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2026-01-01
Modified
2026-02-28
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<p>The<a>long division</a>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<a>square root</a>using the long division method, step by step:</p>
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<p>The<a>long division</a>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<a>square root</a>using the long division method, step by step:</p>
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<p><strong>Step 1:</strong>To begin with, we need to group the numbers from right to left. In the case of 253, we need to group it as 53 and 2.</p>
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<p><strong>Step 1:</strong>To begin with, we need to group the numbers from right to left. In the case of 253, we need to group it as 53 and 2.</p>
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<p><strong>Step 2:</strong>Now we need to find n whose square is<a>less than</a>or equal to 2. We can say n as ‘1’ because 1 x 1 is lesser than or equal to 2. Now the<a>quotient</a>is 1; after subtracting 1 from 2, the<a>remainder</a>is 1.</p>
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<p><strong>Step 2:</strong>Now we need to find n whose square is<a>less than</a>or equal to 2. We can say n as ‘1’ because 1 x 1 is lesser than or equal to 2. Now the<a>quotient</a>is 1; after subtracting 1 from 2, the<a>remainder</a>is 1.</p>
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<p><strong>Step 3:</strong>Now let us bring down 53, which is the new<a>dividend</a>. Add the old<a>divisor</a>with the same number 1 + 1 to get 2, which will be our new divisor.</p>
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<p><strong>Step 3:</strong>Now let us bring down 53, which is the new<a>dividend</a>. Add the old<a>divisor</a>with the same number 1 + 1 to get 2, which will be our new divisor.</p>
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<p><strong>Step 4:</strong>The new divisor will be the sum of the dividend and quotient. Now we get 2n as the new divisor, and we need to find the value of n.</p>
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<p><strong>Step 4:</strong>The new divisor will be the sum of the dividend and quotient. Now we get 2n as the new divisor, and we need to find the value of n.</p>
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<p><strong>Step 5:</strong>The next step is finding 2n × n ≤ 153. Let us consider n as 7, now 27 x 7 = 189, which is too large, so n is 5.</p>
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<p><strong>Step 5:</strong>The next step is finding 2n × n ≤ 153. Let us consider n as 7, now 27 x 7 = 189, which is too large, so n is 5.</p>
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<p><strong>Step 6:</strong>Subtract 125 (25 x 5) from 153; the difference is 28, and the quotient is 15.</p>
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<p><strong>Step 6:</strong>Subtract 125 (25 x 5) from 153; the difference is 28, and the quotient is 15.</p>
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<p><strong>Step 7:</strong>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 2800.</p>
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<p><strong>Step 7:</strong>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 2800.</p>
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<p><strong>Step 8:</strong>Now we need to find the new divisor that is 310 because 310 x 9 = 2790. Step 9: Subtracting 2790 from 2800, we get the result 10.</p>
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<p><strong>Step 8:</strong>Now we need to find the new divisor that is 310 because 310 x 9 = 2790. Step 9: Subtracting 2790 from 2800, we get the result 10.</p>
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<p><strong>Step 10:</strong>Now the quotient is 15.9</p>
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<p><strong>Step 10:</strong>Now the quotient is 15.9</p>
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<p><strong>Step 11:</strong>Continue doing these steps until we get two numbers after the decimal point. Suppose if there are no decimal values, continue till the remainder is zero So the square root of √253 is approximately 15.90.</p>
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<p><strong>Step 11:</strong>Continue doing these steps until we get two numbers after the decimal point. Suppose if there are no decimal values, continue till the remainder is zero So the square root of √253 is approximately 15.90.</p>
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