Approximating square roots using binomial expansion. Through the binomial expansion of $(1 - 2x)^\frac{1}{2}$, I am required to find an approximation of $\sqrt2$.

Binomial expansion
$ (1 + x)^n = 1 + \frac{n}{1}x + \frac{n(n-1)}{1*2}x^2 + ... $

Thus, the expansion of $(1 - 2x)^\frac{1}{2}$: $$
 = 1 - x -\frac{1}{2}x^2 - \frac{1}{2}x^3 + ... $$
The suggested way, is to choose a value for $x$ so that $(1-2x)$ has the form $2*$'a perfect square'. This can be done by taking $x = 0.01$. Thus, $(1 - 2x)=(1-2*0.01) = 0.98 = 2*0.7^2$
And
$$
(1 - 2x)^\frac{1}{2} = 0.98^\frac{1}{2} = 0.7\sqrt2$$
Which is equal to the previously established expansion, so we can now go ahead and find $\sqrt2$. The problem I am facing, is that there was no mention of how the value of $x=0.01$ was arrived at.
Is there an easy way to determine an appropriate value for $x$?
 A: 
We want to (manually)  approximate $\sqrt{2}$ by using the first few terms of the binomial series expansion of 
  \begin{align*}
\sqrt{1-2x}&=  \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}(-2x)^n\qquad\qquad\qquad\qquad |x|<\frac{1}{2}\\
&= 1-x-\frac{1}{2}x^2-\frac{1}{2}x^3+\cdots\tag{1}
\end{align*}
  Here we look for a way to determine appropriate values of $x$ using the binomial expansion.

In order to apply (1) we are  looking for a number $y$ with
\begin{align*}
\sqrt{1-2x}&=\sqrt{2y^2}=y\sqrt{2}\tag{2}\\
\color{blue}{\sqrt{2}}&\color{blue}{=\frac{1}{y}\sqrt{1-2x}}
\end{align*}
We see it is convenient to choose $y$ to be a square number which can be easily factored out from the root.  We obtain from (2)
\begin{align*}
1-2x&=2y^2\\
\color{blue}{x}&\color{blue}{=\frac{1}{2}-y^2}\tag{3}
\end{align*}

When looking for an appropriate $y$  which fulfills (3) there are some aspects to consider:
  
  
*
  
*We  have  to respect the radius of convergence $|x|<\frac{1}{2}$.
  
*Since we want to calculate an approximation of $\sqrt{2}$ by hand we should take $y\in\mathbb{Q}$ with rather small numbers as numerator and denominator.
  
*Last but not least: We want to find a value $x$ which provides a good approximation for $\sqrt{2}$.
We will see it's not hard to find values which have these properties.

We see in (1) a good approximation is given if $x$ is close to $0$. If $x$ is close to zero we will also fulfill the convergence condition. $x$ close to zero means that in (3) we have to choose $y$ so that
$
y^2
$ 
is close to $\frac{1}{2}$. We have already   (1)  and (3) appropriately considered. Now we want to find  small natural  numbers $a,b$ so that
\begin{align*}
y^2=\frac{a^2}{b^2}\approx \frac{1}{2}
\end{align*}
This can be done easily. When going through small numbers of $a$ and $b$ whose squares are apart by a factor $2$ we might quickly come to $100$ and $49$. These are two small squares and we have $2\cdot 49=98$ close to $100$. That's all.

Now it's time to harvest. We choose $y^2=\frac{49}{100}$ resp. $\color{blue}{y=\frac{7}{10}}$. We obtain for $x$ from (3)
  \begin{align*}
x=\frac{1}{2}-y^2=\frac{1}{2}-\frac{49}{100}=\frac{1}{100}
\end{align*}
  We have now a nice value $\color{blue}{x=\frac{1}{100}}$ and we finally get  from (1) the approximation:
  \begin{align*}
\color{blue}{\sqrt{2}} \approx \frac{10}{7}\left(1- 10^{-2}-\frac{1}{2}\cdot 10^{-4}-\frac{1}{2}\cdot 10^{-6}\right)\color{blue}{=1.414\,213\,5}71\ldots
\end{align*}
We have $\color{blue}{\sqrt{2}=1.414\,213\,5}62\ldots$ with an approximation error $\approx 9.055\times 10^{-9}$. This result is quite impressive when considering that we have used just four terms of the binomial series.

Note: In a section about binomial series expansion in Journey through Genius by W. Dunham the author cites Newton: Extraction of roots are much shortened by this theorem, indicating how valuable this technique was for Newton.
A: Personally, I wouldn't have done it that way.  So here is how I would've done it:
Method 1:
$$\sqrt2=\sqrt{1+1}=1+\frac12-\frac18+\dots\approx1+\frac12-\frac18=\frac{11}8=1.375$$
which is much clearer to me, since it avoids having to take decimals raised to powers and gives you something you can easily do by hand.
$$1.375^2=1.890625$$
Obviously it approaches the correct value as you take more terms.

Method 2:
This is called fixed-point iteration/Newton's method, and it basically goes like this:
$$x=\sqrt2\implies x^2=2$$
$$2x^2=2+x^2$$
Divide both sides by $2x$ and we get
$$x=\frac{2+x^2}{2x}$$
Now, interestingly, I'm going to call the $x$'s on the left $x_{n+1}$ and the $x$'s on the right $x_n$, so
$$x_{n+1}=\frac{2+(x_n)^2}{2x_n}$$
and with $x_0\approx\sqrt2$, we will then have $x=\lim_{n\to\infty}x_n$.  For example, with $x_0=1$,
$x_0=1$
$x_1=\frac{2+1^2}{2(1)}=\frac32=1.5$
$x_2=\frac{2+(3/2)^2}{2(3/2)}=\frac{17}{12}=1.41666\dots$
$x_3=\dots=\frac{577}{408}=1.414215686$
And one can quickly check that $(x_3)^2=2.000006007\dots$, which is pretty much the square root of $2$.
A: In fact you can take any two numbers which can be added to get 2 (not nesserly 0.01 but at least you should know the root of one of them 
So for example $\sqrt{2} = {(1+1)^{1/2}}$ 
Know all what you need is to expand it using bio theorem and for 2 terms you ll get 1.5
$1+(0.5)\cdot 1= 1.5 $
Which is not too bad approximation for $\sqrt{2} = 1.4142135624 $
