What is the square root of 67.98?

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Answer 1 · 2015-10-13T19:57:34

$67.98 = (2*3*11*103)/100$ , so the simplest algebraic form is:

$sqrt(67.98) = sqrt(6798)/10 ~~ 8.245$

To calculate an approximation for $sqrt(67.98)$ , find an approximation for $sqrt(6798)$ and divide by $10$

$80^2 = 6400 < 6798 < 8100 = 90^2$

To find the square root of a number $n$ , you can choose a first approximation $a_0$ and iterate using the formula:

$a_(i+1) = (a_i^2+n)/(2a_i)$

I prefer to work with integers, so I express $a_0 = p_0/q_0$ and iterate using the formulae:

$p_(i+1) = p_i^2 + n q_i^2$

$q_(i+1) = 2 p_i q_i$

Then $p_i/q_i$ is the same as $a_i$ .

If the resulting $p_(i+1)$ and $q_(i+1)$ have a common factor larger than $1$ then divide both by that before the next iteration.

Let $n=6798$ , $p_0 = 80$ , $q_0 = 1$

Then:

$p_1 = p_0^2 + n q_0^2 = 80^2 + 6798 * 1^2 = 6400+6798 = 13198$

$q_1 = 2 p_0 q_0 = 2 * 80 * 1 = 160$

These are both divisible by $2$ , so do that to get:

$p_(1a) = p_1/2 = 6599$

$q_(1a) = q_1/2 = 80$

Next iteration:

$p_2 = p_(1a)^2 + n q_(1a)^2 = 6599^2 + 6798*80^2 = 43546801 + 43507200 = 87054001$

$q_2 = 2 p_(1a) q_(1a) = 2 * 6599 * 80 = 1055840$

Stopping here, we get:

$sqrt(6798) ~~ 87054001 / 1055840 ~~ 82.44999 ~~ 82.45$

So

$sqrt(67.98) ~~ 8.245$