# How do you find the square root of 1242?

Jun 7, 2017

It's $3 \cdot \sqrt{138}$

#### Explanation:

As we can see, 1242 is not a perfect square, and so therefore, will not be able to be simplified into a whole number. What we can do, though, is simplify $\sqrt{1242}$ until it's in the form $a \cdot \sqrt{b}$, where $\sqrt{b}$ cannot be simplified further.

Let's start by dividing $\sqrt{1242}$ into its factors:

$\sqrt{1242} = \sqrt{2} \cdot \sqrt{621} = \sqrt{2} \cdot \sqrt{3} \cdot \sqrt{207} = \sqrt{2} \cdot \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{69} = \sqrt{2} \cdot \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{23}$

That last expression can be rewritten as:

$\sqrt{2} \cdot \sqrt{3} \cdot \sqrt{9} \cdot \sqrt{23}$

And that can be rewritten as:

$3 \cdot \sqrt{2} \cdot \sqrt{3} \cdot \sqrt{23}$

Further simplification:

$3 \cdot \sqrt{138}$

Nov 8, 2017

$\sqrt{1242} \approx \frac{12171889}{345380} \approx 35.2420204$

#### Explanation:

To find rational approximations to $\sqrt{1242}$ we can proceed as follows:

First split $1242$ into pairs of digits starting from the right:

$12 \text{|} 42$

Examining the leftmost group of digits, note that:

${3}^{2} = 9 < 12 < 16 = {4}^{2}$

So:

$3 < \sqrt{12} < 4$

and:

$30 < \sqrt{1242} < 40$

In fact note that $12$ is close to the average of $9$ and $16$, so a reasonable approximation to $\sqrt{12}$ is $3.5$ and to $\sqrt{1242}$ is $35$.

Let's use a variant on the Babylonian method:

Given a rational approximation ${p}_{i} / {q}_{i}$ to $\sqrt{n}$, a better approximation is given by ${p}_{i + 1} / {q}_{i + 1}$ where:

$\left\{\begin{matrix}{p}_{i + 1} = {p}_{i}^{2} + n {q}_{i}^{2} \\ {q}_{i + 1} = 2 {p}_{i} {q}_{i}\end{matrix}\right.$

Starting with ${p}_{0} / {q}_{0}$ repeatedly apply these formulae to get better approximations.

Let $n = 1242$ and ${p}_{0} / {q}_{0} = \frac{35}{1}$

Then:

$\left\{\begin{matrix}{p}_{1} = {p}_{0}^{2} + n {q}_{0}^{2} = {35}^{2} + 1242 \cdot {1}^{2} = 1225 + 1242 = 2467 \\ {q}_{1} = 2 {p}_{1} {q}_{1} = 2 \cdot 35 \cdot 1 = 70\end{matrix}\right.$

$\left\{\begin{matrix}{p}_{2} = {p}_{1}^{2} + n {q}_{1}^{2} = {2467}^{2} + 1242 \cdot {70}^{2} = 6086089 + 6085800 = 12171889 \\ {q}_{2} = 2 {p}_{1} {q}_{1} = 2 \cdot 2467 \cdot 70 = 345380\end{matrix}\right.$

So:

$\sqrt{1242} \approx \frac{12171889}{345380} \approx 35.2420204$

Dec 21, 2017

35.2420...

#### Explanation:

$\sqrt{12 ' 42.00 ' 00 ' 00 ' 00 '} = 35.2420$
$- {3}^{2}$
$\text{------------}$
$\textcolor{w h i t e}{\ldots} 342 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} 65 \times 5 = 325$
$- 325$
$\text{------------}$
$\textcolor{w h i t e}{\ldots . .} 1700 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 702 \times 2 = 1404$
$\textcolor{w h i t e}{.} - 1404$
$\text{------------}$
$\textcolor{w h i t e}{\ldots \ldots . .} 29600 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots . \ldots \ldots} 7044 \times 4 = 28176$
$\textcolor{w h i t e}{\ldots .} - 28176$
$\text{--------------}$
$\textcolor{w h i t e}{\ldots \ldots \ldots .} 142400 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots} 70482 \times 2 = 140964$
$\textcolor{w h i t e}{\ldots . .} - 140964$
$\text{--------------------}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots . .} 143600 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots} 704840 \times 0 = 0$