# How can we find square root of a number by hand?

To be able to evaluate this, recall that  with a square root of N such that `N= AB` , it follows square root of N lies in between the A and B.

A<sqrt(N)< B

This  is our clue that we can apply factoring for the value inside the square root sign.

For a radical `root(n)(x)` ,  the parts are called:

n = index

A square root has an index of 2 which has a radical sign root(2)(x) or  sqrt(x).

Suppose we have sqrt( 30).

Apply factoring on the radicand: `30 =5*6` then we know that

`5 ltsqrt(30)lt6`

To solve it numerically, note that an average of two number `(A+B)/2` will be in between A and B.

Then,  `Alt(A+B)/2 ltB` .

Average value`= (5+6)/2 =11/2` or `5.5`

Apply the average value to the factoring of the radicand such that:

Divide radicand by the average value:

`sqrt(x)/(((A+B)/2)) = 30/ ((11/2))`

`= 30*(2/11)`

`= 60/11`

Then, factoring of the radicand: `30 = 11/2* 60/11`

and it follows it square root will lie in between:

`60/11 lt sqrt(30)lt11/2`

Note:`60/11 lt11/2 `  since` 60/11~~5.455` and `11/2=5.5`

Note that the boundary values is approximately same as "5.5" then we can estimate the value of the square root:` sqrt(30)~~5.5`

For more accurate estimation, repeat the same procedure with the new set of factors of the radicand:

`30 = 11/2* 60/11`

Then,

average value`= (11/2+ 60/11)/2 = 241/44` or `5.477` rounded off.

`sqrt(x)/(((A+B)/2)) =30/((241/44)) = 1320/241`

then new factoring: `30=(241/44)*(1320/241)`

new range will be:`(1320/241)<sqrt(30)<(241/44)`

or `5.47718 ltsqrt(30)lt5.477273`

Note that the boundary values is approximately same as "5.4772" then we can estimate the value of the square root:  `sqrt(30)~~5.4772`