[Maths Class Notes] on Square Root Questions Pdf for Exam

A square root is a value that, when multiplied by itself, gives the original number.

How to Find the Square Root?

We can use two methods to find square roots – Prime factorization and the Long division method.

What are Squares and Square Roots?

Just the opposite method of squaring a number is the square root. If a number n, such as n², is squared, then the square root of n² is equal to the original number n.

How to Square a Number?

We need to multiply the number by itself to find the square of a number.

For example, 3 multiplied by 3 is equal to 9.

Similarly,

Square of 4 is: 4 multiplied by 4 = 4 × 4 = 16.

Square Root Problems and Answers

Q1: Can you find out the number of numbers lying between the squares of these following pairs of numbers?

(i) 25 and 26

(ii) 99 and 100

Solution – As we know, between n² and (n + 1)², the number of non–perfect square numbers are 2n.

(i) Between 252 and 262 there are 2 × 25 = 50 natural numbers.

(ii) Between 992 and 1002 there are 2 × 99 = 198 natural numbers.

Q2: Write a Pythagorean triplet whose one of the required member is:

(i) 6

(ii) 14

(iii) 16

(iv) 18

Solution –

We know, for any natural number m, 2m, m² – 1, m² + 1 is a Pythagorean triplet.

(i) 2m = 6

⇒ m = 6/2 = 3

m² – 1 = 3² – 1 = 9 – 1 = 8

m² + 1 = 3² + 1 = 9 + 1 = 10

Therefore, (6, 8, 10) is a Pythagorean triplet.

(ii) 2m = 14

⇒ m = 14/2 = 7

m² – 1 = 72 – 1 = 49 – 1 = 48

m² + 1 = 72 + 1 = 49 + 1 = 50

(14, 48, 50) is not a Pythagorean triplet.

(iii) 2m = 16

⇒ m = 16/2 = 8

m² – 1 = 8² – 1 = 64 – 1 = 63

m² + 1 = 8² + 1 = 64 + 1 = 65

Therefore, (16, 63, 65) is a Pythagorean triplet.

(iv) 2m = 18

⇒ m = 18/2 = 9

m² – 1 = 92 – 1 = 81 – 1 = 80

m² + 1 = 92 + 1 = 81 + 1 = 82

Therefore, (18, 80, 82) is a Pythagorean triplet.

Q3: (n + 1)² – n² = ?

Solution –

(n + 1)² – n²

= (n² + 2n + 1) – n²

= 2n + 1

Q4: State that the number 121 is the sum of 11 odd natural numbers.

Solution – As 121 = 112

We recognize that n2 is the sum of the first n odd natural numbers.

It shows that 121 = the sum of the first 11 odd natural numbers.

= 1 + 3 + 5+ 7 + 9 + 11 +13 + 15 + 17 + 19 + 21

Q5: Use the identity and find the square of 189.

(a – b)² = a² – 2ab + b²

Solution – 189 = (200 – 11)²

= 40000 – 2 × 200 × 11 + 112

= 40000 – 4400 + 121

= 35721

Q6: Find out the square root of 625 using the mathematical identity as stated:  (a + b)² = a² + b² + 2ab?

Solution – (625)²

= (600 + 25)²

= 6002 + 2 × 600 × 25 + 252

= 360000 + 30000 + 625

= 390625

Properties of Square Roots

A square root function is defined in mathematics as a one-to-one function that takes as an input a positive number and returns the square root of the given input number.

f(x) = √x

The following are some of the essential properties of the square root:

• If a number is a perfect square number, a perfect square root exists.

• It may have a square root if a number ends with an even number of zeros (0’s).

• It is possible to multiply the two square root values. For instance, √3 can be multiplied by √2, then √6 should be the result.

• If two same square roots are multiplied, then a radical number should be the product. This implies that a non-square root number is a product. When √7 is multiplied by √7, for example, the result obtained is 7.

• It does not define the square root of any negative numbers. And no negative can be the perfect square.

• If a number ends with 2, 3, 7, or 8 (in the digit of the unit), the perfect square root will not exist.

• If in the unit digit, a number ends with 1, 4, 5, 6, or 9, the number would have a square root.

Using the prime factorization process, the square root of a perfect square number is simple to measure. For example –

Square Root By Prime Factorisation

With the help of an example, let us understand this concept:

Example 1: Solve √10 to 2 decimal places.

Solution –

Step 1: Choose any two perfect square roots between which you feel your number may fall.

We know that 22 = 4; 32 = 9, 42 = 16 and 52 = 25

Now, choose 3 and 4 (as √10 lies between these 2 numbers)

Step 2: Divide the given number into one of the square roots chosen.

Divide 10 by 3.

=> 10/3 = 3.33 (round off answer at 2 places)

Step 3: Find the root average and the product of the step above, i.e.

(3 + 3.33)/2 = 3.1667

Verify: 3.1667 × 3.1667 = 10.0279 (Not required)

Repeat step 2 and step 3

Now 10/3.1667 = 3.1579

Average of 3.1667 and 3.1579.

(3.1667+3.1579)/2 = 3.1623

Verify: 3.1623 × 3.1623 = 10.0001 (more accurate)

Stop the process.

Example 2: Find the square roots of whole numbers from 1 to 100 that are perfect squares.

Solution – The perfect squares are – 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

Example 3: What is:

1. The square root of 2.

2. The square root of 3.

3. The square root of 4.

4. The square root of 5

Solution – Use a square root list, we have

1. Value of root 2 i.e. √2 = 1.4142.

2. Value of root 3 i.e. √3 = 1.7321.

3. Value of root 4 i.e. √4 = 2.

4. Value of root 5 i.e. √5 = 2.2361.