How can this binomial expansion result in two different approximations of root 2? I have been working on a problem on approximating $\sqrt{2}$ using the first three terms of the following binomial expansion, and a substitution of $x = -\frac{1}{10}$ :
$$(4 - 5x)^.5  = 2 - \frac{5x}{4} - \frac{25x^2}{64}$$
After substituting, I got to the stage:
$ \frac{3}{\sqrt{2}}= \frac{543}{256}$
Now what is bizarre, is that if I solve this equation for $\sqrt{2}$, there are two routes I could take, but they give slightly different answers.
Route 1:
Reciprocate both sides, and multiply both sides by 3 to get:
$ \sqrt{2}= \frac{256}{181} $
Route 2:
Rationalize the left hand side to make the equation:
$ \frac{3\sqrt{2}}{2}= \frac{543}{256}$
Divide both sides by $\frac{3}{2}$:
$ \sqrt{2}= \frac{181}{128}$
So we end up with two approximations of $ \sqrt{2}$: $\frac{256}{181}$ and $\frac{181}{128}$
I am really struggling to understand how this has happened?
Why is the same equation leading to two different solutions?
 A: If we denote the approximation with $\ c\ $, we have two equations that we can solve for $\ x\ $ :


*

*$\ \frac{3}{x}=c\ $ giving $\ x=\frac{3}{c}\ $

*$\ \frac{3x}{2}=c\ $ giving $\ x=\frac{2c}{3}\ $
If $\ c\ $ were exactly $\ \sqrt{2}\ $, both solutions would coincide. But $\ c\ $ is only an approximation, hence the values cannot coincide excactly. Their product , however , is exactly $\ 2\ $ , so one approximation is too small and the other too large.
A: Let $f(x) = 2 - \frac{5x}{4} - \frac{25x^2}{64}$.
Let $c$ denote the value $f(-\frac{1}{10}) = \frac{543}{256}$.
We can write
$$\tag 1 \frac{3}{\sqrt 2} = c + \varepsilon$$
If we take the OP's route 1 (but keeping $\varepsilon$), then
$$\tag 2 \sqrt 2 = \frac{3}{c+\varepsilon}$$
We can check that the 'route 1 approximation', $\frac{3}{c}$, is strictly greater than  $\sqrt 2$. With $\varepsilon$ playing a part in the denominator of the rhs of $\text{(2)}$, to 'fix things up' it must be true that $\varepsilon \gt 0$.
If we take the OP's route 2 (but keeping $\varepsilon$), then
$$\tag 3 \sqrt 2 = \frac{2}{3}\,(c + \varepsilon)$$
Since $\varepsilon \gt 0$, the 'route 2 approximation' $\frac{2c}{3}$ is strictly less than $\sqrt 2$.
