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	<title>DICTIONARY Square root - Revision history</title>
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	<updated>2026-05-21T11:00:26Z</updated>
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		<id>https://www.completenoobs.com/noobs/index.php?title=DICTIONARY_Square_root&amp;diff=434&amp;oldid=prev</id>
		<title>Noob: Noob moved page Square root to DICTIONARY Square root without leaving a redirect</title>
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		<updated>2023-05-16T11:42:47Z</updated>

		<summary type="html">&lt;p&gt;Noob moved page &lt;a href=&quot;/noobs/index.php?title=Square_root&amp;amp;action=edit&amp;amp;redlink=1&quot; class=&quot;new&quot; title=&quot;Square root (page does not exist)&quot;&gt;Square root&lt;/a&gt; to &lt;a href=&quot;/noobs/DICTIONARY_Square_root&quot; title=&quot;DICTIONARY Square root&quot;&gt;DICTIONARY Square root&lt;/a&gt; without leaving a redirect&lt;/p&gt;
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				&lt;td colspan=&quot;1&quot; style=&quot;background-color: #fff; color: #202122; text-align: center;&quot;&gt;← Older revision&lt;/td&gt;
				&lt;td colspan=&quot;1&quot; style=&quot;background-color: #fff; color: #202122; text-align: center;&quot;&gt;Revision as of 11:42, 16 May 2023&lt;/td&gt;
				&lt;/tr&gt;&lt;tr&gt;&lt;td colspan=&quot;2&quot; class=&quot;diff-notice&quot; lang=&quot;en&quot;&gt;&lt;div class=&quot;mw-diff-empty&quot;&gt;(No difference)&lt;/div&gt;
&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;</summary>
		<author><name>Noob</name></author>
	</entry>
	<entry>
		<id>https://www.completenoobs.com/noobs/index.php?title=DICTIONARY_Square_root&amp;diff=347&amp;oldid=prev</id>
		<title>imported&gt;AwesomO: Created page with &quot; The square root of a number is a value that, when multiplied by itself, gives the original number. In mathematical terms, the square root of a number x is denoted as √x or x^(1/2). For example, the square root of 9 is 3 because 3 * 3 = 9.  In this tutorial, we will cover the concept of square roots, different methods to find the square root of a number, examples, and exam-style questions with answers. Methods to Find the Square Root  ==1. Prime Factorization==  To fin...&quot;</title>
		<link rel="alternate" type="text/html" href="https://www.completenoobs.com/noobs/index.php?title=DICTIONARY_Square_root&amp;diff=347&amp;oldid=prev"/>
		<updated>2023-04-20T02:27:42Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;quot; The square root of a number is a value that, when multiplied by itself, gives the original number. In mathematical terms, the square root of a number x is denoted as √x or x^(1/2). For example, the square root of 9 is 3 because 3 * 3 = 9.  In this tutorial, we will cover the concept of square roots, different methods to find the square root of a number, examples, and exam-style questions with answers. Methods to Find the Square Root  ==1. Prime Factorization==  To fin...&amp;quot;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;&lt;br /&gt;
The square root of a number is a value that, when multiplied by itself, gives the original number. In mathematical terms, the square root of a number x is denoted as √x or x^(1/2). For example, the square root of 9 is 3 because 3 * 3 = 9.&lt;br /&gt;
&lt;br /&gt;
In this tutorial, we will cover the concept of square roots, different methods to find the square root of a number, examples, and exam-style questions with answers.&lt;br /&gt;
Methods to Find the Square Root&lt;br /&gt;
&lt;br /&gt;
==1. Prime Factorization==&lt;br /&gt;
&lt;br /&gt;
To find the square root of a number using prime factorization, follow these steps:&lt;br /&gt;
&lt;br /&gt;
:*    Find the prime factors of the given number.&lt;br /&gt;
:*    Pair the prime factors in groups of two identical factors.&lt;br /&gt;
:*    Multiply one factor from each pair to find the square root.&lt;br /&gt;
&lt;br /&gt;
Example:&lt;br /&gt;
&lt;br /&gt;
Find the square root of 36.&lt;br /&gt;
&lt;br /&gt;
:*    Prime factors of 36: 2 * 2 * 3 * 3&lt;br /&gt;
:*    Pair the prime factors: (2 * 2) * (3 * 3)&lt;br /&gt;
:*    Multiply one factor from each pair: 2 * 3 = 6&lt;br /&gt;
&lt;br /&gt;
So, the square root of 36 is 6.&lt;br /&gt;
&lt;br /&gt;
==2. Long Division Method==&lt;br /&gt;
&lt;br /&gt;
The long division method is a technique used to find the square root of a number with decimal places.&lt;br /&gt;
Example:&lt;br /&gt;
&lt;br /&gt;
Find the square root of 50.&lt;br /&gt;
&lt;br /&gt;
:*    Pair the digits from the right: (50)&lt;br /&gt;
:*    Find the largest number whose square is less than or equal to 50: 7 (7 * 7 = 49)&lt;br /&gt;
:*    Subtract the result from 50: 50 - 49 = 1&lt;br /&gt;
:*    Bring down the next pair of digits (if any) and repeat the process.&lt;br /&gt;
&lt;br /&gt;
So, the square root of 50 is approximately 7.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;div class=&amp;quot;toccolours mw-collapsible mw-collapsed&amp;quot;&amp;gt;&lt;br /&gt;
Finding the square root of 50 using the long division method:&lt;br /&gt;
&amp;lt;div class=&amp;quot;mw-collapsible-content&amp;quot;&amp;gt;&lt;br /&gt;
:* Step 1: Write the number 50 and separate the digits into pairs starting from the right. Since there are no decimal places, we only have one pair (50).&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
50&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
:* Step 2: Find the largest number whose square is less than or equal to the first pair (50). In this case, the largest number is 7, as 7 * 7 = 49, which is less than or equal to 50. Write 7 above the pair and write 49 below the pair.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
  7&lt;br /&gt;
-------&lt;br /&gt;
| 50&lt;br /&gt;
  49&lt;br /&gt;
-------&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:* Step 3: Subtract 49 from 50 and write the remainder below.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  7&lt;br /&gt;
-------&lt;br /&gt;
| 50&lt;br /&gt;
  49&lt;br /&gt;
-------&lt;br /&gt;
   1&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:* Step 4: Since there are no more digits to bring down, we can proceed to calculate the decimal places. Add a decimal point to the quotient (7), and add a pair of zeros to the remainder. Bring down the pair of zeros.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
  7.&lt;br /&gt;
-------&lt;br /&gt;
| 50.00&lt;br /&gt;
  49&lt;br /&gt;
-------&lt;br /&gt;
   1 00&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:* Step 5: Double the quotient (7) and write it to the left of the remainder. Treat this doubled value as a single entity (in this case, 14). Find a digit (x) that, when combined with the doubled value (14), creates a number (14x) that can be multiplied by x to get a product less than or equal to the remainder (100). In this case, x = 1, as 141 * 1 = 141, which is less than or equal to 100. Write the 1 in the quotient after the decimal point and write 141 below the 100.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
  7.1&lt;br /&gt;
-------&lt;br /&gt;
| 50.00&lt;br /&gt;
  49&lt;br /&gt;
-------&lt;br /&gt;
   1 00&lt;br /&gt;
    141&lt;br /&gt;
-------&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:* Step 6: Subtract 141 from 100 and write the remainder below.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  7.1&lt;br /&gt;
-------&lt;br /&gt;
| 50.00&lt;br /&gt;
  49&lt;br /&gt;
-------&lt;br /&gt;
   1 00&lt;br /&gt;
    141&lt;br /&gt;
-------&lt;br /&gt;
     59&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:* Step 7: Bring down the next pair of zeros (if necessary) and continue the process to find more decimal places. In this example, we will stop at one decimal place.&lt;br /&gt;
&lt;br /&gt;
So, the square root of 50 is approximately 7.1 using the long division method. You can continue this process to find more decimal places if needed.&lt;br /&gt;
&amp;lt;/div&amp;gt;&lt;br /&gt;
&amp;lt;/div&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;div class=&amp;quot;toccolours mw-collapsible mw-collapsed&amp;quot;&amp;gt;&lt;br /&gt;
Finding the square root of 33 using the long division method:&lt;br /&gt;
&amp;lt;div class=&amp;quot;mw-collapsible-content&amp;quot;&amp;gt;&lt;br /&gt;
:* Step 1: Write the number 33 and separate the digits into pairs starting from the right. Since there are no decimal places, we only have one pair (33).&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
33&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
:* Step 2: Find the largest number whose square is less than or equal to the first pair (33). In this case, the largest number is 5, as 5 * 5 = 25, which is less than or equal to 33. Write 5 above the pair and write 25 below the pair.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  5&lt;br /&gt;
-------&lt;br /&gt;
| 33&lt;br /&gt;
  25&lt;br /&gt;
-------&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
:* Step 3: Subtract 25 from 33 and write the remainder below.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  5&lt;br /&gt;
-------&lt;br /&gt;
| 33&lt;br /&gt;
  25&lt;br /&gt;
-------&lt;br /&gt;
   8&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
:* Step 4: Since there are no more digits to bring down, we can proceed to calculate the decimal places. Add a decimal point to the quotient (5), and add a pair of zeros to the remainder. Bring down the pair of zeros.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  5.&lt;br /&gt;
-------&lt;br /&gt;
| 33.00&lt;br /&gt;
  25&lt;br /&gt;
-------&lt;br /&gt;
   8 00&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
:* Step 5: Double the quotient (5) and write it to the left of the remainder. Treat this doubled value as a single entity (in this case, 10). Find a digit (x) that, when combined with the doubled value (10), creates a number (10x) that can be multiplied by x to get a product less than or equal to the remainder (800). In this case, x = 7, as 107 * 7 = 749, which is less than or equal to 800. Write the 7 in the quotient after the decimal point and write 749 below the 800.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  5.7&lt;br /&gt;
-------&lt;br /&gt;
| 33.00&lt;br /&gt;
  25&lt;br /&gt;
-------&lt;br /&gt;
   8 00&lt;br /&gt;
     749&lt;br /&gt;
-------&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
:* Step 6: Subtract 749 from 800 and write the remainder below.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  5.7&lt;br /&gt;
-------&lt;br /&gt;
| 33.00&lt;br /&gt;
  25&lt;br /&gt;
-------&lt;br /&gt;
   8 00&lt;br /&gt;
     749&lt;br /&gt;
-------&lt;br /&gt;
      51&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
:* Step 7: Bring down the next pair of zeros and continue the process to find more decimal places. In this example, we will find 2 more decimal places.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  5.7&lt;br /&gt;
-------&lt;br /&gt;
| 33.00&lt;br /&gt;
  25&lt;br /&gt;
-------&lt;br /&gt;
   8 00&lt;br /&gt;
     749&lt;br /&gt;
-------&lt;br /&gt;
      51 00&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
:* Step 8: Double the quotient without the decimal (57) and write it to the left of the remainder. Treat this doubled value as a single entity (in this case, 114). Find a digit (x) that, when combined with the doubled value (114), creates a number (114x) that can be multiplied by x to get a product less than or equal to the remainder (5100). In this case, x = 4, as 1144 * 4 = 4576, which is less than or equal to 5100. Write the 4 in the quotient after the 7 and write 4576 below the 5100.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  5.74&lt;br /&gt;
-------&lt;br /&gt;
| 33.00&lt;br /&gt;
  25&lt;br /&gt;
-------&lt;br /&gt;
   8 00&lt;br /&gt;
     749&lt;br /&gt;
-------&lt;br /&gt;
      51 00&lt;br /&gt;
        4576&lt;br /&gt;
-------&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
:* 9: Subtract 4576 from 5100 and write the remainder below.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  5.74&lt;br /&gt;
-------&lt;br /&gt;
| 33.00&lt;br /&gt;
  25&lt;br /&gt;
-------&lt;br /&gt;
   8 00&lt;br /&gt;
     749&lt;br /&gt;
-------&lt;br /&gt;
      51 00&lt;br /&gt;
        4576&lt;br /&gt;
-------&lt;br /&gt;
         524&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
:* Step 10: Bring down the next pair of zeros and continue the process to find more decimal places. In this example, we will find 1 more decimal place.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  5.74&lt;br /&gt;
-------&lt;br /&gt;
| 33.00&lt;br /&gt;
  25&lt;br /&gt;
-------&lt;br /&gt;
   8 00&lt;br /&gt;
     749&lt;br /&gt;
-------&lt;br /&gt;
      51 00&lt;br /&gt;
        4576&lt;br /&gt;
-------&lt;br /&gt;
         524 00&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
:* Step 11: Double the quotient without the decimal (574) and write it to the left of the remainder. Treat this doubled value as a single entity (in this case, 1148). Find a digit (x) that, when combined with the doubled value (1148), creates a number (1148x) that can be multiplied by x to get a product less than or equal to the remainder (52400). In this case, x = 4, as 11484 * 4 = 45936, which is less than or equal to 52400. Write the 4 in the quotient after the second 4 and write 45936 below the 52400.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  5.744&lt;br /&gt;
-------&lt;br /&gt;
| 33.00&lt;br /&gt;
  25&lt;br /&gt;
-------&lt;br /&gt;
   8 00&lt;br /&gt;
     749&lt;br /&gt;
-------&lt;br /&gt;
      51 00&lt;br /&gt;
        4576&lt;br /&gt;
-------&lt;br /&gt;
         524 00&lt;br /&gt;
           45936&lt;br /&gt;
-------&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
:* Step 12: Subtract 45936 from 52400 and write the remainder below.&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
  5.744&lt;br /&gt;
-------&lt;br /&gt;
| 33.00&lt;br /&gt;
  25&lt;br /&gt;
-------&lt;br /&gt;
   8 00&lt;br /&gt;
     749&lt;br /&gt;
-------&lt;br /&gt;
      51 00&lt;br /&gt;
        4576&lt;br /&gt;
-------&lt;br /&gt;
         524 00&lt;br /&gt;
           45936&lt;br /&gt;
-------&lt;br /&gt;
            6464&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
Now we have calculated the square root of 33 with three decimal places (5.744).&lt;br /&gt;
&lt;br /&gt;
&amp;lt;/div&amp;gt;&lt;br /&gt;
&amp;lt;/div&amp;gt;&lt;/div&gt;</summary>
		<author><name>imported&gt;AwesomO</name></author>
	</entry>
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