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Squares and Square Roots - Finding Square Roots by Long Division Method

Grade 8ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Concept 1: Grouping with Bars (Pairing): To begin the process, we group the digits of the number in pairs starting from the units place and moving towards the left. For the decimal part, we group digits from the decimal point moving towards the right. Each bar, such as in 15‾21‾\overline{15}\overline{21}, represents one digit in the resulting square root quotient.

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Concept 2: Identifying the First Divisor: We look at the leftmost period (the first group under a bar). We find the largest number whose square is less than or equal to this period. For example, if the first period is 1515, we choose 33 because 32=93^2 = 9, which is less than 1515, while 42=164^2 = 16 is too large. This number 33 is written as both the first divisor and the first digit of the quotient.

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Concept 3: Forming the New Dividend: After subtracting the square of the quotient digit from the current period, we bring down the next pair of digits from the original number. These digits are placed to the right of the remainder to form the new dividend. If our remainder was 66 and the next pair is 2121, our new dividend becomes 621621.

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Concept 4: Doubling the Quotient for the Next Divisor: To find the next divisor, we double the current quotient and write it with a blank space for a new digit at the end. Visually, if the quotient is 33, the new divisor starts as 6x6\text{x}. We then find a digit xx such that the product (60+x)×x(60 + x) \times x is the largest possible value less than or equal to the current dividend.

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Concept 5: Decimal Point Placement: When finding the square root of a decimal number, the decimal point is placed in the quotient immediately after we have finished using the integer part of the number and are about to bring down the first pair of digits from the decimal part.

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Concept 6: Non-Perfect Squares and Precision: If the remainder is not zero after all pairs have been used, the number is not a perfect square. We can add pairs of zeros after a decimal point (e.g., 5.00005.0000) to continue the process and calculate the square root to a required number of decimal places, such as two or three decimal points.

📐Formulae

Number of digits in N=n2 (if n is even)\text{Number of digits in } \sqrt{N} = \frac{n}{2} \text{ (if } n \text{ is even)}

Number of digits in N=n+12 (if n is odd)\text{Number of digits in } \sqrt{N} = \frac{n+1}{2} \text{ (if } n \text{ is odd)}

New Divisor Step: (20×Q+x)×x≤Dividend (where Q is the current quotient)\text{New Divisor Step: } (20 \times Q + x) \times x \leq \text{Dividend} \text{ (where } Q \text{ is the current quotient)}

x2=∣x∣\sqrt{x^2} = |x|

💡Examples

Problem 1:

Find the square root of 10241024 using the long division method.

Solution:

  1. Group the digits into pairs from the right: 10‾24‾\overline{10}\overline{24}.
  2. The first period is 1010. The largest square ≤10\leq 10 is 32=93^2 = 9. Write 33 in the quotient and divisor. Subtract 99 from 1010 to get remainder 11.
  3. Bring down the next pair 2424. The new dividend is 124124.
  4. Double the current quotient (3×2=63 \times 2 = 6) to get the start of the new divisor: 6x6\text{x}.
  5. Find a digit xx such that 6x×x≤1246x \times x \leq 124. Trying x=2x = 2, we get 62×2=12462 \times 2 = 124.
  6. The remainder is 00, and the quotient is 3232.
  7. Therefore, 1024=32\sqrt{1024} = 32.

Explanation:

We paired the digits, found the square root of the first group (1010), then used the doubling rule to find the digit 22, which completed the perfect square.

Problem 2:

Find the square root of 31.3631.36.

Solution:

  1. Group the digits: 31‾.36‾\overline{31}.\overline{36}.
  2. For the first period 3131, the largest square ≤31\leq 31 is 52=255^2 = 25. Subtract 2525 from 3131 to get 66. The first quotient digit is 55.
  3. Bring down the next pair 3636 and place a decimal point in the quotient because we are crossing the decimal. The quotient is now 5.5. and the new dividend is 636636.
  4. Double the quotient 55 to get 1010. The new divisor is 10x10\text{x}.
  5. Find a digit xx such that 10x×x=63610x \times x = 636. Trying x=6x = 6, we get 106×6=636106 \times 6 = 636.
  6. The remainder is 00, and the quotient is 5.65.6.
  7. Therefore, 31.36=5.6\sqrt{31.36} = 5.6.

Explanation:

The decimal point is handled by placing it in the quotient as soon as we bring down the digits following the decimal point in the dividend.