Prime Factors Fail the Moment a Number Is Not a Perfect Square

Because it only gives an answer when the number is a perfect square.

Factorise $1764$ and the primes pair up neatly, giving $\sqrt{1764}=42$ exactly. Try the same on $7$ and there is nothing to pair — $7$ is prime, and $\sqrt{7}$ is not a whole number at all. No amount of factorising will produce it.

So a second method is needed, one that works on any number and gives as many decimal places as you ask for. That is the long division method, and it is the core of this page — along with prime factorisation for the cases where it does work, and the problems both are used to solve.

Break the number into prime factors, group them in pairs, and take one factor from each pair.

Worked example 1. Find $\sqrt{1764}$.

Dividing repeatedly by primes:

$$1764=2\times 2\times 3\times 3\times 7\times 7$$

Grouping in pairs — $(2\times 2)$, $(3\times 3)$, $(7\times 7)$ — and taking one from each:

$$\sqrt{1764}=2\times 3\times 7=42$$

Checking: $42^2=1764$. Correct.

Worked example 2. Find $\sqrt{5184}$.

$$5184=2^6\times 3^4$$

Halve each exponent:

$$\sqrt{5184}=2^3\times 3^2=8\times 9=72$$

Checking: $72^2=5184$. Correct.

The exponent shortcut. Once the number is written with exponents, the square root simply halves every exponent — which is why a number is a perfect square exactly when every prime exponent is even. That is also the test for whether the method will work at all.

Worked example 3 — showing a failure. Is $500$ a perfect square?

$$500=2^2\times 5^3$$

The exponent of $5$ is **$3$, which is odd**, so one $5$ has no partner and $500$ is not a perfect square. Prime factorisation has answered the question but cannot give the root.

Where factorisation is still the better method. For a perfect square it is exact and often faster than long division, and it is the only method that tells you why a number is or is not a perfect square. For anything else — and for every decimal answer — the next section is required.

Mark off the digits in pairs from the right, then build the root one digit at a time.

The procedure:

- Pair the digits from the right for the whole-number part, and from the decimal point rightwards for the decimal part.
- Find the largest number whose square is not more than the leftmost group. That is the first digit of the root.
- Subtract, bring down the next pair, and form the new dividend.
- Double the root so far to get the start of the new divisor, then find the digit that makes divisor $\times$ digit not exceed the dividend.
- Repeat until the pairs run out, or until you have enough decimal places.

Worked example 1. Find $\sqrt{1369}$.

Pair as $13\mid 69$.

- Largest square not over $13$ is $9=3^2$, so the first digit is **$3$**. Remainder $13-9=4$.
- Bring down $69$, giving $469$.
- Double the root so far: $3+3=6$. Find $x$ with $\overline{6x}\times x\le 469$. Trying $x=7$: $67\times 7=469$ exactly.

So $\sqrt{1369}=37$, and $37^2=1369$ confirms it.

Worked example 2. Find $\sqrt{2025}$.

Pair as $20\mid 25$. Largest square not over $20$ is $16=4^2$, so the first digit is **$4$**, remainder $4$. Bring down $25$ to give $425$. Doubling gives $8$, and $85\times 5=425$ exactly.

So $\sqrt{2025}=45$.

Worked example 3 — a decimal. Find $\sqrt{6.25}$.

Pair the whole part and the decimal part separately: $6\mid 25$. Largest square not over $6$ is $4=2^2$, remainder $2$. Bring down $25$ to give $225$. Doubling gives $4$, and $45\times 5=225$ exactly.

The decimal point in the answer sits above the decimal point in the number, so $\sqrt{6.25}=2.5$. Checking: $2.5^2=6.25$. Correct.

Worked example 4 — a fraction. Find $\sqrt{\frac{529}{1024}}$.

Take the square root of the numerator and the denominator separately:

$$\frac{\sqrt{529}}{\sqrt{1024}}=\frac{23}{32}$$

since $23^2=529$ and $32^2=1024$.

Worked example 5 — to a given number of decimal places. Find $\sqrt{2}$ correct to two decimal places.

Carrying the long division out gives $1.4142\dots$, so to two places $\sqrt{2}=1.41$.

The rule that is easy to get wrong. To give an answer correct to **$n$ places you must work out $n+1$** places and then round. Here the third digit is $4$, so the second place stays as $1$. But for $\sqrt{7}=2.6457\dots$, the third digit is $5$, so two places gives **$2.65$** and not $2.64$ — stopping the division at two places would have given the wrong answer.

Find the square root by long division and look at the remainder, or equivalently find the nearest perfect squares on either side.

To subtract, go down to the perfect square below. To add, go up to the perfect square above.

Worked example 1. What is the least number that must be subtracted from $2000$ to make it a perfect square?

Since $44^2=1936$ and $45^2=2025$, the largest perfect square below $2000$ is $1936$:

$$2000-1936=64$$

So **$64$** must be subtracted, leaving $1936=44^2$.

Worked example 2. What is the least number that must be added to $2000$ to make it a perfect square?

The smallest perfect square above $2000$ is $45^2=2025$:

$$2025-2000=25$$

So **$25$** must be added, giving $2025=45^2$.

Notice that the two answers are quite different — $64$ and $25$ — because $2000$ sits much closer to $2025$ than to $1936$.

Worked example 3. Find the least number to be subtracted from $3250$, and the least to be added.

Here $57^2=3249$ and $58^2=3364$.

$$\text{subtract: }3250-3249=1$$

$$\text{add: }3364-3250=114$$

So one is enough to subtract, while $114$ is needed to add.

The connection to long division. Carrying out the long division of $2000$ leaves a remainder after the last complete step, and that remainder is exactly the number to subtract. To find the number to add, increase the root by $1$, square it, and subtract the original — which is why examiners accept either route.

The trap to avoid. For the add case you must go to the next square up, not simply reuse the remainder. Adding the remainder $64$ to $2000$ gives $2064$, which is not a perfect square. Getting these two directions the wrong way round is the most frequent error on this topic.

Whenever a quantity is squared in the setting — a square field, a square array of objects — the square root undoes it.

Worked example 1 — a square field. The area of a square field is $1764{\text{m}}^2$. Find its side and its perimeter.

$$\text{side}=\sqrt{1764}=42\text{m}$$

$$\text{perimeter}=4\times 42=168\text{m}$$

Worked example 2 — equal rows. A gardener has $2025$ plants and wants to plant them so that the number of rows equals the number of plants in each row. How many rows?

If there are $r$ rows with $r$ plants each, then $r^2=2025$, so

$$r=\sqrt{2025}=45$$

Fortyfive rows of $45$ plants.

Worked example 3 — with a leftover. A commander wants to arrange $8000$ soldiers in a perfect square. What is the least number that must be left out?

Since $89^2=7921$ and $90^2=8100$, the largest square that fits is $7921$:

$$8000-7921=79$$

So **$79$** soldiers are left out, and the rest form a $89\times 89$ square.

Worked example 4 — cost leading to area. The cost of levelling a square field at $\text{₹}5$ per square metre is $\text{₹}8820$. Find the side of the field.

Work backwards. First the area:

$$\text{area}=\frac{8820}{5}=1764{\text{m}}^2$$

Then the side:

$$\text{side}=\sqrt{1764}=42\text{m}$$

How to recognise which problem you are in. If the question asks for a side, a row count or an edge, you are taking a square root. If it gives you one of those and asks for the area or the total, you are squaring. And when the total does not come out to a perfect square, the question is almost always asking for the least number to add or subtract from the previous section.

Always keep the units. A side is in metres and an area in square metres, so $\sqrt{1764{\text{m}}^2}=42\text{m}$ — the square root halves the unit's power as well as the number's exponent.