Prime factorisation method: factor the number into primes, group identical primes into pairs; if all primes pair up the number is a perfect square; the square root is product of one element from each pair.

Example: \(324 = 2\times2\times3\times3\times3\times3 = 2^2\cdot3^4\). Pair factors: \((2^2),(3^4)\) → one of each pair gives \(2\cdot3^2=2\cdot9=18\). So \(\sqrt{324}=18\). (Long answer — full explanation and steps). :contentReference[oaicite:48]{index=48}

Long-division method steps: group digits in pairs from decimal point leftwards; find largest square ≤ left-most group; subtract, bring down next pair; double current quotient as divisor base, find next digit, etc. Example: For 5476, groups: (54)(76). Largest square ≤54 is 7^2=49. Quotient digit 7; remainder 5; bring 76 → 576. Double quotient 7→ 14_. Find digit x so that \( (140 + x)\times x \le 576\). x=4 → (144×4=576). So final quotient 74. Thus \(\sqrt{5476}=74\). This matches the example in the chapter (least number subtract example). :contentReference[oaicite:49]{index=49}

Repeated subtraction: subtract successive odd numbers starting at 1. If after n subtractions you get 0 then original number is \(n^2\); n is its square root. E.g., 81: 81−1−3−5−...−17=0 after 9 steps → root 9. Usefulness: conceptually neat, links square to sum of odds. Limitation: inefficient for large numbers (many subtractions). The chapter demonstrates this to show conceptual basis; for practical large numbers use factorisation or long-division. :contentReference[oaicite:50]{index=50}

If the units digit is 2, 3, 7 or 8, the number is definitely not a perfect square (no square ends in these digits). Examples from chapter: 1057 (ends with 7) cannot be square; 7928 (ends with 8) not square. Conversely, ending in 1,4,5,6,9,0 doesn't guarantee it is square—further checks needed. This quick check helps eliminate many numbers. :contentReference[oaicite:51]{index=51}

Prime-factor the number; for each prime with odd exponent, multiply by that prime to make exponents even. Example: 90 = \(2\cdot3^2\cdot5\) → primes 2 and 5 have exponent 1 (odd) → multiply by \(2\times5=10\) → \(90\times10=900=30^2\). Example: 2352 = \(2^4\cdot3\cdot7^2\) → missing pair for 3 → multiply by 3 → \(2352\times3=7056=84^2\). The method ensures smallest multiplier.

Observe: \(1^2=1\), \(11^2=121\), \(111^2=12321\), \(1111^2=1234321\). This palindromic pattern arises because of place-value expansion of repeated '1's; chapter shows this as an interesting pattern and invites exploration. Two examples: \(111^2=12321\), \(11111^2=123454321\). (Detailed algebraic explanation: expand and group powers of 10). :contentReference[oaicite:53]{index=53}

Greatest 4-digit number is 9999. Using long-division root method we find remainder 198 when trying to find \(\sqrt{9999}\); subtract remainder: \(9999-198=9801\). \(\sqrt{9801}=99\). So greatest 4-digit perfect square is 9801 (i.e., \(99^2\)). The long-division remainder approach finds nearest lower perfect square. :contentReference[oaicite:54]{index=54}

Algebraic identity: \((a+1)(a-1)=a^2-1\) by expansion. This helps compute products of numbers around a central 'a' (e.g., \(44\times46=(45-1)(45+1)=45^2-1\)). It is useful to compute squares or near-square products quickly and spot patterns. Chapter uses this to get patterns like 44×46 etc. :contentReference[oaicite:55]{index=55}

Group integral part (17) and decimal pairs (64). Largest square ≤17 is 4^2=16 → first digit 4; remainder 1; bring down 64 → 164. Double current quotient 4→8_ ; find x so that (80+x)x ≤164 → x=2 since 82×2=164. Put decimal point in quotient → 4.2. So \(\sqrt{17.64}=4.2\). Chapter gives detailed step-by-step illustration. :contentReference[oaicite:56]{index=56}

Sum of first n odd numbers: \(1+3+5+\dots+(2n-1)=n^2\). Proof by induction or pairwise addition. Since every square is such a sum, subtracting odd numbers until remainder 0 yields number of subtractions n → square root n. The book demonstrates with 81 example: 81−1−3−5−...−17=0 in 9 steps. :contentReference[oaicite:57]{index=57}

Choose integer m>1. Then \( (2m)^2+(m^2-1)^2=(m^2+1)^2\). Examples: m=3 → (6,8,10); m=4 → (8,15,17). The chapter walks through derivation and examples. :contentReference[oaicite:58]{index=58}

Find \(\sqrt{1300}\) via long division; remainder shows nearest lower square \(36^2=1296\) and next square \(37^2=1369\). Difference \(1369-1300=69\). So add 69 to get 1369 which is \(37^2\). Chapter provides the long-division steps. :contentReference[oaicite:59]{index=59}

Count digits in integer part. For 25600, group digits in pairs from right: (2)(56)(00) → leftmost group '2' has 1 digit → root will have 3 digits if leftmost group >1 (or compute via powers: \(100^2=10000\) and \(200^2=40000\) so root between 100 and 200 → 3 digits). Chapter gives rules about digits. (Detailed logic shown in text). :contentReference[oaicite:60]{index=60}

Prime-factor a number. If any prime has odd exponent → not a square. If all exponents even → root is product of primes raised to half their exponents. Example: \(6400=2^8\cdot5^2\) → root \(=2^4\cdot5^1=80\). The chapter gives multiple worked examples (6400, 324 etc.). :contentReference[oaicite:61]{index=61}

Repeated subtraction is conceptually instructive (links sums of odd numbers), prime factor method is fastest for integer checks and exact roots, long-division method is systematic and works for decimals and large numbers. For exam speed, prime-factor + pattern recognition is best; for algorithmic precision and non-integers, use long division. The chapter balances conceptual and practical methods with examples and exercises.