Solve quadratic equation when coefficients may be 0 - c

Question

I was given a problem to write a C program which would solve the equation ax2+bx+c=0, where a, b and c are coefficients with double type. Any of the coefficients may be zero. In this problem it is unclear to me how to handle the double variables.
Here is my code. As for now, I know that my program can't distinguish between two roots and infinitely many roots. It also doesn't detect the "linear equation situation". How can I make it detect an infinite number of solutions? I was also advised in the comments to calculate the root with the minus before the discriminant if b > 0 and then use the Viet's theorem. I understand that it is because it is always more accurate to sum two numbers. I also guess I should do the exact opposite with b < 0. But what if b == 0 ? In this case, the program will not do anything. Or should I just include b == 0 in b < 0 and have b <= 0 ?
#include <stdio.h>
#include <math.h>
#include <float.h>
int main() {
double a, b, c, x1, x2;
scanf("%lf", &a);
scanf("%lf", &b);
scanf("%lf", &c); // just reading variables
//ax^2+bx+c=0
if ((b * b - 4 * a * c) < 0) {
printf("no");
} else {
x1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a); //calculating roots
x2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);
if ((fabs((a * x1 * x1 + b * x1 + c)) < DBL_EPSILON) & (fabs((a * x2 * x2 + b * x2 + c)) < DBL_EPSILON)) { //plugging the roots in
if (fabs((x1 - x2)) < DBL_EPSILON) { //checking if the roots are equal
printf("%lf", &x1); // if they are equal, we print only one of them
} else {
printf("%lf", &x1); // if they are not equal, we print both.
printf("\n %lf", &x2);
}
} else { // if there are no two valid roots
if ((fabs((a * x1 * x1 + b * x1 + c)) < DBL_EPSILON)) // we try to find one root.
printf("%lf", &x1);
if (fabs((a * x2 * x2 + b * x2 + c)) < DBL_EPSILON)
printf("%lf", &x2);
if ((fabs((a * x1 * x1 + b * x1 + c)) > DBL_EPSILON) & (fabs((a * x2 * x2 + b * x2 + c)) > DBL_EPSILON)) // if both of the plugged roots don't satisfy the equation
printf("no");
}
}
return 0;
}

Answer 1

Solve quadratic equation when coefficients may be 0
How can I make it detect an infinite number of solutions?
When a==0 && b == 0 && c == 0.
No DBL_EPSILON needed really anywhere in this code. See also #Eric Postpischil.
But what if b == 0 ?
if (b == 0) { // y = a*x*x + c
if (a) {
double dd = -c/a;
if (dd >= 0) {
double d = sqrt(d);
printf_roots("+/- roots", d,-d);
} else {
printf_roots("Complex roots", NAN, NAN); // Note NAN may not exist
}
} else if (c) { // y = 0*x*x + c, c != 0
printf_roots("No roots", NAN, NAN);
} else { // y = 0*x + 0
printf_roots("Infinite roots", -HUGE_VAL, HUGE_VAL);
}
Or should I just include b == 0 in b < 0 and have b <= 0 ?
Unless the coding goal requires a special output when b==0, I would only vector code on b==0 as a subtest when a==0 occurred.
if (a==0) {
if (b == 0) {
The quadric equation, like much FP code, can readily overflow and hit 0, both cases losing all precision.
Consider the code below: the unnecessary subtraction may cause overflow or truncation to 0 versus the second which may not. It is dependent on many things.
if ((b * b - 4 * a * c) < 0)
//
if (b * b < 4 * a * c)
Further, C allows various calculations to occur using wider math. Research FLT_EVAL_METHOD. Because of this, to prevent sqrt(value_less_than_0), code should calculate the discriminate and then test the object x that is going to be applied to sqrt(x).
//if ((b * b - 4 * a * c) < 0) {
// printf("no");
//} else {
// x1 = (-b + sqrt(b * b - 4 * a * c))
double discriminate = b * b - 4 * a * c;
if (discriminate < 0) {
printf("no");
} else {
double d = sqrt(discriminate);
x1 = (-b + d)
As to the idea of "calculate the root with the minus before the discriminant if b > 0 and then use the Viet's theorem", I'd suggest for improved retained precision the below which does not subtract like signed values.
double d = sqrt(discriminate);
// Note x1*x2 = c/a
if (b < 0) {
x2 = (-b + d)/(2*a);
x1 = c/a/x2;
} else {
x1 = (-b - d)/(2*a);
x2 = c/a/x1;
}
printf_roots("2 roots", x1, x2);
Notes on printf("%lf", &x1);. You are not compiling with all warnings enabled. Save time - enable them. Should be printf("%lf", x1); No &.
Further double is floating point. For FP code development use "%e", "%a" or"%g" to full see significant information.
printf("%g\n", some_double);
// or better
printf("%.*e\n", DBL_DECIMAL_DIG -1, some_double);

Answer 2

Since division by zero is not allowed, you have to split the problem into 4 cases :
a != 0:
this is case you treated in your code.
a == 0 && b != 0 :
This is a linear equation where the solution is x = -c/b
a == 0 && b == 0 && c != 0 : There's no possible value for x.
In this last case, a, b and c are equals to 0 : there's infinitly many solutions for x.
EDIT: comparisons with epsilon removed since they seem to be useless

Answer 3

There are some problems in your code:
you should check the return values of scanf() to avoid undefined behavior on invalid input.
you should use local variables for intermediary results to improve code readability
your printf statements are incorrect: you should pass the values of the double variables instead of their addresses: printf("%lf", &x1); should read:
printf("%f", x1);
Regarding the degenerate cases, you should just test those before trying to resolve the second degree equation.
Here is a corrected version:
#include <stdio.h>
#include <math.h>
int main() {
double a, b, c, delta, x1, x2;
if (scanf("%lf%lf%lf", &a, &b, &c) != 3) {
printf("invalid input\n");
return 1;
}
if (a == 0) {
// not a quadratic equation
if (b != 0) {
printf("one solution: %g\n", -c / b);
} else {
if (c != 0) {
printf("no solution\n");
} else {
printf("all real values are solutions\n");
}
}
} else {
delta = b * b - 4 * a * c;
if (delta < 0) {
printf("no real solution\n");
} else
if (delta == 0) {
printf("one double solution: %g\n", -b / (2 * a));
} else {
x1 = (-b + sqrt(delta)) / (2 * a);
x2 = (-b - sqrt(delta)) / (2 * a);
printf("two solutions: %g, %g\n", x1, x2);
}
}
return 0;
}