How do you find the square root of 4783?

2 Answers
Jan 31, 2017

Answer:

#sqrt4783~~69.159#

Explanation:

As #4783# is not a perfect square, to find the square root of #4783#, we should do a special long division, where we pair, the numbers in two, starting from decimal point in either direction.

Here first pair is #47# and the number whose square is just less than it is #6#, so we write #36# below #47#. The difference is #11# and now we bring down next two digits #83#. As a divisor we first write double of #6# i.e. #12# and then find a number #x# so that #12x# (here #x# stands for single digit in units place) multiplied by #x# is just less than the number, here #1183#. We find for #x=9#, we have #129xx9=1161# and get the difference as #22#.

Now as we still have a remainder of #22#, we bring #00# after decimal point. Also put decimal after #69# and this makes it #6900#.

Now recall we had brought as divisor #6xx2=12#, but this time we have #69# so we make the divisor as #138x# and identify an #x# so that #138x# multiplied by #x# is just less than #2200#. This number is just #1#, as making it #2# will make the product #1382xx2=2764>2200#.

We continue to do this till we have desired accuracy.

#color(white)(xx)6color(white)(xx)9color(white)(xx).1color(white)(xx)5color(white)(xx)9#
#ul6|bar(47)color(white)(.)bar(83).color(white)(.)bar(00)color(white)(.)bar(00)color(white)(.)bar(00)color(white)(.)bar(00)#
#color(white)(X)ul(36)color(white)(X)darr#
#color(red)(12)9|color(white)(X)11color(white)(.)83#
#color(white)(xxxxx)ul(11color(white)(.)61)#
#color(white)(x)color(red)(138)1|color(white)(X)22color(white)(.)00#
#color(white)(xxxxxxx)ul(13color(white)(.)81)#
#color(white)(xx)color(red)(1382)5|color(white)(.)8color(white)(.)19color(white)(.)00#
#color(white)(xxxxxxxx)ul(6color(white)(.)91color(white)(.)25)#
#color(white)(xx)color(red)(13830)9|color(white)()1color(white)(.)27color(white)(.)75color(white)(.)00#
#color(white)(xxxxxxxx)ul(1color(white)(.)24color(white)(.)47color(white)(.)81)#
#color(white)(xxxxxxxxx)color(white)(.)32color(white)(.)71color(white)(.)9#

Hence #sqrt4783~~69.159#

Feb 5, 2017

Answer:

#sqrt(4783) = 69+22/(138+22/(138+22/(138+...)))#

Explanation:

Here's a method using generalised continued fractions.

First the theory...

Suppose we can find numbers #a# and #b# such that:

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

Then we have:

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

Multiplying both ends by #(a+sqrt(n))# we get:

#a^2+color(red)(cancel(color(black)(asqrt(n))))+b = color(red)(cancel(color(black)(asqrt(n))))+n#

Hence:

#b = n-a^2#

In our example, first note that #4783 < 4900 = 70^2#, so let's see what we get when we square #69#:

#69^2 = (70-1)^2 = 70^2-2*70+1 = 4900-140+1 = 4761#

So putting #n=4783# and #a=69# we get:

#b = n-a^2 = 4783-4761 = 22#

So:

#sqrt(4783) = 69+22/(138+22/(138+22/(138+...)))#

We can truncate after any number of terms to get a rational approximation for #sqrt(4783)#:

#sqrt(4783) ~~ 69#

#sqrt(4783) ~~ 69+22/138 = 4772/69 ~~ 69.1594#

#sqrt(4783) ~~ 69+22/(138+22/138) = 69+22/(9533/69) = 659295/9533 ~~ 69.1592363#

A calculator tells me:

#sqrt(4873) ~~ 69.1592365487#