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Last updated on April 9th, 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 8640.
The square root is the inverse of the square of the number. 8640 is not a perfect square. The square root of 8640 is expressed in both radical and exponential form. In the radical form, it is expressed as √8640, whereas 8640^(1/2) in the exponential form. √8640 ≈ 92.878, 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 useful for perfect square numbers. However, for non-perfect square numbers like 8640, the long division method and approximation method are more effective. 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 8640 is broken down into its prime factors:
Step 1: Finding the prime factors of 8640 Breaking it down, we get 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 5, or 2^6 x 3^2 x 5^1
Step 2: Now we have found the prime factors of 8640. The second step is to make pairs of those prime factors. Since 8640 is not a perfect square, the digits of the number can’t be grouped in complete pairs.
Therefore, calculating √8640 using prime factorization involves taking the square root of these factors: √(2^6 x 3^2 x 5) = 2^3 x 3 x √5 = 24√5, approximately 92.878.
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: Begin by grouping the numbers from right to left. For 8640, we group it as 86 and 40.
Step 2: Now find n whose square is closest to 86. We can say n is 9 because 9² = 81 which is less than 86. Now the quotient is 9, and after subtracting 81 from 86, the remainder is 5.
Step 3: Bring down 40 to make the new dividend 540. Add the old divisor with the same number 9 + 9 to get 18, which will be our new divisor.
Step 4: The next step is finding 18n × n ≤ 540. Let us consider n as 3, now 183 × 3 = 549, which is too large. Thus, n should be 2 because 182 × 2 = 364.
Step 5: Subtract 364 from 540, the difference is 176, and the quotient is 92.
Step 6: 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 17600.
Step 7: Continue the process until you get the desired precision.
The result is approximately √8640 = 92.878.
The approximation method is another way to find 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 8640 using the approximation method.
Step 1: Find the closest perfect square to 8640.
The smallest perfect square less than 8640 is 8281 (91²), and the largest perfect square greater than 8640 is 8836 (94²).
√8640 falls between 91 and 94.
Step 2: Now we need to apply the interpolation formula:
(Given number - smaller perfect square) / (Larger perfect square - smaller perfect square)
Applying the formula, (8640 - 8281) / (8836 - 8281) = 0.678.
Using the formula, we identified the decimal point of our square root.
Adding this to the lower bound, 91 + 0.678 = 91.678, but the approximation by long division gives us approximately 92.878, indicating a more accurate decimal adjustment.
Can you help Max find the area of a square box if its side length is given as √8640?
A square-shaped building measuring 8640 square feet is built; if each of the sides is √8640, what will be the square feet of half of the building?
Calculate √8640 × 5.
What will be the square root of (8640 + 360)?
Find the perimeter of the rectangle if its length ‘l’ is √8640 units and the width ‘w’ is 40 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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