**N**, the undertaking is to write a C program to find the square solution of the given number N.

**Examples:**

Input:N = 12

Output:3.464102Input:N = 16

Output:4

**Method 1: Using inbuilt ** **sqrt()** ** function:** The sqrt ( ) routine returns the sqrt of any count N. Below is the implementation of the above approach :

## C

`#include ` `#include ` `double` `findSQRT(` `double` `N) { ` `return` `sqrt` `(N); }` `int` `main()` `{` ` ` ` ` `int` `N = 12;` ` ` ` ` `printf` `(` `"%f "` `, findSQRT(N));` ` ` `return` `0;` `}` |

**Output:**

3.464102

**Method 2: Using ** **Binary Search** **:** This approach is used to find the squarely root of the given count N with preciseness upto 5 decimal places .

- The square root of number N lies in range 0 ≤ squareRoot ≤ N. Initialize start = 0 and end = number.
- Compare the square of the mid integer with the given number. If it is equal to the number, then we found our integral part, else look for the same in the left or right side of mid depending upon the condition.
- After finding an integral part, we will find the fractional part.
- Initialize the increment variable by 0.1 and iteratively calculate the fractional part upto
**5 decimal places**. - For each iteration, change increment to
**1/10th**of its previous value. - Finally, return the answer computed.

Below is the implementation of the above approach :

## C

`#include ` `#include ` `float` `findSQRT(` `int` `number)` `{` ` ` `int` `start = 0, end = number;` ` ` `int` `mid;` ` ` ` ` `float` `ans;` ` ` ` ` ` ` `while` `(start <= end) {` ` ` ` ` `mid = (start + end) / 2;` ` ` ` ` ` ` `if` `(mid * mid == number) {` ` ` `ans = mid;` ` ` `break` `;` ` ` `}` ` ` ` ` ` ` ` ` `if` `(mid * mid < number) {` ` ` ` ` `ans=start;` ` ` ` ` `start = mid + 1;` ` ` `}` ` `
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**Output:**

3.464099

**Method 3: Using ** **log2()** **:** The square-root of a numeral N can be calculated using log2 ( ) as :

Let five hundred be our answer for input signal number N, then Apply log2 ( ) both sides therefore, five hundred = prisoner of war ( 2, 0.5*log2 ( newton ) )

Below is the execution of the above approach :

## C

`#include ` `#include ` `double` `findSQRT(` `double` `N) { ` `return` `pow` `(2, 0.5 * log2(N)); }` `int` `main()` `{` ` ` ` ` `int` `N = 12;` ` ` ` ` `printf` `(` `"%f "` `, findSQRT(N));` ` ` `return` `0;` `}` |

**Output:**

3.464102

My Personal Notes

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