Write a C Program to Determine the roots of the quadratic equation (ax^2 + bx + c = 0) [Code and Explanation]

Determine the roots of the quadratic equation

Using the well-known quadratic formula

Allow for the possibility that one of the constants has a value of zero, and that the quantity

is less than or equal to zero. Test the program using the following sets of data:

(Code explanation at the bottom of the post)

#include "stdafx.h"
#include <stdio.h>
#include <cmath>

int main()
{
double a = 0, b = 0, c = 0, d, sqD, x1, x2;

printf("Enter a: ");                                    // Enter the value for a
scanf_s("%lf" , &a);
printf("\n");

printf("Enter b: ");                                    // Enter the value for b
scanf_s("%lf", &b);
printf("\n");

printf("Enter c: ");                                    // Enter the value for c
scanf_s("%lf", &c);
printf("\n");

d = pow(b, 2) - (4 * a * c); // we assign (b^2) - 4ac to variable d

if (d < 0) // we can't have imaginary number, so if d < 0, it gives an error
{
printf("Undetermined, imaginary number");
}

else
{
sqD = sqrt(d); // Square root of d

x1 = ((-b) + (sqD)) / (2 * a); // Root 1

x2 = ((-b) - (sqD)) / (2 * a); // Root 2

printf("Root 1 = %lf\n", x1);
printf("Root 2 = %lf\n", x2);
}



    return 0;
}


Once you run the program in Visual Studio, you will get something like this:

After you input a number for a and press enter, you will be asked to input a number for b, then same process for c:

After you enter the number, if d is imaginary, then an error message will show stating the number is undetermined:

However, if d is real then the program will display the roots just fine:

Ok! let's breakdown the code line by line:


We include the header <cmath> because we are dealing with double\floating numbers , in addition to  using the mathematical functions pow and sqrt. We can't use sqrt or pow without the cmath header.

--------------------------------------------------------------------------

printf("Enter a: ");                                    // Enter the value for a
scanf_s("%lf" , &a);
printf("\n");

printf("Enter b: ");                                    // Enter the value for b
scanf_s("%lf", &b);
printf("\n");

printf("Enter c: ");                                    // Enter the value for c
scanf_s("%lf", &c);
printf("\n");

In this part, we are simply asking the user to enter the values that will be stored in a, b, and c.

--------------------------------------------------------------------------

d = pow(b, 2) - (4 * a * c);

Here, we will calculate the equation below and store the value to a variable named d

pow(b,2) means that we are rising b to the power of 2, b^2

--------------------------------------------------------------------------

if (d < 0) // we can't have imaginary number, so if d < 0, it gives an error
{
printf("Undetermined, imaginary number");
}

After d is calculated, we are gonna compare it with 0. If d < 0, then we can't determine the roots because we will have to take square root of d, which will be an imaginary number since d < 0.

If that's the case, then we will have the program prints out a message stating that you can't determine the roots.

--------------------------------------------------------------------------

else
{
sqD = sqrt(d); // Square root of d

x1 = ((-b) + (sqD)) / (2 * a); // Root 1

x2 = ((-b) - (sqD)) / (2 * a); // Root 2

printf("Root 1 = %lf\n", x1);
printf("Root 2 = %lf\n", x2);
}


If the calculated value of d is not less than 0, then we proceed to finding the roots using the below equation

as we know [(b^2) - 4ac] is assigned to d, to find the square root of it we use a math function sqrt (stands for Square Root), sqrt(d). We will take that and find the roots and print them out with this code:

x1 = ((-b) + (sqD)) / (2 * a); // Root 1
x2 = ((-b) - (sqD)) / (2 * a); // Root 2
printf("Root 1 = %lf\n", x1);                                  // Print out Root 1
printf("Root 2 = %lf\n", x2);                                  // Print out Root 2



and that's it....