C program to Find the Roots of a Quadratic Equation

January 31, 2023

How to write a C program to find the roots of a Quadratic Equation?

Writing a C program to Find the Roots of a Quadratic Equation is pretty easy if you know the basic knowledge about what is it and how its works. A Quadratic equation is a second degree equation in x in Mathematics.
C-program-to-find-the-roots-of-a-quadratic-equation image

General representation of a Quadratic Equation:

ax2 + bx + c = 0, where
a, b and c are real constant numbers and
a != 0

Syntax:

discriminant = b * b - 4 * a * c;
root1 = (-b + sqrt(discriminant)) / (2 * a);
root2 = (-b - sqrt(discriminant)) / (2 * a);

Working of Program :

In the program, we will require some numbers from the user or pre-defined variables depending on the program requirement to display the roots of a Quadratic Equation.

Steps to write C program to Find the Roots of a Quadratic Equation :

  • Step 1: Start
  • Step 2: Declare the variable num1=2, num2=4 and num3.
  • Step 3: Read input num1, num2 and num3 from the user or predefine it according to the need.
  • Step 4: Find the discriminant using the formula provided.
  • Step5: Using the discriminant find the roots of the Quadratic Equation.
  • Step 6: Print the result of the program compiled.
  • Step 7:  Program end.

Problem 1:

Write a program to calculate the the roots of a Quadratic Equation in which the variables are pe-defined.

Code:

Run 
#include<math.h>
#include<stdio.h>
int main() {
    double a=1, b=-45, c=324, discriminant, root1, root2, real, imag;
    discriminant = b * b - 4 * a * c;
    // condition for real and different roots
    if (discriminant > 0) {
        root1 = (-b + sqrt(discriminant)) / (2 * a);
        root2 = (-b - sqrt(discriminant)) / (2 * a);
        printf("root1 = %.2lf and root2 = %.2lf", root1, root2);
    }
    // condition for real and equal roots
    else if (discriminant == 0) {
        root1 = root2 = -b / (2 * a);
        printf("root1 = root2 = %.2lf;", root1);
    }
    // if roots are not real
    else {
        real = -b / (2 * a);
        imag = sqrt(-discriminant) / (2 * a);
        printf("root1 = %.2lf+%.2lfi and root2 = %.2f-%.2fi", real, imag, real, imag);
    }
    return 0;
}

Output

root1 = 36.00 and root2 = 9.00
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Problem 2:

In the following program we will display the roots of a Quadratic Equation using user defined variables.

Code:

Run 
#include<math.h>
#include<stdio.h>
int main() {
    double a, b, c, discriminant, root1, root2, real, imag;
    printf("Enter coefficients a, b and c: ");
    scanf("%lf %lf %lf", &a, &b, &c);
    discriminant = b * b - 4 * a * c;
    // condition for real and different roots
    if (discriminant > 0) {
        root1 = (-b + sqrt(discriminant)) / (2 * a);
        root2 = (-b - sqrt(discriminant)) / (2 * a);
        printf("root1 = %.2lf and root2 = %.2lf", root1, root2);
    }
    // condition for real and equal roots
    else if (discriminant == 0) {
        root1 = root2 = -b / (2 * a);
        printf("root1 = root2 = %.2lf;", root1);
    }
    // if roots are not real
    else {
        real = -b / (2 * a);
        imag = sqrt(-discriminant) / (2 * a);
        printf("root1 = %.2lf+%.2lfi and root2 = %.2f-%.2fi", real, imag, real, imag);
    }
    return 0;
}

Input:

Enter coefficients a, b and c: 1
-45
365

Output:

root1 = 34.38 and root2 = 10.62
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Problem 3:

Write a program to calculate the roots of a Quadratic Equations using functions.

Code:

Run 
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
void Roots(int a, int b, int c)
{
    if (a == 0) 
    {
        printf("Invalid");
        return;
    }
    int d = b * b - 4 * a * c;
    double sqrt_val = sqrt(abs(d));
    if (d > 0) 
    {
        printf("Roots are real and different ");
        printf("%f%f", 
              (double)(-b + sqrt_val) / 
                      (2 * a),
              (double)(-b - sqrt_val) / 
                      (2 * a));
    }
    else if (d == 0) 
    {
        printf("Roots are real and same ");
        printf("%f", 
                -(double)b / (2 * a));
    }
    else // d < 0
    {
        printf("Roots are complex ");
        printf("%f + i%f%f - i%f", 
               -(double)b / (2 * a),
               sqrt_val/(2 * a), 
               -(double)b / (2 * a), 
               sqrt_val/(2 * a));
    }
}
int main()
{
    int num1 = 1, num2 = -7, num3 = 12;
    Roots(num1, num2, num3);
    return 0;
}

Output

Roots are real and different 4.0000003.000000

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