Squares Square Roots and Sensible Estimates

What you will learn

A square-root answer needs both a correct value and a correct meaning. In this lesson you will:

• Distinguish squaring, the principal square root, and the solutions of a squared equation
• Reject impossible integer squares and confirm possible ones using prime-factor pairs
• Find the smallest positive integer multiplier or exact divisor that produces an integer square
• Find integer and decimal square roots by digit-pair long division
• Find exact fraction roots, bound other roots, and label an estimate honestly

Square root is named in the Mathematics syllabus in RRB CEN 09/2025 §14.1. This foundation lesson does not predict exam weightage. It uses the prime factorisation and fraction/decimal skills from earlier lessons. Advanced surds, rationalisation and cube roots are outside this lesson.

1 Before you begin

Try these retrieval checks before reading the feedback.

1. Evaluate 4² and (−4)².
2. Write 180 as a product of prime factors.
3. Express 0.49 as a fraction with an integer numerator and denominator.

Retrieval feedback

1. Both values are 16: 4 × 4 = 16 and (−4) × (−4) = 16. A square means multiplying a number by itself, not by 2. Retry: 7² = 49 and (−7)² = 49. Remember that −7², without brackets, is −49.
2. 180 = 18 × 10 = 2² × 3² × 5. Count repeated occurrences of each prime. Retry: 72 = 2³ × 3².
3. 0.49 = 49/100 because the final digit is in the hundredths place. Retry: 0.09 = 9/100 and 0.009 = 9/1000; the zeros determine the places.

If a check was difficult, revisit the related earlier lesson before starting long division. There is no time limit or pass cutoff.

2 Squares and the meaning of the root sign

For a real number a, its square is a² = a × a. Every real square is nonnegative: a positive or negative nonzero number squares to a positive number, and 0² = 0. In this lesson, an integer perfect square is a nonnegative integer of the form n², where n is an integer. Thus 0 and 1 are included.

For a ≥ 0, √a means the unique nonnegative number whose square is a. This is the principal square root. The symbol √ does not ask for both signs. A negative number has no real square root, because no real number squares to a negative value; complex numbers are outside our scope.

Worked example 1 One principal root and two possible solutions

Compare √169 with all real solutions of x² = 169. Also find √0.

1. 13 × 13 = 169, so √169 = 13.
2. (−13) × (−13) = 169 as well. Therefore x² = 169 has solutions x = 13 and x = −13. The shorthand x = ±13 means these two alternatives.
3. √0 = 0. The equation x² = 0 has only the solution x = 0; −0 and 0 are the same number.

Error check: √169 = ±13 is incorrect. The equation has two solutions, while the principal-root symbol has one value. Similarly, √((−6)²) = √36 = 6, not −6.

3 Use the last digit as a rejection test

Square each possible units digit 0 through 9. The units digits of the resulting squares are, in that order, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1. Tens and higher places cannot change the units digit of the product. Therefore an integer square can end only in 0, 1, 4, 5, 6 or 9.

An integer ending in 2, 3, 7 or 8 is not a perfect square. An allowed ending is necessary, but it is not sufficient: it only means this quick test has not rejected the number. You still need a square, factorisation, or consecutive-square comparison to confirm it.

Worked example 2 An allowed ending is not a certificate

Can 746 and 728 be perfect squares?

1. 728 ends in 8, so it is not a perfect square.
2. 746 ends in 6, which is allowed. This alone gives no final answer.
3. 27² = 729 and 28² = 784. Since 729 < 746 < 784, no integer can square to 746. It is not a perfect square either.

The squares of nonnegative integers increase as the integers increase. Between two consecutive integers there is no other integer whose square could fit the gap.

Error check: An allowed ending does not mean “definitely a square”. Do not apply this integer-ending test directly to the last written digit of a decimal.

4 Pair prime factors to find an integer root

When a positive integer is squared, its whole prime factorisation is repeated. Every prime therefore occurs an even number of times. Conversely, if every prime occurrence can be paired, taking one prime from each pair gives an integer whose square is the original number. This is both a test and a method. The integer 1 is a square with root 1; handle 0 separately rather than trying to prime-factorise it.

Worked example 3 Take one factor from each pair

Find √1764 by prime factorisation.

1. 1764 = 4 × 441 = (2 × 2) × (3 × 3) × (7 × 7).
2. Each prime has a partner: 1764 = 2² × 3² × 7².
3. Take one from each pair: √1764 = 2 × 3 × 7 = 42.
4. Check by multiplication: 42 × 42 = 1764.

Error check: Take one factor from each pair, not one factor from the entire list. An unpaired prime means the positive integer is not an integer square. The root still exists as a real number, but is not an integer.

5 Repair odd prime counts with the smallest change

Here the multiplier and divisor must be positive integers. A divisor must divide the original number exactly, and the quotient must be an integer square.

For a positive integer N, identify the primes with odd counts in its factorisation. Multiply one copy of each such prime together.

• Multiplying N by this product adds one to each odd count and makes every count even.
• Dividing N by this product removes one from each odd count and also makes every count even.

Why is this the smallest positive choice? Each odd count needs an odd number of copies added or removed; at least one copy of that prime is unavoidable. Every qualifying multiplier or divisor must therefore contain the product just described. Including extra copies or other primes cannot produce a smaller positive integer. If N is already a square, the smallest multiplier and divisor are both 1.

Worked example 4 The smallest multiplier and the smallest divisor

Find both for 540.

1. 540 = 54 × 10 = 2² × 3³ × 5.
2. The count of 2 is even. The counts of 3 and 5 are odd, so the required product is 3 × 5 = 15.
3. Multiply: 540 × 15 = 2² × 3⁴ × 5² = 8100 = 90².
4. Divide: 540 ÷ 15 = 2² × 3² = 36 = 6².
5. Both answers are 15, but the resulting squares differ. The multiplication and division checks confirm each result.

Error check: Dividing by 540 gives 1, a square, but 540 is not the smallest qualifying divisor. Using only 3 or only 5 leaves the other odd count unresolved.

Pause and check

What is the smallest positive multiplier for 98? What is the smallest positive divisor of 98 that leaves an integer square?

Feedback: 98 = 2 × 7². Both answers are 2: 98 × 2 = 196 = 14² and 98 ÷ 2 = 49 = 7². “Smallest” refers to the multiplier or divisor, not the resulting square.

6 Square-root long division one digit at a time

Long division finds root digits without first finding every prime factor. Each pair of input digits produces one root digit. All the steps below are written out; no picture is needed.

Pair the digits before calculating

Start at the decimal point, or at the units digit if no decimal point is written.

• In the integer part, make pairs towards the left. Only the leftmost group may have one digit: 4624 becomes 46 | 24, and 10404 becomes 1 | 04 | 04.
• In the fractional part, make pairs towards the right: 5.76 becomes 5 . 76. The dot is still the decimal point; the vertical bars only separate pairs, not division operations.
• If one fractional digit is left over, put a zero on its right: 354.7 = 354.70, grouped as 3 | 54 . 70. Never add that zero to the integer part.
• To continue beyond the given decimal digits, append pairs of zeros after the decimal point. For example, 70 = 70.00 00 … . Each extra pair supplies one more decimal place of its root.

The repeatable procedure

1. From the leftmost integer group, choose the largest digit whose square does not exceed the group. Write this first root digit and subtract its square. For a number below 1, the integer group is 0 and the first root digit is 0.
2. Bring down the next pair. If the old remainder is r and the pair is p, the new working number is 100 × r + p. Bringing down 04 means adding 4 after multiplying the remainder by 100; the leading zero preserves the two places.
3. Let q be the root digits already found, read as a whole number without their decimal point. Try a next digit d from 0 to 9. Choose the largest d for which (20 × q + d) × d does not exceed the new working number.
4. Subtract that product. Append d to the root digits, so the new prefix is 10 × q + d. Use this entire updated prefix for the next step. Repeat until all required pairs have been processed.
5. Put the root's decimal point immediately after the digits supplied by the integer groups, before processing the first fractional pair. A zero digit must still be written; do not skip its place.

The multiplier in step 3 is the new single digit d, not the entire updated root. If q = 6 and d = 8, the product to subtract is 128 × 8, not 128 × 68.

Why the trial product works

Suppose the next whole-number prefix is 10 × q + d. Multiplying it by itself gives:

(10 × q + d)² = 100 × q² + 20 × q × d + d² = 100 × q² + (20 × q + d) × d.

Bringing down two input digits multiplies the previous processed number by 100 before adding the pair. The old square contribution is therefore 100 × q². The extra amount needed for the new digit is exactly (20 × q + d) × d. This explains both the digit pairing and the trial product, rather than treating “double and attach” as an unexplained trick.

Worked example 5 Integer and decimal long division

A Find √4624

Groups: 46 | 24.

1. First group 46: 6² = 36 ≤ 46, while 7² = 49 > 46. First root digit = 6; remainder = 46 − 36 = 10.
2. Bring down 24: working number = 100 × 10 + 24 = 1024.
3. Current prefix q = 6, so try (120 + d) × d. Digit 8 gives 128 × 8 = 1024; digit 9 gives 129 × 9 = 1161, too large. Choose 8.
4. New prefix = 68; remainder = 1024 − 1024 = 0. All pairs are used, so √4624 = 68 exactly. Check: 68² = 4624.

B Find √5.76

Groups: 5 . 76.

1. First group 5: 2² = 4 ≤ 5 < 3². First root digit = 2; remainder = 1.
2. All integer groups are used, so put the decimal point after 2.
3. Bring down 76: working number = 100 × 1 + 76 = 176. The current digit prefix is q = 2.
4. Choose d = 4 because (40 + 4) × 4 = 176, while (40 + 5) × 5 = 225 is too large. The remainder is 0.
5. The fractional pair supplies the tenths digit 4, so √5.76 = 2.4. Check: 2.4 × 2.4 = 5.76.

C Keep a zero digit and update the whole prefix

For √10404, use groups 1 | 04 | 04.

1. First group 1 gives root prefix 1 and remainder 0.
2. Bring down 04 to get 4. Digit 1 would require (20 + 1) × 1 = 21 > 4, so choose digit 0. Write root prefix 10 and keep remainder 4.
3. Bring down 04: working number = 100 × 4 + 4 = 404. Now q = 10, not 1 and not the last digit 0.
4. Choose digit 2: (200 + 2) × 2 = 404. Digit 3 would give 609, too large. Root = 102 and remainder = 0; check 102² = 10404.

A zero remainder is not permission to stop when unused nonzero pairs remain. In part C the first remainder was zero, but both 04 groups still mattered. If the remainder is zero and all remaining pairs are 00, continue them or fill the corresponding root places with zeros. Stop with an exact result only when the required places are accounted for and the remainder is zero.

Check the decimal position

Try grouping 0.0009 before reading the answer.

Feedback: 0 . 00 | 09. The integer group gives root 0. The first fractional pair 00 gives tenths digit 0. For the next pair 09, q = 0 and d = 3 gives (0 + 3) × 3 = 9, so the hundredths digit is 3. Hence √0.0009 = 0.03; (0.03)² = 0.0009. Omitting the zero in the tenths place would give the wrong answer 0.3.

If the remainder stays nonzero, more pairs give more digits; a stopped decimal is not automatically the exact root. For rounding to a chosen decimal place, find at least the following digit and apply the rounding rule from the previous lesson.

7 Exact fractions and sensible bounds

For a ≥ 0 and b > 0, √(a/b) = √a/√b. In elementary exact problems, check whether numerator and denominator are squares after simplifying the fraction. The fraction rule works because the nonnegative quotient on the right squares to a/b. In particular, √(49/100) = 7/10 = 0.7, so √0.49 = 0.7, not 0.07.

Worked example 6 Exact value or bounded estimate

A Find √(81/196)

81 = 9² and 196 = 14². Therefore √(81/196) = 9/14. Check: (9/14) × (9/14) = 81/196. Both numerator and denominator contribute to the root.

B Locate √70 and refine the estimate

1. 8² = 64 < 70 < 81 = 9², so 8 < √70 < 9. These are consecutive-integer bounds.
2. The interval does not imply that the root is its midpoint 8.5. Indeed 8.5² = 72.25, not 70.
3. For a tighter interval, 8.3² = 68.89 < 70 and 8.4² = 70.56 > 70, so 8.3 < √70 < 8.4.
4. To round to the nearest tenth, check the midpoint 8.35: 8.35² = 69.7225 < 70, so √70 is above 8.35. Hence √70 ≈ 8.4 to the nearest tenth. Write ≈ for the approximation, not =.

A short rational and irrational distinction

A rational number can be written as p/q with integers p and q ≠ 0. Its decimal representation terminates or eventually repeats a fixed digit block. An irrational number cannot be written in that form; its decimal neither terminates nor eventually repeats.

For a nonnegative integer N, √N is rational exactly when N is a perfect square. Thus √81 = 9 is rational, while √70 is irrational. A brief reason uses the prime counts you already know: if √N = p/q in lowest terms with q > 0, then p² = N × q². Any prime dividing q would divide p² and hence p, contradicting lowest terms. Therefore q = 1; the root must be an integer, and N must be its square. For N = 0, the root is simply 0.

A calculator's finite display of √70 is an approximation, not evidence of a terminating exact decimal. Not every radical is irrational: √0.49 = 0.7 and √(81/196) = 9/14 are rational.

8 Products and sums behave differently

For nonnegative real a and b, √(a × b) = √a × √b. For division, also require b > 0. These rules follow by squaring the proposed nonnegative result. They do not supply a corresponding rule for a sum.

Worked example 7 Disprove a root of a sum shortcut

Is √(16 + 9) = √16 + √9?

1. Left side: √(16 + 9) = √25 = 5.
2. Right side: √16 + √9 = 4 + 3 = 7.
3. Since 5 ≠ 7, the claimed equality is false. The entire sum under the root must be evaluated first.

Error check: One counterexample disproves a general rule. The statement is not “the two sides can never be equal”: if one term is zero, they do agree. The point is that there is no rule allowing this split for every pair of nonnegative numbers.

9 Independent practice

Attempt all twelve questions before the answer key. Choose one option for each question and write a reason or calculation. This is original formative practice without timing, negative marking or a pass cutoff; it is not a full RRB CBT simulation.

1. What is 23²?

A. 46 B. 69 C. 529 D. 523

2. What is the principal root √225?

A. −15 B. 15 C. ±15 D. 25

3. What is √0?

A. −1 B. 1 C. Undefined D. 0

4. Which units digit immediately rules out an integer perfect square?

A. 2 B. 4 C. 6 D. 9

5. Use prime factorisation to find √2025.

A. 40 B. 45 C. 55 D. 90

6. What is the smallest positive integer by which 588 must be multiplied to obtain an integer perfect square?

A. 2 B. 6 C. 12 D. 3

7. What is the smallest positive integer divisor of 720 that leaves an integer perfect-square quotient?

A. 5 B. 10 C. 15 D. 20

8. Use digit-pair long division to find √10.24.

A. 0.32 B. 32 C. 3.2 D. 5.12

9. Find √(121/144).

A. 11/144 B. 121/12 C. 12/11 D. 11/12

10. Which consecutive-integer bounds contain √150?

A. 10 < √150 < 11 B. 12 < √150 < 13 C. 14 < √150 < 15 D. 150 < √150 < 151

11. Which correctly gives √64 and all real solutions of x² = 64?

A. √64 = ±8; x = 8 or −8 B. √64 = −8; x = 8 only C. √64 = 8; x = 8 or −8 D. √64 = 8; x = 8 only

12. Which evaluation and conclusion correctly check the claim √(25 + 144) = √25 + √144?

A. False: left side 13, right side 17 B. True: both sides 17 C. True: both sides 13 D. False: left side 169, right side 17

10 Explained answer key

1. C. 23² means 23 × 23. Multiply 23 × 20 = 460 and 23 × 3 = 69, then add: 460 + 69 = 529. Multiplying by 2 gives 46, which is not squaring.

2. B. 15² = 225 and 15 is nonnegative, so √225 = 15. Both signs solve x² = 225, but the radical denotes only the principal root.

3. D. 0² = 0, so √0 = 0. A root of zero is defined. Do not confuse it with division by zero, which is undefined.

4. A. Squaring the digits 0 through 9 produces only units digits 0, 1, 4, 5, 6 and 9. An ending of 2 is impossible. The allowed endings 4, 6 and 9 alone do not prove a square.

5. B. 2025 = 81 × 25 = 3⁴ × 5². Taking one factor from each pair gives 3² × 5 = 45. Check: 45² = 2025.

6. D. 588 = 2² × 3 × 7². Only the prime 3 has an odd count, so the smallest multiplier is 3. Then 588 × 3 = 1764 = 42². The multiplier 12 also works but is not the smallest.

7. A. 720 = 2⁴ × 3² × 5. Remove one factor 5 to make all prime counts even: 720 ÷ 5 = 144 = 12². Any qualifying divisor must include 5, so 5 is the smallest.

8. C. Group 10 . 24. First digit 3 gives 3² = 9 and remainder 1. Put the root decimal after 3. Bring down 24 to get 124; choose 2 because (60 + 2) × 2 = 124. Thus √10.24 = 3.2; check 3.2² = 10.24.

9. D. The numerator is 11² and the denominator is 12², so the nonnegative root is 11/12. Squaring both numerator and denominator checks the result. The root is less than 1, as 121/144 is less than 1.

10. B. 12² = 144 < 150 < 169 = 13², so 12 < √150 < 13. These bounds do not claim the root equals 12.5; that number would square to 156.25.

11. C. The radical selects the nonnegative root 8. Both 8² and (−8)² equal 64, so the equation has both solutions. A list containing only 8 omits a real solution.

12. A. The left side is √169 = 13, whereas the right side is 5 + 12 = 17. Since 13 ≠ 17, the claim is false. On the left, adding inside the root is not the last step; take the root of the sum.

Before moving on

Explain aloud why an allowed final digit is not enough, why a trial digit is multiplied by the trial divisor rather than the whole new root, and why a decimal point or zero digit cannot be skipped. If a practice error remains, redo the matching worked block and check your result by squaring.

11 Sources and scope note

• RRB CEN 09/2025 §14.1, Mathematics syllabus: https://www.rrbthiruvananthapuram.gov.in/assets/pdf/FinalDetailed_CEN_092025_Level1updated_on_30012026.pdf#page=28. Used for topic scope; the official indexed text was inspected, but the PDF binary was not verified in this source check.
• NCERT Class VIII Mathematics Chapter 5 via IIT Kanpur SATHEE: §§5.1–5.2, 5.5.1, 5.5.3–5.5.4 and 5.6, for squares, root meaning, factorisation and digit-pair methods: https://sathee.iitk.ac.in/sathee-railway-exams/student-corner/ncert-books/class-08/mathematics/chapter-5-squares-and-square-roots/. Relevant official indexed text was inspected; the official-host printed edition and page layout were not authenticated.
• NIOS Secondary Mathematics 211 Chapter 1, rational and irrational numbers, particularly §1.10, printed p.28: https://digital.nios.ac.in/content/211en/Chapter-1.pdf.
• Visual cross-check only: a third-party-hosted NCERT-labelled Reprint 2024–25 chapter at https://ncert24.com/assets/ots/pdf/ncert-books-class-8-maths-chapter-5.pdf, §§5.5.4–5.6, printed pp.65–68. Its byte identity with an official-host NCERT PDF remains unverified. No source page, figure or worked layout is reproduced here.

The explanations, solutions and text layouts in this lesson were authored independently. Standard mathematical target numbers may also occur in textbook tables or exercises. The trial-digit recurrence and numerical steps were checked independently; source wording is not an algorithm specification.

Analogy

From a square arrangement back to one side

Imagine 49 identical tiles arranged in a square grid. The same number of tiles must run across each row and down each column. Seven rows of seven tiles use 7 × 7 = 49 tiles, so the number along one side is √49 = 7.

The count along a side cannot be −7. This helps explain why the principal square root is nonnegative. In a separate number equation, however, both 7 and −7 solve x² = 49 because both square to 49. The physical tile model does not represent the negative solution.

If there are 50 tiles, they cannot all make a completely filled square grid with an integer number of tiles along each side. That does not mean √50 has no real value; it lies between 7 and 8. The tile comparison explains integer squares and the principal-root choice, not the full long-division procedure.

Quick reference

Meaning and signs

• a² = a × a; every real square is nonnegative
• For a ≥ 0, √a is the nonnegative principal root; √0 = 0
• For a > 0, x² = a has solutions x = √a and x = −√a; x² = 0 has only x = 0
• A negative number has no real square root; √((−6)²) = 6

Integer square checks

• Allowed last digits: 0, 1, 4, 5, 6, 9; an allowed ending does not prove a square
• Last digit 2, 3, 7 or 8 rules out an integer square
• A positive integer is a square exactly when every prime count is even; take one factor per pair for its root
• 0 and 1 are squares; do not prime-factorise 0
• Smallest positive integer multiplier or exact divisor producing an integer square: multiply one copy of each prime with an odd count; if none, use 1

Digit-pair long division

1. Pair integer digits to the left and decimal digits to the right from the decimal point. Pad a lone final fractional digit with a zero on its right
2. First group: choose the largest digit whose square fits, and subtract that square
3. Bring down the next pair p: new working number = 100 × old remainder + p
4. With current root digits read as the integer q, choose the largest digit d from 0 to 9 such that (20 × q + d) × d fits
5. Subtract that product; append d to give new prefix 10 × q + d. Use the whole new prefix next time, ignoring its decimal point in the trial calculation
6. Put the root decimal after the integer groups are finished. Keep zero root digits. Additional 00 pairs after the decimal give additional root places

Multiply by the new digit, not the whole new root. A zero remainder gives an exact answer only when all required places are accounted for. Otherwise continue or label the stated-precision result with ≈.

Fractions estimates and error checks

• For a ≥ 0 and b > 0, √(a/b) = √a/√b; check by squaring
• For a,b ≥ 0, √(a × b) = √a × √b; there is no general rule √(a + b) = √a + √b
• If k ≥ 0 and k² < N < (k + 1)², then k < √N < k + 1; bounds do not make the root their midpoint
• For nonnegative integer N, √N is rational exactly when N is a perfect square; a finite calculator display is not proof of exactness
• Preserve exact values until a rounding instruction requires an estimate; report the requested precision