How do you find the square root of $6724$ $6724$ ?

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How do you find the square root of $6724$ $6724$ ?

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Answer 1 · 2017-05-03T19:54:37

$\sqrt{6724} = 82$ $sqrt(6724) = 82$

Explanation:

We can try factoring $6724$ $6724$ and identifying square factors:

$0000006724$ $color(white)(000000)6724$
$000000/00\$ $color(white)(000000)"/"color(white)(00)"\"$
$0000020003362$ $color(white)(00000)2color(white)(000)3362$
$000000000/00\$ $color(white)(000000000)"/"color(white)(00)"\"$
$0000000020001681$ $color(white)(00000000)2color(white)(000)1681$
$000000000000/00\$ $color(white)(000000000000)"/"color(white)(00)"\"$
$00000000000410041$ $color(white)(00000000000)41color(white)(00)41$

So:

$\sqrt{6724} = \sqrt{2^{2} \cdot 41^{2}} = 2 \cdot 41 = 82$ $sqrt(6724) = sqrt(2^2*41^2) = 2*41=82$

That "worked", but I cheated slightly in knowing that $1681 = 41^{2}$ $1681 = 41^2$

Let's try another approach...

Given $6724$ $6724$ , split off pairs of digits starting from the right to get:

$67 ∣ 24$ $67|24$

Looking at the most significant pair of digits, note that:

$67 > 64 = 8^{2}$ $67 > 64 = 8^2$

Hence the square root of $6724$ $6724$ is a little larger than $80$ $80$

We can use the Babylonian method to find a better approximation:

Given a number $n$ $n$ of which you want the square root and an approximation $a_{i}$ $a_i$ to that root, a better approximation $a_{i + 1}$ $a_(i+1)$ is given by the formula:

$a_{i + 1} = \frac{a_{i}^{2} + n}{2 a_{i}}$ $a_(i+1) = (a_i^2+n)/(2a_i)$

So in our case, with $n = 6724$ $n=6724$ and $a_{0} = 80$ $a_0 = 80$ we find:

$a_{1} = \frac{80^{2} + 6724}{2 \cdot 80} = \frac{6400 + 6724}{160} = 82.025$ $a_1 = (80^2+6724)/(2*80) = (6400+6724)/160 = 82.025$

Hmmm - that's suspiciously close to $82$ $82$ . Does $82$ $82$ work?

$82^{2} = {(80 + 2)}^{2} = 80^{2} + 2 \cdot 80 \cdot 2 + 2^{2} = 6400 + 320 + 4 = 6724$ $82^2 = (80+2)^2 = 80^2+2*80*2+2^2 = 6400+320+4 = 6724$

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How do you find the square root of $6724$ $6724$ ?

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How do you find the square root of $6724$ $6724$ ?