 # How to Find Square Root of a Number  View Notes

The square root of a number x is the value which when multiplied by itself gives the original number x as its product. The square root of a number x is denoted by $\sqrt{x}$.

Mathematically, if y is the square root of  a number x, i.e., $\sqrt{x}$ = y, then  multiplying y with itself gives x as its product, i.e., y × y = $y^2$ = x.

For example, the square root of 4 = $\sqrt{4}$ = 2, the square root of 25 = $\sqrt{25}$ = 5, etc.

In this article, you will learn different methods to find the square root of a given number.

There are two methods which are generally used for finding the square root of a number. These two methods are:

(1) Prime Factorisation Method

(2) Long-Division Method

METHOD 1

Finding Square Root of a Perfect square number by Prime Factorisation Method

When a given number is a perfect square, the following steps are used to find its square root:

Step 1: Resolve the given number into its prime factors.

Step 2: Make pairs of similar prime factors.

Step 3: Separate out one factor from every pair.

Step 4: Multiply these separate out factors, their product will give the required square root.

For Example: Find the square root of 256 by Prime factorisation method.

We will follow the above steps to find the square root of 256:

Step 1: Resolve the given number 256 into its prime factors.

So, 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Step 2: Make pairs of similar prime factors.

256 = $\underline{2 × 2}$ ×  $\underline{2 × 2}$ ×  $\underline{2 × 2}$ ×  $\underline{2 × 2}$

Step 3: Separate out one factor from every pair.

Step 4: Multiplying these separate out factors, we get their product as; 2 × 2 × 2 × 2 = 16

Therefore, the square root of 256 = $\sqrt{256}$ = 2 × 2 × 2 × 2 = 16.

METHOD 2

Finding Square Root of a Perfect square number by Long Division Method

To learn the long division method for finding the square root of a number, let us consider an example, Find the square root of 529 by long division method.

Step 1: Place a bar over every pair of digits starting from the digit at the unit's place. If the number of digits in the given number is odd, then the left-most single digit too will have a bar. Thus we have, $\overline{5}$ $\overline{29}$.

Step 2: Find the largest number whose square is less than or equal to the number under the left-most bar (22 < 5 < 32 ). Take this number as the divisor and the number under the extreme left bar as the dividend (here 5) for division and get the remainder.

Step 3: Write down the number under the next bar, which is 29 in this case, to the right of the remainder. So the new dividend is 129.

Step 4: Double the quotient and write it with a blank on its right.

Step 5: Choose a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product obtained is less than or equal to the dividend. In this case 42 × 2 = 84. As 43 × 3 = 129. So we choose the new digit as 3. Perform the division with new divisor 43 and dividend 129 to get the remainder.

Step 6: Since the remainder obtained is 0 and no digits are left in the given number, therefore, the square root of 529 = $\sqrt{529}$  = 23.

How to Solve the Square Root Equations?

Let us consider an example to learn how to solve the square root equations. Consider a square root equation below and find the value of x:

$\sqrt{x-2}$ = 5.

To solve the above equation, we will first remove the square root from LHS of the equation by squaring both LHS and RHS.

$(\sqrt{x-2})^2$ = (5)2

⇒ x - 2 = 25

⇒ x = 25 + 2

⇒ x  = 27

Therefore, the required value of x for the given square root equation is 27.

How to Find Cube Root?  Square Root of 289  Square Root of 576  Square Root of 144  Square Root of 8  Square Root Tricks  Square Root Symbol  How to Calculate the Percentage of Marks?  Square Root Prime Factorization  How to Make a Model of Seasons?  NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots  NCERT Solutions for Class 12 Accountancy Chapter 4 - Reconstitution of a Partnership Firm:Retirement/Death of a Partner  NCERT Solutions for Class 12 Accountancy Chapter 3 - Reconstitution of a Partnership Firm: Admission of a Partner  NCERT Solutions for Class 10 English First Flight Chapter 1 A Letter to God  NCERT Solutions Class 11 English Woven Words Poem Chapter 11 Ode to a Nightingale  NCERT Solutions of Class 6 English Chapter 1 - A Tale of Two Birds  NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (Ex 1.4) Exercise 1.4  NCERT Solutions for Class 9 Maths Chapter 1 Number System  NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots Exercise 6.4  NCERT Solutions for Class 7th English Chapter 2 A Gift of Chappals  Area of Square Formula  Number of Moles Formula  Cube Root Formula  Square Formula  Area of a Sector of a Circle Formula  Perfect Square Formula  CBSE Class 8 Maths Chapter 6 - Squares and Square Roots Formulas  Area of a Rhombus Formula  Perimeter of a Trapezoid Formula  Class 9 Maths Revision Notes for Number Systems of Chapter 1  