# How do you find the square root of 74889?

##### 3 Answers

#### Answer:

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.

#### Answer:

#### Explanation:

As **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

Here for **bring down next two digits**

Next we **bring down next two digits**

Now as we still have a remainder of

We continue in similar way by bringing down

Hence

#### Answer:

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#