#include <math.h>
#include <stdio.h>
main() {
int a, b, c, x, x1, x2;
printf("enter the values of a,b,c:");
scanf("%d%d%d", &a, &b, &c);
printf("The quadratic equation is %d*pow(x,2) %d*x %d=0", a, b, c);
if (pow(b, 2) - 4 * a * c >= 0) {
x1 = (-b sqrt(pow(b, 2) - 4 * a * c)) / 2 * a;
x2 = (-b - sqrt(pow(b, 2) - 4 * a * c)) / 2 * a;
printf("the roots of the equation are x1=%d,x2=%d", x1, x2);
}
else
printf("roots of the equation in the form of x iy and x-iy");
return 0;
}
Is this code alright for the given question, i had a bit confusion at that printing imaginary roots. could you please help
CodePudding user response:
- Use proper
main
prototype. - Use floating point numbers instead of integers
- pow(b,2) == b*b
/ 2 * a
->/ (2 * a)
int main(void) {
double a, b, c, x, x1, x2;
printf("enter the values of a,b,c:");
if(scanf("%lf %lf %lf", &a, &b, &c) != 3) { /* handle error */}
printf("\nThe quadratic equation is %f*x^2 %f*x %f=0\n", a, b, c);
if (b*b - 4 * a * c >= 0) {
x1 = (-b sqrt(b*b - 4 * a * c)) / (2 * a);
x2 = (-b - sqrt(b*b - 4 * a * c)) / (2 * a);
printf("the roots of the equation are x1=%f,x2=%f\n", x1, x2);
}
else
{
double r = -b / (2*a);
double z = sqrt(fabs(b*b - 4 * a * c));
printf("the roots of the equation are x1 = %f i%f, x2 = %f - i%f", r,z,r,z);
}
}