求方程ax2+bx+c=0的根,用3个函数分别求当b2-4ac大于0,等于0,和小于0时的根并输出结果.从主函数a,b,c输用c语言编程,
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![求方程ax2+bx+c=0的根,用3个函数分别求当b2-4ac大于0,等于0,和小于0时的根并输出结果.从主函数a,b,c输用c语言编程,](/uploads/image/z/3960160-16-0.jpg?t=%E6%B1%82%E6%96%B9%E7%A8%8Bax2%2Bbx%2Bc%3D0%E7%9A%84%E6%A0%B9%2C%E7%94%A83%E4%B8%AA%E5%87%BD%E6%95%B0%E5%88%86%E5%88%AB%E6%B1%82%E5%BD%93b2-4ac%E5%A4%A7%E4%BA%8E0%2C%E7%AD%89%E4%BA%8E0%2C%E5%92%8C%E5%B0%8F%E4%BA%8E0%E6%97%B6%E7%9A%84%E6%A0%B9%E5%B9%B6%E8%BE%93%E5%87%BA%E7%BB%93%E6%9E%9C.%E4%BB%8E%E4%B8%BB%E5%87%BD%E6%95%B0a%2Cb%2Cc%E8%BE%93%E7%94%A8c%E8%AF%AD%E8%A8%80%E7%BC%96%E7%A8%8B%2C)
求方程ax2+bx+c=0的根,用3个函数分别求当b2-4ac大于0,等于0,和小于0时的根并输出结果.从主函数a,b,c输用c语言编程,
求方程ax2+bx+c=0的根,用3个函数分别求当b2-4ac大于0,等于0,和小于0时的根并输出结果.从主函数a,b,c输
用c语言编程,
求方程ax2+bx+c=0的根,用3个函数分别求当b2-4ac大于0,等于0,和小于0时的根并输出结果.从主函数a,b,c输用c语言编程,
#include
#include
// b^2-4ac == 0
void fun1(double &a,double &b,double &c,double &d){
double ans = -b/(2*a);
printf("b^2-4ac == 0 , x1 = x2 = %lf.\n",ans);
}
// b^2-4ac > 0
void fun2(double &a,double &b,double &c,double &d){
double ans1,ans2;
ans1 = (-b+sqrt(d)) / (2*a);
ans2 = (-b-sqrt(d)) / (2*a);
printf("b^2-4ac > 0 , x1 = %lf , x2 = %lf.\n",ans1,ans2);
}
// b^2-4ac < 0
void fun3(double &a,double &b,double &c,double &d){
double real,imar;
real = -b/(2*a);
imar = sqrt(-d) / (2*a);
printf("b^2-4ac < 0 , x1 = %lf+%lfi , x2 = %lf-%lfi.\n",real,imar,real,imar);
}
int main(){
double a,b,c,d;
printf("please input a,b,c.\n");
while(scanf("%lf%lf%lf",&a,&b,&c)!=EOF){
d = b*b-4*a*c;
if(d==0) fun1(a,b,c,d);
else if(d>0) fun2(a,b,c,d);
else fun3(a,b,c,d);
printf("please input a,b,c.\n");
}
}
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