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

## C Programming Mathematics: Exercise-19 with Solution

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

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