Question

What is the square root of 133?

Answer 1

Hint: First we will check if we can use the prime factorization method to find the square root or not. If we will be able to write all the prime factors as exponent equal to 2 then we will be able to find the root easily. If this method fails then we will apply the long division method to get our answer in decimal up to two places.

Complete step by step solution:
Here we have been asked to find the square root of 133.
Now, 133 is a prime number so we will not be able to write the factors with their exponent equal to 2. Therefore, we need to apply the long division process to find the square root. So, let us see each step while finding the root.
Step (1): Starting from the rightmost digit we will form pairs of two digits till all the digits are paired or a single digit is left at the leftmost place, here we have three digits so 33 will be paired and 1 will be left.
Step (2): We will select a number whose square will be less than or equal to 1. Here we have to select 1. Now, we will write 1 as the divisor and the quotient and subtract the product from 1. The remainder will be 0. Now, we will write the initially formed pair 33 beside this remainder.
\[\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,1 \\
 & \begin{matrix}
   \,\,\,\,\,\,\,1 \\
   \,\,\,+1 \\
\end{matrix}\left| \!{\overline {\,
 \begin{align}
  & \,\,\,\,\,\,\,1\overline{33} \\
 & \,\,\,-1 \\
 & \overline{\,\,\,\,\,\,\,0\,33} \\
\end{align} \,}} \right. \\
\end{align}\]
Step (3): The new dividend thus becomes 33, at the divisor place we will get the sum 1 + 1 = 2. Now, we need to select a digit after 2 such that when we will take the product of this selected digit with the new divisor we will get a number less than 33. So here we will select 1. The new divisor will become 21 + 1 = 22, the remainder will be 12 and the new quotient will become 11.
\[\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11 \\
 & \begin{matrix}
   \,\,\,\,\,\,\,\,1 \\
   \,\,\,\,+1 \\
   \overline{\,\,\,\,\,\,\,\,\,21} \\
   \,\,\,\,\,\,\,+1 \\
   \overline{\,\,\,\,\,\,\,\,22} \\
\end{matrix}\left| \!{\overline {\,
 \begin{align}
  & \,\,\,\,\,\,\,1\overline{33} \\
 & \,\,\,-1 \\
 & \overline{\,\,\,\,\,\,\,\,\,033} \\
 & \,\,\,\,\,\,\,\,-21 \\
 & \overline{\,\,\,\,\,\,\,\,\,\,\,\,12} \\
\end{align} \,}} \right. \\
\end{align}\]
Step (4): Now, there are no more paired digits to be placed besides the remainder 12 so we will take the decimal point in the quotient and place two 0’s after 12. The dividend thus becomes 1200. Further we need to select a digit after 22 such that the product of this digit with the new divisor will be less than 1200, so we will select 5.
\[\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11.5 \\
 & \begin{matrix}
   \,\,\,\,\,\,\,1 \\
   \,\,\,+1 \\
   \overline{\,\,\,\,\,\,\,21} \\
   \,\,\,\,\,\,+1 \\
   \overline{\,\,\,\,\,\,\,\,\,225} \\
   \,\,\,\,\,\,\,\,\,\,+5 \\
   \overline{\,\,\,\,\,\,\,\,230} \\
\end{matrix}\left| \!{\overline {\,
 \begin{align}
  & \,\,\,\,\,\,\,133 \\
 & \,\,\,-1 \\
 & \overline{\,\,\,\,\,\,\,\,\,\,033} \\
 & \,\,\,\,\,\,\,\,\,-21 \\
 & \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,1200} \\
 & \,\,\,\,\,\,\,\,\,-1125 \\
 & \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,75} \\
\end{align} \,}} \right. \\
\end{align}\]
Step (5): Now, we will get the divisor 225 + 5 = 230 and the remainder will be 75. The quotient is now 11.5. Again two 0’s will be placed after the remainder so that it becomes 7500 due to the decimal point. Now, we need to select 3 beside the divisor so that the product of 2303 and 3 will be less than 7500. Therefore, the quotient will become 11.53.
\[\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11.53 \\
 & \begin{matrix}
   \,\,\,\,\,\,1 \\
   \,\,+1 \\
   \overline{\,\,\,\,\,\,\,21} \\
   \,\,\,\,\,\,+1 \\
   \overline{\,\,\,\,\,\,\,\,\,225} \\
   \,\,\,\,\,\,\,\,\,+5 \\
   \,\,\overline{\,\,\,\,\,\,\,2303} \\
   \,\,\,\,\,\,\,\,\,\,\,\,+3 \\
   \,\,\,\overline{\,\,\,\,\,2306} \\
\end{matrix}\left| \!{\overline {\,
 \begin{align}
  & \,\,\,\,\,\,\,1\overline{33} \\
 & \,\,\,-1 \\
 & \overline{\,\,\,\,\,\,033} \\
 & \,\,\,\,\,-21 \\
 & \overline{\,\,\,\,\,\,\,\,\,1200} \\
 & \,\,\,\,\,-1125 \\
 & \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,7500} \\
 & \,\,\,\,\,\,\,\,\,\,-6909 \\
 & \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,591} \\
\end{align} \,}} \right. \\
\end{align}\]
Hence, the square root of 133 is 11.53 correct up to two places of decimal.

Note: You must remember the long division method to find the square root because this method is very useful in determining the square roots of numbers which are not perfect squares. In case we have only two digits whose square root is to be found then in such case both the digits will be paired in the first step and we will proceed in the similar manner as above.