# How to root single and double digit non perfect square?

I want to know a simple process to root single or double digit non-perfect square for a competitive exam where time is very limited. For example, $$\sqrt 3$$ $$\sqrt 13$$

You can take any number to show the process.


\begin{align} \root{13} & = \root{325 \over 25} = {1 \over 5}\root{324 + 1} = {18 \over 5}\root{1 + {1 \over 324}}\quad \pars{~follow\ the\ above\ procedure~} \end{align}

I understand you want a simple way to approximate the sqaure root of a number below $100$. An easy way may be to use Taylor expansion.

You have $\sqrt{a^2+h}=a \cdot \sqrt{1+h/a^2} \approx a\cdot(1+{h \over 2a^2}) \approx a + {h \over 2a}$.

You need to know the squares from $1$ to $10$ (not too hard), and pick the closest one to the input number ($h$ can be negative).

For example, let's take $73$, then we choose $a=9$ and $h=-8$, since $73 = 9^2-8$.

$\sqrt{73} \approx 9-8/18 \approx 9-0.444 \approx 8.556$ and $\sqrt{73} = 8.544 \dots$. Our approximation is not too bad.

If you have more time you could add the second order term which is $a \cdot {-h^2 \over 8a^4}={-h^2 \over 8a^3}$ but that begins to be tedious to compute.

For small numbers ($2, 3, 5, 6, 7, 8$) you can learn the roots by heart, they are often useful.