What is the square root of #48569834# ?

Question

Hazel Anglin · Accepted Answer

The square root is irrational.

#6969^2 = 48566961# and #6970^2 = 48580900#. Since 48,569,384 lies between two consecutive perfect squares, its square root is irrational. If you need an approximation for it every computer comes with a calculator that will provide one. Alternatively, use Google. Type sqrt(48569384), and Google's calculator will approximate the answer. [Every positive number has two square roots, of course; the square root of principle is positive, and the square root of the other is negative.]

Isaiah Anthony · Answer

#sqrt(48569384) = 6969+2423/(13938+2423/(13938+2423/(13938+...)))#

To attempt to find the square root of #48569384#, we can first split off pairs of digits starting from the right to get:

#48"|"56"|"93"|"84#

Note that #48 < 49 = 7^2#

Hence the square root is a little under #7000#

To obtain a more accurate approximation, we can try utilizing a technique based on the Babylonian method.

Starting with #p_0/q_0 = 7000/1#, to find better approximations for #sqrt(n)# where #n=48569384#, apply the following formulae:

#{ (p_(i+1) = p_i^2+nq_i^2), (q_(i+1) = 2p_i q_i) :}#

Thus:

#{ (p_1 = p_0^2+n q_0^2 = 7000^2+48569384*1^2 = 97569384), (q_1 = 2p_0 q_0 = 2*7000*1 = 14000) :}#

After just one step, let's see where we are at with this:

#p_1/q_1 = 97569384/14000 ~~ 6969.2#

Try this:

#6969^2 = 48566961 < 48569384#

#6970^2 = 48580900 > 48569384#

So the square root is somewhere strictly between #6969# and #6970#. It cannot be rational, so it's an irrational number.

Keep in mind that generally speaking:

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

where #0 < a < sqrt(n)# and #b = n-a^2#

Right now:

#48569384-6969^2 = 2423#

additionally:

#2*6969 = 13938#

Thus:

#sqrt(48569384) = 6969+2423/(13938+2423/(13938+2423/(13938+...)))#

Eli Assouline · Answer

undefined The square root of 48569834 is approximately 6973.