Question

Find the square root of the following numbers by long division method:
(a) 531441, (b) 4401604.

Answer 1

Hint: We start solving the problem by recalling the procedure and steps that should be followed in order to get square root through long division method. We first follow those steps one by one from dividing into periods up to getting the absolute value of the square root of 531441. We then follow those steps one by one from dividing into periods up to getting the absolute value of the square root of 4401604.

Complete step by step answer:
According to the problem, we need to find the square root of 531441, 4401604 by using a long division method.
Before finding the square root, let us recall the procedure of the long division method. We need to follow the steps as shown below in order to find a square root.
(i) We first start with pairing the digits present in the number starting from the unit place. Each pair (single digits if) obtained is called a period.
(ii) Now we need to divide the first period with the number whose square is equal or less than the period value. Here divisor and quotient will be the same number.
(iii) Now, we subtract the first period with the value we just obtained and the digits in the next period will bring down to the next of remainder to get our new dividend.
(iv) We now multiply our quotient with 2 to get the first digits of new divisor and we annex with a digit next to the digits that already present in divisor in such a way that the product of divisor with the digit we just annexed is equal or less than the new dividend.
(v) Now, we repeat steps (ii), (iii), (iv) till all the periods complete and we get a zero remainder, then the quotient will be required to be a square root.

Now, we follow these steps to find the square root of 531441 and 4401604.
Let us first find the square root of 531441. We first pair the digits to form periods.
$$|\underset{―}{\overline{53}\overline{14}\overline{41}}$$. So, we can see that we have got three periods and the first period is 53. We know that the square of the number 7 is 49 which is less than 53. So, the divisor we need to start is 7.
$$\begin{array}{rl}& 7|\underset{―}{\begin{array}{rl}& \overline{53}\overline{14}\overline{41}(7\\ & 49\end{array}}\\ & |\underset{―}{4}\end{array}$$.
We have got the remainder 4 we bring down the period 14 to get our new dividend. So, we get 414 as a new dividend.
$$\begin{array}{rl}& 7|\underset{―}{\begin{array}{rl}& \overline{53}\overline{14}\overline{41}(7\\ & 49\end{array}}\\ & |\underset{―}{414}\end{array}$$.
We get the first digits of our new divisor by multiplying the quotient with 2. So, the first digits of our new divisor are $(7\times 2)=14$.
$$\begin{array}{rl}& 7|\underset{―}{\begin{array}{rl}& \overline{53}\overline{14}\overline{41}(72\\ & 49\end{array}}\\ & 142|\underset{―}{\begin{array}{rl}& 414\\ & 284\end{array}}\\ & |\underset{―}{130}\end{array}$$.
We have got the remainder 130 we bring down the period 41 to get our new dividend. So, we get 13041 as a new dividend.
$$\begin{array}{rl}& 7|\underset{―}{\begin{array}{rl}& \overline{53}\overline{14}\overline{41}(72\\ & 49\end{array}}\\ & 142|\underset{―}{\begin{array}{rl}& 414\\ & 284\end{array}}\\ & |\underset{―}{13041}\end{array}$$.
We get the first digits of our new divisor by multiplying the quotient with 2. So, first digits of our new divisor are $(7\times 2)=144$.
$$\begin{array}{rl}& 7|\underset{―}{\begin{array}{rl}& \overline{53}\overline{14}\overline{41}(729\\ & 49\end{array}}\\ & 142|\underset{―}{\begin{array}{rl}& 414\\ & 284\end{array}}\\ & 1449|\underset{―}{\begin{array}{rl}& 13041\\ & \text{13041}\end{array}}\\ & |\underset{―}{0}\end{array}$$.
We can see that there is no period left and the remainder we obtained is zero. So, we have got out square root as 729.
∴ The square root of 531441 is 729.

Now, let us first find the square root of 4401604. We first pair the digits to form periods.
$$|\underset{―}{\bar{4}\overline{40}\overline{16}\overline{04}}$$. So, we can see that we have got three periods and the first period is 4. We know that the square of the number 2 is 4 which is equal to 4. So, the divisor we need to start is 2.
$$\begin{array}{rl}& 2|\underset{―}{\begin{array}{rl}& \bar{4}\overline{40}\overline{16}\overline{04}(2\\ & 4\end{array}}\\ & |\underset{―}{0}\end{array}$$.
We have got the remainder 0 we bring down the period 40 to get our new dividend. So, we get 40 as new dividend.
$$\begin{array}{rl}& 2|\underset{―}{\begin{array}{rl}& \bar{4}\overline{40}\overline{16}\overline{04}(2\\ & 4\end{array}}\\ & |\underset{―}{40}\end{array}$$.
We get the first digits of our new divisor by multiplying the quotient with 2. So, the first digits of our new divisor is $(2\times 2)=4$.
$$\begin{array}{rl}& 2|\underset{―}{\begin{array}{rl}& \bar{4}\overline{40}\overline{16}\overline{04}(20\\ & 4\end{array}}\\ & 40|\underset{―}{\begin{array}{rl}& 40\\ & 00\end{array}}\\ & |\underset{―}{40}\end{array}$$.
We have got the remainder 40 we bring down the period 16 to get our new dividend. So, we get 4016 as new dividend.
$$\begin{array}{rl}& 2|\underset{―}{\begin{array}{rl}& \bar{4}\overline{40}\overline{16}\overline{04}(20\\ & 4\end{array}}\\ & 40|\underset{―}{\begin{array}{rl}& 40\\ & 00\end{array}}\\ & |\underset{―}{4016}\end{array}$$.
We get the first digits of our new divisor by multiplying the quotient with 2. So, the first digits of our new divisor are $(20\times 2)=40$.
$$\begin{array}{rl}& 2|\underset{―}{\begin{array}{rl}& \bar{4}\overline{40}\overline{16}\overline{04}(209\\ & 4\end{array}}\\ & 40|\underset{―}{\begin{array}{rl}& 40\\ & 00\end{array}}\\ & 409|\underset{―}{\begin{array}{rl}& 4016\\ & 3681\end{array}}\\ & |\underset{―}{335}\end{array}$$.
We have got the remainder 335 we bring down the period 04 to get our new dividend. So, we get 33504 as a new dividend.
$$\begin{array}{rl}& 2|\underset{―}{\begin{array}{rl}& \bar{4}\overline{40}\overline{16}\overline{04}(209\\ & 4\end{array}}\\ & 40|\underset{―}{\begin{array}{rl}& 40\\ & 00\end{array}}\\ & 409|\underset{―}{\begin{array}{rl}& 4016\\ & 3681\end{array}}\\ & |\underset{―}{33504}\end{array}$$.
We get the first digits of our new divisor by multiplying the quotient with 2. So, the first digits of our new divisor are $(209\times 2)=418$.
$$\begin{array}{rl}& 2|\underset{―}{\begin{array}{rl}& \bar{4}\overline{40}\overline{16}\overline{04}(2098\\ & 4\end{array}}\\ & 40|\underset{―}{\begin{array}{rl}& 40\\ & 00\end{array}}\\ & 409|\underset{―}{\begin{array}{rl}& 4016\\ & 3681\end{array}}\\ & 4188|\underset{―}{\begin{array}{rl}& 33504\\ & 33504\end{array}}\\ & |\underset{―}{0}\end{array}$$.
We can see that there is no period left and the remainder we obtained is zero. So, we have got our square root as of 2098.
∴ The square root of 4401604 is 2098.

Note:
We should know that the period will contain one or two digits. We should take periods for digits that were present right to left in a given number. We should not make any calculation mistakes while solving this problem. In this problem, we have got the natural number as a square root. This may not be the case always so; we may get the decimal values while square roots for some numbers. We should keep a limit in the decimal places while calculating those limits. We can also find the approximate values of irrational numbers using this technique.