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Last updated on April 28th, 2025
If a number is multiplied by the same number, the result is a square. The inverse of the square is a square root. The square root is used in the field of vehicle design, finance, etc. Here, we will discuss the square root of 153.
The square root is the inverse of the square of the number. 153 is not a perfect square. The square root of 153 is expressed in both radical and exponential forms. In the radical form, it is expressed as √153, whereas (153)¹/² in the exponential form. √153 ≈ 12.3693, which is an irrational number because it cannot be expressed in the form of p/q, where p and q are integers and q ≠ 0.
The prime factorization method is used for perfect square numbers. However, the prime factorization method is not used for non-perfect square numbers where the long-division method and approximation method are used. Let us now learn the following methods:
The product of prime factors is the prime factorization of a number. Now let us look at how 153 is broken down into its prime factors.
Step 1: Finding the prime factors of 153 Breaking it down, we get 3 x 3 x 17.
Step 2: Now we found out the prime factors of 153. The second step is to make pairs of those prime factors. Since 153 is not a perfect square, the digits of the number can’t be grouped in pairs.
Therefore, calculating 153 using prime factorization is impossible.
The long division method is particularly used for non-perfect square numbers. In this method, we should check the closest perfect square number for the given number. Let us now learn how to find the square root using the long division method, step by step.
Step 1: To begin with, we need to group the numbers from right to left. In the case of 153, we need to group it as 53 and 1.
Step 2: Now we need to find n whose square is 1. We can say n as ‘1’ because 1 x 1 is lesser than or equal to 1. Now the quotient is 1, after subtracting 1-1, the remainder is 0.
Step 3: Now let us bring down 53, which is the new dividend. Add the old divisor with the same number 1 + 1, we get 2, which will be our new divisor.
Step 4: The new divisor will be the sum of the dividend and quotient. Now we get 2n as the new divisor, we need to find the value of n.
Step 5: The next step is finding 2n × n ≤ 53. Let us consider n as 2, now 2 x 2 x 2 = 44.
Step 6: Subtract 53 from 44, the difference is 9, and the quotient is 12.
Step 7: Since the dividend is less than the divisor, we need to add a decimal point. Adding the decimal point allows us to add two zeroes to the dividend. Now the new dividend is 900.
Step 8: Now we need to find the new divisor that is 24, because 248 x 3 = 744.
Step 9: Subtracting 744 from 900, we get the result 156. Step 10: Now the quotient is 12.3.
Step 11: Continue doing these steps until we get two numbers after the decimal point. Suppose if there is no decimal value, continue till the remainder is zero.
So the square root of √153 is approximately 12.37.
The approximation method is another method for finding the square roots. It is an easy method to find the square root of a given number. Now let us learn how to find the square root of 153 using the approximation method.
Step 1: Now we have to find the closest perfect square of √153. The smallest perfect square less than 153 is 144, and the largest perfect square greater than 153 is 169. √153 falls somewhere between 12 and 13.
Step 2: Now we need to apply the formula: (Given number - smallest perfect square) ÷ (Greater perfect square - smallest perfect square). Going by the formula (153 - 144) ÷ (169 - 144) = 0.36. Using the formula, we identified the decimal point of our square root. The next step is adding the value we got initially to the decimal number, which is 12 + 0.36 = 12.36.
Therefore, the square root of 153 is approximately 12.36.
Can you help Max find the area of a square box if its side length is given as √153?
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Calculate √153 x 5.
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Find the perimeter of the rectangle if its length ‘l’ is √153 units and the width ‘w’ is 30 units.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.