Square roots by long division

Some square roots are gifts — $144 = 12^2$ falls apart into primes in three easy steps. But hand a student $\sqrt{7921}$ or $\sqrt{15.21}$ and prime factorisation turns into a slog of trial division. The long-division method solves this by working digit by digit, the way ordinary long division builds a quotient one place at a time, except each new digit must satisfy a growing "divisor" condition instead of a fixed one. This article stays inside the ICSE Class 8 syllabus: whole numbers up to 4 digits and decimals up to 2 places. We rebuild the method from a clean algebraic reason — why doubling the quotient-so-far is exactly the right move, not a rule to memorise — then work through the two syllabus examples ( $\sqrt{7921}=89$ and $\sqrt{15.21}=3.9$ ) end to end, and close with the traps that cost the most marks.

The core insight

Long division finds a square root one digit at a time by pairing the number's digits from the decimal point outward and, at each stage, choosing the largest digit $d$ for which (double the quotient so far, with $d$ appended) $\times\, d$ does not exceed the digits brought down so far.

The core idea

Why pairing matters. A 2-digit root like $89$ has a 4-digit square ( $7921$ ); a 3-digit root like $123$ has a 5- or 6-digit square. In general an $n$ -digit root corresponds to a square with $2n$ or $2n-1$ digits — so grouping the square's digits into pairs, from the units place outward, lines each pair up with exactly one digit of the root.

Why the divisor keeps changing. At every stage the method is really solving "square minus what I've already accounted for equals the next chunk." As shown in the derivation below, that leftover chunk always factors as (double the quotient so far, then this new digit) times (this new digit) — which is precisely the "find $d$ " rule.

Whole numbers vs decimals. For a whole number, pair digits starting from the units digit and moving left. For the decimal part, pair digits starting right after the decimal point and moving right; the decimal point is written into the quotient at the exact moment the first decimal pair is brought down. Padding with extra zero-pairs after the decimal point lets the process continue to the syllabus limit of 2 decimal places.

x = \overline{P_1\,P_2\,P_3\,\cdots} \quad\Longrightarrow\quad n = \sqrt{x} = \overline{d_1 d_2 d_3 \cdots}

where: $x$ is the number under the root, $P_1, P_2, \ldots$ are its digit-pairs read left to right (the leftmost group may have only one digit), and $d_1, d_2, \ldots$ are the digits of the root produced one pair at a time.

Figure

Digit pairing for 7921 showing the groups 79 and 21 feeding the quotient digits 8 and 9

Derivation AWhy doubling the quotient-so-far gives the correct next divisor

The goal is to show algebraically, using nothing beyond expanding a square, why the long-division rule "double the digits found so far, append a trial digit $d$ , and require (that number) $\times d \le$ the current dividend" is exactly the condition for $d$ to be the true next digit of the square root — not an arbitrary shortcut but a forced consequence of how a two-digit number squares.

Step 1. Set up the algebra for a 2-digit root. Suppose the square root of a 4-digit number $x$ is the two-digit number $n = 10a + b$ , where $a$ is the tens digit (found first) and $b$ is the units digit (found second). Squaring gives $x = n^2 = (10a+b)^2 = 100a^2 + 20ab + b^2$ .

Step 2. Interpret the first term as the first step of long division. The long-division method's first move is to find the largest $a$ with $a^2 \le$ (first digit-group of $x$ ), which corresponds to comparing against $100a^2$ — the part of $x$ that depends on $a$ alone. Subtracting $100a^2$ from $x$ (after the first group has been matched) and bringing down the next pair leaves exactly $x - 100a^2 = 20ab + b^2$ .

Step 3. Factor the remainder to expose the divisor rule. Factor out $b$ : $20ab + b^2 = b(20a + b) = (20a + b)\times b$ . This is precisely "double the quotient found so far ( $2a$ , written as $20a$ in place value) with a new digit $b$ appended, multiplied by $b$ " — the exact rule used in Step 5 of the procedure. So the divisor's leading part, $20a$ , is not an arbitrary doubling; it falls straight out of expanding $(10a+b)^2$ .

Step 4. Generalise and confirm. The same expansion applies at every later stage with $a$ standing for "all quotient digits found so far" and $b$ for "the next digit," because each new digit is just the next place value being appended to the existing root. This is why the rule is always double the current quotient, append the trial digit, multiply by the trial digit, and require the result to stay at or below the current dividend — it is the algebraic remainder of a squared number, uncovered one digit at a time.

Worked example

Find $\sqrt{7921}$ by long division, then extend the method to $\sqrt{15.21}$ .

Step 1. Pair the digits of $7921$ from the right: $\overline{79}\ \overline{21}$ .

Step 2. Match the first group. The largest digit $a$ with $a^2 \le 79$ is $a = 8$ , since $8^2 = 64 \le 79$ but $9^2 = 81 > 79$ . Write $8$ as the first quotient digit and subtract: $79 - 64 = 15$ .

Step 3. Bring down the next pair. Bringing down $21$ turns the remainder into a new dividend, $1521$ .

Step 4. Build the new divisor and find the next digit. Double the quotient so far ( $8\to16$ ) to start the divisor " $16\_$ ", then find $d$ with $(160+d)\times d \le 1521$ . Trying $d=9$ gives $169 \times 9 = 1521$ , an exact match, so the quotient becomes $89$ with remainder $0$ .

Step 5. Check and extend to the decimal case. $\sqrt{7921} = \mathbf{89}$ , and indeed $89^2 = 7921$ . For $\sqrt{15.21}$ , pair as $\overline{15}.\overline{21}$ : $3^2=9\le15$ gives first digit $3$ (remainder $6$ ); bringing down $21$ gives dividend $621$ , and the decimal point is written into the quotient right here. Doubling $3$ gives divisor start " $6\_$ ", and $d=9$ satisfies $69\times9=621$ exactly. So $\sqrt{15.21} = \mathbf{3.9}$ , confirmed by $3.9^2 = 15.21$ .

Competitive stretch

For a number with an odd number of digits, such as $\sqrt{529}$ , the leftmost group has only one digit: $\overline{5}\ \overline{29}$ . The largest $a$ with $a^2 \le 5$ is $a=2$ (remainder $1$ ); bringing down $29$ gives dividend $129$ , doubling $2$ gives " $4\_$ ", and $d=3$ works since $43\times3=129$ exactly. So $\sqrt{529} = 23$ — the odd digit-count only changes how the first group is sized, not the rest of the algorithm.

Common traps

Pairing whole-number digits from the wrong end. Always start from the units digit and move left; the leftmost group can have one digit if the total count is odd.
Placing the decimal point in the quotient too early or too late. It must appear at the exact step where the first decimal digit-pair is brought down, not before and not after.
Doubling the previous divisor instead of doubling the quotient digits found so far — Step 4 of the derivation shows why it must be the quotient, not the divisor, that gets doubled.
Accepting a trial digit $d$ that makes the product exceed the dividend, or stopping at a digit that is not the largest valid one — always test whether $d+1$ would also work before settling.

Ask the tutor: stuck on where to place the decimal point or how to double the quotient — ask the AI tutor to walk through your own number step by step.

Why does the divisor grow a new digit at every step instead of staying fixed

Because the divisor is built from the quotient digits discovered so far, and those grow by one digit at each stage. As the derivation shows, the remainder after each step algebraically equals (double the quotient so far, with a new digit appended) times that new digit — so as the quotient grows longer, the divisor's fixed part grows with it. It is not a separate rule; it falls directly out of expanding the square of a multi-digit number.

What if, after pairing, the trial digit makes the product come out slightly more than the dividend

Then that trial digit is too large — try one digit smaller. The rule requires (divisor with digit $d$ appended) $\times d$ to be less than or equal to the current dividend, never more. Once you find a digit that satisfies this, it is worth checking that the next digit up would fail, confirming you picked the largest valid digit rather than stopping too early.

Does the long-division method work for numbers that are not perfect squares

Yes, and this is actually where it becomes most useful. If a number is not a perfect square, the process never reaches a remainder of zero; instead you keep bringing down pairs of zeros after the decimal point and continue finding digits, giving the square root correct to as many decimal places as needed. ICSE Class 8 restricts practice to 2 decimal places, but the same digit-by-digit logic extends further with more zero-pairs.

Square roots by long division

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads