求大佬帮忙看看，为什么复数i不能正确使用？整个程序好像还有点什么错误

[weixin_45003333]

#include <stdio.h> #include<math.h> int main() { char i=sqrt(-1); float a, b, c,x1,x2,D; printf("请输入a,b,c"); scanf_s("%f %f %f", &a, &b, &c); D = b * b - 4 * a*c; if (D < 0) { x1 = (-b + sqrt(-D)*i ) / (2 * a); x2 = (-b - sqrt(-D)*i )/ (2 * a); printf("该函数有两个不相等的虚数根x1=%f,x2=%f", x1, x2); } else { if (D = 0) { x1 = x2 = -b / (2 * a); printf("该函数有两个相等的实数根x1=x2=%f", x1=x2); } else { x1 = (-b + sqrt(D)) / (2 * a); x2 = (-b - sqrt(D)) / (2 * a); printf("该函数有两个不相等的实数根x1=%f,x2=%f", x1, x2); } } return 0; }

[lunhui2016]

-1开啥方啊，sqrt()要求的参数为非负数

[吖如]

#include <stdio.h>
#include <math.h>
void main()
{
float a,b,c,x1,x2,judge;
printf("please input three coefficient:");
scanf("%f,%f,%f",&a,&b,&c);
if(a==0)
printf("not one yuan quadratic equation");
else
{
judge=b*b-4*a*c;
if(judge<0)
printf("no roots");
else if (judge==0)
printf("x1=x2=%f",(-b)/(2*a));
else
printf("x1=%f,x2=%f",((-b)+sqrt(b*b-4*a*c))/(2*a),((-b)+sqrt(b*b-4*a*c))/(2*a));
}
}

[weixin_45003333]

引用 2 楼 吖如的回复:

#include <stdio.h>
#include <math.h>
void main()
{
float a,b,c,x1,x2,judge;
printf("please input three coefficient:");
scanf("%f,%f,%f",&a,&b,&c);
if(a==0)
printf("not one yuan quadratic equation");
else
{
judge=b*b-4*a*c;
if(judge<0)
printf("no roots");
else if (judge==0)
printf("x1=x2=%f",(-b)/(2*a));
else
printf("x1=%f,x2=%f",((-b)+sqrt(b*b-4*a*c))/(2*a),((-b)+sqrt(b*b-4*a*c))/(2*a));
}
}

当b²-4ac<0，该怎么弄呢？

[自信男孩]

if (D = 0)

这里应该是判断语句的，所以改成如下试试：

if (D == 0)