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Last updated on March 22nd, 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 943.
The square root is the inverse of the square of the number. 943 is not a perfect square. The square root of 943 can be expressed in both radical and exponential form. In the radical form, it is expressed as √943, whereas (943)(1/2) in the exponential form. √943 ≈ 30.7083, 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 943 is broken down into its prime factors:
Step 1: Finding the prime factors of 943 Breaking it down, we get 23 and 41: 943 = 23 × 41
Step 2: We found out the prime factors of 943. Since 943 is not a perfect square, the digits of the number can’t be grouped in pairs.
Therefore, calculating 943 using prime factorization is not feasible for finding its square root directly.
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 943, we need to group it as 43 and 9.
Step 2: Now we need to find n whose square is ≤ 9. We can say n as 3 because 3 × 3 = 9. Now the quotient is 3 after subtracting 9 - 9 the remainder is 0.
Step 3: Now let us bring down 43, which is the new dividend. Add the old divisor with the same number 3 + 3, we get 6, which will be our new divisor.
Step 4: The new divisor will be the sum of the dividend and quotient. Now we get 6n as the new divisor, we need to find the value of n.
Step 5: The next step is finding 6n × n ≤ 43. Let us consider n as 7, now 6 × 7 = 42, and 42 × 7 = 294, which is greater than 43. So we take n as 6.
Step 6: Subtract 36 from 43, the difference is 7, and the quotient is 36.
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 700.
Step 8: Now we need to find the new divisor. The new divisor is 66 because 66 × 10 = 660, which is less than 700.
Step 9: Subtracting 660 from 700 we get the result 40.
Step 10: Continue doing these steps until we get a suitable approximation or two numbers after the decimal point.
So the square root of √943 is approximately 30.70.
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 943 using the approximation method.
Step 1: Now we have to find the closest perfect square of √943. The smallest perfect square less than 943 is 900 and the largest perfect square greater than 943 is 961. √943 falls somewhere between 30 and 31.
Step 2: Now we need to apply the formula: (Given number - smallest perfect square) / (Greater perfect square - smallest perfect square).
Going by the formula (943 - 900) ÷ (961 - 900) = 43 ÷ 61 ≈ 0.70. Using the formula, we identified the decimal point of our square root. The next step is to add the value we got initially to the decimal number, which is 30 + 0.70 = 30.70.
So the square root of 943 is approximately 30.70.
Can you help Max find the area of a square box if its side length is given as √943?
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Calculate √943 × 4.
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Find the perimeter of the rectangle if its length ‘l’ is √943 units and the width ‘w’ is 43 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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