Square Root Formula

Algebra is an important topic of mathematics. A square root is an operation that is used in many formulas and different fields of mathematics. This article is about square root and square root formula. The square root of a number is a number squaring which gives the original number. There are multiple square root formulas that are discussed in this article with their problems.

Square Root 

The square root of a number is a number squaring which gives the original number. It is that factor of the number that when squared gives the original number. It is the value of power 1/2 of that number. The square root of a number is represented as √.

Example: Square root of 9 ⇒ √9 = ± 3

                 Here, 3² = 9

                       (-3)² = 9

The number inside the square root is called radicand and the square root symbol is called radical.

Methods for finding the square root of a number

There are two methods for finding the square roots of a number that are highly used in mathematics. These are discussed below.

1. Prime Factorization Method
2. Long Division Method

Prime Factorization Method

Prime factorization is a method in which a number is represented in the form of the product of prime numbers.  Then the square root of the number is fined according to the given concept.

To find the square root using the prime factorization method:

Step 1: Represent the number in its prime factors using prime factorization method.

Step 2: Form the pair of the same factors.

Step 3:Take one factor from each pair and then, find the products of all the factors obtained by taking one factor from each pair.

Step 4:The resultant product is the square root of the number.

Example for Prime Factorization Method:

Prime Factorization method for finding the square root
2	196
2	98
7	49
7	7
	1
Prime factors of 196 = (2×2) × (7×7)
For square root, we take the product of 2 from (2×2)  and 7 from (7×7)
The square root of 196 = 2×7 The square root of 196 = 14

Long Division Method

Long Division Method is a very useful method for finding the square root of the number. The square root of imperfect squares such as 14, 15,18, etc can be found using the long division method. For finding the square root using the long division method we have to follow the specific steps which are described below.

Steps to find the square root of a number using the long division method:

Example: Find the square root of 256 using the long division method.

Step 1. Divide the number into pairs starting from one place. For example, pairs starting from one place:= 2, 56

Step 2. After dividing the digits into pairs, start from the leftmost pair or digit. The largest number whose square is just less than or equal to the first pair or digit is taken as the divisor and also the quotient. In the above example, the largest number whose square is just less than 2 is 1. So, the divisor is 1 and the quotient is also 1.

Step 3. Subtract the square of the divisor from the first pair or digit and bring the next pair down to the right of the reminder to get the new dividend. In the above example 2 – 1 = 1 then, we bring the next pair i.e. 56 down and the new dividend becomes 156.

Step 4. Now, the new divisor is obtained by adding the previous divisor and the previous quotient digit and concatenating it with a suitable digit which is also taken as the next digit of the quotient, chosen in a way that the product of the new divisor and this new digit in quotient is equal to or just less than the new dividend. In the above example, the previous divisor is 1 and the previous quotient digit is 1 and their addition gives 2 which is the new divisor. Now, we have to choose a digit so that the product of the new divisor and the new digit in the quotient is equal to or less than the new dividend i.e. 26 is the new divisor and 6 in the new digit which is concatenated with the previous quotient. Now, the current quotient is 16.

Step 5. Repeat steps 2, 3, and 4 till all the pairs have been taken. Now, the resultant quotient is the square root of the given number. In the above example, all pairs have been taken, and hence, the square root of the number 256 is 16. 

Table for Square and Square root of first ten natural numbers

Number	Square	Square root
1	1	√1 = 1
2	4	√2 = 1.414
3	9	√3 = 1.732
4	16	√4 = 2
5	25	√5 = 2.236
6	36	√6 = 2.4494
7	49	√7 = 2.645
8	64	√8 = 2.828
9	81	√9 = 3
10	100	√10 = 3.162

Square Root Formula

The square root of a number has the exponent 1/2. The square root formula is used to find the square root of a number.

Square Root Formula :      √x = x^1/2

Square Root Formula
√x = x1/2
√x . √x = x
x√y . x√y = x2y
√(x × y) = √x × √y
√(x / y) = √x / √y
x / √y = (x / √y) × ( √y / √y) = (x .  √y)/y
x. √y ±  z. √y = (x ± z) √y
x / (y + z. √p) = [x / (y + z√p)]×[(y – z√p)/ (y – z√p)]                       = [x(y – z√p)]/[y2 – z2p] Rationalize denominator by multiplying by [(y – z√p)/ (y – z√p)]
x / (y – z. √p) = [x / (y – z√p)]×[(y + z√p)/ (y + z√p)]                       = [x(y + z√p)]/[y2 – z2p] Rationalize denominator by multiplying by [(y + z√p)/ (y + z√p)]

Sample Problems

Question 1: Find the sum: 5√3 + 6√12

Solution:

5√3 + 6√12 = 5√3 + 6(√(4 × 3)

                              = 5√3 + 6 × √4 × √3

                              = 5√3 + 6 ×2√3

                              = 5√3 + 12√3

                              = 17√3

Question 2: Evaluate:  √64 – √25

Solution:

 √64 – √25 = √(8 × 8)  – √(5 × 5 )

                          = 8 – 5 = 3

Question 3: Evaluate: √63 / √28

Solution:

 √63 / √28 = √(7 × 9 )/ √(7 × 4 )

                          = √[(7 × 9 )/ (7 × 4 )]

                          = √(9 /4)

                          = √9 / √4 

                          = 3/2

Question 4: Evaluate: 5 /√15 

Solution: 

5 /√15 = (5 /√15)×(√15 /√15 ) 

             = (5√15)/15

             = (√15)/3

Question 5: Evaluate: 4 / (5 + √6)

Solution:

4 / (5 + √6) = [4 / (5 + √6)] × [(5 – √6) / (5 – √6)]

Above step is rationalizing denominator by multiplying by [(5 – √6) / (5 – √6)]

                    = [4×(5 – √6)] / [(5 + √6)× (5 – √6)]

                    = [4×(5 – √6)] / [5² – (√6)²]

                    = (20 – 4√6) / (25 – (√6 × √6)]

                    = (20 – 4√6) / (25 – 6)

                    = (20 – 4√6) / 19

Question 6: Evaluate : 7 / (8 – √10)

Solution:

7 / (8 – √10) = [7 / (8 – √10)] × [(8 + √10) / (8 + √10)]

Above step is rationalizing denominator by multiplying by [(8 + √10) / (8 + √10)]

                   = [7×(8 + √10)] / [(8 – √10) × (8 + √10)]

                   = [7×(8 + √10)] / [8² – (√10)²]

                   = (56 + 7√10) / (64 – (√10 × √10)]

                   = (56 + 7√10) / (64 – 10)

                   = (56 + 7√10) / 54

Question 7: Evaluate : (2 +√5) / (4 – √2)

Solution:

(2 +√5) / (4 – √2) = [(2 +√5) / (4 – √2)] × [(4 + √2) / (4 + √2)]

Above step is rationalizing denominator by multiplying by [(4 + √2) / (4 + √2)]

                            = [(2 +√5) × (4 + √2)] / [(4 – √2) × (4 + √2)]

                            = [(2 +√5) × (4 + √2)] / [4² – (√2)²]

                            = [8 + 2√2 + 4√5 + (√5 × √2)] / [16 –  (√2 × √2)]

= [8 + 2√2 + 4√5 + √(5×2)] / [16 – 2]

                            = [8 + 2√2 + 4√5 + √10] / 14