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C program to find the roots of a quadratic equation

Comprehensive Collection of C Programming Examples

In this example, you will learn how to find the roots of a quadratic equation in C programming.

To understand this example, you should understand the followingC programmingTopic:

The standard form of a quadratic equation is:

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

b2-4acThe term is called the discriminant of a quadratic equation. It describes the nature of the roots.

  • If the discriminant is greater than 0, the roots are different real numbers

  • If the discriminant is equal to 0, then the roots are real and equal.

  • If the discriminant is less than 0, then the roots are different complex numbers.

Program to find the roots of a quadratic equation

#include <math.h>
#include <stdio.h>
int main() {
    double a, b, c, discriminant, root1, root2, realPart, imagPart;
    printf("Input coefficients a, b and c: ");
    scanf("%lf %lf %lf", &a, &b, &c);
    discriminant = b * b - 4 * a * c;
    // The condition for unequal real roots
    if (discriminant > 0) {
        root1 = (-b + sqrt(discriminant)) / (2 * a);
        root2 = (-b - sqrt(discriminant)) / (2 * a);
        printf("root1 = %.2lf and root2 = %.2lf", root1, root2);
    }
    // The condition for equal real roots
    else if (discriminant == 0) {
        root1 = root2 = -b / (2 * a);
        printf("root1 = root2 = %.2lf;", root1);
    }
    // If the root is not a real number
    else {
        realPart = -b / (2 * a);
        imagPart = sqrt(-discriminant) / (2 * a);
        printf("root1 = %.2lf+%.2lfi and root2 = %.2f-%.2fi
    }
    return 0;
}

Output Result

Enter coefficients a, b, and c: 2.3
4
5.6
root1 = -0.87+1.30i and root2 = -0.87-1.30i

In this program, the sqrt() library function is used to find the square root of a number. For more information, please visit:sqrt() Function.

Comprehensive Collection of C Programming Examples