## Approximate Square Roots

We can find quite an accurate estimate of a square root using simple algebra. The square root of
$1300$
is approximately 36 -
$36^2=1296$
, so the square root is just a little bit more.
We can find a better estimate by writing
$\sqrt{1300}=36+x$
where
$x$
is small.
Then
$1300=(36+x)^2=1296+72x+x^2$
(1)
Since
$x$
is small,
$x^2$
is very small and we can ignore it, so
$1300 \simeq 1296+72x$
.
$4 \simeq 72x$
.
$x \simeq \frac{4}{72} =0.0555555...$
.
Then
$\sqrt{1300} \simeq 36.055555...$
.
We can improve this estimate by taking
$x^2$
into account. From (1) we have
$x^2+72x-4=0 \rightarrow x=\frac{4-x^2}{72}$
.
We can use this as an iteration expression
$x_{n+1}=\frac{4-x_n^2}{72}$
, and by putting
$x_1=\frac{1}{18}$
on the right hand side we get
$x_2=\frac{4-(1/18)^2}{72} \simeq 0.055513$
, so that
$\sqrt{1300} \simeq 37.055513$
.
This is correct to 8 significant figures.