Show $\sqrt[4]{2}$ on horizontal axis It is very basic question ,to show $\sqrt[2]{2}$ on $x$ axis .we do like below 
but,how can we show $\sqrt[4]{2}$ on $x$ axis ?(like we do for $\sqrt2)$
I am thankful if you guide me .
  $\bf{Remark}:$ may be a question for many people .
 A: It's possible to repeatedly square root any constructible number. So far, you know how to construct the point located at $A(\sqrt 2, 0)$. Now construct the point $B(-1, 0)$ and connect the two points (note that they pass through the origin $O(0, 0)$). Construct a semi-circle above the $x$-axis whose diameter is $AB$ and let $C(0, k)$ be its $y$-intercept. We claim that $OC = k = \sqrt[4]{2}$.
Indeed, observe that triangles $AOC$ and $COB$ are similar, so we have:
$$
\frac{k}{1} = \frac{\sqrt 2}{k} \iff k^2 = \sqrt 2 \iff k = \sqrt[4]{2}
$$
A: The following steps will work . . .
\begin{align*}
1.\;\,&\text{Construct a segment of length$\,\sqrt{2}$.}\\[4pt]
2.\;\,&\text{Construct a segment $AB$ of length$\,\sqrt{2} + {\small{\frac{1}{4}}}$.}\\[4pt]
3.\;\,&\text{Construct a circle with $AB$ as a diameter.}\\[4pt]
4.\;\,&\text{Construct a point $P$ on $AB$ such that $PB$ has length$\,\sqrt{2} - {\small{\frac{1}{4}}}$.}\\[4pt]
5.\;\,&\text{Construct a circle with center $B$, and radius $BP$.}\\[4pt]
6.\;\,&\text{Let $C$ be one of the two points where the circles intersect.}\\[10pt]
\end{align*}
\begin{align*}
\text{Then}\;\;|AC|^2 &= |AB|^2 - |BC|^2
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,
\\[4pt]
&=\left(\sqrt{2} + {\small{\frac{1}{4}}}\right)^2 -\left(\sqrt{2} - {\small{\frac{1}{4}}}\right)^2\\[4pt]
&=
\left(2+{\small{\frac{1}{2}}}\sqrt{2}+{\small{\frac{1}{16}}}\right)
-
\left(2-{\small{\frac{1}{2}}}\sqrt{2}+{\small{\frac{1}{16}}}\right)\\[4pt]
&=\sqrt{2}\\[8pt]
\text{hence}\;\;|AC|\;&= \sqrt[4]{2},\;\text{as required.}\\[4pt]
\end{align*}
