How do you find the square root of 1414?

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Answer 1 · 2016-06-26T17:39:44

$sqrt(1414)$ cannot be simplified, but you can calculate approximations.

For example:

$sqrt(1414) ~~ 35347/940 ~~ 37.60319$

Factorising $1414$ into prime factors, we find:

$1414 = 2 xx 7 xx 101$

which has no square factors. So its square root has no simpler form.

You can find rational approximations for the square root using a kind of Newton Raphson method.

Given an initial approximation $a_0$ for $sqrt(n)$ , iterate using the formula:

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

In order to make the arithmetic less messy, I prefer to split $a_i = p_i/q_i$ and iterate using the formulae:

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

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

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

In our example $n=1414$ .

Note that $37^2 = 1369$ and $38^2 = 1444$

So linearly interpolating, choose:

$a_0 = 37+(1414-1369)/(1444-1369) = 37.6 = 188/5$

So $p_0=188$ , $q_0=5$

Then:

$p_1 = p_0^2+n q_0^2 = 188^2+1414*5^2 = 35344+35350 = 70694$

$q_1 = 2 p_0 q_0 = 2*188*5 = 1880$

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

$p_(1a) = 70694/2 = 35347$

$q_(1a) = 1880/2 = 940$

If we stopped at this stage we would have:

$sqrt(1414) ~~ 35347/940 ~~ 37.603191489$

Let's try another iteration:

$p_2 = p_(1a)^2 + n q_(1a)^2 = 35347^2+ 1414*940^2 = 1249410409 + 1249410400 = 2498820809$

$q_2 = 2 p_(1a) q_(1a) = 2*35347*940 = 66452360$

So:

$sqrt(1414) ~~ 2498820809/66452360 ~~ 37.60319135392633158551$

Actually:

$sqrt(1414) ~~ 37.60319135392633134161$