C Program to Find the Roots of a Quadratic Equation

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

A quadratic equation will have two roots and a standard quadratic equation is defined as ax² + bx + c = 0 where a,b and c are real numbers and a!=0.

The nature of the quadratic equation is determined by the discriminant value (b² – 4ac).

• If discriminant value is greater than 0, then the roots are real and different.
• If discriminant value is equal to 0, then the roots are real and equal.
• If discriminant value is less than 0, then the roots are complex and different.

C Program to Find the Roots of a Quadratic Equation

#include <stdio.h>
#include <math.h>

int main() {
    float a, b, c, root1, root2, discriminant, realPart, imagPart;
    printf("Enter coefficients a, b and c: ");
    scanf("%f %f %f", &a, &b, &c);
    
    // calculate discriminant
    discriminant = b*b - 4*a*c;

    // condition for roots are real and different
    if (discriminant > 0) {
        root1 = (-b + sqrt(discriminant)) / (2 * a);
        root2 = (-b - sqrt(discriminant)) / (2 * a);
        printf("Root1 = %.3f and Root2 = %.3f", root1, root2);
    }
    // condition for roots are real and equal
    else if (discriminant == 0) {
        root1 = root2 = -b / (2 * a);
        printf("Root1 = Root2 = %.3f;", root1);
    }
    // if roots are complex
    else {
        realPart = -b / (2 * a);
        imagPart = sqrt(-discriminant) / (2 * a);
        printf("Root1 = %.3f+%.3fi and Root2 = %.3f-%.3fi", realPart, imagPart, realPart, imagPart);
    }

    return 0;
} 

Output

Enter coefficients a, b and c: 1 1 4
Root1 = -0.500+1.936i and Root2 = -0.500-1.936i

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