How do you find the square root of 74889?

3 Answers
Aug 21, 2017

Answer:

The simplest form of the square root is #3sqrt(8321)#

We can find approximations such as:

#sqrt(74889) ~~ 273.659#

Explanation:

Given:

#74889#

Notice that the sum of the digits is divisible by #9#. That is:

#7+4+8+8+9 = 36 = 4*9#

So #74889# is divisible by #9# (#=3^2#) too:

#74889 = 9 * 8321#

#8321# is less easy to factorise, but eventually you might find:

#8321=53*157#

In fact, to check for square factors we only needed to look for prime factors up to #19#, since #20# and #21# are composite and:

#21^3 = 9261 > 8321#

Since there are no more square factors, the simplest form of the square root is given by:

#sqrt(74889) = sqrt(3^2*8321) = sqrt(3^2)*sqrt(8321) = 3sqrt(8321)#

This is an irrational number, not expressible as a fraction, but we can find rational approximations:

Given:

#74889#

First split into pairs of digits from the right:

#7"|"48"|"89"#

Note that:

#2^2 = 4 < 7 < 9 = 3^2#

Hence:

#2 < sqrt(7) < 3#

and:

#200 < sqrt(74889) < 300#

For a better estimate, if we know a few more square roots we can include the next two digits and note that:

#27^2 = 729 < 748 < 784 = 28^2#

Hence:

#270 < sqrt(74889) < 280#

We can linearly interpolate between these limits to find:

#sqrt(74889) ~~ 270+10*(74889 - 72900)/(78400-72900) = 270+10*1989/5500 ~~ 273.6#

Let us choose #274# as an approximation.

Given an approximation #a# for the square root of a number #n#, a better approximation is given by:

#(a^2+n)/(2a)#

So in our case, putting #n=74889# and #a=274# we find:

#sqrt(74889) ~~ (274^2+74889)/(2*274) = (75076+74889)/548 = 149965/548 ~~ 273.659#

If we want more accuracy, then repeat with this new approximation. Each iteration will roughly double the number of significant digits which are correct.

Aug 21, 2017

Answer:

#sqrt74889~~273.658#

Explanation:

As #74889# is not a perfect square, to find the square root of #74889#, we should do a special long division, where we pair, the numbers in two, starting from decimal point in either direction. When we group them we do so starting from #89#, then #48# and then #7#.

Here for #7#, the number whose square is just less than it, is #2#, whose square is #4# and so we write #4# below #7#. The difference is #3# and now we bring down next two digits #48#. As a divisor we first write double of #2# i.e. #4# and then find a number #x# so that #4x# (here #x# stands for single digit in units place) multiplied by #x# is just less than the number, here #348#. We find for #x=7#, we have #47xx7=329# and get the difference as #19#.

Next we bring down next two digits #89# and we have #1989#. Recall we had brought as divisor #2xx2=4#, but this time we have #27# double of which is #54#, so we make the divisor as #54x#, selecting an #x# so that #54x# multiplied with #x#, just goes in #1989# and this is #3#, for which we get #543xx3=1629# and have a remainder #360#.

Now as we still have a remainder of #360#, for more accuracy, we bring #00# after decimal point. Also put decimal after #273#.

We continue in similar way by bringing down #00#, which makes it #36000# and continue as shown till we have desired accuracy. You can see the details below, where we have gone up to three decimal places.

#color(white)(xX)2color(white)(xx)7color(white)(xx)3color(white)(xx).6color(white)(xx)5color(white)(xx)8#
#ul2|bar(7)color(white)(.)bar(48)color(white)(.)bar(89).color(white)(.)bar(00)color(white)(.)bar(00)color(white)(.)bar(00)#
#color(white)(XX)ul(4)color(white)(.)darr#
#color(red)(4)7|3color(white)(X)48#
#color(white)(xxx)ul(3color(white)(..)29)#
#color(white)(x)color(red)(54)3|color(white)(.)19color(white)(x)89#
#color(white)(xxx.xx)ul(16color(white)(.)29)#
#color(white)(x)color(red)(546)6|color(white)(.)3color(white)(.)60color(white)(.)00#
#color(white)(xxxxxx)ul(3color(white)(.)27color(white)(.)96)#
#color(white)(xx)color(red)(5472)5|color(white)()32color(white)(.)04color(white)(.)00#
#color(white)(xxXxxxx)ul(27color(white)(.)36color(white)(.)25#
#color(white)(xx)color(red)(54730)8|4color(white)(.)67color(white)(.)75color(white)(.)00#
#color(white)(xxxxxxxx)ul(4color(white)(.)37color(white)(.)84color(white)(.)64#
#color(white)(xxxxxxxxx.)29color(white)(.)90color(white)(.)36#

Hence #sqrt74889~~273.658#

Aug 21, 2017

Answer:

Here's another method for finding rational approximations...

Explanation:

For interest, here's another idea for finding rational approximations to #sqrt(74889)#.

Start by noting that:

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

Given that #274# is the closest integer approximation, with #274^2 = 75076 = 74889 + 187#, we can write a generalised continued fraction for the square root, putting #a=274# and #b=-187# to get:

#sqrt(74889) = 274-187/(548-187/(548-187/(548-187/(548-...))))#

This is related to #274+sqrt(74889)# being one of the zeros of:

#(x-274-sqrt(74889))(x-274+sqrt(74889)) = x^2-548x+187#

Now consider a sequence defined recursively as follows:

#{ (a_0 = 0), (a_1 = 1), (a_(n+2) = 548a_(n+1)-187a_n" for "n >= 0):}#

The first few terms are:

#0, 1, 548, 300117, 164361640, 90014056841, 49296967522188#

Because of the way it is constructed, the ratio between pairs of successive terms tends to #274+sqrt(74889)#. In fact each successive ratio corresponds to truncating the continued fraction after one more term.

So we can use this sequence to get successively better approximations to #sqrt(74889)#, such as:

#sqrt(74889) ~~ 164361640/300117 - 274 ~~ 273.6585465#