C Quadratic Roots: Cases
C Quadratic Roots Program
A science exercise asks for roots of ax² + bx + c. Check whether the equation is actually quadratic before dividing by 2a.
For a nonzero quadratic coefficient, the discriminant b²−4ac determines whether the roots are two real roots, a repeated real root or a complex-conjugate pair.
Before this exercise, review arithmetic and conversions and numeric types and their limits. The worked program below applies those ideas to this specific problem.
Explain
Explain why quadratic roots: cases uses this sequence of operations.
Trace
Trace the sample values and predict the printed result.
Apply
Use a=1, b=0 and c=1 and identify why the result is complex.
Quadratic Roots: Cases: follow the data
Given data
A science exercise asks for roots of ax² + bx + c. Check whether the equation is actually quadratic before dividing by 2a.
Rule
For a nonzero quadratic coefficient, the discriminant b²−4ac determines whether the roots are two real roots, a repeated real root or a complex-conjugate pair.
Observed result
Roots: -1.00 -2.00
Understand the operation
The direct formula uses (-b ± sqrt(discriminant))/(2a). A negative discriminant uses the square root of its magnitude for the imaginary part instead of asking sqrt to evaluate a negative real value.
When a is zero, use the linear equation if b is nonzero; otherwise c determines whether every value or no value is a solution. The complete C example includes those branches even though its first fixture is quadratic.
Read the complete C program
The input values are written into this complete C17 example so the calculation can be reproduced without an unavailable service. Change the named data, rebuild and compare the actual output with your prediction.
C17 · complete program
#include <stdio.h>
#include <math.h>
int main(void) {
double a = 1.0, b = 3.0, c = 2.0;
if (a == 0.0) {
if (b != 0.0) printf("Linear root: %.2f\n", -c / b);
else puts(c == 0.0 ? "Every value is a solution" : "No solution");
} else {
double discriminant = b * b - 4.0 * a * c;
if (discriminant >= 0.0) {
double root = sqrt(discriminant);
printf("Roots: %.2f %.2f\n", (-b + root) / (2.0 * a), (-b - root) / (2.0 * a));
} else {
double real = -b / (2.0 * a), imaginary = sqrt(-discriminant) / fabs(2.0 * a);
printf("Roots: %.2f + %.2fi and %.2f - %.2fi\n", real, imaginary, real, imaginary);
}
}
return 0;
}
Expected output
Roots: -1.00 -2.00
Follow the values through the code
C17 · code used in this explanation
#include <stdio.h>
#include <math.h>
int main(void) {
double a = 1.0, b = 3.0, c = 2.0;
if (a == 0.0) {
if (b != 0.0) printf("Linear root: %.2f\n", -c / b);
else puts(c == 0.0 ? "Every value is a solution" : "No solution");
} else {
double discriminant = b * b - 4.0 * a * c;
if (discriminant >= 0.0) {
double root = sqrt(discriminant);
printf("Roots: %.2f %.2f\n", (-b + root) / (2.0 * a), (-b - root) / (2.0 * a));
} else {
double real = -b / (2.0 * a), imaginary = sqrt(-discriminant) / fabs(2.0 * a);
printf("Roots: %.2f + %.2fi and %.2f - %.2fi\n", real, imaginary, real, imaginary);
}
}
return 0;
}
For a=1, b=3 and c=2, the discriminant is 9−8=1.
Its square root is one. The two numerators are -2 and -4, divided by two to give -1 and -2.
Follow the worked calculation
Start with the sample
For a=1, b=3 and c=2, the discriminant is 9−8=1.
Apply the rule
Its square root is one. The two numerators are -2 and -4, divided by two to give -1 and -2.
Check the result
Roots: -1.00 -2.00
Roots: -1.00 -2.00
Try a small variation
C17 · code used in this explanation
#include <stdio.h>
#include <math.h>
int main(void) {
double a = 1.0, b = 3.0, c = 2.0;
if (a == 0.0) {
if (b != 0.0) printf("Linear root: %.2f\n", -c / b);
else puts(c == 0.0 ? "Every value is a solution" : "No solution");
} else {
double discriminant = b * b - 4.0 * a * c;
if (discriminant >= 0.0) {
double root = sqrt(discriminant);
printf("Roots: %.2f %.2f\n", (-b + root) / (2.0 * a), (-b - root) / (2.0 * a));
} else {
double real = -b / (2.0 * a), imaginary = sqrt(-discriminant) / fabs(2.0 * a);
printf("Roots: %.2f + %.2fi and %.2f - %.2fi\n", real, imaginary, real, imaginary);
}
}
return 0;
}
Use a=1, b=0 and c=1 and identify why the result is complex.
Worked answer
The discriminant is -4. The real part is zero and the imaginary magnitude is one. The sample may display signed negative zero from floating-point arithmetic; it represents the same zero real part here.
C17 · complete program
#include <stdio.h>
#include <math.h>
int main(void) {
double a = 1.0, b = 0.0, c = 1.0;
if (a == 0.0) {
if (b != 0.0) printf("Linear root: %.2f\n", -c / b);
else puts(c == 0.0 ? "Every value is a solution" : "No solution");
} else {
double discriminant = b * b - 4.0 * a * c;
if (discriminant >= 0.0) {
double root = sqrt(discriminant);
printf("Roots: %.2f %.2f\n", (-b + root) / (2.0 * a), (-b - root) / (2.0 * a));
} else {
double real = -b / (2.0 * a), imaginary = sqrt(-discriminant) / fabs(2.0 * a);
printf("Roots: %.2f + %.2fi and %.2f - %.2fi\n", real, imaginary, real, imaginary);
}
}
return 0;
}
Expected output
Roots: -0.00 + 1.00i and -0.00 - 1.00i
Recognize the important boundaries
The direct formula can lose precision when nearly equal values are subtracted. This small-coefficient teaching example is not a numerically robust general solver.
On toolchains requiring a separate math library, link with -lm. Exact zero tests here describe exact literal fixtures, not a universal tolerance policy for measured coefficients.
Check your understanding
Choose the answer, then compare the reasoning for every option.
Check your understanding
Choose the answer, then compare the reasoning for every option.
Use the idea independently
For a nonzero quadratic coefficient, the discriminant b²−4ac determines whether the roots are two real roots, a repeated real root or a complex-conjugate pair.
Continue with floating-point calculations.
Language reference: WG14 C11 committee draft. These examples use the C17-compatible subset of the explained language features.
Reaching the recap and completing the quizzes are separate milestones.