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Class 8th Chapters
1. Rational Numbers 2. Linear Equations in One Variable 3. Understanding Quadrilaterals
4. Practical Geometry 5. Data Handling 6. Squares and Square Roots
7. Cubes and Cube Roots 8. Comparing Quantities 9. Algebraic Expressions and Identities
10. Visualising Solid Shapes 11. Mensuration 12. Exponents and Powers
13. Direct and Inverse Proportions 14. Factorisation 15. Introduction to Graphs
16. Playing with Numbers

Content On This Page
Square Numbers or Perfect Squares Properties of Square Numbers Patterns
Pythagorean Triplets Square Roots Finding Square Roots through Repeated Subtraction and Prime Factorisation
Finding Square Root by Division Method Estimating the Number of Digits of Square Root Square Roots of Decimals
Square Root of a Fraction

Chapter 6 Squares and Square Roots (Concepts)

Embark on this numerical exploration as we delve into the fundamental concepts of Squares and Square Roots. This chapter builds upon basic arithmetic operations by introducing the idea of squaring a number – a process of self-multiplication – and its crucial inverse operation, finding the square root. Understanding squares and square roots is essential not only for advancing in algebra and geometry but also for solving practical problems involving area, distance, and various scientific calculations. We will investigate the properties of numbers generated by squaring and learn systematic methods to reverse this process.

The square of a number is simply the result obtained when the number is multiplied by itself. If we represent a number by the variable $n$, its square is denoted mathematically as $n^2$, which is equivalent to $n \times n$. For example, the square of $5$ is $5^2 = 5 \times 5 = 25$, and the square of $-3$ is $(-3)^2 = (-3) \times (-3) = 9$. We will practice calculating squares of various numbers, including integers and rational numbers (e.g., $(\frac{2}{3})^2 = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$). Numbers obtained by squaring integers (like 1, 4, 9, 16, 25, ...) are known as perfect squares or square numbers.

Perfect squares exhibit several interesting and useful properties that help in identifying them or understanding their nature:

Intriguing patterns also emerge, such as the sum of the first $n$ odd natural numbers being equal to $n^2$ (e.g., $1 = 1^2$; $1+3 = 4 = 2^2$; $1+3+5 = 9 = 3^2$). We also encounter Pythagorean triplets: sets of three positive integers $(a, b, c)$ that satisfy the Pythagorean theorem relation $a^2 + b^2 = c^2$. A common formula used to generate such triplets (for an integer $m > 1$) is $(2m, m^2 - 1, m^2 + 1)$. For $m=2$, this yields $(4, 3, 5)$ since $4^2 + 3^2 = 16+9 = 25 = 5^2$.

The inverse operation to squaring is finding the square root, denoted by the radical symbol $\sqrt{\phantom{x}}$. The square root of a given number $x$ is a number $y$ such that when $y$ is squared, the result is $x$ (i.e., $y^2 = x$). Finding square roots is a major focus, and we will learn two primary methods:

  1. Prime Factorization Method: This involves finding the prime factors of the given number and grouping identical factors into pairs. For each pair, one factor is taken out of the square root. This method works perfectly for finding the exact square root of perfect squares. For example, $\sqrt{144} = \ $$ \sqrt{2 \times 2 \times 2 \times 2 \times 3 \times 3} = \ $$ \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3)} = \ $$ 2 \times 2 \times 3 = \ $$ 12$.
  2. Long Division Method: This is a systematic, step-by-step algorithm, similar in structure to arithmetic long division, used to find the square root of any positive number. It is particularly useful for finding the square roots of large numbers and for finding approximate decimal values for non-perfect squares.

It's important to note that every positive number technically has two square roots: a positive one and a negative one (e.g., both $3^2 = 9$ and $(-3)^2 = 9$, so the square roots of 9 are $+3$ and $-3$, often written as $\pm 3$). However, the symbol $\sqrt{x}$ by convention usually denotes the principal (positive) square root. We will also practice estimating the square roots of numbers that are not perfect squares. The skills learned here have direct applications in geometry, such as finding the side length of a square given its area ($side = \sqrt{Area}$) and applying the Pythagorean theorem ($c = \sqrt{a^2 + b^2}$). This chapter significantly enhances numerical fluency by introducing this vital inverse operation.

Square Numbers or Perfect Squares

In geometry, we often calculate the area of a square by multiplying its side length by itself. For example, if a square has a side of length 3 units, its area is $3 \times 3 = 9$ square units.

Let's look at the areas of squares with different side lengths:

The numbers we get as the area (1, 4, 9, 16, 25, ...) are special. They are obtained by multiplying a whole number by itself.

Definition of Square Number or Perfect Square

When a number is multiplied by itself, the resulting product is called the square of that number.

For example, the square of 7 is $7 \times 7 = 49$.

We use a special notation to represent the square of a number. We write the number and put a small '2' as a superscript above and to the right of it. This '2' indicates that the number is multiplied by itself (raised to the power of 2).

The square of a number $n$ is written as $n^2 = n \times n$.

... (i)

For example:

A natural number (or a positive integer) is called a square number or a perfect square if it is the square of some natural number.

In other words, a natural number $N$ is a perfect square if there exists a natural number $n$ such that $N = n^2$.

Examples of perfect squares:

Examples of numbers that are NOT perfect squares:

The numbers 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, etc., are not perfect squares because they cannot be expressed as the square of any natural number. There is no natural number $n$ such that $n^2$ equals any of these numbers.

List of First Few Square Numbers:

It's helpful to know the squares of the first few natural numbers by heart. Here is a table showing the squares of natural numbers from 1 to 20:

Number (n) Square ($n^2$)
1$1^2 = 1$
2$2^2 = 4$
3$3^2 = 9$
4$4^2 = 16$
5$5^2 = 25$
6$6^2 = 36$
7$7^2 = 49$
8$8^2 = 64$
9$9^2 = 81$
10$10^2 = 100$
11$11^2 = 121$
12$12^2 = 144$
13$13^2 = 169$
14$14^2 = 196$
15$15^2 = 225$
16$16^2 = 256$
17$17^2 = 289$
18$18^2 = 324$
19$19^2 = 361$
20$20^2 = 400$

You can extend this table further to find the squares of larger numbers.

Checking if a Number is a Perfect Square

One way to check if a number is a perfect square is to see if it can be written as the product of two equal integers. We will learn more systematic methods, like prime factorisation, later in this chapter.

For now, let's look at the unit digits (last digit) of the square numbers in the table above: 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0.

Notice a pattern: The unit digit of a square number is always one of these digits: 0, 1, 4, 5, 6, or 9.

This gives us a simple way to identify numbers that are NOT perfect squares:

Any number ending with the digits 2, 3, 7, or 8 at the unit place is never a perfect square.

Example: Is 152 a perfect square? No, because its unit digit is 2. Is 237 a perfect square? No, because its unit digit is 7.

However, if a number ends with 0, 1, 4, 5, 6, or 9, it *might* be a perfect square, but it's not guaranteed. For example, 14 is not a perfect square, but it ends in 4. 36 is a perfect square and ends in 6. So, the unit digit rule can only tell you if a number is definitely *not* a perfect square.

Example 1. Which of the following numbers are perfect squares? 121, 132, 289, 1000.

Answer:

  • 121: We know that $11 \times 11 = 121$, or $11^2 = 121$. Since 121 is the square of the natural number 11, 121 is a perfect square.
  • 132: The unit digit of 132 is 2. Numbers ending in 2 are not perfect squares. So, 132 is not a perfect square.
  • 289: We know that $17 \times 17 = 289$, or $17^2 = 289$. Since 289 is the square of the natural number 17, 289 is a perfect square.
  • 1000: The unit digit is 0. This doesn't immediately rule it out. Let's think about numbers ending in 0. $10^2 = 100$, $20^2 = 400$, $30^2 = 900$. When you square a number ending in 0, the square ends in an even number of zeros (00, 0000, etc.). 1000 ends in three zeros, which is an odd number. Therefore, 1000 is not a perfect square.

Properties of Square Numbers

Now that we know what square numbers (or perfect squares) are, let's explore some interesting properties they possess. These properties can help us identify perfect squares and understand their characteristics without necessarily calculating their square roots.

Property 1: Unit Digits of Square Numbers

Let's revisit the unit digits of the squares of the first few natural numbers:

Observe the pattern of the unit digits of these square numbers: 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0.

The digits that appear in the unit's place of a perfect square are always one of these: 0, 1, 4, 5, 6, or 9.

This leads to a useful property: Any number ending with the digits 2, 3, 7, or 8 at its unit's place cannot be a perfect square.

This is a quick check to eliminate numbers that are definitely not perfect squares. However, remember that if a number ends in 0, 1, 4, 5, 6, or 9, it is *not necessarily* a perfect square (e.g., 14 ends in 4 but is not a perfect square; 20 ends in 0 but is not a perfect square). This rule only helps to identify numbers that are *not* perfect squares.

Example 1. Without calculating the squares, determine which of the following numbers are definitely not perfect squares:

1024, 1025, 1026, 1027, 1028, 1133, 1257, 1348

Answer:

We examine the unit digit of each number:

  • 1024: Unit digit is 4. A square number *can* end in 4 (e.g., $2^2=4, 8^2=64$). This number *might* be a perfect square. (In fact, $32^2 = 1024$).
  • 1025: Unit digit is 5. A square number *can* end in 5 (e.g., $5^2=25, 15^2=225$). This number *might* be a perfect square.
  • 1026: Unit digit is 6. A square number *can* end in 6 (e.g., $4^2=16, 6^2=36$). This number *might* be a perfect square.
  • 1027: Unit digit is 7. Numbers ending in 7 are never perfect squares. So, 1027 is definitely not a perfect square.
  • 1028: Unit digit is 8. Numbers ending in 8 are never perfect squares. So, 1028 is definitely not a perfect square.
  • 1133: Unit digit is 3. Numbers ending in 3 are never perfect squares. So, 1133 is definitely not a perfect square.
  • 1257: Unit digit is 7. Numbers ending in 7 are never perfect squares. So, 1257 is definitely not a perfect square.
  • 1348: Unit digit is 8. Numbers ending in 8 are never perfect squares. So, 1348 is definitely not a perfect square.

The numbers that are definitely not perfect squares are 1027, 1028, 1133, 1257, and 1348.

Property 2: Number of Zeros at the End of a Perfect Square

Consider the squares of numbers ending in zero:

If a number ends in one zero, its square ends in two zeros ($10^2 = 100$). If a number ends in two zeros, its square ends in four zeros ($100^2 = 10000$). If a number ends in $k$ zeros, its square ends in $2k$ zeros.

This observation gives us another property: A perfect square can only have an even number of zeros at its end. A number ending with an odd number of zeros (like one zero, three zeros, five zeros, etc.) is never a perfect square.

Example 2. Which of the following numbers are perfect squares?

400, 8000, 250000, 36000, 100000, 490000

Answer:

We examine the number of zeros at the end of each number:

  • 400: Ends in 2 zeros (even number). Check if the non-zero part (4) is a perfect square. Yes, $2^2=4$. So, $400 = 4 \times 100 = 2^2 \times 10^2 = (2 \times 10)^2 = 20^2$. It is a perfect square.
  • 8000: Ends in 3 zeros (odd number). It is not a perfect square.
  • 250000: Ends in 4 zeros (even number). Check if the non-zero part (25) is a perfect square. Yes, $5^2=25$. So, $250000 = 25 \times 10000 = 5^2 \times 100^2 = (5 \times 100)^2 = 500^2$. It is a perfect square.
  • 36000: Ends in 3 zeros (odd number). It is not a perfect square.
  • 100000: Ends in 5 zeros (odd number). It is not a perfect square.
  • 490000: Ends in 4 zeros (even number). Check if the non-zero part (49) is a perfect square. Yes, $7^2=49$. So, $490000 = 49 \times 10000 = 7^2 \times 100^2 = (7 \times 100)^2 = 700^2$. It is a perfect square.

The perfect squares among the given numbers are 400, 250000, and 490000.

Property 3: Square of an Even Number and an Odd Number

Let's look at the squares of some even and odd numbers:

Observation: The square of an even number is always an even number. The square of an odd number is always an odd number.

This property is useful: if a number is odd, its square must be odd. If a number is even, its square must be even. This also means that if a perfect square is odd, its square root must be odd, and if a perfect square is even, its square root must be even.

Property 4: Numbers Between Consecutive Squares

Consider the squares of consecutive natural numbers and count the numbers between them:

Do you see a pattern? Between the squares of $n$ and $(n+1)$, the number of non-perfect square numbers is 2 times $n$.

Property: There are $2n$ non-perfect square numbers between the squares of two consecutive natural numbers, $n^2$ and $(n+1)^2$.

Let's think why this is true. The natural numbers between $n^2$ and $(n+1)^2$ are $n^2+1, n^2+2, \dots, (n+1)^2-1$. The number of integers from $A$ to $B$ (inclusive) is $B - A + 1$. The number of integers *between* $A$ and $B$ is $B - A - 1$. Here, $A = n^2$ and $B = (n+1)^2$. The number of integers between $n^2$ and $(n+1)^2$ is $(n+1)^2 - n^2 - 1$. $(n+1)^2 - n^2 - 1 = (n^2 + 2n + 1) - n^2 - 1 = n^2 + 2n + 1 - n^2 - 1 = 2n$. Since $n^2$ and $(n+1)^2$ are consecutive perfect squares, any integer between them cannot be a perfect square. Thus, there are exactly $2n$ non-perfect square numbers between $n^2$ and $(n+1)^2$.

Example 3. How many non-perfect square numbers lie between the squares of 15 and 16?

Answer:

We need to find the number of non-perfect square numbers between $15^2$ and $16^2$.

Here, we are looking at the numbers between the squares of two consecutive natural numbers, $n$ and $n+1$. In this case, $n=15$ and $n+1=16$.

The number of non-perfect square numbers between $n^2$ and $(n+1)^2$ is given by the formula $2n$.

Number of non-perfect square numbers $= 2 \times n$

[Using Property 4]

Substitute $n=15$:

Number $= 2 \times 15$

$= 30$

So, there are 30 non-perfect square numbers lying between $15^2$ and $16^2$.

Check: $15^2 = 225$ and $16^2 = 256$. The numbers between 225 and 256 are 226, 227, ..., 255. The number of integers from 226 to 255 (inclusive) is $255 - 226 + 1 = 29 + 1 = 30$. This matches the formula.


Patterns in Square Numbers

Square numbers are not just the result of multiplying a number by itself; they reveal several fascinating patterns. Discovering these patterns can make working with squares more interesting and can sometimes help in identifying perfect squares or performing calculations more quickly.

Pattern 1: Sum of Consecutive Odd Numbers

Consider the sums of the first few consecutive odd numbers starting from 1:

The pattern is clear: The sum of the first $n$ consecutive odd numbers is equal to $n^2$.

Sum of first $n$ odd numbers $= 1 + 3 + 5 + \dots + (2n-1) = n^2$

... (i)

This property provides a method to check if a natural number is a perfect square. You can try subtracting consecutive odd numbers (1, 3, 5, ...) from the given number, starting with 1. If the number becomes 0 after subtracting a certain number of consecutive odd numbers, then the original number is a perfect square, and the number of subtractions performed is its square root.

Example 1. Is 25 a perfect square? Use the method of subtracting consecutive odd numbers to check.

Answer:

We will subtract consecutive odd numbers starting from 1 from 25:

Step 1: $25 - 1 = 24$

Step 2: $24 - 3 = 21$

Step 3: $21 - 5 = 16$

Step 4: $16 - 7 = 9$

Step 5: $9 - 9 = 0$

We reached 0 after performing 5 consecutive subtractions of odd numbers starting from 1.

Therefore, 25 is a perfect square, and its square root is the number of subtractions, which is 5.

Pattern 2: Squares of Numbers Ending in 5

There is a simple trick to quickly calculate the square of a number ending in 5.

Consider a number whose unit digit is 5. We can write such a number as $(10 \times N + 5)$, where $N$ is the number formed by the digits before 5. For example, for 25, $N=2$; for 45, $N=4$; for 105, $N=10$.

The square of $(10N + 5)$ is $(10N + 5)^2 = (10N + 5)(10N + 5)$

$(10N + 5)^2 = (10N)(10N) + (10N)(5) + (5)(10N) + 5^2$

$= 100N^2 + 50N + 50N + 25$

$= 100N^2 + 100N + 25$

$= 100N(N + 1) + 25$

... (ii)

This formula tells us the pattern: To square a number ending in 5:

  1. Identify the number $N$ formed by the digits before the 5.
  2. Multiply $N$ by the next consecutive integer, $(N+1)$.
  3. Write '25' at the end of the result obtained in step 2.

Examples:

Pattern 3: Difference of Squares

Consider the difference between the squares of two consecutive natural numbers. Let the two consecutive natural numbers be $n$ and $(n+1)$.

The difference between their squares is $(n+1)^2 - n^2$.

Using the algebraic identity $a^2 - b^2 = (a+b)(a-b)$, where $a = (n+1)$ and $b = n$:

$(n+1)^2 - n^2 = ((n+1) + n)((n+1) - n)$

$= (n+1+n)(n+1-n)$

$= (2n+1)(1)$

$= 2n+1$

... (iii)

Also, the sum of the two consecutive numbers $n$ and $(n+1)$ is $n + (n+1) = 2n+1$.

So, the difference between the squares of two consecutive natural numbers is equal to the sum of the two consecutive numbers.

Examples:

Pattern 4: Sum of Two Consecutive Natural Numbers

We just saw that the difference between the squares of two consecutive natural numbers is an odd number ($2n+1$). This also means that any odd number ($k$) can be expressed as the difference of the squares of two consecutive natural numbers. If $k = 2n+1$, then $k = (n+1)^2 - n^2$.

If an odd number $k$ can be expressed as the sum of two consecutive natural numbers, say $n$ and $n+1$, then $k = n + (n+1) = 2n+1$. The numbers $n$ and $n+1$ can be found if $k$ is known: $2n = k-1 \implies n = \frac{k-1}{2}$, and $n+1 = \frac{k-1}{2} + 1 = \frac{k+1}{2}$.

So, any odd number $k$ can be written as the sum of two consecutive natural numbers $\frac{k-1}{2}$ and $\frac{k+1}{2}$.

Example: Express 15 as the sum of two consecutive natural numbers. $k=15$ (odd). $n = \frac{15-1}{2} = \frac{14}{2} = 7$. $n+1 = 7+1=8$. So, $15 = 7 + 8$.

If an odd number is also a perfect square, say $m^2$, then we can express $m^2$ as the sum of two consecutive natural numbers. $m^2 = n + (n+1)$, where $n = \frac{m^2-1}{2}$ and $n+1 = \frac{m^2+1}{2}$.

This gives us a relationship: for an odd perfect square $m^2$, we have $m^2 + n^2 = (n+1)^2$, which means $m^2 + (\frac{m^2-1}{2})^2 = (\frac{m^2+1}{2})^2$. This is a way to generate Pythagorean Triplets (discussed later). For any odd number $m$, $(m, \frac{m^2-1}{2}, \frac{m^2+1}{2})$ forms a Pythagorean triplet.

Example 2. Express the perfect square 81 as the sum of two consecutive natural numbers.

Answer:

The given perfect square is 81. It is an odd number.

We can express any odd number as the sum of two consecutive natural numbers using the formula $k = n + (n+1)$, where $n = \frac{k-1}{2}$. Here $k=81$.

$n = \frac{81-1}{2} = \frac{80}{2} = 40$

The two consecutive natural numbers are $n=40$ and $n+1=41$.

Check: $40 + 41 = 81$. This is correct.

Also, related to the difference of squares: $41^2 - 40^2 = (41-40)(41+40) = 1 \times 81 = 81$.

So, the perfect square 81 can be expressed as the sum of the consecutive natural numbers 40 and 41.


Pythagorean Triplets

In geometry, you might recall the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (legs). If the legs have lengths $a$ and $b$, and the hypotenuse has length $c$, then $a^2 + b^2 = c^2$.

Definition of a Pythagorean Triplet

A collection of three positive integers $a, b, c$ such that $a^2 + b^2 = c^2$ is called a Pythagorean triplet. These three integers represent the lengths of the sides of a right-angled triangle.

The most famous example is the triplet (3, 4, 5), because $3^2 + 4^2 = 9 + 16 = 25$, and $5^2 = 25$. Since $3^2 + 4^2 = 5^2$, (3, 4, 5) is a Pythagorean triplet.

Other examples of Pythagorean triplets include (6, 8, 10), (5, 12, 13), (8, 15, 17), etc. You can verify these:

Generating Pythagorean Triplets

While you can find Pythagorean triplets by trial and error, there is a useful formula to generate them. For any natural number $m > 1$, the triplet $(2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet.

Let's verify this formula by substituting these expressions into the Pythagorean theorem $a^2 + b^2 = c^2$. Let $a = 2m$, $b = m^2 - 1$, and $c = m^2 + 1$.

We need to check if $(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2$.

Calculate the LHS:

LHS $= (2m)^2 + (m^2 - 1)^2$

Using the identity $(a-b)^2 = a^2 - 2ab + b^2$ for $(m^2 - 1)^2$:

$= (2m \times 2m) + ((m^2)^2 - 2(m^2)(1) + 1^2)$

$= 4m^2 + (m^4 - 2m^2 + 1)$

$= m^4 + 4m^2 - 2m^2 + 1$

$= m^4 + 2m^2 + 1$

Calculate the RHS:

RHS $= (m^2 + 1)^2$

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$:

$= (m^2)^2 + 2(m^2)(1) + 1^2$

$= m^4 + 2m^2 + 1$

Since LHS = RHS ($m^4 + 2m^2 + 1 = m^4 + 2m^2 + 1$), the equation $(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2$ is true for any value of $m$.

Thus, for any natural number $m > 1$, the integers $2m$, $m^2 - 1$, and $m^2 + 1$ form a Pythagorean triplet. Note that the condition $m>1$ is used to ensure that $m^2-1$ is a positive integer (if $m=1$, $m^2-1 = 0$, which is not a side length of a triangle).

Using the Formula to Generate Triplets:

Choose a natural number $m > 1$ and substitute it into the formula $(2m, m^2 - 1, m^2 + 1)$:

This formula generates many, but not all, Pythagorean triplets. For example, (7, 24, 25) is a Pythagorean triplet ($7^2 + 24^2 = 49 + 576 = 625 = 25^2$), which can be generated by $2m=...$, $m^2-1=...$, $m^2+1=...$. If $m^2-1=7$, $m^2=8$, $m=\sqrt{8}$ not natural. If $2m=7$, $m=3.5$ not natural. If $m^2+1=25$, $m^2=24$, $m=\sqrt{24}$ not natural. However, the triplet (7, 24, 25) can be generated using a more general formula $(m^2-n^2, 2mn, m^2+n^2)$ for natural numbers $m>n$. With $m=4, n=3$, $(4^2-3^2, 2 \times 4 \times 3, 4^2+3^2) = (16-9, 24, 16+9) = (7, 24, 25)$. The formula $(2m, m^2-1, m^2+1)$ is a special case of this more general formula when $n=1$. For Class 8, the formula $(2m, m^2 - 1, m^2 + 1)$ is usually sufficient.

Example 1. Write a Pythagorean triplet whose smallest member is 8.

Answer:

Given: Smallest member of the Pythagorean triplet is 8.

To Find: The other two members of the triplet.

Solution:

We use the general form of a Pythagorean triplet generated by the formula $(2m, m^2 - 1, m^2 + 1)$ for $m > 1$. The smallest member could be $2m$ or $m^2 - 1$. (Note that $m^2+1$ is always greater than $m^2-1$ for $m>1$ and greater than $2m$ for most $m>1$, but not always, e.g., for $m=2$, $2m=4$, $m^2-1=3$, $m^2+1=5$; for $m=3$, $2m=6$, $m^2-1=8$, $m^2+1=10$).

Let's consider the possibilities for the smallest member being 8 using the formula $(2m, m^2 - 1, m^2 + 1)$:

Case 1: Let $2m$ be the smallest member.

$2m = 8$

$m = \frac{8}{2} = 4$

Since $m=4$ is a natural number and $4 > 1$, this is a valid value for $m$ using this formula.

Substitute $m=4$ into the other parts of the triplet:

$m^2 - 1 = 4^2 - 1 = 16 - 1 = 15$

$m^2 + 1 = 4^2 + 1 = 16 + 1 = 17$

The triplet is (8, 15, 17).

Let's check if 8 is indeed the smallest member of this triplet: 8, 15, 17. Yes, 8 is the smallest.

Case 2: Let $m^2 - 1$ be the smallest member.

$m^2 - 1 = 8$

$m^2 = 8 + 1 = 9$

$m = \sqrt{9} = 3$

(Since $m$ must be a natural number)

Since $m=3$ is a natural number and $3 > 1$, this is a valid value for $m$ using this formula.

Substitute $m=3$ into the other parts of the triplet:

$2m = 2(3) = 6$

$m^2 - 1 = 3^2 - 1 = 9 - 1 = 8$

$m^2 + 1 = 3^2 + 1 = 9 + 1 = 10$

The triplet is (6, 8, 10). The members are 6, 8, and 10. The smallest member in this triplet is 6, not 8. So, this is not the required triplet.

Case 3: Let $m^2 + 1$ be the smallest member.

$m^2 + 1 = 8$

$m^2 = 8 - 1 = 7$

$m = \sqrt{7}$

Since $m=\sqrt{7}$ is not a natural number, this case is not possible using the formula $(2m, m^2 - 1, m^2 + 1)$ to generate a triplet with 8 as $m^2+1$.

From the cases considered, the only Pythagorean triplet generated by the formula $(2m, m^2 - 1, m^2 + 1)$ that has 8 as its smallest member is (8, 15, 17).

Check the triplet (8, 15, 17): $8^2 + 15^2 = 64 + 225 = 289$. $17^2 = 289$. $8^2 + 15^2 = 17^2$. It is a Pythagorean triplet. The smallest member is 8.

The required Pythagorean triplet is (8, 15, 17).


Square Roots

We have learned about squaring a number, which is multiplying a number by itself. Now, we will explore the inverse operation of squaring, which is finding the square root. Finding the square root of a number is like asking: "What number, when multiplied by itself, gives the original number?"

Definition of Square Root

If the square of a number $m$ is equal to another number $n$, i.e., $m^2 = n$, then $m$ is called the square root of $n$.

In simpler terms, the square root of a number $n$ is the number $m$ such that when $m$ is squared ($m \times m$), you get $n$.

The symbol used to denote the square root is $\sqrt{}$. This symbol is called the radical sign.

So, if $m^2 = n$, we write $\sqrt{n} = m$.

Examples:

Finding the square root is essentially looking for the base number when the exponent is 2.

Positive and Negative Square Roots

Consider the number 16. We know that $4^2 = 4 \times 4 = 16$. So, 4 is a square root of 16. However, let's also consider $(-4)^2 = (-4) \times (-4)$. Remember the rule for multiplying negative numbers: negative times negative is positive. So, $(-4) \times (-4) = 16$. This means that -4 is also a square root of 16.

Every positive number has two square roots: one positive and one negative.

The positive square root of a number is called the principal square root and is denoted by the symbol $\sqrt{}$.

Example: $\sqrt{16} = 4$ (This refers specifically to the positive square root). $\sqrt{25} = 5$. $\sqrt{100} = 10$.

The negative square root of a number is denoted by $-\sqrt{}$.

Example: $-\sqrt{16} = -4$. $-\sqrt{25} = -5$. $-\sqrt{100} = -10$.

When we refer to both square roots of a positive number, we use the symbol $\pm \sqrt{}$.

Example: The square roots of 16 are $\pm \sqrt{16} = \pm 4$ (meaning +4 and -4).

In this chapter, unless mentioned otherwise, when we talk about the square root of a positive number, we will generally be referring to its positive (principal) square root.

Example: Find the square root of 49. We are looking for the positive number whose square is 49. We know $7^2 = 49$. So, $\sqrt{49} = 7$. Note that $(-7)^2 = 49$ as well, but $\sqrt{49}$ specifically means the positive root.

Square Root of Zero:

The square of 0 is $0^2 = 0 \times 0 = 0$. The only number whose square is 0 is 0 itself. So, the square root of 0 is 0.

$\sqrt{0} = 0$

Square Root of a Negative Number:

Can we find the square root of a negative number, like $\sqrt{-9}$? We are looking for a real number $m$ such that $m^2 = -9$.

The square of any real number (positive, negative, or zero) is always greater than or equal to zero ($m^2 \geq 0$). Therefore, there is no real number whose square is a negative number.

Conclusion: Negative numbers do not have real square roots. You will learn about imaginary numbers in higher classes, which are used to find square roots of negative numbers.

Example 1. What is the principal square root of 64?

Answer:

We need to find the positive number whose square is 64.

We know that $8^2 = 8 \times 8 = 64$.

The principal (positive) square root of 64 is 8.

$\sqrt{64} = 8$

Example 2. Find the square roots of 144.

Answer:

We need to find the numbers whose square is 144.

We know that $12^2 = 12 \times 12 = 144$. So, 12 is a square root.

Also, $(-12)^2 = (-12) \times (-12) = 144$. So, -12 is also a square root.

The square roots of 144 are 12 and -12.

We can write this using the $\pm$ symbol:

The square roots of 144 are $\pm \sqrt{144} = \pm 12$


Finding Square Roots through Repeated Subtraction and Prime Factorisation

We have defined square roots and identified some properties of perfect squares. Now, let's learn systematic methods to find the square root of a number, particularly when the number is a perfect square.

Method 1: Repeated Subtraction of Consecutive Odd Numbers

This method is based on the property that the sum of the first $n$ consecutive odd numbers is $n^2$. If a number is a perfect square, say $N = n^2$, then subtracting the first $n$ consecutive odd numbers from $N$ will result in 0.

To find the square root of a number using this method, repeatedly subtract consecutive odd numbers (1, 3, 5, 7, ...) from the number, in order, until you reach 0. If you reach 0, the number is a perfect square, and the number of subtractions you performed is its square root.

Example 1. Find the square root of 81 by the method of repeated subtraction.

Answer:

We start with 81 and subtract consecutive odd numbers:

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 reached 0 after exactly 9 steps of subtracting consecutive odd numbers starting from 1.

Therefore, 81 is a perfect square, and its square root is 9. $\sqrt{81} = 9$.

This method is simple conceptually, but it becomes very lengthy and impractical for finding the square roots of large numbers.

Method 2: Prime Factorisation Method

The prime factorisation method is a more systematic and efficient way to find the square root of a perfect square. It is based on the fact that if a number is a perfect square, then each of its prime factors appears an even number of times in its prime factorisation.

Steps to find the square root of a number using prime factorisation:

  1. Find the prime factorisation of the given number. Express the number as a product of its prime factors.
  2. Group the prime factors in pairs of identical factors. For a number to be a perfect square, every prime factor must form a pair.
  3. For each pair of identical prime factors, take one factor.
  4. Multiply the factors selected in Step 3. The product is the square root of the given number.

If, after prime factorisation, any prime factor is left unpaired, it means the original number is not a perfect square.

Example 2. Find the square root of 144 by prime factorisation.

Answer:

We need to find the square root of 144.

Step 1: Find the prime factorisation of 144.

$\begin{array}{c|cc} 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorisation of 144 is $2 \times 2 \times 2 \times 2 \times 3 \times 3$.

Step 2: Group the identical prime factors in pairs.

$144 = (2 \times 2) \times (2 \times 2) \times (3 \times 3)$

Each prime factor (2 and 3) forms a pair. This confirms that 144 is a perfect square.

Step 3: Take one factor from each pair.

From $(2 \times 2)$, take 2.

From $(2 \times 2)$, take 2.

From $(3 \times 3)$, take 3.

Step 4: Multiply the selected factors to find the square root.

$\sqrt{144} = 2 \times 2 \times 3$

$= 4 \times 3 = 12$

The square root of 144 is 12.

Example 3. Is 2904 a perfect square? If not, find the smallest natural number by which it must be multiplied so that the product is a perfect square. Also, find the square root of the resulting perfect square.

Answer:

To check if 2904 is a perfect square, we find its prime factorisation.

Step 1: Prime factorisation of 2904.

$\begin{array}{c|cc} 2 & 2904 \\ \hline 2 & 1452 \\ \hline 2 & 726 \\ \hline 3 & 363 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

The prime factorisation of 2904 is $2 \times 2 \times 2 \times 3 \times 11 \times 11$.

Step 2: Group the identical prime factors in pairs.

$2904 = (2 \times 2) \times 2 \times 3 \times (11 \times 11)$

We see that the prime factors 2 and 3 are left unpaired. Since not all prime factors can be grouped into pairs, 2904 is not a perfect square.

To make 2904 a perfect square, we need to multiply it by the factors that are unpaired, so that all factors form pairs. The unpaired factors are 2 and 3. To make them pairs, we need to multiply by $2 \times 3 = 6$.

The smallest natural number by which 2904 must be multiplied to get a perfect square is 6.

The new number is $2904 \times 6$. Its prime factorisation will be:

$2904 \times 6 = (2 \times 2 \times 2 \times 3 \times 11 \times 11) \times (2 \times 3)$

$= (2 \times 2 \times 2 \times 2) \times (3 \times 3) \times (11 \times 11)$

Now, all prime factors are in pairs. The number is $17424$.

Step 3: Find the square root of the resulting perfect square (17424) by taking one factor from each pair.

$\sqrt{17424} = \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (11 \times 11)}$

$= 2 \times 2 \times 3 \times 11$

$= 4 \times 33 = 132$

The square root of the resulting perfect square is 132.

Example 4. Is 128 a perfect square? If not, find the smallest natural number by which it must be divided so that the quotient is a perfect square. Also, find the square root of the resulting perfect square.

Answer:

To check if 128 is a perfect square, we find its prime factorisation.

Step 1: Prime factorisation of 128.

$\begin{array}{c|cc} 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

The prime factorisation of 128 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$.

Step 2: Group the identical prime factors in pairs.

$128 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times 2$

We see that one factor 2 is left unpaired. Therefore, 128 is not a perfect square.

To make 128 a perfect square by division, we must remove the unpaired factor. The unpaired factor is 2. Dividing 128 by 2 will eliminate this unpaired factor.

The smallest natural number by which 128 must be divided to get a perfect square is 2.

The resulting number (quotient) is $128 \div 2 = 64$.

The prime factorisation of the resulting number (64) is $(2 \times 2) \times (2 \times 2) \times (2 \times 2)$. All prime factors are now in pairs.

Step 3: Find the square root of the resulting perfect square (64) by taking one factor from each pair.

$\sqrt{64} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2)}$

$= 2 \times 2 \times 2 = 8$

The square root of the resulting perfect square is 8.


Finding Square Root by Division Method

The prime factorisation method is efficient for smaller numbers and for determining if a number is a perfect square. However, for larger numbers, or when dealing with decimals, or when you need to find the square root of a non-perfect square up to a certain number of decimal places, the division method is generally more practical and faster.

Steps for Finding Square Root by Division Method (for Whole Numbers):

This method involves a process similar to long division. Here are the steps:

  1. Pair the Digits: Write the number down. Place a bar over every pair of digits starting from the unit's digit (the rightmost digit) and moving towards the left. If the number of digits is odd, the leftmost single digit will also have a bar over it. Each section under a bar is called a 'period'.
  2. Find the First Digit of the Root: Consider the number under the leftmost bar. Find the largest number whose square is less than or equal to this number. This number is the first digit of the square root. Write this digit above the leftmost period (as the quotient) and also to the left (as the divisor).
  3. First Subtraction and Bring Down: Subtract the square of the first digit (found in step 2) from the number under the leftmost bar. Bring down the next period (the pair of digits under the next bar) to the right of the remainder. This combined number is the new dividend.
  4. Prepare the New Divisor: Double the quotient obtained so far and write it down, leaving a blank space to its right. This forms the new partial divisor (e.g., if the quotient is 'a', the partial divisor is '2a_').
  5. Find the Next Digit of the Root: Find the largest possible digit (0 to 9) to fill the blank space in the partial divisor such that when the new divisor (e.g., '2ax') is multiplied by the same digit (x), the product is less than or equal to the current dividend. Write this digit (x) as the next digit in the quotient (above the next period) and also in the blank space of the divisor.
  6. Subtract and Repeat: Subtract the product (new divisor $\times$ selected digit) from the current dividend. Bring down the next pair of digits (the next period) to the right of this remainder to form the new dividend.
  7. Repeat: Repeat steps 4, 5, and 6 until all periods have been brought down and the remainder is 0 (if the original number is a perfect square). If the remainder is not 0, you can add pairs of zeros after the decimal point and continue the process to find the square root of a non-perfect square or a perfect square with decimal places.

Example 1. Find the square root of 529 by the division method.

Answer:

Given number: 529.

Step 1: Place bars over pairs of digits from right to left: $\overline{5}\overline{29}$. The leftmost period is 5.

Step 2: Find the largest number whose square is less than or equal to 5. $1^2 = 1$, $2^2 = 4$, $3^2 = 9$. The largest square $\leq 5$ is 4, and its square root is 2. Write 2 as the first digit of the quotient and the divisor.

Step 3: Subtract $2^2 = 4$ from 5. Remainder is $5 - 4 = 1$. Bring down the next period (29). The new dividend is 129.

Step 4: Double the current quotient (2) to get 4. Write it to the left, leaving a blank space: 4_.

Step 5: Find the largest digit to fill the blank such that $4\_ \times \_$ is $\leq 129$. Try 1: $41 \times 1 = 41$ (less than 129). Try 2: $42 \times 2 = 84$ (less than 129). Try 3: $43 \times 3 = 129$ (equal to 129). The digit is 3. Write 3 as the next digit in the quotient and in the blank space of the divisor. The new divisor is 43.

Step 6: Subtract $43 \times 3 = 129$ from the dividend 129. Remainder is $129 - 129 = 0$.

All periods have been brought down, and the remainder is 0.

The calculation can be shown as:

$\begin{array}{c|cc} & 2 \ 3 & \\ \hline \phantom{()} 2 & \overline{5} \ \overline{29} & \\ + \; 2 & 4\phantom{29)} \\ \hline \phantom{()} 4 \; 3 & 1 \; 29 \\ \phantom{()} \times 3 & 1 \; 29 \\ \hline \phantom{()} 46 & 0 \end{array}$

The quotient is 23. The remainder is 0.

Therefore, the square root of 529 is 23. $\sqrt{529} = 23$.

Example 2. Find the square root of 4096 by the division method.

Answer:

Given number: 4096.

Step 1: Place bars over pairs of digits from right to left: $\overline{40}\overline{96}$. The leftmost period is 40.

Step 2: Find the largest number whose square is less than or equal to 40. $6^2 = 36$, $7^2 = 49$. The largest square $\leq 40$ is 36, and its square root is 6. Write 6 as the first digit of the quotient and the divisor.

Step 3: Subtract $6^2 = 36$ from 40. Remainder is $40 - 36 = 4$. Bring down the next period (96). The new dividend is 496.

Step 4: Double the current quotient (6) to get 12. Write it to the left, leaving a blank space: 12_.

Step 5: Find the largest digit to fill the blank such that $12\_ \times \_$ is $\leq 496$. Try 1: $121 \times 1 = 121$. Try 2: $122 \times 2 = 244$. Try 3: $123 \times 3 = 369$. Try 4: $124 \times 4 = 496$. The digit is 4. Write 4 as the next digit in the quotient and in the blank space of the divisor. The new divisor is 124.

Step 6: Subtract $124 \times 4 = 496$ from the dividend 496. Remainder is $496 - 496 = 0$.

All periods have been brought down, and the remainder is 0.

The calculation can be shown as:

$\begin{array}{c|cc} & 6 \ 4 & \\ \hline \phantom{()} 6 & \overline{40} \ \overline{96} & \\ + \; 6 & 36\phantom{96)} \\ \hline \phantom{()} 12 \; 4 & 4 \; 96 \\ \phantom{()} \times 4 & 4 \; 96 \\ \hline \phantom{()} 128 & 0 \end{array}$

The quotient is 64. The remainder is 0.

Therefore, the square root of 4096 is 64. $\sqrt{4096} = 64$.


Estimating the Number of Digits of Square Root

Before find ing the square root of a large perfect square, it can be helpful to know how many digits its square root will have. This can give you an idea of the magnitude of the result and help in checking your answer when using calculation methods like the division method. There's a simple rule based on the number of digits in the original number.

Rule for Estimating the Number of Digits

Let $N$ be a perfect square number. Let $n$ be the number of digits in $N$. The number of digits in the square root of $N$ can be estimated based on whether $n$ is an even or an odd number.

This rule is directly related to the process of placing bars over pairs of digits in the division method. The number of bars (or periods) you get over the number when pairing from the right indicates the number of digits in the square root.

Let's verify this rule with some examples:

Example 1. Without calculating the square root, estimate the number of digits in the square root of 65536.

Answer:

Given number: 65536.

Count the number of digits in the number. 65536 has 5 digits.

The number of digits is $n = 5$.

Since $n=5$ is an odd number, we use the formula for the number of digits in the square root for an odd number of digits: $\frac{n+1}{2}$.

Number of digits in $\sqrt{65536} = \frac{5+1}{2}$

[Using the rule for odd n]

$= \frac{6}{2} = 3$

So, the square root of 65536 will have 3 digits.

Verification (Optional, not needed for the answer): Using the division method or calculator, $\sqrt{65536} = 256$. The number 256 has 3 digits, which matches our estimate.

Example 2. Estimate the number of digits in the square root of the largest 4-digit perfect square.

Answer:

The largest 4-digit number is 9999. The largest 4-digit perfect square is the square of the largest integer whose square is less than or equal to 9999. We know $100^2 = 10000$. So, we check $99^2$. $99^2 = 9801$. The largest 4-digit perfect square is 9801.

The number 9801 has 4 digits.

The number of digits is $n = 4$.

Since $n=4$ is an even number, we use the formula for the number of digits in the square root for an even number of digits: $\frac{n}{2}$.

Number of digits in $\sqrt{9801} = \frac{4}{2}$

[Using the rule for even n]

$= 2$

So, the square root of the largest 4-digit perfect square will have 2 digits.

Verification: $\sqrt{9801} = 99$, which indeed has 2 digits.

Making a Number a Perfect Square

Sometimes we need to find the smallest number to multiply, divide, add, or subtract from a given number to make it a perfect square. The methods for these operations differ.

Finding the Least Number to be Multiplied or Divided

To find the least number to multiply or divide a number by to make it a perfect square, we use the prime factorization method. A number is a perfect square if all its prime factors occur in pairs.

Example 3. Find the smallest whole number by which 252 must be multiplied to get a perfect square. Also, find the square root of the new number.

Answer:

First, we find the prime factors of 252.

$\begin{array}{c|cc} 2 & 252 \\ \hline 2 & 126 \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

The prime factorization of 252 is $2 \times 2 \times 3 \times 3 \times 7$.

Let's group the factors into pairs: $(2 \times 2) \times (3 \times 3) \times 7$.

Here, the prime factor 7 does not have a pair. To make 252 a perfect square, we need to complete the pair for 7. Therefore, we must multiply 252 by 7.

The smallest number to be multiplied is 7.

New number = $252 \times 7 = 1764$.

Now, find the square root of the new number.

$\sqrt{1764} = \sqrt{(2 \times 2) \times (3 \times 3) \times (7 \times 7)}$

From each pair, we take one factor:

$\sqrt{1764} = 2 \times 3 \times 7 = 42$.

So, the square root of the new number is 42.

Finding the Least Number to be Subtracted or Added

To find the least number to subtract or add to a number to make it a perfect square, we use the long division method for finding square roots.

Example 4. Find the least number which must be subtracted from 402 to obtain a perfect square. Also, find the square root of this perfect square.

Answer:

We use the long division method to find the square root of 402.

$\begin{array}{c|cc} & 2\ 0 & \\ \hline \phantom{()} 2 & \overline{4} \; \overline{02} \\ + \; 2 & 4\phantom{(...)} \\ \hline \phantom{()} 4 \; 0 & 0 \; 02 \phantom{(.)} \\ \phantom{()} +0 & 00 \phantom{(.)} \\ \hline \phantom{()} & 2 \\ \end{array}$

From the long division, we see that $20^2 = 400$, and when we try to find the square root of 402, we get a remainder of 2. This means that 402 is 2 more than the perfect square $20^2$.

Therefore, the least number that must be subtracted from 402 to make it a perfect square is the remainder, which is 2.

New number = $402 - 2 = 400$.

The square root of the new number is the quotient from the division.

$\sqrt{400} = 20$.

Example 5. Find the least number which must be added to 1300 so as to get a perfect square. Also, find the square root of the perfect square so obtained.

Answer:

We use the long division method to find an estimate for the square root of 1300.

$\begin{array}{c|cc} & 3\ 6 & \\ \hline \phantom{()} 3 & \overline{13} \; \overline{00} \\ + \; 3 & 9\phantom{(...)} \\ \hline \phantom{()} 6 \; 6 & 4 \; 00 \phantom{(.)} \\ \phantom{()} +6 & 396 \phantom{(.)} \\ \hline \phantom{()} & 4 \\ \end{array}$

The quotient is 36, and there is a remainder of 4. This tells us that $36^2 < 1300$.

The next perfect square after $36^2$ is $37^2$.

Let's calculate $36^2$ and $37^2$.

$36^2 = 1296$

$37^2 = 1369$

The given number, 1300, lies between these two perfect squares. To make 1300 a perfect square by adding to it, we must make it equal to the next perfect square, which is 1369.

The number to be added is:

Number to be added $= 37^2 - 1300$

$ = 1369 - 1300 = 69$

So, the least number to be added is 69.

New perfect square = $1300 + 69 = 1369$.

The square root of the new number is 37.

$\sqrt{1369} = 37$.


Square Roots of Decimals

We can also find the square root of a decimal number using the division method. The process is very similar to finding the square root of a whole number, with one important addition: handling the decimal point and pairing the digits in the decimal part.

Steps for Finding Square Root of a Decimal Number by Division Method:

Here's how to apply the division method to a decimal number:

  1. Place Bars on Integral Part: Place bars over every pair of digits in the integral part (the whole number part) of the decimal number, starting from the unit's digit and moving towards the left. If the integral part has an odd number of digits, the leftmost single digit will have a bar.
  2. Place Bars on Decimal Part: Place bars over every pair of digits in the decimal part, starting from the first digit after the decimal point and moving towards the right. If the decimal part has an odd number of digits, add a zero at the end to make it even, and then place the bars.
  3. Start Division: Find the largest number whose square is less than or equal to the number under the leftmost bar (in the integral part). This number is the first digit of the square root. Write it as the quotient and the divisor for the first step, just like with whole numbers.
  4. Handle the Decimal Point in Quotient: Continue the division process. As soon as you bring down the first pair of digits from the decimal part, place a decimal point in the quotient before writing the next digit.
  5. Continue Division Process: Bring down the next pairs of digits (periods) from the decimal part one by one. For each new dividend, double the quotient obtained so far (ignoring the decimal point for the purpose of doubling) and proceed as in the division method for whole numbers (find the blank digit, multiply, subtract).
  6. Repeat: Repeat the steps until the remainder is 0 (for a perfect square decimal) or until you have obtained the square root up to the desired number of decimal places. If you need the square root up to, say, two decimal places, you should calculate it up to three decimal places and then round off. To do this, add pairs of zeros (00) after the last digit of the decimal part of the original number as needed to get enough periods for the calculation.

Example 1. Find the square root of 12.25.

Answer:

Given number: 12.25.

Step 1 & 2: Place bars. For the integral part (12), place a bar over 12: $\overline{12}$. For the decimal part (.25), place a bar over 25: $\overline{.25}$. The number with bars is $\overline{12}\overline{.25}$. The leftmost period is 12.

Step 3: Find the largest number whose square is $\leq 12$. $3^2 = 9$, $4^2 = 16$. The largest is 3. Write 3 as the first digit of the quotient and the divisor.

Step 4: Subtract $3^2 = 9$ from 12. Remainder is $12 - 9 = 3$. Bring down the next period (25). This is the first pair from the decimal part (.25), so place a decimal point in the quotient after 3. The new dividend is 325.

Step 5: Double the current quotient (3) to get 6. Write it to the left, leaving a blank space: 6_.

Step 6: Find the largest digit to fill the blank such that $6\_ \times \_$ is $\leq 325$. Try 1: $61 \times 1 = 61$. Try 2: $62 \times 2 = 124$. Try 3: $63 \times 3 = 189$. Try 4: $64 \times 4 = 256$. Try 5: $65 \times 5 = 325$. The digit is 5. Write 5 as the next digit in the quotient and in the blank space of the divisor. The new divisor is 65.

Step 7: Subtract $65 \times 5 = 325$ from the dividend 325. Remainder is $325 - 325 = 0$.

All periods have been brought down, and the remainder is 0.

The calculation is shown as:

$\begin{array}{c|cc} & 3 \ . \ 5 & \\ \hline \phantom{()} 3 & \overline{12} \ \overline{.25} & \\ + \; 3 & 9\phantom{.25)} \\ \hline \phantom{()} 6 \; 5 & 3 \; 25 \\ \phantom{()} \times 5 & 3 \; 25 \\ \hline \phantom{()} 70 & 0 \end{array}$

The quotient is 3.5. The remainder is 0.

Therefore, the square root of 12.25 is 3.5. $\sqrt{12.25} = 3.5$.

Check: $3.5 \times 3.5 = 12.25$. Correct.

Example 2. Find the square root of 2.89.

Answer:

Given number: 2.89.

Step 1 & 2: Place bars: $\overline{2}\overline{.89}$. Integral part is 2, decimal part is 89.

Step 3: Largest square $\leq 2$. $1^2 = 1$, $2^2 = 4$. Largest is 1. Write 1 as the first digit of the quotient and the divisor.

Step 4: Subtract $1^2 = 1$ from 2. Remainder is $2 - 1 = 1$. Bring down the next period (89). This is the first pair from the decimal part, so place a decimal point in the quotient after 1. The new dividend is 189.

Step 5: Double the current quotient (1) to get 2. Write it to the left, leaving a blank space: 2_.

Step 6: Find the largest digit to fill the blank such that $2\_ \times \_$ is $\leq 189$. Try 1: $21 \times 1 = 21$. ... Try 7: $27 \times 7 = 189$. The digit is 7. Write 7 as the next digit in the quotient and in the blank space of the divisor. The new divisor is 27.

Step 7: Subtract $27 \times 7 = 189$ from the dividend 189. Remainder is $189 - 189 = 0$.

All periods have been brought down, and the remainder is 0.

The calculation is shown as:

$\begin{array}{c|cc} & 1 \ . \ 7 & \\ \hline \phantom{()} 1 & \overline{2} \ \overline{.89} & \\ + \; 1 & 1\phantom{.89)} \\ \hline \phantom{()} 2 \; 7 & 1 \; 89 \\ \phantom{()} \times 7 & 1 \; 89 \\ \hline \phantom{()} 34 & 0 \end{array}$

The quotient is 1.7. The remainder is 0.

Therefore, the square root of 2.89 is 1.7. $\sqrt{2.89} = 1.7$.

Check: $1.7 \times 1.7 = 2.89$. Correct.

Example 3. Find the square root of 0.0009.

Answer:

Given number: 0.0009.

Step 1 & 2: Place bars. Integral part is 0, place a bar over it: $\overline{0}$. For the decimal part (.0009), pair digits from left to right: $\overline{00}\overline{09}$. The number with bars is $\overline{0.}\overline{00}\overline{09}$. The leftmost period is 0.

Step 3: Find the largest number whose square is $\leq 0$. This is 0. Write 0 as the first digit of the quotient. Since we are starting with the integral part which is 0, immediately place the decimal point in the quotient after this 0. Subtract $0^2 = 0$ from 0. Remainder is 0. Bring down the next period (00). The new dividend is 00.

Step 4: Double the current quotient (0, ignoring the decimal for doubling) to get 0. Write it to the left, leaving a blank space: 0_.

Step 5 & 6: Find the largest digit to fill the blank such that $0\_ \times \_$ is $\leq 00$. Try 0: $00 \times 0 = 00$. The digit is 0. Write 0 as the next digit in the quotient and in the blank space of the divisor. The new divisor is 00. Subtract $00 \times 0 = 00$ from the dividend 00. Remainder is 0. Bring down the next period (09). The new dividend is 09.

Step 4 (contd.): Double the current quotient (00, ignoring the decimal for doubling) to get 00. Write it to the left, leaving a blank space: 00_.

Step 5 & 6 (contd.): Find the largest digit to fill the blank such that $00\_ \times \_$ is $\leq 09$. Try 1: $001 \times 1 = 1$. Try 2: $002 \times 2 = 4$. Try 3: $003 \times 3 = 9$. The digit is 3. Write 3 as the next digit in the quotient and in the blank space of the divisor. The new divisor is 003 or simply 3.

Step 7: Subtract $003 \times 3 = 9$ from the dividend 09. Remainder is $09 - 9 = 0$.

All periods have been brought down, and the remainder is 0.

The calculation is shown as:

$\begin{array}{c|cc} & 0 \ . \ 0 \ 3 & \\ \hline \phantom{()} 0 & \overline{0.} \ \overline{00} \ \overline{09} \\ + \; 0 & 0\phantom{.0009} \\ \hline \phantom{()} 0 \; 0 & 0 \; 0 \phantom{09} \\ \phantom{()} \times 0 & 0 \; 0 \phantom{09} \\ \hline \phantom{()} 00 \; 3 & \phantom{.} 0 \; 9 \\ \phantom{()} \times 3 & \phantom{.} 0 \; 9 \\ \hline \phantom{()} 006 & 0 \end{array}$

The quotient is 0.03. The remainder is 0.

Therefore, the square root of 0.0009 is 0.03. $\sqrt{0.0009} = 0.03$.

Check: $0.03 \times 0.03 = 0.0009$. Correct.

Example 4. Find the square root of 2 correct to two decimal places.

Answer:

Given number: 2. We need the square root correct to two decimal places. This means we must calculate the square root up to three decimal places and then round the result to two decimal places. To calculate up to three decimal places, we need three pairs of zeros after the decimal point in the number.

Write 2 as 2.000000.

Step 1 & 2: Place bars. For the integral part (2), place a bar over 2: $\overline{2}$. For the decimal part (.000000), pair digits from left to right: $\overline{00}\overline{00}\overline{00}$. The number with bars is $\overline{2.}\overline{00}\overline{00}\overline{00}$. The leftmost period is 2.

Step 3: Find the largest number whose square is $\leq 2$. $1^2 = 1$, $2^2 = 4$. Largest is 1. Write 1 as the first digit of the quotient and the divisor. Subtract $1^2 = 1$ from 2. Remainder is $2 - 1 = 1$. Bring down the next period (00). This is the first pair from the decimal part, so place a decimal point in the quotient after 1. The new dividend is 100.

Step 4: Double the current quotient (1) to get 2. Write it to the left, leaving a blank space: 2_.

Step 5 & 6: Find the largest digit to fill the blank such that $2\_ \times \_$ is $\leq 100$. Try 1: $21 \times 1 = 21$. Try 2: $22 \times 2 = 44$. Try 3: $23 \times 3 = 69$. Try 4: $24 \times 4 = 96$. Try 5: $25 \times 5 = 125$ (too large). The digit is 4. Write 4 as the next digit in the quotient and in the blank space of the divisor. The new divisor is 24. Subtract $24 \times 4 = 96$ from 100. Remainder is $100 - 96 = 4$. Bring down the next period (00). The new dividend is 400.

Step 4 (contd.): Double the current quotient (14, ignoring decimal for doubling) to get 28. Write it to the left, leaving a blank space: 28_.

Step 5 & 6 (contd.): Find the largest digit to fill the blank such that $28\_ \times \_$ is $\leq 400$. Try 1: $281 \times 1 = 281$. Try 2: $282 \times 2 = 564$ (too large). The digit is 1. Write 1 as the next digit in the quotient and in the blank space of the divisor. The new divisor is 281. Subtract $281 \times 1 = 281$ from 400. Remainder is $400 - 281 = 119$. Bring down the next period (00). The new dividend is 11900.

Step 4 (contd.): Double the current quotient (141, ignoring decimal for doubling) to get 282. Write it to the left, leaving a blank space: 282_.

Step 5 & 6 (contd.): Find the largest digit to fill the blank such that $282\_ \times \_$ is $\leq 11900$. Try 1: $2821 \times 1 = 2821$. Try 2: $2822 \times 2 = 5644$. Try 3: $2823 \times 3 = 8469$. Try 4: $2824 \times 4 = 11296$. Try 5: $2825 \times 5 = 14125$ (too large). The digit is 4. Write 4 as the next digit in the quotient and in the blank space of the divisor. The new divisor is 2824. Subtract $2824 \times 4 = 11296$ from 11900. Remainder is $11900 - 11296 = 604$.

We have calculated the square root up to three decimal places: 1.414... The quotient is 1.414.

The calculation is shown as:

$\begin{array}{c|cc} & 1\ . \ 4 \ 1 \ 4 & \\ \hline \phantom{()} 1 & \overline{2.} \ \overline{00} \ \overline{00} \ \overline{00} & \\ + \; 1 & 1\phantom{.000000)} \\ \hline \phantom{()} 2 \; 4 & 1 \; 00 \phantom{0000)} \\ \phantom{()} \times 4 & 96 \phantom{000)} \\ \hline \phantom{()} 28 \; 1 & 4 \; 00 \phantom{00)} \\ \phantom{()} \times 1 & 281 \phantom{00)} \\ \hline \phantom{()} 282 \; 4 & \phantom{.} 119 \; 00 \\ \phantom{()} \times 4 & \phantom{.} 11296 \\ \hline \phantom{()} 2828 & \phantom{....} 604 \end{array}$

The square root of 2 is approximately 1.414.

To round this to two decimal places, we look at the third decimal digit, which is 4. Since 4 is less than 5, we round down, meaning the second decimal digit remains unchanged.

Square root of 2 correct to two decimal places is 1.41.


Square Root of a Fraction

We have learned how to find the square root of whole numbers and decimal numbers. Now, let's consider how to find the square root of a fraction. The process is quite simple and relies on finding the square roots of the numerator and the denominator separately.

Rule for Finding the Square Root of a Fraction

To find the square root of a fraction $\frac{a}{b}$, where $a$ is a non-negative number and $b$ is a positive number, we find the square root of the numerator ($a$) and divide it by the square root of the denominator ($b$).

The formula is:

$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$

... (i)

This property holds because $(\frac{\sqrt{a}}{\sqrt{b}})^2 = \frac{(\sqrt{a})^2}{(\sqrt{b})^2} = \frac{a}{b}$, by the definition of a square root.

Note: For $\sqrt{a}$ to be a real number, $a$ must be non-negative ($a \ge 0$). For the fraction $\frac{a}{b}$ to be defined and $\sqrt{b}$ to be a real number in the denominator, $b$ must be positive ($b > 0$).

Example 1. Find the square root of $\frac{16}{25}$.

Answer:

Given fraction: $\frac{16}{25}$. Both numerator and denominator are perfect squares ($16=4^2$, $25=5^2$).

Using the formula for the square root of a fraction:

$\sqrt{\frac{16}{25}} = \frac{\sqrt{16}}{\sqrt{25}}$

[Using Formula (i)]

Find the square root of the numerator and the denominator separately:

$\sqrt{16} = 4$ ($4^2 = 16$)

$\sqrt{25} = 5$ ($5^2 = 25$)

Substitute these values back into the fraction:

$\sqrt{\frac{16}{25}} = \frac{4}{5}$

The square root of $\frac{16}{25}$ is $\frac{4}{5}$.

Check: Square the result: $(\frac{4}{5})^2 = \frac{4^2}{5^2} = \frac{16}{25}$. This matches the original fraction, so the answer is correct.

Example 2. Find the square root of $\frac{196}{289}$.

Answer:

Given fraction: $\frac{196}{289}$. Both numerator and denominator are perfect squares ($196=14^2$, $289=17^2$).

Using the formula $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$:

$\sqrt{\frac{196}{289}} = \frac{\sqrt{196}}{\sqrt{289}}$

Find the square root of the numerator and the denominator:

$\sqrt{196} = 14$ ($14^2 = 196$)

$\sqrt{289} = 17$ ($17^2 = 289$)

Substitute the values:

$\sqrt{\frac{196}{289}} = \frac{14}{17}$

The square root of $\frac{196}{289}$ is $\frac{14}{17}$.

Square Root of a Mixed Number

If you need to find the square root of a mixed number, the first step is always to convert the mixed number into an improper fraction. Then, apply the rule for finding the square root of a fraction.

Example: Find the square root of $1\frac{9}{16}$.

First, convert to an improper fraction: $1\frac{9}{16} = \frac{(1 \times 16) + 9}{16} = \frac{16 + 9}{16} = \frac{25}{16}$.

Now, find the square root of $\frac{25}{16}$: $\sqrt{\frac{25}{16}} = \frac{\sqrt{25}}{\sqrt{16}} = \frac{5}{4}$.

Example 3. Find the square root of $2\frac{14}{25}$.

Answer:

Given mixed number: $2\frac{14}{25}$.

Step 1: Convert the mixed number to an improper fraction.

$2\frac{14}{25} = \frac{(2 \times 25) + 14}{25}$

$= \frac{50 + 14}{25} = \frac{64}{25}$

Step 2: Find the square root of the resulting improper fraction $\frac{64}{25}$.

Using the formula $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$:

$\sqrt{\frac{64}{25}} = \frac{\sqrt{64}}{\sqrt{25}}$

Find the square root of the numerator and the denominator:

$\sqrt{64} = 8$ ($8^2 = 64$)

$\sqrt{25} = 5$ ($5^2 = 25$)

Substitute the values:

$\sqrt{2\frac{14}{25}} = \frac{8}{5}$

The square root of $2\frac{14}{25}$ is $\frac{8}{5}$. You can leave the answer as an improper fraction or convert it back to a mixed number ($1\frac{3}{5}$) or a decimal (1.6).

Check: $(\frac{8}{5})^2 = \frac{8^2}{5^2} = \frac{64}{25} = 2\frac{14}{25}$. Correct.