What is the square root of #48569834# ?

2 Answers
Apr 4, 2017

Answer:

The square root is irrational.

Explanation:

#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.

[Of course, every positive number has two square roots. The principle square root is positive, and the other square root is negative.]

Apr 4, 2017

Answer:

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

Explanation:

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#

We can try using a method based on the Babylonian method to get a better approximation.

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) :}#

So:

#{ (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) :}#

Let's see where this has got us after just one step:

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

Try:

#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.

Note that in general:

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

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

Now:

#48569384-6969^2 = 2423#

and:

#2*6969 = 13938#

So:

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