What is the square root of 30?

2 Answers
Sep 14, 2015

You can't get an exact answer by hand, but you can estimate it using some tricks.

#sqrt(25) = 5#
#sqrt(36) = 6#

so it must be between #5# and #6#.

#30# is slightly less than halfway between #5# and #6#.

What you can begin with is assuming that because:

#30 - 25 = 5#
#36 - 25 = 11#
#5/11 = 45.45%#

... the square root of #30# is probably near #0.45 + 5 = 5.45#. What you are assuming then is:

#x + 0.45*[(x+1)-x] ~~ x^2 + 0.45*[(x+1)^2 - x^2]#

or more specifically,

#5 + 0.45*(6-5) ~~ 25 + 0.45*(36 - 25)#

Actually, it's not a bad bet. The actual square root is about #5.477#.

Sep 14, 2015

Answer:

#30 = 2*3*5# has no square factors, so it is not possible to simplify #sqrt(30)#.

You can calculate an approximation by hand as shown below...

Explanation:

I explained my favourite method (a sort of Newton Raphson method) for approximating square roots of integers in an answer to the following question:

How do you find the square root 28?

Given an integer #n#, choose integers #p_0# and #q_0# so #p_0/q_0# is a reasonable first approximation to #sqrt(n)#.

Then iterate using the formulas:

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

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

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

The successive pairs #p_i#, #q_i# provide a sequence of rational approximations #p_i/q_i# to #sqrt(n)# that converge quite rapidly.

For our example, let #n = 30#, #p_0 = 11#, #q_0 = 2# (using an initial approximation of #5.5# since #30# is about halfway between #5^2 = 25# and #6^2 = 36#).

#p_1 = p_0^2 + n q_0^2 = 11^2 + 30*2^2 = 121 + 120 = 241#

#q_1 = 2 p_0 q_0 = 2 * 11 * 2 = 44#

This would give #sqrt(30) ~~ 241/44 = 5.47dot(7)dot(2)#

#p_2 = p_1^2 + n q_1^2 = 241^2 + 30*44^2 = 58081 + 58080 = 116161#

#q_2 = 2 p_1 q_1 = 2*241*44 = 21208#

This gives #sqrt(30) ~~ 116161/21208 ~~ 5.477225575#