{3}=x_2+y_2+2\sqrt{x_2y_2}$
$x_2+y_2=2$; $2\sqrt{x_2y_2}=\sqrt{3} \Rightarrow x_2y_2=\dfrac{3}{4}$
$(x_2-y_2)^2=(x_2+y_2)^2-4x_2y_2=4-4\left(\dfrac{3}{4}\right)=1 \Rightarrow x_2-y_2=1$
$x_2=\dfrac{3}{2}, y_2=\dfrac{1}{2}$
Hence $\sqrt[4]{X}=\sqrt[4]{14+8\sqrt{3}}=\sqrt{2\sqrt{2}+\sqrt{6}} =\sqrt[4]{2}\left(\sqrt{\dfrac{3}{2}}+\sqrt{\dfrac{1}{2}}\right)=\sqrt[4]{2}\left(\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{2}}{2}\right)=Y$
The solution however is given as $\sqrt[4]{X}=\dfrac{1}{\sqrt[4]{2}}(\sqrt{3}+1)=Z$, but I feel my calculations are correct.
Why I am getting $\sqrt[4]{X}=Y$ while text book is giving $\sqrt[4]{X}=Z$ ?
So where did I go wrong? Thanks.