The following Topics and Sub-Topics are covered in this chapter and are available on MSVgo:
You must have come across real-life scenarios when squares and square roots are used in the form of the Pythagorean theorem by architects, engineers and carpenters. This forms an important chapter for students from an exam point of view as it is used in different chapters in higher education. Let us see more details about squares and square roots, their properties, and how to find these in case of a whole number and decimal in this article.
i. Perfect squares: When whole numbers are squared, it results in perfect squares.
Example: 82 = 16.
ii. The squares of numbers always end with 0, 1, 4, 5, 6, or 9. You will never find the square ending with 2, 3, 7, or 8 as in the following cases:
iii. If numbers are ending with 1 or 9, i.e., if these digits are in the unit’s place, then their square will end with 1.
iv. The square of numbers with 4 or 6 in the unit’s place always ends with 6.
v. If ‘n’ is a natural number and ‘n + 1’ is the next natural number then:
(n + 1)2 – n2 = (n2 + 2n + 1) – n2 = 2n + 1.
So, we can tell that between the squares of 2 consecutive numbers, there are 2n non-perfect square numbers.
Example: How many numbers are there between 42 and 52?
Here, n = 4 and n + 1 = 5
So there are (2 × 4) = 8 non square numbers between 42 and 52, i.e., 17, 18, 19, 20, 21, 22, 23, 24.
vi. A natural number is not a perfect square if you can’t express it as a sum of successive numbers beginning from 1. This can be used for finding if a number is a perfect square. In other words, for knowing if you need to check if n2 is a perfect square, add n numbers from 1 and see if the result is n2.
Example: 62 = 36
So let us see the sum of 1st 6 odd numbers, which is = 1 + 3 + 5 + 7 + 9 + 11 = 36.
vii. Square of an odd number is the sum of two consecutive positive integers.
Example: 92 = 81, which can be written as 40 + 41.
viii. The product of two consecutive odd or even numbers can be written as n2 – 1, where ‘n’ is a natural number.
We can also write (n + 1)(n – 1) = n2 – 1.
Example: Let n = 5, so 6 × 4 = 24 = 25 – 1.
ix. Pythagorean triplets:
If we take any natural number ‘n’ greater than 1, then:
(2n)2 + (n2 – 1)2 = (n2 + 1)2
Here, 2n, n2 – 1 and n2 + 1 are termed as pythagorean triplets.
Let n = 3
So, (2 * 3)2 + (32 – 1)2 = (32 + 1)2
= 36 + 64 = 100 = 102
i.e., 62 + 82 = 102
Hence, 6, 8, and 10 are Pythagorean triplets.
i. Repeated subtraction method
We have already seen that every perfect square can be written as the sum of successive odd natural numbers beginning from one.
Let us consider an example 64 and go in the reverse order, i.e., subtract the successive odd numbers from the given number, starting from 1 till you get a zero. If you get zero at the nth step, then the square root of the given number is ‘n’.
So, 64 – 1 = 63
63 – 3 = 60
60 – 5 = 55
55 – 7 = 48
48 – 9 = 39
39 – 11 = 28
28 – 13 = 15
15 – 15 = 0
It took us 8 steps to reach zero, so √64 = 8.
ii. Prime factorisation
We will understand this by an example:
Let us take a number 324. Its prime factorisation comes to 2 × 2 × 3 × 3 × 3 × 3. We can pair the prime factors, which gives 22 x 32 x 32.
Hence, we can tell that square root of 324 is:
√324 = 2 × 3 × 3 = 18
What if the prime factors are not in pairs?
Example: √48 = 2 × 2 × 2 × 2 × 3.
Here, since we are unable to form pairs, this is not a perfect square.
iii. Division method
The prime factorisation method can be time taking in case of larger numbers. Hence the division method can be used in such cases.
We will understand this by an example: √529.
Put a bar on the pair of digits from the extreme right end. In case there are an odd number of digits, a single digit will be left in the left end without a bar. 3 will be without a bar in our case.
Take the largest number whose square is less than that of the number under the extreme left bar. So it will be 22 < 5 < 32.
And we will take 2 as the divisor, and the number under the extreme left bar will be the quotient. You will get reminder 1 after dividing.
Then the number under the next bar, i.e., 29, has to be brought down right to the existing reminder, which forms 129. This has to be divided by 4 (doubling the divisor).
Now the new digit in the quotient will be the largest possible number to fill the blank and it will adhere to the formula of new divisor multiplied to new quotient, which results in a number less than or equal to the dividend.
Here, it will be 42 x 4, i.e., 84. Since 43 x 3 = 129, the new digit chosen is 3 and the remainder is obtained which is 0. So, the process ends, and √529 = 23.
What is the relationship between square roots and squares?
The square root of a number is the number which when multiplied with itself results in the original number.
So, if z2 = a,
Then, √a = z.
Does a square root cancel out a square?
Yes, when the number is a perfect square, a square root will cancel out the square.
How do you simplify square roots with square roots?
Consider an example to understand this: √64 = √82.
Here, square root will cancel out the square, so the answer will be 8.
Does a negative have a square root?
No. You can find the square root of non-negative numbers only.
This was a brief explanation of various aspects of squares and square roots. You can refer to MSVgo app or website for learning the topic in a much better way. It has numerous examples along with visual illustrations to clarify concepts so that you don’t face issues while solving even the most complex problems. Once the concept is clear, practice the different types of questions well to get a good grasp on the topic.