C语言编程问题

Question

输入一个方程ax^2+bx+c=0
判断方程的根,并写出方程的根

Answer 1

#include<math.h>
main()
{float a,b,c,d,disc,x1,x2,realpart,imagpart;
scanf("%f,%f,%f",&a,&b,&c);
printf("The equation");
if(fabs(a)<=le-6)
printf("is not a quadratic");
else
{disc=b*b-4*a*c;
if(fabs(disc)<=le-6)
printf("has two equal roots:%8.4\n",-b/(2*a));
else if(disc>le-6)
{x1=(-b+sqrt(dise))/(2*a);
x2=(-b-sqrt(dise))/(2*a);
printf("has distinct real roots:%8.4f and %8.4f\n",x1,x2);
}
else
{realpart=-b/(2*a);
imagpart=sqrt(-disc)/(2*a);
printf("has complex roots:\n");
printf("%8.4f+%8.4fi\n",realpart,imagpart);
printf("%8.4f-%8.4fi\n",realpart,imagpart);
} }
}

Answer 2

这个我编译过了.肯定可以.而且简单.
#include"stdio.h"
#include"math.h"
void main()
{
float a,b,c;
printf("%s\n","Please Input a,b,c");
scanf("%f%f%f",&a,&b,&c);
if((b*b-4*a*c) < 0)
printf("%s\n","Error:b^2-4*a*c Must > 0");
else
{
printf("X1=%2f\n",(-b+sqrt(b*b-4*a*c))/2*a);
printf("X2=%2f\n",(-b-sqrt(b*b-4*a*c))/2*a);
}
}

Answer 3

main()
{float a,b,c,x;
float x1,x2;
scanf("%f,%f,%f,%f",&a,&b,&c,&x);
{if(b^2<4*a*c)
printf("error!!!");
else
x2+

待续!!!