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#