Last updated on May 26th, 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 fields such as vehicle design, finance, etc. Here, we will discuss the square root of 15129.
The square root is the inverse of the square of a number. 15129 is a perfect square. The square root of 15129 is expressed in both radical and exponential forms. In the radical form, it is expressed as √15129, whereas 15129(1/2) is the exponential form. √15129 = 123, which is a rational number because it can 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. The long division method and approximation method can also be applied. Let us now learn the following methods: -
The product of prime factors is the prime factorization of a number. Here is how 15129 is broken down into its prime factors:
Step 1: Finding the prime factors of 15129. By inspection or trial division, we find that 15129 = 123 × 123. Thus, the prime factorization is 32 × 412
Step 2: Since 15129 is a perfect square, the digits of the number can be grouped into pairs: (3 × 41) × (3 × 41).
Hence, calculating the square root of 15129 using prime factorization gives us 123.
The long division method can also be used for non-perfect square numbers, but let's apply it to 15129 for illustration:
Step 1: Start by grouping the digits from right to left. For 15129, group as 15 | 129.
Step 2: Find the largest number whose square is less than or equal to 15. This number is 3, because 3 × 3 = 9.
Step 3: Subtract 9 from 15 to get a remainder of 6. Bring down the next pair of digits, 29, to get the new dividend 629.
Step 4: Double the quotient obtained so far (3) and write it as 6_.
Step 5: Find a digit n such that 6n × n is less than or equal to 629. The digit is 3, because 63 × 3 = 189.
Step 6: Subtract 189 from 629 to get 440. Bring down the next pair of zeros to get 4400.
Step 7: Continue the process. The whole number part of the quotient is the square root. Through this method, we find that the square root of 15129 is 123.
The approximation method involves finding the closest perfect squares. Since 15129 is a perfect square, this method is straightforward:
Step 1: Identify the closest perfect squares around 15129.
Step 2: Since 15129 is a perfect square, we already know its square root is exactly 123.
Step 3: If it were not a perfect square, you would use the formula (Given number - smaller perfect square) / (larger perfect square - smaller perfect square) for a more refined approximation.
Students make mistakes while finding square roots, such as forgetting negative square roots or skipping the long division method. Here are some common mistakes students make and how to avoid them.
Can you help Max find the area of a square box if its side length is given as √15129?
The area of the square is 15129 square units.
The area of a square = side².
The side length is given as √15129.
Area of the square = side² = √15129 × √15129 = 123 × 123 = 15129.
Therefore, the area of the square box is 15129 square units.
A square-shaped building measuring 15129 square feet is built. If each of the sides is √15129, what will be the square feet of half of the building?
7564.5 square feet
To find half of the building's square feet, divide the given area by 2.
Dividing 15129 by 2 = 7564.5
So half of the building measures 7564.5 square feet.
Calculate √15129 × 5.
615
First, find the square root of 15129, which is 123.
Then multiply 123 by 5. So, 123 × 5 = 615.
What will be the square root of (15129 + 0)?
The square root is 123.
To find the square root, calculate 15129 + 0 = 15129, and then √15129 = 123.
Therefore, the square root of (15129 + 0) is ±123.
Find the perimeter of the rectangle if its length ‘l’ is √15129 units and the width ‘w’ is 38 units.
The perimeter of the rectangle is 322 units.
Perimeter of the rectangle = 2 × (length + width)
Perimeter = 2 × (√15129 + 38) = 2 × (123 + 38) = 2 × 161 = 322 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.
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