While finding root of a square number in dividing method why we make double of the first root number and why we take the numbers in pair?

1 Answer
Mar 10, 2017

Please see below

Explanation:

Let a number be #kpqrstm#. Observe that square of a single digit number can have up to two digits, square of a two digit number can have up to four digits, square of a three digit number can have up to six digits and square of a four digit number can have up to eight digits. You may have already got a hint now that why we take the numbers in pairs.

As the number has seven digits, so square root will have four digits. And making them in pairs we get #ulk" "ul(pq)" "ul(rs)" "ul(tm)# and as#k# is single digit, square root could start from #3,2# or #1#.

The numerical value of number is

#kxx1000000+pxx100000+qxx10000+rxx1000+sxx100+txx10+m#

we an also write it the following way, which we say (A)

#kxx1000000+(10p+q)xx10000+(10r+s)xx100+(10t+m)#

Let us consider a two digit number #abc# and let its square root be #fg#. Actually numerical value of these numbers is #100a+10b+c# and #10f+g# and hence we must have

#100a+10b+c=(10f+g)^2=100f^2+20fg+g^2#

or #100a+10b+c=100f^2+ul(2(10f+g))g#

Hence, in division method we first search for some #f#, whose square is equal or just less than #a#. Naturally #f# comes in the place for quotient and remainder would be #(a-f^2)#, with place value #100(a-f^2)#.

For next digit, we choose divisor as double of #f# (note that its place value is #10f# and choose a #g#, which makes it #10f+g#.

I hope this makes this clear. Would have gone for a bigger number like #kpqrstm#, but things get too complicated.