To find a square root by the division (long-division) method , you split the number into pairs of digits and then build the root digit by digit using a special kind of division.

What this method does (in simple words)

The division method (also called the digit‑by‑digit or long‑division method) is a manual way to find N\sqrt{N}N​ that works like ordinary long division but with a pattern: divide, multiply, subtract, bring down, and repeat.

It works for big numbers and for numbers whose square roots are not easy to factor.

Step‑by‑step rule (for whole numbers)

Here is the general procedure.

  1. Group digits in pairs from right to left
    • Write the number and put bars over pairs of digits starting from the units place (right‑most).
    • Example:
      • 1225→1225\rightarrow 1225→ 12‾overline25\overline{12}\\overline{25}12overline25
   * 7396→7396\rightarrow 7396→ 73‾overline96\overline{73}\\\overline{96}73overline96
  1. Start with the leftmost group
    • Take the first (leftmost) group as your starting “segment” or “period”.
    • Find the largest integer whose square is ≤ this group.
    • This integer is:
      • Your first digit of the square root (quotient) , and
      • Your first divisor.
  1. Subtract and bring down the next pair
    • Square this first digit and subtract it from the first group.
    • Bring down the next pair of digits to the right of the remainder to form a new “dividend”.
  1. Double the current quotient to form a new divisor base
    • Take the current quotient and double it ; write it as something like 2Q__2Q\{}\ 2Q__.
    • You will fill the blank with a digit you are going to test (0–9).
  1. Choose the next digit (trial digit)
    • Let the new divisor be (2Qd)(2Qd)(2Qd) where ddd is a digit from 0 to 9.
    • Choose the largest digit ddd such that
      (2Qd)×d≤(2Qd)\times d\le (2Qd)×d≤ current dividend.
 * This ddd becomes:
   * The **next digit of the quotient** , and
   * The **last digit that completes your new divisor**.
  1. Repeat: subtract, bring down, extend divisor
    • Subtract (2Qd)×d(2Qd)\times d(2Qd)×d from the current dividend to get a new remainder.
    • Bring down the next pair of digits (if any).
    • Form the new divisor base again by doubling the updated quotient , and repeat the process.
  1. Stop when all pairs are used
    • When there are no more digit pairs to bring down, the quotient you have is the square root (for a perfect square).
    • If the number is not a perfect square, you can continue by adding decimal pairs of zeros to get more digits after the decimal point.

Worked example 1: 1225\sqrt{1225}1225​

We’ll apply the method to 1225.

Step 1: Group digits

  • 1225→12‾overline251225\rightarrow \overline{12}\\overline{25}1225→12overline25

Step 2: First group 12‾\overline{12}12

  • Largest perfect square ≤ 12 is 32=93^2=932=9.
  • So first quotient digit = 3, first divisor = 3.
  • Write:
    • Quotient: 3
    • Divisor: 3
    • Subtract: 12−9=312-9=312−9=3.

Step 3: Bring down next pair

  • Bring down “25” next to 3 → new dividend = 325.

Step 4: Double the quotient

  • Current quotient = 3 → double = 6.
  • Write new divisor base as 6_6\_ 6_.

Step 5: Find the next digit ddd

We want (6d)×d≤325(6d)\times d\le 325(6d)×d≤325.

Try digits:

  • d=4⇒64×4=256d=4\Rightarrow 64\times 4=256d=4⇒64×4=256 (≤ 325).
  • d=5⇒65×5=325d=5\Rightarrow 65\times 5=325d=5⇒65×5=325 (perfect match).

So choose d=5d=5d=5:

  • New divisor = 65.
  • New quotient = 35.
  • Subtract: 325−325=0325-325=0325−325=0.

No more pairs and remainder is 0 → 1225=35\sqrt{1225}=351225​=35.

Worked example 2: 169\sqrt{169}169​

This is a classic textbook example.

Step 1: Group digits

  • 169→1‾overline69169\rightarrow \overline{1}\\overline{69}169→1overline69.

Step 2: First group 1‾\overline{1}1

  • Largest square ≤ 1 is 12=11^2=112=1.
  • Quotient first digit = 1, divisor = 1.
  • Subtract: 1−1=01-1=01−1=0.

Step 3: Bring down next pair

  • Bring down “69” → new dividend = 69.

Step 4: Double quotient

  • Quotient = 1 → double = 2; divisor base = 2_2\_ 2_.

Step 5: Choose next digit ddd

Find largest ddd such that (2d)×d≤69(2d)\times d\le 69(2d)×d≤69.

Try digits:

  • d=3⇒23×3=69d=3\Rightarrow 23\times 3=69d=3⇒23×3=69 (exact).

So:

  • New divisor = 23.
  • New quotient = 13.
  • Subtract: 69−69=069-69=069−69=0.

No more pairs, remainder 0 → 169=13\sqrt{169}=13169​=13.

Quick pattern to remember

You can think of the method as these five repeating actions.

  • Divide : Use the largest square in the leftmost group.
  • Multiply : Square the current quotient digit and later multiply (2Qd)(2Qd)(2Qd) by ddd.
  • Subtract : Subtract these products from the current dividend.
  • Bring down : Bring down the next pair of digits.
  • Repeat : Double the quotient, choose a new digit, and go again.

Using it for non-perfect squares (idea only)

For numbers that are not perfect squares , the process is the same, but after you finish all the original digits, you:

  • Put a decimal point in the answer.
  • Attach pairs of zeros (00, 00, 00, …) to the number, bringing them down one pair at a time.
  • Keep repeating the same “double, test digit, multiply, subtract, bring down” cycle to get more decimal places.

This way you can approximate roots like 2\sqrt{2}2​ or 10\sqrt{10}10​ to as many decimal digits as you like.

Mini HTML table: Steps at a glance

html

<table>
  <thead>
    <tr>
      <th>Step</th>
      <th>What to do</th>
      <th>Key idea</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>1</td>
      <td>Group digits in pairs from right (e.g., 1225 → 12 | 25).</td>
      <td>Each pair is one stage of the process.[web:1][web:2]</td>
    </tr>
    <tr>
      <td>2</td>
      <td>Take the leftmost group, find largest integer whose square ≤ it.</td>
      <td>This gives the first digit of the root and initial divisor.[web:1][web:2]</td>
    </tr>
    <tr>
      <td>3</td>
      <td>Subtract the square, bring down next digit pair.</td>
      <td>Forms the new dividend.[web:1][web:2]</td>
    </tr>
    <tr>
      <td>4</td>
      <td>Double current quotient to form new divisor base (2Q_).</td>
      <td>Prepares for choosing the next digit.[web:1][web:2]</td>
    </tr>
    <tr>
      <td>5</td>
      <td>Find digit d (0–9) so (2Qd) × d ≤ current dividend.</td>
      <td>Largest possible d keeps remainder non-negative.[web:1][web:2]</td>
    </tr>
    <tr>
      <td>6</td>
      <td>Subtract, update quotient by adding d, repeat with next pair.</td>
      <td>Continue until all pairs used (and zeros for decimals if needed).[web:1][web:4]</td>
    </tr>
  </tbody>
</table>

Tiny story to lock it in

Imagine you’re building the square root like a house number , one digit at a time.

  • The street gives you “blocks” (pairs of digits).
  • At each block, you double the house number you already have, try a new digit in front, and check whether the “rent” (2Qd)×d(2Qd)\times d(2Qd)×d is not more than the money you have (current dividend).
  • If it fits, you keep that digit and move to the next block; if not, you try a smaller digit.

Following this pattern carefully turns any big, scary number into a neat, step‑by‑step division game that gives you its square root. Meta description (SEO):
Learn how to find square root by division method with clear steps, examples (like 1225 and 169), and an easy pattern you can use for both perfect and non‑perfect squares.

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