\[1300\]

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.