What is the square root of 26?

2 Answers
Sep 6, 2015

Answer:

You can only have an approximation: 5.09901951...

Explanation:

The square root of a number #x# is a number #y# such that #y^2=x#. So, we're looking for a number #y# such that #y^2=26#. Since #5^2=25# and #6^2=36#, the square root of #26# is between #5# and #6#. Since there is no algorithm to compute it exactly, you can only have an approximation. A possible way is the following: we know that #sqrt(26)# is between #5# and #6#. So, since #5^2=25# and #5.1^2=26.01#, #sqrt(26)# must be between #5# and #5.1#.

Iterating this process gives you all the decimal digits you need.

Sep 6, 2015

Answer:

#sqrt(26)# does not simplify, but you can calculate an approximation efficiently using Newton Raphson method as:

#sqrt(26) ~~ 54100801 / 10610040 ~~ 5.099019513592786#

Explanation:

#26 = 2 * 13# has no square factors, so #sqrt(26)# cannot be simplified.

If you want to calculate an approximation by hand, then I would recommend a form of Newton Raphson method, starting with first approximation #a_0 = 5#.

To iterate you can use the formula:

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

where #n = 26# is the number you are approximating the square root of.

Personally, I like to deal with these approximations as rational approximations in the form #p_i/q_i = a_i# where #p_i# and #q_i# are integers as follows:

#n = 26#
#p_0 = 5#
#q_0 = 1#

Iterate using:

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

So:

#p_1 = 5^2 + 26*1^2 = 25+26 = 51#
#q_1 = 2*5*1 = 10#

#p_2 = 51^2 + 26*10^2 = 2601 + 2600 = 5201#
#q_2 = 2*51*10 = 1020#

#p_3 = 5201^2 + 26*1020^2 = 27050401 + 27050400 = 54100801#
#q_3 = 2*5201*1020 = 10610040#

Stop when you think you have enough significant digits (typically about the number of significant digits of #p_i# + the number of significant digits of #q_i#).

#sqrt(26) ~~ 54100801 / 10610040 ~~ 5.099019513592786#

Actually #sqrt(26) ~~ 5.099019513592785#