How do you find the square root of 43?

2 Answers
Feb 14, 2017

Answer:

See below

Explanation:

If you are looking at approximation methods that you can employ using pen, paper and some mental arithmetic, you can try a Binomial Expansion.

If you start at 36, a square number, so that you are looking for #sqrt(36 +7)#, you can now play with that a little:

#sqrt(36 +7) = sqrt 36 sqrt(1 +7/36) = 6 sqrt(1 +7/36)#

You can then use the Binomial Expansion , ie:

#(1+x)^alpha = 1 + alpha x + (alpha (alpha - 1))/(2!)x^2 + ...#:

In this case:

#6 (1 +7/36)^(1/2)#

#=6 (color(green)(1 +1/2 * 7/36) + 1/2 (-1/2)(1/(2!)) (7/36)^2 + ...)#

Even just the first two terms give #79/12 approx 6.583# and #6.583^2 approx 43.34#

We could get a little closer by using a different square number. If you start at 49, another square number, you are now looking at:

#sqrt(49 -6) = sqrt 49 sqrt(1 - 6/49) = 7 sqrt(1 - 6/49)#

Using just the first 2 terms of the Binomial Expansion:

#= 7 (1 - 1/2 * 6/49 ) = 46/7 approx 6.571#

And #6.571^2 = 43.18#

Feb 17, 2017

Answer:

#sqrt(43) = 13/2+(3/4)/(13+(3/4)/(13+(3/4)/(13+...)))#

Explanation:

#43# is a prime number, so its square root is irrational.

We can find approximations to it as follows...

Note that #43# is roughly half way between #36=6^2# and #49=7^2#

So a good first approximation for #sqrt(43)# would be #13/2#.

We find:

#(13/2)^2 = 169/4 = 42.25#

Given #n > 0# and #0 < a < sqrt(n)# we can write #sqrt(n)# as a generalised continued fraction:

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

where #b = n-a^2#

So in our example:

#n = 43#, #a = 13/2# and #b=43-169/4 = 3/4#

So:

#sqrt(43) = 13/2+(3/4)/(13+(3/4)/(13+(3/4)/(13+...)))#

We can truncate this to get rational approximations.

For example:

#sqrt(43) ~~ 13/2+(3/4)/13 = 341/52 ~~ 6.5577#

#sqrt(43) ~~ 13/2+(3/4)/(13+(3/4)/13) = 8905/1358 ~~ 6.557437#

A calculator tells me:

#sqrt(43) ~~ 6.5574385243#

See https://socratic.org/s/aCh3Xasm for another example and explanation of this method.