关于c语言求b^2-4ac的根#include "stdio.h"#include "conio.h"main(){double a,b,c,d,x1,x2,p,q;scanf("%lf,%lf,%lf",&a,&b,&c);d=sqrt(b*b-4*a*c);*a);x1=p+q;x2=p-q;if(d>=0){ if(d>0) printf("real roots:\nx1=%7.2f\nx2=%7.2f\n",x1,x2);else printf("has tw
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/25 06:58:52
![关于c语言求b^2-4ac的根#include](/uploads/image/z/15148510-70-0.jpg?t=%E5%85%B3%E4%BA%8Ec%E8%AF%AD%E8%A8%80%E6%B1%82b%5E2-4ac%E7%9A%84%E6%A0%B9%23include+%22stdio.h%22%23include+%22conio.h%22main%28%29%7Bdouble+a%2Cb%2Cc%2Cd%2Cx1%2Cx2%2Cp%2Cq%3Bscanf%28%22%25lf%2C%25lf%2C%25lf%22%2C%26a%2C%26b%2C%26c%29%3Bd%3Dsqrt%28b%2Ab-4%2Aa%2Ac%29%3B%2Aa%29%3Bx1%3Dp%2Bq%3Bx2%3Dp-q%3Bif%28d%3E%3D0%29%7B+if%28d%3E0%29+printf%28%22real+roots%3A%5Cnx1%3D%257.2f%5Cnx2%3D%257.2f%5Cn%22%2Cx1%2Cx2%29%3Belse+printf%28%22has+tw)
关于c语言求b^2-4ac的根#include "stdio.h"#include "conio.h"main(){double a,b,c,d,x1,x2,p,q;scanf("%lf,%lf,%lf",&a,&b,&c);d=sqrt(b*b-4*a*c);*a);x1=p+q;x2=p-q;if(d>=0){ if(d>0) printf("real roots:\nx1=%7.2f\nx2=%7.2f\n",x1,x2);else printf("has tw
关于c语言求b^2-4ac的根
#include "stdio.h"
#include "conio.h"
main()
{
double a,b,c,d,x1,x2,p,q;
scanf("%lf,%lf,%lf",&a,&b,&c);
d=sqrt(b*b-4*a*c);
*a);
x1=p+q;
x2=p-q;
if(d>=0)
{ if(d>0) printf("real roots:\nx1=%7.2f\nx2=%7.2f\n",x1,x2);
else printf("has two same soulution:\nx1=x2=%7.2f\n",x1,x2);
}
else
printf("no real roots\n");
getch();
}
我不知道到底是哪儿出问题了.得出的结论错的.
关于c语言求b^2-4ac的根#include "stdio.h"#include "conio.h"main(){double a,b,c,d,x1,x2,p,q;scanf("%lf,%lf,%lf",&a,&b,&c);d=sqrt(b*b-4*a*c);*a);x1=p+q;x2=p-q;if(d>=0){ if(d>0) printf("real roots:\nx1=%7.2f\nx2=%7.2f\n",x1,x2);else printf("has tw
你先写了d=sqrt(b*b-4*a*c)然后再判断d>=0,当然错了,要先判断正负再开根号的,还有你的p,q是什么东西,定义了没赋值就x1=p+q; x2=p-q; 这怎么能运行?