题目：求方程ax平方+bx+c=o的根，用三个函数分别求档b平方-4ac大于0，等于0，小于0时的根，并输出结果。从主函数输入a,b,c.
下面是我个人所编写的，但是错误，请问为什么，并希望高人能给出正确答案，谢谢！



#include <iostream.h>
#include <cmath.h>
int main()
{ float a,b,c,x1,x2;
  cout<<"please enter three numbers:"<<endl;
  cin>>a>>b>>c;
  if((-b+sqrt(b*b-4*a*c)/(2*a)<0)  &&  (-b-sqrt(b*b-4*a*c)/(2*a)<0)
      { cout<<"there id no gen"<<endl; }
  if((-b+sqrt(b*b-4*a*c)/(2*a)>=0)  &&  (-b-sqrt(b*b-4*a*c)/(2*a)>=0)
      {    x1=(-b+sqrt(b*b-4*a*c)/(2*a));
    x2=(-b-sqrt(b*b-4*a*c)/(2*a));  }
  cout<<"x1="<<x1<<endl;
  cout<<"x2="<<x2<<endl;
       return 0;





   }

我写了一个供楼主参考，请多指教：

#include<iostream>
#include<cmath>
using namespace std;

int main(int argc, char* argv[])
{
    float a,b,c;
    float x1,x2;
    cout<<"请输入方程ax^2+bx+x=0的各项系数："<<endl;
    cout<<"a=";
    cin>>a;
    cout<<"b=";
    cin>>b;
    cout<<"c=";
    cin>>c;

    if(a==0)
        cout<<"方程只有一个实根：x="<<-(c/b)<<endl;
    float temp=b*b-4.0*a*c;
    if(temp<0)
        cout<<"方程没有实根！"<<endl;
    else if(temp<1.0e-6)        //这句是判断temp是否等于0，浮点数是不精确存储的，所以不直接跟0做比较。
        cout<<"方程有两个相等的实根："<<"x1=x2="<<-b/2.0*a<<endl;
    else
    {
        temp=sqrt(temp);
        x1=(-b+temp)/2.0*a;
        x1=(-b-temp)/2.0*a;
        cout<<"方程有两个实根："<<"x1="<<x1<<ends<<ends<<"x2="<<x2<<endl;

    }

 

#include <iostream.h>
#include <cmath.h>
int main()
{
  float a,b,c,d，x1,x2;
  cout<<"please enter three numbers:"<<endl;
  cin>>a>>b>>c;
  d=sqrt(b*b-4*a*c);
   x1=(-b+d)/(2*a));
   x2=(-b-d)/(2*a));
    if(sqrt(b*b-4*a*c)<0)  
       cout<<"there id no gen"<<endl;
  esle
  if(sqrt(b*b-4*a*c)==0)
{
   cout<<"there is one gen"<<endl;
  cout<<"x1=x2"<<x1<<endl;
}
 else
{
 cout<<"there is two  gen"<<endl;

   
  cout<<"x1="<<x1<<endl;
  cout<<"x2="<<x2<<endl;
}
}
      

看看那个行不行.........

#include <iostream.h>
#include <cmath.h>
int main()
{ float a,b,c,x1,x2;
  cout<<"please enter three numbers:"<<endl;
  cin>>a>>b>>c;
  if((-b+sqrt(b*b-4*a*c)/(2*a)<0)  &&  (-b-sqrt(b*b-4*a*c)/(2*a)<0) ( 不是这样判断的)
      { cout<<"there id no gen"<<endl; }
  if((-b+sqrt(b*b-4*a*c)/(2*a)>=0)  &&  (-b-sqrt(b*b-4*a*c)/(2*a)>=0)
      {    x1=(-b+sqrt(b*b-4*a*c)/(2*a));
    x2=(-b-sqrt(b*b-4*a*c)/(2*a));  }
  cout<<"x1="<<x1<<endl;
  cout<<"x2="<<x2<<endl;
       return 0;
   }
数学公式要吃透点，帮你改了改看是不是这东东
#include <iostream.h>
#include <math.h>
#include"stdlib.h"
int main()
{ float a,b,c;
double x1,x2;
cout<<"please enter three numbers:"<<endl;
  cin>>a>>b>>c;
    if(b*b-4*a*c<0){cout<<"there id no gen"<<endl;}
      else if(b*b-4*a*c==0){x1=x2=((-b)/2*a);}
        else{
             x1=(-b-sqrt(b*b-4*a*c))/(2*a);
             x2=(-b+sqrt(b*b-4*a*c))/(2*a);
             }
       cout<<"x1="<<x1<<endl;
  cout<<"x2="<<x2<<endl;
       return 0;
    }