Squares And Square Roots

1.0Square Number Or Perfect Square

If a natural number ' $m$ ' can be expressed as $n$ 2, where ' $n$ ' is also a natural number, then ' $m$ ' is a square number or a perfect square, i.e. if $m=n^2$ where $m,n$ are natural numbers, then $m$ is a square number or a perfect square. E.g., (i) $36=2\times 2\times 3\times 3$ $=2^2\times 3^2=(2\times 3{)}^2=6^2$ Hence, 36 is a perfect square of 6 . (ii) $50=2\times 5\times 5$ Hence, 2 is unpaired, so, 50 is not a perfect square. Properties of square numbers

The square of a number, either positive or negative is always positive. $(-5{)}^2=(-5)\times (-5)=25$ $(5{)}^2=5\times 5=25$. So, for an integer $n,(-n{)}^2=n^2$

Following table shows the square of numbers from 1 to 20.

Study the square numbers in the above table. All these numbers end with $0,1,4,5,6$ or 9 at unit's place. None of these end with $2,3,7$ or 8 at unit's place. So, we can say that a number that ends in $2,3,7$ or 8 is never a perfect square. Thus, by just looking at the numbers $152,453,1657$ and 2348 , we can say that these are not perfect squares.

• Q. Which of the following numbers are perfect squares? (i) 1057 (ii) 7928 (iii) 784 Explanation (i) 1057 The number ends with 7 . $\Rightarrow$ It is not a perfect square. (ii) 7928 The number ends with 8. $\Rightarrow$ It is not a square number. (iii) 784 The number ends with 4. $\Rightarrow$ It can be a square number. So, we check it. $784=2\times 2\times 2\times 2\times 7\times 7$ $=(2\times 2\times 7{)}^2=(28{)}^2$ Hence, 784 is a perfect square.

• If number ends with $0,1,4,5,6$ or 9 , then it may or may not be a square number but a square number must end with $0,1,4,5,6$ or 9 .

One's digit in square of a number

(i) The ones digit in the square of a number can be determined if the ones digit of the number is known.

(ii) The number of zeros at the end of a perfect square is always even and is double the number of zeros at the end of the number. For example,

(iii) The square of an even number is always an even number and square of an odd number is always an odd number.

2.0Some More Interesting Patterns

1. Adding triangle numbers: The sum of two consecutive triangular numbers is a square number.
   
   Note: Triangular numbers are numbers whose dot patterns can be arranged as triangles.
2. Number between square numbers

There are ' 2 n ' non perfect square numbers between the squares of two consecutive natural numbers n and ( $\mathrm{n}+1$ ). Between $1^2(=1)$ and $2^2(=4)\to 2,3$ $2\times 1=2$ non square numbers exist. Between $2^2(=4)$ and $3^2(=9)\to 5,6,7,8$ $2\times 2=4$ non square numbers exist. Between $3^2(=9)$ and $4^2(=16)\to 10,11,12,13,14,15$ $2\times 3=6$ non square numbers exist.

3. Adding odd numbers

Consider the following

So, we can say that the sum of first ' $n$ ' odd natural numbers is $n^2$ or we can say if the number is a square number, it has to be the sum of successive odd numbers starting from 1.

• Q. Find whether each of the following numbers is a perfect square or not by using successive subtraction? (i) 25 (ii) 38 Explanation: (i) 25 Successively subtract $1,3,5,7,9,\dots ...$. from 25. $25-1=24$ $24-3=21$ $21-5=16$ $16-7=9$ $9-9=0$ Thus $25=1+3+5+7+9$ 25 is a perfect square. (ii) 38 Successively subtract 1, 3, 5, 7, 9,.... from 38. $38-1=37$ $37-3=34$ $34-5=29$ $29-7=22$ $22-9=13$ $13-11=2$ $2-13=-11$ This shows that we are not able to express 38 as the sum of consecutive odd numbers starting with 1. $\Rightarrow 38$ is not a perfect square.

4. A sum of consecutive natural numbers

The square of an odd natural number ' $n$ ' can be expressed as the sum of two consecutive natural numbers $\frac{{\mathrm{n}}^2-1}{2}$ and $\frac{{\mathrm{n}}^2+1}{2}$.

Then, ${\mathrm{n}}^2=\frac{{\mathrm{n}}^2-1}{2}+\frac{{\mathrm{n}}^2+1}{2}$ $3^2=9=4+5\mathrm{s}$ $\to$ $\frac{3^2-1}{2}\frac{3^2+1}{2}$ $5^2=25=12+13$ $\to$ $\frac{5^2-1}{2}\frac{5^2+1}{2}$ $11^2=121=60+61$ $\to$ $\frac{11^2-1}{2}\frac{11^2+1}{2}$

5. Product of two consecutive even or odd natural numbers

If $(n+1)$ and $(n-1)$ are two consecutives even or odd numbers, then their product i.e. $(n+1)(n-1)$ is $n^2-1$. $11\times 13=143=12^2-1$ Also, $11\times 13=(12-1)(12+1)$ $\Rightarrow 11\times 13=(12-1)(12+1)=12^2-1$ Similarly, $13\times 15=(14-1)(14+1)=14^2-1$

6. Finding the square of a number

It is easy to find the square of small numbers like $2,3,6,7,9$ etc. but we cannot find square of large number so quickly. For example, There is another way also to find $47\times 47$ without multiplying. $47=40+7$ $47^2=(40+7)(40+7)$ $=40(40+7)+7(40+7)$ $=1600+280+280+49=2209$

Avoid these common mistakes

• $(100+5{)}^2\ne 100^2+5^2$ $(100+5{)}^2=(100{)}^2+(5{)}^2+2\times (100)\times (5)$
• $(100-5{)}^2\ne 100^2-5^2$ $(100-5{)}^2=(100{)}^2+(5{)}^2-2\times 100\times 5$

• $\sqrt{4}+\sqrt{6}\ne \sqrt{10}$ $\sqrt{4}+\sqrt{6}=2+\sqrt{6}$

• $\sqrt{7}-\sqrt{3}\ne \sqrt{4}$ (This cannot be further simplified.)

3.0Some More Pattern In Square Numbers

Observe the squares of numbers $1,11,111,\dots$. etc.

This pattern is restricted up to natural numbers having nine 1's. Another interesting pattern

1. Other patterns in squares

Consider the following pattern $25^2=625=(2\times 3)$ hundreds +25 $35^2=1225=(3\times 4)$ hundreds +25 $75^2=5625=(7\times 8)$ hundreds +25 Generalising it, consider a number with unit digit 5 i.e., a5 $(a5{)}^2=(10a+5{)}^2$ $=10\mathrm{a}(10\mathrm{a}+5)+5(10\mathrm{a}+5)$ $=100a^2+50\mathrm{a}+50\mathrm{a}+25$ $=100{\mathrm{a}}^2+100\mathrm{a}+25$ $=100\mathrm{a}(\mathrm{a}+1)+25$ $=a(a+1)$ hundred +25

• Area of square $=(\text{ side }{)}^2$
• Side of a square $=\sqrt{\text{ Area of a square }}$
• In a right angled triangle,
  
  ${\mathrm{H}}^2={\mathrm{B}}^2+{\mathrm{P}}^2$ [Pythagoras theorem] Where, H = Hypotenuse B = Base P= Perpendicular

4.0Pythagorean Triplets

Consider the following, $3^2+4^2=9+16=25=5^2$ The collection of numbers 3,4 and 5 is known as Pythagorean triplet. $6,8,10$ is also a Pythagorean triplet Since, $6^2+8^2=36+64=100=10^2$ Generalising it, for any natural number $m>1$, we have, $(2m{)}^2+{\left(m^2-1\right)}^2={\left(m^2+1\right)}^2$ So, $2m,m^2-1$ and $m^2+1$ form a Pythagorean triplet.

• Q. Find a Pythagorean triplet where one of the numbers is 12. Solution: If we take ${\mathrm{m}}^2-1=12$ $\Rightarrow {\mathrm{m}}^2=12+1=13$ then the value of $m$ will not be an integer. So, we try to take, ${\mathrm{m}}^2+1=12$ $\Rightarrow {\mathrm{m}}^2=12-1=11$ Again, the value of $m$ will not be an integer. So, let us take $2\mathrm{m}=12$ $\Rightarrow \mathrm{m}=6$ Thus, the other members of Pythagorean triplet are as follows $m^2-1=6^2-1=3-1=35$ $m^2+1=6^2+1=36+1=37$ Thus, the required triplet is 12,35 and 37 .

5.0Square Roots

Finding the square root is the inverse operation of squaring. If $4\times 4=16$ i.e., $4^2=16$, then the square root of 16 is 4 . In other words, if $q=p^2$, then $p$ is called the square root of $q$. The square root of a number is the number which when multiplied by itself gives the number as the product. The square root of a number ' $n$ ' is denoted by the symbol $\sqrt{n}$ or $\sqrt[2]{n}$ or ( n${)}^{1/2}$. For example, $\sqrt{16}=4$ Note: When we look for division of a number, we need to look only up to its square root, as, one number of the pair will be less than the square root of the number and the other will be more. (If the number is square number, then its square root will be its own partner).

6.0Some Properties Of Square Roots

(i) The square root of an even perfect square is even and that of an odd perfect square is odd. For example, $\sqrt{4}=2,\sqrt{16}=4$, $\sqrt{9}=3,\sqrt{25}=5$. (ii) Since there is no number whose square is negative, the square root of a negative number is not defined. (iii) If a number ends in an odd number of zeroes, then it cannot have a square root which is a natural number. (iv) If the unit's digit of a number is $2,3,7$ or 8 , then square root of that number (in natural numbers) is not possible. (v) If $m$ is not a perfect square, then there is no integer ' $n$ ' such that square root of $m$ is $n$.

7.0Methods Of Finding Square Root

1. Finding square root through repeated subtraction

Recall the pattern formed while adding consecutive odd numbers. $1+3+5+7+9=5^2=25$ $1+3+5+7+9+11=6^2=36$ Sum of first n odd numbers $={\mathrm{n}}^2$ The above pattern can be used to find the square root of the given number. (i) Obtain the given perfect square whose square root is to be calculated. Let the number be a. (ii) Subtract from it successively $1,3,5,7,9,\dots .$. till you get zero. (iii) Count the number of times the subtraction is performed to arrive at zero. Let the number of times be $n$. (iv) Write $\sqrt{\mathrm{a}}=\mathrm{n}$.

• Q. Find the square root of 36 by successive subtractions. Explanation We have, 36-1 = 35 $35-3=32$ $32-5=27$ $27-7=20$ $20-9=11$ $11-11=0$ Clearly, we have performed subtraction six times. $\therefore \sqrt{36}=6$

This method is the simplest method for finding square root but this method is convenient for calculating square root of small numbers only, because for large numbers it will be lengthy and time taking. So, we have more methods for finding square roots which are, prime factorisation method and long division method.

2. Finding square root through prime factorisation

In order to find the square root of a perfect square by prime factorisation, we follow the following steps. (i) Obtain the given number. (ii) Resolve the given number into prime factors by successive division. (iii) Make pairs of prime factors such that both the factors in each pair are equal. (iv) Take one factor from each pair and find their product. (v) The product obtained is the required square root.

• Q. Find the square root of $7744$ by prime factorisation. Explanation: Resolving 7744 into prime factors, | 2 | 7744 | |----|------| | 2 | 3872 | | 2 | 1936 | | 2 | 968 | | 2 | 484 | | 2 | 242 | | 11 | 121 | | 11 | 11 | | | 1 | $\Rightarrow 7744=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{11\times 11}$ Taking one factor from each pair $=2\times 2\times 2\times 11$ $\sqrt{7744}=8\times 11=88$
• Q. Find the smallest number by which 162 should be multiplied to make it a perfect square. Explanation: $162=2\times \underline{3\times 3}\times \underline{3\times 3}$ All the prime factors can be paired except 2 . Thus, to make 162 a perfect square, we multiply it by 2 . | 2 | 162 | |----|-----| | 3 | 81 | | 3 | 27 | | 3 | 9 | | 3 | 3 | | | 1 |

8.0Square Root Of Decimals

Consider a number 176.341. Put bars on both integral part and decimal part. In what way is putting bars on decimal part different from integral part? Notice, for 176 we start from the unit's place close to the decimal and move towards left. The first bar is over 76 and the second bar over 1. For 341, we start from the decimal and move towards right. First bar is on 34 and for the second bar we put 0 after 1 and make $\overline{34}\cdot \overline{10}$.

Consider the following steps to find the square root of 17.64.

Step-1 : To find the square root of a decimal number we put bars on the integral part (i.e. 17) of the number in the usual manner. And place bars on the decimal part (i.e. 64) on every pair of digits beginning with the first decimal place. Proceed as usual. We get $\overline{17}\overline{64}$.

Step-2 :Now proceed in a similar manner. The left most bar is on 17 and $4^2<17<5^2$. Take this number as the divisor and the number under the left-most bar as the dividend (i.e. 17). Divide and get the remainder.

Step-3: The remainder is 1 . Write the number under the next bar (i.e. 64) to the right of this remainder, to get 164 .

Step-4 : Double the divisor and enter it with a blank on its right.

Step-5: We know $82\times 2=164$, therefore, the new digit is 2 . Divide and get the remainder.

Step 6 : Since the remainder is 0 and no bar left,

9.0Estimating Square Root

We can also estimate square roots of the numbers by finding the number whose square is closest to the given number. It can be illustrated by the following example.

Estimate the square root of 300 . We know that, $100<300<400$. Since, $\sqrt{100}=10$ and $\sqrt{400}=20$ So, $10<\sqrt{300}<20$ We know that, $17^2=289$ and $18^2=324$. Thus, $17<\sqrt{300}<18$. But, 300 is closer to 289 as compared to 324 . Therefore, $\sqrt{300}$ is approximately equal to 17 .