How do you find the square root of 1242?

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How do you find the square root of 1242?

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Answer 1 · 2017-06-07T21:20:57

It's $3 \cdot \sqrt{138}$ $3*sqrt138$

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 $\sqrt{1242}$ $sqrt1242$ until it's in the form $a \cdot \sqrt{b}$ $a*sqrtb$ , where $\sqrt{b}$ $sqrtb$ cannot be simplified further.

Let's start by dividing $\sqrt{1242}$ $sqrt1242$ into its factors:

$\sqrt{1242} = \sqrt{2} \cdot \sqrt{621} = \sqrt{2} \cdot \sqrt{3} \cdot \sqrt{207} = \sqrt{2} \cdot \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{69} = \sqrt{2} \cdot \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{23}$ $sqrt1242=sqrt2*sqrt621=sqrt2*sqrt3*sqrt207=sqrt2*sqrt3*sqrt3*sqrt69=sqrt2*sqrt3*sqrt3*sqrt3*sqrt23$

That last expression can be rewritten as:

$\sqrt{2} \cdot \sqrt{3} \cdot \sqrt{9} \cdot \sqrt{23}$ $sqrt2*sqrt3*sqrt9*sqrt23$

And that can be rewritten as:

$3 \cdot \sqrt{2} \cdot \sqrt{3} \cdot \sqrt{23}$ $3*sqrt2*sqrt3*sqrt23$

Further simplification:

$3 \cdot \sqrt{138}$ $3*sqrt138$

Answer 2 · 2017-11-08T00:19:32

$\sqrt{1242} \approx \frac{12171889}{345380} \approx 35.2420204$ $sqrt(1242) ~~ 12171889/345380 ~~ 35.2420204$

Explanation:

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

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

$12 | 42$ $12"|"42$

Examining the leftmost group of digits, note that:

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

So:

$3 < \sqrt{12} < 4$ $3 < sqrt(12) < 4$

and:

$30 < \sqrt{1242} < 40$ $30 < sqrt(1242) < 40$

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

Let's use a variant on the Babylonian method:

Given a rational approximation $\frac{p_{i}}{q_{i}}$ $p_i/q_i$ to $\sqrt{n}$ $sqrt(n)$ , a better approximation is given by $\frac{p_{i + 1}}{q_{i + 1}}$ $p_(i+1)/q_(i+1)$ where:

${ (p_(i+1) = p_i^2+n q_i^2), (q_(i+1) = 2 p_i q_i) :}$

Starting with $p_0/q_0$ repeatedly apply these formulae to get better approximations.

Let $n=1242$ and $p_0/q_0 = 35/1$

Then:

${ (p_1 = p_0^2+n q_0^2 = 35^2+1242 * 1^2 = 1225+1242 = 2467), (q_1 = 2 p_1 q_1 = 2 * 35 * 1 = 70) :}$

${ (p_2 = p_1^2+n q_1^2 = 2467^2 + 1242 * 70^2 = 6086089 + 6085800 = 12171889), (q_2 = 2 p_1 q_1 = 2 * 2467 * 70 = 345380) :}$

So:

$sqrt(1242) ~~ 12171889/345380 ~~ 35.2420204$

Answer 3 · 2017-12-21T15:37:21

35.2420...

Explanation:

$sqrt(12'42.00'00'00'00')=35.2420$
$-3^2$
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$color(white)(...)342color(white)(..........................)65xx5=325$
$-325$
$"------------"$
$color(white)(.....)1700 color(white)(........................)702xx2=1404$
$color(white)(.)-1404$
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$color(white)(........)29600 color(white)(............. ......)7044xx4=28176$
$color(white)(....)-28176$
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$color(white)(..........)142400 color(white)(..................)70482xx2=140964$
$color(white)(.....)-140964$
$"--------------------"$
$color(white)(..............)143600color(white)(...............) 704840xx0=0$

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How do you find the square root of 1242?

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