Question

用C++,计算一元二次方程ax�0�5+bx+c=0；注意不同的根有不同的输出。请高手帮帮忙啊，，尽量简单点？？？

Answer 1

#include<stdio.h>
#include<math.h>
int main()
{
float a,b,c,d;
printf("请输入一元二次方程的三个参数a b c(中间用空格隔开)：");
scanf("%f %f %f",&a,&b,&c);
d=b*b-4*a*c;
if(a==0)
{
if(b==0)
{
if(c==0)
printf("参数都为零，方程无意义！\n");
else
printf("a和b为0，c不为0，方程不成立\n");
}
else
printf("x1=x2=%.2f\n",-c/b);
}
else
if(d>=0)
{
printf("x1=%.2f\n",(-b+sqrt(d))/(2*a));
printf("x2=%.2f\n",(-b-sqrt(d))/(2*a));
}
else
{
printf("x1=%.2f+%0.2fi\n",-b/(2*a),sqrt(-d)/(2*a));
printf("x2=%.2f-%0.2fi\n",-b/(2*a),sqrt(-d)/(2*a));
}
fflush(stdin);
getchar();
return 0;
} 望采纳

Answer 2

#include <stdio.h>
#include <math.h>
#include <stdlib.h>int main()
{
double a,b,c,d;
scanf("%lf %lf %lf",&a,&b,&c);
if(fabs(a)<1e-6)
{
printf("x=%lf\n",-c/b);
return 0;
}
d=b*b-4*a*c;
if(fabs(d)<1e-6)
{
printf("x=%lf\n",-b/(2*a));
return 0;
}
if(d<0)
{
printf("no root\n");
return 0;
}
printf("x1=%lf\n",(-b+sqrt(d))/(2*a));
printf("x2=%lf\n",(-b-sqrt(d))/(2*a));
return 0;
}