How do you find the square root of 74889?
3 Answers
The simplest form of the square root is
We can find approximations such as:
#sqrt(74889) ~~ 273.659#
Explanation:
Given:
#74889#
Notice that the sum of the digits is divisible by
#7+4+8+8+9 = 36 = 4*9#
So
#74889 = 9 * 8321#
#8321=53*157#
In fact, to check for square factors we only needed to look for prime factors up to
#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
Given an approximation
#(a^2+n)/(2a)#
So in our case, putting
#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.
Explanation:
As
Here for
Next we bring down next two digits
Now as we still have a remainder of
We continue in similar way by bringing down
Hence
Here's another method for finding rational approximations...
Explanation:
For interest, here's another idea for finding rational approximations to
Start by noting that:
#sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+b/(2a+...))))#
Given that
#sqrt(74889) = 274-187/(548-187/(548-187/(548-187/(548-...))))#
This is related to
#(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
So we can use this sequence to get successively better approximations to
#sqrt(74889) ~~ 164361640/300117 - 274 ~~ 273.6585465#