SQUARE AND SQUARE ROOTS

1.1 Square Numbers

• A square number or perfect square is a natural number obtained by multiplying an integer by itself.
• If a natural number \(m\) can be expressed as \(n^2\) for some natural number \(n\), then \(m\) is a square number.
• Examples:
• \(1 = 1 \times 1 = 1^2\)
• \(4 = 2 \times 2 = 2^2\)
• \(9 = 3 \times 3 = 3^2\)
• \(16 = 4 \times 4 = 4^2\)

1.2 Identifying Perfect Squares

• To check if a number is a perfect square, find its prime factorization.
• If all prime factors appear in pairs, then the number is a perfect square.
• Example: Is 324 a perfect square?
• Prime factorization of \(324 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 = 2^2 \times 3^4 = (2 \times 3^2)^2 = 18^2\).
• Since all prime factors (2 and 3) are in pairs, 324 is a perfect square.

2.1 Unit Digits of Square Numbers

• Observe the unit digits of squares of numbers from 1 to 10:
• \(1^2 = 1\) (unit digit 1)
• \(2^2 = 4\) (unit digit 4)
• \(3^2 = 9\) (unit digit 9)
• \(4^2 = 16\) (unit digit 6)
• \(5^2 = 25\) (unit digit 5)
• \(6^2 = 36\) (unit digit 6)
• \(7^2 = 49\) (unit digit 9)
• \(8^2 = 64\) (unit digit 4)
• \(9^2 = 81\) (unit digit 1)
• \(10^2 = 100\) (unit digit 0)
• Conclusion: A number ending in 2, 3, 7, or 8 can never be a perfect square.
• Perfect squares can only end in 0, 1, 4, 5, 6, or 9.

2.2 Number of Zeros at the End of a Perfect Square

• If a number ends with zero, its square ends with an even number of zeros.
• Examples:
• \(10^2 = 100\) (2 zeros)
• \(20^2 = 400\) (2 zeros)
• \(100^2 = 10000\) (4 zeros)
• A number ending in an odd number of zeros is never a perfect square (e.g., 10, 1000, 250).

2.3 Squares of Even and Odd Numbers

• The square of an even number is always an even number.
• \(2^2 = 4\), \(4^2 = 16\), \(12^2 = 144\)
• The square of an odd number is always an odd number.
• \(3^2 = 9\), \(5^2 = 25\), \(11^2 = 121\)

2.4 Sum of Consecutive Odd Numbers

• The sum of the first \(n\) odd natural numbers is \(n^2\).
• Examples:
• \(1 = 1^2\)
• \(1 + 3 = 4 = 2^2\)
• \(1 + 3 + 5 = 9 = 3^2\)
• \(1 + 3 + 5 + 7 = 16 = 4^2\)
• This property can be used to check if a number is a perfect square by repeatedly subtracting consecutive odd numbers starting from 1. If the result is 0, it's a perfect square.

2.5 Numbers Between Consecutive Squares

• Between any two consecutive square numbers \(n^2\) and \((n+1)^2\), there are \(2n\) non-perfect square numbers.
• Example: Between \(2^2 = 4\) and \(3^2 = 9\):
• \(2n = 2 \times 2 = 4\) non-square numbers (5, 6, 7, 8).
• Between \(5^2 = 25\) and \(6^2 = 36\):
• \(2n = 2 \times 5 = 10\) non-square numbers (26, 27, ..., 35).

2.6 Adding Triangular Numbers

• Triangular numbers are numbers that can be represented as dots arranged in a triangle (1, 3, 6, 10, 15, ...).
• The sum of two consecutive triangular numbers is a perfect square.
• Examples:
• \(T_1 = 1, T_2 = 3 \implies T_1 + T_2 = 1 + 3 = 4 = 2^2\)
• \(T_2 = 3, T_3 = 6 \implies T_2 + T_3 = 3 + 6 = 9 = 3^2\)
• \(T_3 = 6, T_4 = 10 \implies T_3 + T_4 = 6 + 10 = 16 = 4^2\)

3.1 Direct Multiplication

• The most straightforward method is to multiply the number by itself.
• Example: \(23^2 = 23 \times 23 = 529\)

3.2 Using Algebraic Identity for Two-Digit Numbers

• For a two-digit number \(ab\) (where \(a\) is the tens digit and \(b\) is the units digit), we can write it as \((10a + b)\).
• Using the identity \((x+y)^2 = x^2 + 2xy + y^2\):
• \((10a + b)^2 = (10a)^2 + 2(10a)(b) + b^2 = 100a^2 + 20ab + b^2\)
• Example: Find \(23^2\)
• Here, \(a=2, b=3\).
• \(23^2 = (20+3)^2 = 20^2 + 2 \times 20 \times 3 + 3^2\)
• \(= 400 + 120 + 9 = 529\)

3.3 Column Method (for two-digit numbers)

• This method is based on the identity \((10a + b)^2 = 100a^2 + 20ab + b^2\).
• Steps:

1. Divide into three columns: Column I (\(a^2\)), Column II (\(2ab\)), Column III (\(b^2\)).
2. Write the values in each column.
3. Underline the unit digit of Column III and carry over the tens digit to Column II.
4. Add the carry-over to Column II. Underline the unit digit of the new Column II value and carry over the remaining digits to Column I.
5. Add the carry-over to Column I. Underline the entire resulting number.
6. The underlined digits, read from left to right, form the square.

• Example: Find \(23^2\)
• \(a=2, b=3\)
• Column I (\(a^2\)) | Column II (\(2ab\)) | Column III (\(b^2\))
• \(2^2 = 4\) | \(2 \times 2 \times 3 = 12\) | \(3^2 = 9\)
• Carry over from III to II: 0. Underline 9.
• \(4\) | \(12 + 0 = 12\) | \(\underline{9}\)
• Carry over from II to I: 1. Underline 2.
• \(4 + 1 = 5\) | \(\underline{2}\) | \(\underline{9}\)
• Underline 5.
• \(\underline{5}\) | \(\underline{2}\) | \(\underline{9}\)
• Result: 529

3.4 Diagonal Method (for numbers with more digits)

• This method is useful for finding squares of numbers with 2 or more digits.
• Steps:

1. Draw a square and divide it into smaller squares (cells) equal to the number of digits squared (e.g., 2 digits = 2x2 grid, 3 digits = 3x3 grid).
2. Draw diagonals in each small square.
3. Write the digits of the number horizontally above the cells and vertically to the left of the cells.
4. Multiply each digit on the left by each digit on the top and write the product in the corresponding cell. The tens digit goes above the diagonal, and the units digit goes below.
5. Add the numbers along the diagonals, starting from the bottom right. Carry over any tens digit to the next diagonal sum.
6. The digits obtained form the square.

• Example: Find \(23^2\)

1. Draw a 2x2 grid.
2. Write 2 above the first column, 3 above the second. Write 2 to the left of the first row, 3 to the left of the second.
3. Cell (1,1): \(2 \times 2 = 04\). Cell (1,2): \(2 \times 3 = 06\).

Cell (2,1): \(3 \times 2 = 06\). Cell (2,2): \(3 \times 3 = 09\).

1. Sum diagonals:

• Bottom-most: 9
• Next: \(6 + 0 + 6 = 12\) (write 2, carry 1)
• Next: \(0 + 4 + 0 + 1 \text{ (carry)} = 5\)
• Top-most: 0

1. Result: 0529 = 529

4.1 Pattern of \(1, 11, 111, ...\)

• \(1^2 = 1\)
• \(11^2 = 121\)
• \(111^2 = 12321\)
• \(1111^2 = 1234321\)
• Observation: The square of a number consisting only of ones (up to 9 ones) follows a pattern of ascending digits up to the number of ones, then descending back to 1.

4.2 Squares of Numbers Ending in 5

• Consider a number ending in 5, say \(N5\), where \(N\) is the digit(s) before 5.
• \((10N + 5)^2 = 100N^2 + 100N + 25 = 100N(N+1) + 25\)
• Rule: To find the square of a number ending in 5:

1. Multiply the digit(s) before 5 (let's call it \(N\)) by \((N+1)\).
2. Append 25 to the result.

• Examples:
• \(25^2\): \(N=2\). \(2 \times (2+1) = 2 \times 3 = 6\). Append 25. Result: 625.
• \(35^2\): \(N=3\). \(3 \times (3+1) = 3 \times 4 = 12\). Append 25. Result: 1225.
• \(105^2\): \(N=10\). \(10 \times (10+1) = 10 \times 11 = 110\). Append 25. Result: 11025.

4.3 Sum of Consecutive Natural Numbers

• Some square numbers can be expressed as the sum of two consecutive natural numbers.
• Rule: This is possible for squares of odd numbers.
• \(n^2 = \frac{n^2-1}{2} + \frac{n^2+1}{2}\)
• Examples:
• \(3^2 = 9 = 4 + 5\)
• \(5^2 = 25 = 12 + 13\)
• \(7^2 = 49 = 24 + 25\)

4.4 Product of Two Consecutive Even or Odd Numbers

• \((a-1)(a+1) = a^2 - 1^2 = a^2 - 1\)
• This means \(a^2 = (a-1)(a+1) + 1\).
• Rule: The square of a number is 1 more than the product of its preceding and succeeding numbers.
• Examples:
• \(11^2 = (10 \times 12) + 1 = 120 + 1 = 121\)
• \(24^2 = (23 \times 25) + 1 = 575 + 1 = 576\)
• This is particularly useful for squaring numbers close to a multiple of 10 or 5.

4.5 Pythagorean Property (covered in detail in t5)

• A set of three positive integers \((a, b, c)\) such that \(a^2 + b^2 = c^2\) is called a Pythagorean triplet.
• Example: \((3, 4, 5)\) is a Pythagorean triplet because \(3^2 + 4^2 = 9 + 16 = 25 = 5^2\).

5.1 Definition

• A Pythagorean triplet consists of three positive integers \((m, n, p)\) such that \(m^2 + n^2 = p^2\).
• These numbers represent the sides of a right-angled triangle, where \(p\) is the hypotenuse.
• Examples: \((3, 4, 5)\), \((6, 8, 10)\), \((5, 12, 13)\).

5.2 General Form of Pythagorean Triplets

• For any natural number \(m > 1\), the numbers \(2m, m^2-1, m^2+1\) form a Pythagorean triplet.
• Verification:
• We need to check if \((2m)^2 + (m^2-1)^2 = (m^2+1)^2\).
• LHS: \(4m^2 + (m^4 - 2m^2 + 1) = m^4 + 2m^2 + 1\)
• RHS: \((m^2+1)^2 = m^4 + 2m^2 + 1\)
• Since LHS = RHS, the formula is correct.

5.3 Finding a Triplet when One Member is Given

• If one member of the triplet is given, say \(x\), we can equate it to one of \(2m, m^2-1, m^2+1\) to find \(m\), and then find the other two members.
• Case 1: Given member is even.
• If the given member \(x\) is an even number, it is usually equated to \(2m\).
• Example: Find a Pythagorean triplet whose one member is 6.
• Let \(2m = 6 \implies m = 3\).
• Other members: \(m^2-1 = 3^2-1 = 9-1 = 8\).
• \(m^2+1 = 3^2+1 = 9+1 = 10\).
• The triplet is \((6, 8, 10)\).
• Check: \(6^2 + 8^2 = 36 + 64 = 100 = 10^2\).
• Case 2: Given member is odd.
• If the given member \(x\) is an odd number, it is usually equated to \(m^2-1\) or \(m^2+1\).
• Example: Find a Pythagorean triplet whose one member is 5.
• Let \(m^2-1 = 5 \implies m^2 = 6\). \(m = \sqrt{6}\) (not a natural number, so this assignment is not suitable).
• Let \(m^2+1 = 5 \implies m^2 = 4 \implies m = 2\).
• Other members: \(2m = 2 \times 2 = 4\).
• \(m^2-1 = 2^2-1 = 4-1 = 3\).
• The triplet is \((4, 3, 5)\) or \((3, 4, 5)\).
• Check: \(3^2 + 4^2 = 9 + 16 = 25 = 5^2\).

• Important Note: The formula \(2m, m^2-1, m^2+1\) generates primitive Pythagorean triplets (where the numbers have no common factors other than 1) only if \(m\) is chosen appropriately. However, for Class 8, this general form is sufficient.

6.1 Introduction to Square Roots

• The square root of a number is the inverse operation of squaring a number.
• If \(n^2 = m\), then \(n\) is the square root of \(m\).
• The symbol for square root is \(\sqrt{}\) (radical sign).
• Example: Since \(5^2 = 25\), the square root of 25 is 5. We write \(\sqrt{25} = 5\).
• Important: Every positive number has two square roots: a positive one and a negative one. For example, \(\sqrt{25} = \pm 5\) because \(5^2 = 25\) and \((-5)^2 = 25\). However, in this chapter, we primarily focus on the positive square root of a natural number.

6.2 Methods for Finding Square Roots

• There are several methods to find the square root of a number:

1. Repeated Subtraction Method (for perfect squares)
2. Prime Factorization Method (for perfect squares)
3. Division Method (for perfect squares and non-perfect squares)
4. Estimation Method (for perfect squares)

• We will explore each method in detail.

7.1 Concept

• This method is based on the property that the sum of the first \(n\) odd natural numbers is \(n^2\).
• Conversely, if we subtract consecutive odd numbers (1, 3, 5, 7, ...) from a perfect square until we reach 0, the number of subtractions will be its square root.
• This method is practical only for small perfect squares.

7.2 Steps

1. Start with the given number.
2. Subtract the first odd number (1).
3. From the result, subtract the next odd number (3).
4. Continue subtracting consecutive odd numbers (5, 7, 9, ...) from the previous result.
5. Count the number of steps (subtractions) until the result is 0. This count is the square root.

7.3 Example: Find \(\sqrt{81}\)

1. \(81 - 1 = 80\)
2. \(80 - 3 = 77\)
3. \(77 - 5 = 72\)
4. \(72 - 7 = 65\)
5. \(65 - 9 = 56\)
6. \(56 - 11 = 45\)
7. \(45 - 13 = 32\)
8. \(32 - 15 = 17\)
9. \(17 - 17 = 0\)

• We performed 9 subtractions. Therefore, \(\sqrt{81} = 9\).

8.1 Concept

• This method relies on the fact that if a number is a perfect square, its prime factors can be grouped into pairs.
• To find the square root, we take one factor from each pair.

8.2 Steps

1. Find the prime factorization of the given number.
2. Group the identical prime factors in pairs.
3. If the number is a perfect square, all prime factors will form pairs. If any factor is left unpaired, the number is not a perfect square.
4. Take one prime factor from each pair.
5. Multiply these selected prime factors. The product is the square root of the given number.

8.3 Example: Find \(\sqrt{324}\)

1. Prime factorization of 324:

• \(324 = 2 \times 162\)
• \(162 = 2 \times 81\)
• \(81 = 3 \times 27\)
• \(27 = 3 \times 9\)
• \(9 = 3 \times 3\)
• So, \(324 = 2 \times 2 \times 3 \times 3 \times 3 \times 3\)

1. Group factors in pairs:

• \(324 = (2 \times 2) \times (3 \times 3) \times (3 \times 3)\)

1. Take one factor from each pair:

• \(2 \times 3 \times 3\)

1. Multiply them:

• \(2 \times 3 \times 3 = 18\)

• Therefore, \(\sqrt{324} = 18\).

8.4 Finding the Smallest Multiplier/Divisor to Make a Perfect Square

• If a number is not a perfect square, its prime factorization will have some unpaired factors.
• To make it a perfect square:
• Multiply: Multiply the number by the product of all unpaired prime factors.
• Divide: Divide the number by the product of all unpaired prime factors.
• Example: Is 252 a perfect square? If not, by what smallest number should it be multiplied/divided to make it a perfect square?
• Prime factorization of \(252 = 2 \times 2 \times 3 \times 3 \times 7\)
• Grouping pairs: \((2 \times 2) \times (3 \times 3) \times 7\)
• The factor 7 is unpaired. So, 252 is not a perfect square.
• To make it a perfect square, multiply by 7: \(252 \times 7 = 1764 = (2 \times 3 \times 7)^2 = 42^2\).
• To make it a perfect square, divide by 7: \(252 \div 7 = 36 = (2 \times 3)^2 = 6^2\).

9.1 Concept

• The division method is a general method for finding the square root of any number (perfect square or not, integers or decimals) to any desired number of decimal places.
• It's a systematic procedure similar to long division.

9.2 Steps for Integers

1. Pairing Digits: Draw lines over every pair of digits starting from the unit's place (right to left). If the number of digits is odd, the leftmost single digit will have a bar.
2. First Digit/Pair: Find the largest number whose square is less than or equal to the first pair (or single digit). This number is the first digit of the square root. Write it as the divisor and the quotient. Subtract its square from the first pair.
3. Bring Down Next Pair: Bring down the next pair of digits to the right of the remainder to form the new dividend.
4. New Divisor: Double the quotient obtained so far and write it with a blank digit to its right. This forms the new partial divisor.
5. Find Next Digit: Find the largest digit (let's say \(x\)) to fill the blank in the new partial divisor such that when the new divisor (e.g., \(2q x\)) is multiplied by \(x\), the product is less than or equal to the new dividend. This digit \(x\) is the next digit of the square root. Write \(x\) in the quotient and subtract the product (new divisor \(\times x\)) from the new dividend.
6. Repeat: Repeat steps 3-5 until all pairs of digits have been brought down and the remainder is 0 (for perfect squares) or to the desired number of decimal places.

9.3 Example: Find \(\sqrt{529}\)

` 2 3 _______ 2| 5 29 | -4 |_____ 43| 1 29 | -1 29 |______ | 0 `

1. Pair digits: \(\overline{5}\ \overline{29}\).
2. First digit: Largest square \(\le 5\) is \(2^2 = 4\). Write 2 in quotient. \(5-4 = 1\).
3. Bring down 29. New dividend = 129.
4. Double quotient (2) = 4. New partial divisor = \(4\_\).
5. Find \(x\) such that \(4x \times x \le 129\). If \(x=3\), \(43 \times 3 = 129\). Write 3 in quotient. \(129-129 = 0\).
6. Remainder is 0. So, \(\sqrt{529} = 23\).

9.4 Steps for Decimals

1. Pairing Digits: For the integral part, pair digits from right to left. For the decimal part, pair digits from left to right. Add zeros if needed to make even pairs in the decimal part.
2. Proceed as with integers. When you bring down the first pair of decimal digits, place a decimal point in the quotient.

9.5 Example: Find \(\sqrt{22.09}\)

`

1. 7

_______ 4| 22.09 | -16 |_____ 87| 6 09 | -6 09 |______ | 0 `

1. Pair digits: \(\overline{22}.\overline{09}\).
2. First digit: Largest square \(\le 22\) is \(4^2 = 16\). Write 4 in quotient. \(22-16 = 6\).
3. Bring down 09 (decimal part). Place decimal point in quotient. New dividend = 609.
4. Double quotient (4) = 8. New partial divisor = \(8\_\).
5. Find \(x\) such that \(8x \times x \le 609\). If \(x=7\), \(87 \times 7 = 609\). Write 7 in quotient. \(609-609 = 0\).
6. Remainder is 0. So, \(\sqrt{22.09} = 4.7\).

10.1 Concept

• This method is useful for quickly estimating the square root of large perfect squares.
• It combines knowledge of unit digits of squares and powers of 10.

10.2 Steps

1. Unit Digit: Look at the unit digit of the number. This tells you the possible unit digit(s) of its square root.

• If number ends in 1, root ends in 1 or 9.
• If number ends in 4, root ends in 2 or 8.
• If number ends in 5, root ends in 5.
• If number ends in 6, root ends in 4 or 6.
• If number ends in 9, root ends in 3 or 7.
• If number ends in 00, root ends in 0.

1. Tens Digit (or higher): Group the digits into pairs from the right. Ignore the unit's place pair for now.
2. Consider the remaining digits (the leftmost group). Find the largest perfect square less than or equal to this group. The square root of this perfect square gives the first digit(s) of the estimated square root.
3. Combine the possibilities to find the exact square root.

10.3 Example: Find \(\sqrt{1764}\) by estimation

1. Unit Digit: The number 1764 ends in 4. So, its square root must end in 2 or 8.
2. Tens Digit: Group digits from right: \(17\ \overline{64}\). Ignore \(64\).
3. Consider the leftmost group: 17.

• Largest perfect square \(\le 17\) is \(4^2 = 16\).
• So, the tens digit of the square root is 4.

1. Combine: The square root could be 42 or 48.

• To decide, we can check a number in between, like \(45^2\).
• Using the 'ending in 5' rule: \(4 \times (4+1) = 20\). Append 25. So, \(45^2 = 2025\).
• Since \(1764 < 2025\), the square root must be less than 45.
• Therefore, \(\sqrt{1764} = 42\).

10.4 Example: Find \(\sqrt{7921}\) by estimation

1. Unit Digit: Ends in 1. Root ends in 1 or 9.
2. Tens Digit: Group digits: \(79\ \overline{21}\). Ignore \(21\).
3. Consider 79.

• Largest perfect square \(\le 79\) is \(8^2 = 64\) (since \(9^2 = 81\) is too large).
• So, the tens digit is 8.

1. Combine: The square root could be 81 or 89.

• Check \(85^2\): \(8 \times (8+1) = 72\). Append 25. So, \(85^2 = 7225\).
• Since \(7921 > 7225\), the square root must be greater than 85.
• Therefore, \(\sqrt{7921} = 89\).