Suppose x is the square root of y, then it is represented as x=√y, or we can express the same equation as x

For example, the square of 3 is 9, 3

Let us say m is a positive integer, such that √

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

For example, if x=4, then the function returns the output value as 2.

Suppose √-n = i√n, where i is the imaginary number.

where x is the number. The number under the

y = √a |

Since, y.y = y^{2} = a; where ‘a’ is the square of a number ‘y’.

- If a number is a perfect square number, then there exists a perfect square root.
- If a number ends with an even number of zeros (0’s), then it can have a square root.
- The two square root values can be multiplied. For example, √3 can be multiplied by √2, then the result should be √6.
- When two same square roots are multiplied, then the result should be a radical number. It means that the result is a non-square root number. For instance, when √7 is multiplied by √7, the result obtained is 7.
- The square root of any negative numbers is not defined because the perfect square cannot be negative.
- If a number ends with 2, 3, 7 or 8 (in the unit digit), then the perfect square root does not exist.
- If a number ends with 1, 4, 5, 6 or 9 in the unit digit, then the number may have a perfect square root.

Hence, the methods to find the square root of numbers are:

- Square Root by Prime Factorisation
- Square Root by Repeated Subtraction Method
- Square Root by Long Division Method
- Square Root by Estimation Method

Number |
Prime Factorization |
Square Root |

16 | 2x2x2x2 | √16 = 2×2 = 4 |

144 | 2x2x2x2x3x3 | √144 = 2x2x3 = 12 |

169 | 13×13 | √169 = 13 |

256 | 256 = 2×2×2×2×2×2×2×2 | √256 = (2x2x2x2) = 16 |

576 | 576 = 2x2x2x2x2x2x3x3 | √576 = 2x2x2x3 = 24 |

- Repeatedly subtracting consecutive odd numbers from it
- Subtract till the difference is zero
- Number of times we subtract is the required square root

For example, let us find the square root of 25.

- 25 – 1 = 24
- 24 – 3 = 21
- 21 – 5 = 16
- 16 – 7 = 9
- 9 – 9 = 0

Since, the subtraction is done for 5 times, hence the square root of 25 is 5.

Thus, the square root of 436 is 20.880 (rounding to 3 decimals).

For example, the square root of 4 is 2 and the square root of 9 is 3, thus, we can guess the square root of 5 will lie between 2 and 3.

But, we need to check the value of √5 is nearer to 2 or 3. Let us find the square of 2.2 and 2.8.

- 2.2
^{2}= 4.84 - 2.8
^{2}= 7.84

Since the square of 2.2 gives an approximate value of 5, thus we can estimate the square root of 5 is equal to 2.2 approximately.

- 1
^{2}= 1 ⇔ √1 = 1 - 2
^{2}= 4 ⇔ √4 = 2 - 3
^{2}= 9 ⇔ √9 = 3 - 4
^{2}= 16 ⇔ √16 = 4 - 5
^{2}= 25 ⇔ √25 = 5 - 6
^{2}= 36 ⇔ √36 = 6 - 7
^{2}= 49 ⇔ √49 = 7 - 8
^{2}= 64 ⇔ √64 = 8 - 9
^{2}= 81 ⇔ √81 = 9 - 10
^{2}= 100 ⇔ √100 = 10

**Hence, 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100 are the "Perfect Squares" here.**

√n |
Value | √n |
Value | √n |
Value |

√1 |
1 | √18 |
4.2426 | √35 |
5.9161 |

√2 |
1.4142 | √19 |
4.3589 | √36 |
6 |

√3 |
1.7321 | √20 |
4.4721 | √37 |
6.0828 |

√4 |
2 | √21 |
4.5826 | √38 |
6.1644 |

√5 |
2.2361 | √22 |
4.6904 | √39 |
6.2450 |

√6 |
2.4495 | √23 |
4.7958 | √40 |
6.3246 |

√7 |
2.6458 | √24 |
4.8990 | √41 |
6.4031 |

√8 |
2.8284 | √25 |
5 | √42 |
6.4807 |

√9 |
3 | √26 |
5.0990 | √43 |
6.5574 |

√10 |
3.1623 | √27 |
5.1962 | √44 |
6.6332 |

√11 |
3.3166 | √28 |
5.2915 | √45 |
6.7082 |

√12 |
3.4641 | √29 |
5.3852 | √46 |
6.7823 |

√13 |
3.6056 | √30 |
5.4772 | √47 |
6.8557 |

√14 |
3.7417 | √31 |
5.5678 | √48 |
6.9282 |

√15 |
3.8730 | √32 |
5.6569 | √49 |
7 |

√16 |
4 | √33 |
5.7446 | √50 |
7.0711 |

√17 |
4.1231 | √34 |
5.8310 | √51 |
7.1414 |

Taking root on both sides.

N = ±√0.09

As we know,

0.3 x 0.3 = (0.3)

Therefore,

N = ±√(0.3)

N = ±(0.3)

where, a+ib is a complex number.

To solve the radical equation, we need to follow the below steps:

- Isolate the square root to one of the sides (L.H.S or R.H.S)
- Square both the sides of the given equation
- Now solve the rest equation

**Example: Solve the radical equation √(4a+9) – 5 = 0**
**Solution:** Given, √(4a+9) – 5 = 0

Isolate the square root term first. Then the equation becomes,

√(4a+9) = 5

Now on squaring both the sides, we get; 4a+9 = 5^{2}

4a + 9 = 25

4a = 16

a = 16/4

a = 4
**Note:** If the equation has more than one radical or square root, then repeat the above given steps for each square root.
**Example: Solve √(2x−5) − √(x−1) = 1**
**Solution:** Let us isolate one of the square root.

√(2x−5) = 1 + √(x−1)

Now squaring both the sides

2x – 5 = (1 + √(x−1)) ^{2 }

Apply the algebraic identity, (a+b) ^{2 } = a ^{2 } + b ^{2 } + 2ab.

2x−5 = 1 + 2√(x−1) + (x−1)

2x−5 = 2√(x−1) + x

x−5 = 2√(x−1)

Now again isolate the square root.

√(x−1) = (x−5)/2

x−1 = ((x−5)/2) ^{2 }

x−1 = (x 2 − 10x + 25)/4

4x−4 = x ^{2 } − 10x + 25

4x − 4 − x ^{2 } + 10x − 25 = 0

−x ^{2 } + 14x − 29 = 0

x ^{2 } − 14x + 29 = 0

Using the quadratic formula, we can solve the above equation.

x = 2.53 and x = 11.47

- Quadratic Equations
- Algebra
- Geometry
- Calculus

We know that 2

Now, choose 3 and 4 (as

Divide 10 by 3.

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

(3 + 3.33)/2 = 3.1667

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

Square Root |
Result |

√1 |
1 |

√4 |
2 |

√9 |
3 |

√16 |
4 |

√25 |
5 |

√36 |
6 |

√49 |
7 |

√64 |
8 |

√81 |
9 |

√100 |
10 |

√2 = 1.4142

√3 = 1.7321

√5 = 2.2361

Squaring both the sides, we get;

x+2 = √4

x+2 = ±4

x = ±4 – 2

Therefore,

x = 2 or x = -6

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