0 added
0 removed
Original
2026-01-01
Modified
2026-02-28
1
<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>
1
<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>
2
<p><strong>Step 1:</strong>To begin with, we need to group the numbers from right to left. In the case of 230, we need to group it as 30 and 2.</p>
2
<p><strong>Step 1:</strong>To begin with, we need to group the numbers from right to left. In the case of 230, we need to group it as 30 and 2.</p>
3
<p><strong>Step 2:</strong>Now we need to find n whose square is 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 2-1, the<a>remainder</a>is 1.</p>
3
<p><strong>Step 2:</strong>Now we need to find n whose square is 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 2-1, the<a>remainder</a>is 1.</p>
4
<p><strong>Step 3:</strong>Now let us bring down 30, which is the new<a>dividend</a>. Add the old<a>divisor</a>with the same number 1 + 1, we get 2, which will be our new divisor.</p>
4
<p><strong>Step 3:</strong>Now let us bring down 30, which is the new<a>dividend</a>. Add the old<a>divisor</a>with the same number 1 + 1, we get 2, which will be our new divisor.</p>
5
<p><strong>Step 4:</strong>The new divisor will be the<a>sum</a>of the dividend and quotient. Now we get 2n as the new divisor; we need to find the value of n.</p>
5
<p><strong>Step 4:</strong>The new divisor will be the<a>sum</a>of the dividend and quotient. Now we get 2n as the new divisor; we need to find the value of n.</p>
6
<p><strong>Step 5:</strong>The next step is finding 2n × n ≤ 130. Let us consider n as 5, now 25 x 5 = 125.</p>
6
<p><strong>Step 5:</strong>The next step is finding 2n × n ≤ 130. Let us consider n as 5, now 25 x 5 = 125.</p>
7
<p><strong>Step 6:</strong>Subtract 130 from 125, the difference is 5, and the quotient is 15.</p>
7
<p><strong>Step 6:</strong>Subtract 130 from 125, the difference is 5, and the quotient is 15.</p>
8
<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 500.</p>
8
<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 500.</p>
9
<p><strong>Step 8:</strong>Now we need to find the new divisor, which is 31, because 310 x 1 = 310.</p>
9
<p><strong>Step 8:</strong>Now we need to find the new divisor, which is 31, because 310 x 1 = 310.</p>
10
<p><strong>Step 9:</strong>Subtracting 310 from 500, we get the result 190.</p>
10
<p><strong>Step 9:</strong>Subtracting 310 from 500, we get the result 190.</p>
11
<p><strong>Step 10:</strong>Continue doing these steps until we get two numbers after the decimal point. Suppose if there is no decimal values continue till the remainder is zero. So the square root of √230 is approximately 15.16.</p>
11
<p><strong>Step 10:</strong>Continue doing these steps until we get two numbers after the decimal point. Suppose if there is no decimal values continue till the remainder is zero. So the square root of √230 is approximately 15.16.</p>
12
12