How to root single and double digit non perfect square?

Question

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.

Answer 1

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.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \root{3} & = \root{147 \over 49} = {1 \over 7}\root{144 + 3} = {12 \over 7}\root{1 + {3 \over 144}} = {12 \over 7}\root{1 + {1 \over 48}} \\[5mm] & \approx {12 \over 7}\bracks{1 + {1 \over 2}\,{1 \over 48} + {1 \over 2}\,{1 \over 2}\pars{-\,{1 \over 2}}\pars{1 \over 48}^{2}} = \underbrace{\overbrace{\underbrace{12 \over 7}_{\ds{1.7\color{#f00}{1}\ldots}} + {1 \over 56}}^{\ds{1.732\color{#f00}{1}\ldots}} - {1 \over 10752}} _{\ds{1.7320\color{#f00}{4}\ldots}} \end{align}

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\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}