Overview: Squares and Square Roots

Square Numbers

When a whole number is multiplied by itself, the result is called a square number or a perfect square. If a natural number p can be written as y² where y is a natural number, then p is a square number.

• Example:
  \[4 = 2^2\]
  \[9 = 3^2\]
  \[16 = 4^2\]
• Exponential form shows a number multiplied by itself: for instance,
  \[3 \times 3 = 3^2\]
• Squares of negative integers are positive because a negative times a negative is positive. For example,
  \[(-4)^2 = 16\]
  and
  \[\sqrt{16} = 4\]
• Some commonly recognised square numbers are 1, 4, 9, 16, 25, 36, 49, 64, ...
• If a number cannot be expressed as a number multiplied by itself (for example 45 = 9 × 5), it is not a perfect square.

List of Square Numbers

Odd and Even Square Numbers

• Square of an even number is even. For any integer n,
  \[(2n)^2 = 4n^2\]
• Square of an odd number is odd. For any integer n,
  \[(2n+1)^2 = 4(n^2 + n) + 1\]
• Every odd square is of the form 4k + 1. Therefore numbers of the form 4k + 3 cannot be perfect squares.

Properties of Square Numbers

The following properties are useful when recognising or working with square numbers:

• No perfect square ends in the digits 2, 3, 7 or 8 in base ten.
• If the number of trailing zeros in a number is odd, the number is not a perfect square; if the number of trailing zeros is even, the number may be a square (for example 100 = 10²).
• Squaring preserves parity: even → even, odd → odd.
• For any natural number (other than 1), its square is either a multiple of 3 or leaves remainder 1 when divided by 3.
• For any natural number (other than 1), its square is either a multiple of 4 or leaves remainder 1 when divided by 4.
• The unit digit of the square of a number depends only on the unit digit of the original number.
• If a number nis squared, the difference between consecutive squares satisfies
  \[(n+1)^2 - n^2 = (n+1) + n = 2n + 1.\]
• The square of a number equals the sum of the first nodd natural numbers.
  \[1 + 3 + 5 + \dots + (2n-1) = n^2.\]
• For any natural number n > 1, the triple \((2n, n^2 - 1, n^2 + 1)\) is a Pythagorean triplet because
  \[(2n)^2 + (n^2 - 1)^2 = (n^2 + 1)^2.\]
• If a number ends with 1 or 9 then its square ends with 1.
• If a number ends with 4 or 6 then its square ends with 6.
• If a number ends with 0 then its square ends with an even number of zeros.

Finding the Square of a Number

To compute the square of a number efficiently, split the number into convenient parts and use the identity for expansion:

Example: Find the square of 53.

Solution: Divide the number in two parts.
53 = 50 + 3

\[53^2 = (50 + 3)^2\]

\[53^2 = 50^2 + 2 \times 50 \times 3 + 3^2\]

\[53^2 = 2500 + 300 + 9\]

\[53^2 = 2809\]

Pattern for Numbers Ending with 5

For any integer whose last digit is 5, write the number as \((a5)\) where \(a\) denotes the integer formed by the digits before 5. Then

Examples:

• \[25^2 = (2 \times 3)\,100 + 25 = 600 + 25 = 625.\]
• \[45^2 = (4 \times 5)\,100 + 25 = 2000 + 25 = 2025.\]
• \[95^2 = (9 \times 10)\,100 + 25 = 9000 + 25 = 9025.\]
• \[125^2 = (12 \times 13)\,100 + 25 = 15600 + 25 = 15625.\]

Pythagorean Triplets

If the sum of the squares of two integers equals the square of a third integer, those three integers form a Pythagorean triplet. Using the formula above, for any natural p > 1:

Thus, \((2p,\, p^2 - 1,\, p^2 + 1)\) is a triplet.

Example: Write a Pythagorean triplet having 22 as one its members.

Solution:

Let 2p = 6

P = 3

p² + 1 = 10

p² - 1 = 8.

Thus, the Pythagorean triplet is 6, 8 and 10.

6² + 8² = 10²

36 + 64 = 100

Square Root

Taking the square root is the inverse operation of squaring. If \(x^2 = y\), then \(x\) is a square root of \(y\) and we write \(\sqrt{y} = x\) (for the principal positive square root).

The symbol used for the positive square root is the radical sign √.

How to Find the Square Root of an Integer?

Two standard methods are commonly used: prime factorisation and the long-division method.

(i) Prime Factorisation Method

When a number is a perfect square, express it as a product of prime factors. Pair identical prime factors and take one from each pair to form the square root.

• Step I: Resolve the given number into prime factors.
• Step II: Group the same prime factors into pairs.
• Step III: Take one factor from each pair and multiply them; this product is the square root.

Example 1: Find \(\sqrt{484}\) by prime factorisation.

Solution:
Resolving 484 as the product of primes, we get

484 = 2 × 2 × 11 × 11 
√484 = √(2 × 2 × 11 × 11) 
= 2 × 11 
Therefore, √484 = 22

Example 2: Find \(\sqrt{324}\) by prime factorisation.

Solution: The square root of 324 by prime factorization, we get 

324 = 2 × 2 × 3 × 3 × 3 × 3
√324 = √(2 × 2 × 3 × 3 × 3 × 3) 
= 2 × 3 × 3 
Therefore, √324 = 18 

(ii) Long Division Method

This method works well for large numbers or when prime factors are not easy to determine. It is also useful for finding square roots of decimals. The steps are:

1. Group digits in pairs (periods) from the units place to the left, and from the decimal point to the right.
2. Find the largest digit whose square is less than or equal to the leftmost period; this is the first digit of the quotient (the root).
3. Subtract the square just found from the leftmost period and bring down the next period to form the new dividend.
4. Double the current quotient, write it as a tentative divisor, and choose a digit to append to this tentative divisor so that the product of the new divisor and that digit is ≤ new dividend; this digit is the next digit of the quotient.
5. Repeat until all periods are used. The quotient is the required square root.

Example 1: Find \(\sqrt{784}\) using the long division method.

\[\sqrt{784} = 28.\]

Example 2: Evaluate \(\sqrt{5329}\) using the long division method.

\[\sqrt{5329} = 73.\]

Properties of Squares

• In the prime factorisation of a perfect square, each prime occurs an even number of times.
• The difference between consecutive squares is an odd number equal to the sum of those consecutive numbers:
  \[(n+1)^2 - n^2 = 2n + 1.\]
• The unit digit of a square determines possible unit digits of its root: if a square ends in 1, the root ends in 1 or 9; if 4, root ends in 2 or 8; if 9, root ends in 3 or 7; if 6, root ends in 4 or 6.
• The square of any number ending in 5 ends with 25, and preceding digits are given by \(a(a+1)\) where \(a\) are the digits left of 5.
• A perfect square must end in 1, 4, 5, 6, 9, or an even number of zeros; it cannot end in 2, 3, 7, 8 or an odd number of zeros.
• The sum of squares of the first nnatural numbers is
  \[\frac{n(n+1)(2n+1)}{6}.\]
• Squaring increases the magnitude for integers other than 0 and 1; for numbers between 0 and 1 squaring reduces the value (for example, \(0.5^2 < />

Square Root Notation and Examples

If \(x^2 = y\), then \(x\) is a square root of \(y\) and we write

Examples:

• \[\sqrt{4} = 2.\]
• \[\sqrt{9} = 3.\]
• \[\sqrt{196} = 14.\]

Important Points to Remember

• 
• 
• Numbers ending in 8 can never be perfect squares.
• Memorise squares and cubes of 2 to 10 to speed up calculations.

Square Roots of Decimals

To find the square root of a decimal number using the long-division method, place bars over pairs of digits starting from the decimal point outward. Treat each bar as a period and proceed as in the integer long-division method. Insert the decimal point in the root at the appropriate stage when you cross the decimal point during the division.

Example: Find \(\sqrt{7.29}\) using the division method.

Solution:

Thus, √7.29 = 2.7

Remark: To put the bar on a number like 174.241, we will put a bar on 74 and a bar on 1 as it is a single digit left. And in the numbers after decimal, we will put a bar on 24 and put zero after 1 to make it double-digit.

174. 24 10

Estimating Square Roots

When an exact square root is not required or not easy to compute, estimate by locating the number between two consecutive perfect squares.

Solution:

We know that, 300 comes between 100 and 400 i.e. 100 < 300 < 400.
Now, √100 = 10 and √400 = 20.

 So, we can say that

10 < √300 < 20.

We can further estimate the numbers as we know that 17² = 289 and 18² = 324.
Thus, we can say that the square root of √300 = 17 as 289 is much closer to 300 than 324.

Solved Questions on Squaring and Square Roots

Q1: Simplify the following expression.

Sol.

Q2: Simplify the following expression.

Sol.

Q3: Assume x to be positive. Multiply the eighth power of the fourth root of x by the fourth power of the eighth root of x. What is the product?

Sol.