Square roots

What a square root means. For a number $n$, its square root $\sqrt{n}$ is defined as the non-negative number $m$ that satisfies:

where: $n$ is the number whose root is sought, $m$ is the square root, and the restriction $m \ge 0$ is a convention — even though $(-12)\times(-12)=144$ too, $\sqrt{144}$ is agreed to mean $12$, not $-12$.

Exact roots by prime factorisation. Every perfect square, when broken into prime factors, has every prime appearing an even number of times — the primes come in matched pairs. Taking one factor from each pair and multiplying those chosen factors together gives the square root exactly:

This pairing trick only works when $n$ truly is a perfect square. If even one prime in the factorisation has an odd exponent, the primes cannot all be paired, and $\sqrt{n}$ is not a whole number — a signal to switch to estimation instead of forcing an answer.

Estimating roots of non-perfect squares. Most numbers between one perfect square and the next are not perfect squares themselves, yet their square roots can still be located precisely between two whole numbers. If a number $n$ satisfies

then $\sqrt{n}$ must lie strictly between $k$ and $k+1$, because squaring is increasing — larger numbers have larger squares. Comparing how much closer $n$ sits to $k^2$ than to $(k+1)^2$ then gives a reasonable one-decimal estimate. A useful side fact for hunting nearby perfect squares: a perfect square can only end in the digits $0, 1, 4, 5, 6$ or $9$ — never in $2, 3, 7$ or $8$ — which quickly rules out wrong guesses.

Forcing a perfect square by multiplying or dividing. Write $n$ in prime-factorised form and look at the exponents. Any prime whose exponent is odd is "unpaired." Multiplying $n$ by that unpaired prime once more raises its exponent by one, making it even and pairing it off — the smallest such multiplier is the product of all the currently-unpaired primes, each taken once. Dividing works the opposite way: removing an unpaired prime entirely can also leave a perfect square, but only if every other exponent in what remains is already even.

Estimate $\sqrt{130}$ to one decimal place by bracketing, and separately find the smallest number by which $150$ must be multiplied to become a perfect square.

Step 1. List nearby perfect squares around $130$: $11^2 = 121$ and $12^2 = 144$. Since $121 < 130 < 144$, it follows that $11 < \sqrt{130} < 12$.

Step 2. Compare distances: $130$ is $9$ away from $121$ but $14$ away from $144$, so $130$ sits closer to $121$. That means $\sqrt{130}$ is only a little more than $11$ — a reasonable estimate is $\sqrt{130} \approx 11.4$ (a calculator confirms $\sqrt{130} \approx 11.40$).

Step 3. Now prime-factorise $150$: $150 = 2 \times 3 \times 5^2 = 2^1 \times 3^1 \times 5^2$. The exponents of $2$ and $3$ are both odd (each equals $1$), so both are unpaired; the exponent of $5$ is already even and needs nothing.

Step 4. Multiply $150$ by the product of the unpaired primes, $2 \times 3 = 6$: $150 \times 6 = 900$. Check by refactorising: $900 = 2^2 \times 3^2 \times 5^2 = (2\times3\times5)^2 = 30^2$, every exponent now even. The smallest multiplier is $6$, and $\sqrt{900} = 30$.

Competitive stretch

Notice the shortcut: once $150 \times 6 = 900$ is known to be a perfect square, its square root can be read straight off the pairing, $2\times3\times5=30$, without redoing the whole factorisation of $900$ from scratch — the multiplier calculation and the root calculation share the same prime list.

• Forcing the pairing method on a non-perfect square. If any prime's exponent is odd, $\sqrt{n}$ is not a whole number; switch to bracketing between consecutive perfect squares instead of trying to pair an odd count of primes.
• Bracketing with squares that are not consecutive. Always use the perfect square immediately below $n$ and the one immediately above it; skipping to squares further away gives an estimate too wide to be useful.
• Mixing up "multiply to fix" with "divide to fix." For $150 = 2^1\times3^1\times5^2$, multiplying by $6$ works because it pairs both unpaired primes at once; dividing by $6$ would remove both primes but also strip away part of the already-paired $5^2$, so the divisor method only applies when every unpaired prime can be cleanly removed.
• Forgetting the non-negative convention. $\sqrt{n}$ always denotes the non-negative root by definition, even though a negative number squared also gives $n$.

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