How do you simplify #sqrt(1414)#?

1 Answer
Jun 3, 2017

Answer:

#sqrt(1414)# is already in simplest form.

Explanation:

The prime factorisation of #1414# is:

#1414 = 2*7*101#

This contains no square factors, so the square root is already in simplest form.

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Notes

Should the #1414# in the question have been #1444#?

If it was, then we would find:

#1444 = 2*2*19*19 = 38^2#

So:

#sqrt(1444) = 38#

We also find that:

#37^2 = 1369 < 1414 < 1444 = 38^2#

So #sqrt(1414)# is an irrational number somewhere between #37# and #38#.

If you would like a rational approximation, we can start by linearly interpolating between #37# and #38#, finding:

#sqrt(1414) ~~ 37+(1414-1369)/(1444-1369) = 37+45/75 = 37+3/5 = 188/5#

This will be slightly less than #sqrt(1414)#.

We find:

#(188/5^2) = 35344/25 = 35350/25-6/25 = 1414-6/25#

That's not bad, but if we want greater accuracy, we can use a generalised continued fraction based on this approximation.

In general we have:

#sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+b/(2a+...))))#

Putting #a=188/5# and #b=6/25# we get:

#sqrt(1414) = 188/5 + (6/25)/(376/5+(6/25)/(376/5+(6/25)/(376/5+...)))#

You can terminate this continued fraction to get rational approximations, such as:

#sqrt(1414) ~~ 188/5 + (6/25)/(376/5+(6/25)/(376/5)) = 13291036/353455 ~~ 37.603191353921#

A calculator tells me that #sqrt(1414)# is a little closer to:

#37.60319135392633134161#