VTU BPOPS103 · Lab Program 2

Roots of a Quadratic Equation in C

Accept the coefficients a, b and c of ax² + bx + c = 0 and print the roots, handling the linear case and the real, equal and complex root cases.

Algorithm

Read the coefficients a, b and c as decimal numbers.
If a is zero, the equation is linear: print the single root -c/b, or report that there is no unique solution.
Otherwise compute the discriminant D = b*b - 4*a*c.
If D is positive, print two distinct real roots; if D is zero, print one repeated root.
If D is negative, the roots are complex: the real part is -b/(2a) and the imaginary part is sqrt(-D)/(2a).

C program

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

int main(void) {
    double a, b, c, d, r1, r2, real, imag;

    printf("Enter coefficients a, b, c:\n");
    if (scanf("%lf %lf %lf", &a, &b, &c) != 3) {
        printf("Error: invalid input\n");
        return 1;
    }

    if (a == 0) {
        if (b == 0) {
            if (c == 0) {
                printf("Infinitely many solutions\n");
            } else {
                printf("No solution\n");
            }
        } else {
            printf("Linear equation, root = %.2f\n", -c / b);
        }
        return 0;
    }

    d = b * b - 4 * a * c;
    if (d > 0) {
        r1 = (-b + sqrt(d)) / (2 * a);
        r2 = (-b - sqrt(d)) / (2 * a);
        printf("Two distinct real roots: %.2f and %.2f\n", r1, r2);
    } else if (d == 0) {
        r1 = -b / (2 * a);
        printf("Two equal real roots: %.2f\n", r1);
    } else {
        real = -b / (2 * a);
        imag = sqrt(-d) / (2 * a);
        if (imag < 0) {
            imag = -imag;
        }
        printf("Two complex roots: %.2f + %.2fi and %.2f - %.2fi\n", real, imag, real, imag);
    }
    return 0;
}

Sample input and output

Two distinct real roots (x² - 3x + 2 = 0)

Input

1 -3 2

Output

Enter coefficients a, b, c:
Two distinct real roots: 2.00 and 1.00

Complex roots (x² + 2x + 5 = 0)

Input

1 2 5

Output

Enter coefficients a, b, c:
Two complex roots: -1.00 + 2.00i and -1.00 - 2.00i

Edge case: a = 0 (linear equation)

Input

0 2 -4

Output

Enter coefficients a, b, c:
Linear equation, root = 2.00

How the program works

The discriminant D = b² - 4ac decides the nature of the roots. Its sign is checked before taking a square root, because the square root of a negative number is not a real value; the complex branch takes the root of -D instead.

The a == 0 check comes first because the quadratic formula divides by 2a. Without that guard the program would divide by zero and print nan or inf instead of a meaningful message.

Run this code

Edit the program and press Run. It is free and needs no sign-in.

Program input (what the program reads with scanf)

Copy a sample and paste it into the STDIN box in the editor before you press Run.

Two distinct real roots (x² - 3x + 2 = 0)

1 -3 2

Expected output

Enter coefficients a, b, c:
Two distinct real roots: 2.00 and 1.00

Complex roots (x² + 2x + 5 = 0)

1 2 5

Expected output

Enter coefficients a, b, c:
Two complex roots: -1.00 + 2.00i and -1.00 - 2.00i

Edge case: a = 0 (linear equation)

0 2 -4

Expected output

Enter coefficients a, b, c:
Linear equation, root = 2.00

Viva questions

What is the discriminant and what does its sign tell you?

D = b² - 4ac. Positive means two distinct real roots, zero means one repeated real root, and negative means two complex conjugate roots.

Why is sqrt(-d) used in the complex branch?

sqrt needs a non-negative argument for a real result. Since d is negative in that branch, -d is positive and its square root is the imaginary part.

Why do we test a == 0 before using the formula?

The formula divides by 2a, so a zero leading coefficient would divide by zero. The equation is linear in that case.

Which header provides sqrt and how is the program compiled?

math.h declares sqrt. On many systems the program must be linked with the maths library, for example gcc file.c -lm.

Why are the roots printed with %.2f rather than %f?

%f prints six decimal places by default. %.2f limits the output to two places, matching the currency-like precision typical of lab answers.

Common mistakes

Forgetting the -lm link flag, which produces an undefined reference to sqrt on some systems.
Taking sqrt of a negative discriminant and getting nan instead of the complex-root branch.
Testing d == 0 for decimal coefficients: rounding in binary floating point can leave a discriminant that should be zero slightly off, so a tiny tolerance is safer for real input.

Back to all VTU BPOPS103 lab programs.

Algorithm

Read the coefficients a, b and c as decimal numbers.

If a is zero, the equation is linear: print the single root -c/b, or report that there is no unique solution.

Otherwise compute the discriminant D = b*b - 4*a*c.

If D is positive, print two distinct real roots; if D is zero, print one repeated root.

If D is negative, the roots are complex: the real part is -b/(2a) and the imaginary part is sqrt(-D)/(2a).

C program

#include <stdio.h> #include <math.h> int main(void) { double a, b, c, d, r1, r2, real, imag; printf("Enter coefficients a, b, c:\n"); if (scanf("%lf %lf %lf", &a, &b, &c) != 3) { printf("Error: invalid input\n"); return 1; } if (a == 0) { if (b == 0) { if (c == 0) { printf("Infinitely many solutions\n"); } else { printf("No solution\n"); } } else { printf("Linear equation, root = %.2f\n", -c / b); } return 0; } d = b * b - 4 * a * c; if (d > 0) { r1 = (-b + sqrt(d)) / (2 * a); r2 = (-b - sqrt(d)) / (2 * a); printf("Two distinct real roots: %.2f and %.2f\n", r1, r2); } else if (d == 0) { r1 = -b / (2 * a); printf("Two equal real roots: %.2f\n", r1); } else { real = -b / (2 * a); imag = sqrt(-d) / (2 * a); if (imag < 0) { imag = -imag; } printf("Two complex roots: %.2f + %.2fi and %.2f - %.2fi\n", real, imag, real, imag); } return 0; }

Sample input and output

Two distinct real roots (x² - 3x + 2 = 0)

Input

1 -3 2

Output

Enter coefficients a, b, c:
Two distinct real roots: 2.00 and 1.00

Complex roots (x² + 2x + 5 = 0)

Input

1 2 5

Output

Enter coefficients a, b, c:
Two complex roots: -1.00 + 2.00i and -1.00 - 2.00i

Edge case: a = 0 (linear equation)

Input

0 2 -4

Output

Enter coefficients a, b, c:
Linear equation, root = 2.00

How the program works

The a == 0 check comes first because the quadratic formula divides by 2a. Without that guard the program would divide by zero and print nan or inf instead of a meaningful message.

Viva questions

What is the discriminant and what does its sign tell you?

D = b² - 4ac. Positive means two distinct real roots, zero means one repeated real root, and negative means two complex conjugate roots.

Why is sqrt(-d) used in the complex branch?

sqrt needs a non-negative argument for a real result. Since d is negative in that branch, -d is positive and its square root is the imaginary part.

Why do we test a == 0 before using the formula?

The formula divides by 2a, so a zero leading coefficient would divide by zero. The equation is linear in that case.

Which header provides sqrt and how is the program compiled?

math.h declares sqrt. On many systems the program must be linked with the maths library, for example gcc file.c -lm.

Why are the roots printed with %.2f rather than %f?

%f prints six decimal places by default. %.2f limits the output to two places, matching the currency-like precision typical of lab answers.

Common mistakes

Forgetting the -lm link flag, which produces an undefined reference to sqrt on some systems.

Taking sqrt of a negative discriminant and getting nan instead of the complex-root branch.

Testing d == 0 for decimal coefficients: rounding in binary floating point can leave a discriminant that should be zero slightly off, so a tiny tolerance is safer for real input.