C Exercises: Find the square root of a number using Babylonian method

Last update on March 20 2025 12:36:09 (UTC/GMT +8 hours)

19. Babylonian Square Root Variants

Write a C program to find the square root of a number using the Babylonian method.

Example 1:
Input: n = 50
Output: 7.071068
Example 2:
Input: n = 17
Output: 4.123106

Sample Solution:

C Code:

#include <stdio.h>

// Function to calculate the square root of a number using the Babylonian method
float square_Root(float n) { 
    float a = n; // Initial value for 'a' is the number itself
    float b = 1; // Initial value for 'b' is 1
    double e = 0.000001; // Threshold for the difference between 'a' and 'b'

    // Loop to approximate the square root using the Babylonian method
    while (a - b > e) { 
        a = (a + b) / 2; // Update 'a' by averaging 'a' and 'b'
        b = n / a; // Calculate 'b' as the number divided by the updated 'a'
    } 

    return a; // Return the calculated square root
} 

// Main function to test the square_Root function with different values of 'n'
int main(void) { 
    int n = 50; 
    printf("Square root of %d is %f", n, square_Root(n)); // Print the square root of 'n'

    n = 17; 
    printf("\nSquare root of %d is %f", n, square_Root(n)); // Print the square root of 'n'

    return 0;    
}

Sample Output:

Square root of 50 is 7.071068
Square root of 17 is 4.123106

Flowchart:

For more Practice: Solve these Related Problems:

• Write a C program to compute the square root of a number using the Babylonian method with a specified precision.
• Write a C program to implement the Babylonian method iteratively and count the number of iterations needed.
• Write a C program to compare the Babylonian square root approximation with the standard library function.
• Write a C program to calculate square roots using the Babylonian method and handle edge cases like zero and negative inputs.

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