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006. 輸入 A, B, C, 求 AX^2 + BX + C = 0 的解
原發表者：jefferylee

#include<iostream>
#include<math.h>
using namespace std;

void main(void)
{

 double a,b,c,d;

 cout<<"一元二次方程式 AX^2 + BX + C = 0 求解: "<<endl<<"A=";

 cin>>a;

 cout<<"B= ";

 cin>>b;

 cout<<"C= ";

 cin>>c;

 d=pow(b,2)-4*a*c;



 if(d<0)

   cout<<"該方程式無實根. "<<endl;

 else

 {

   if(d==0)

     cout<<"該方程式有一重根: "<<-b/(2*a)<<endl;

   else

     cout<<"該方程式有兩實根, 分別是: "<<endl

         <<(-b+sqrt(d))/(2*a)<<"以及"<<endl

         <<(-b-sqrt(d))/(2*a)<<endl;

 }

}

原發表者:rclrn

//6.輸入 A, B, C，求 AX^2 + BX + C = 0 的解。
//compiled by Borland C++ Builder 6
#include <iostream>
#include <cmath>
using namespace std;

int main()
{
 double a = 0.0, b = 0.0, c = 0.0;

 cout << "輸入 a b c，解AX^2 + BX + C = 0: ";
 cin >> a >> b >> c;

 double d = b*b - 4*a*c;

 if(d > 0)
   cout << "方程式有二相異實根，"
        << (-b + sqrt(d)) / 2 / a << "及"
        << (-b - sqrt(d)) / 2 / a << endl;
 else if(d == 0)
   cout << "方程式有一重根" << -b / 2/ a << endl;
 else
   cout << "方程式無實根" << endl;

 system("pause");
 return 0;
}

原發表者:andyeric

/*006. 輸入 A, B, C, 求 AX^2 + BX + C = 0 的解.*/

#include <iostream.h>
#include <math.h>
#include <conio.h>

int main()
{
   double a,b,c,d;
   cout<<" 輸入 A, B, C, 求 AX^2 + BX + C = 0 的解 "<<endl;
   cout<<"A=";
   cin>>a;
   cout<<"B=";
   cin>>b;
   cout<<"C=";
   cin>>c;
           
   d=b*b-4*a*c;
   
   if(d>0)
   cout<<"有兩根"<<(-b+sqrt(d))/2/a<<","<<(-b-sqrt(d))/2/a<<endl;
   else if(d==0)
        cout<<"有一重根"<<-b/2/a;
        else
        cout<<"沒有實根"<<endl;
         
 
   getch();
   return 0;
}    

原發表者：zoe

/* 輸入 A, B, C，求 AX^2 + BX + C = 0 的解。 */

#include <stdio.h>
#include <math.h>

main()
{
       double A, B, C, D, tmp, X[2];

       scanf("%lf %lf %lf", &A, &B, &C);

       D = B * B - 4 * A * C;

       if ( D >= 0 )
               tmp = sqrt(D);

       if ( D > 0 )                            /* 兩相異實根 */
       {
               X[0] = ( - B + tmp ) / ( 2 * A );
               X[1] = ( - B - tmp ) / ( 2 * A );

               printf("X = %lf, %lf\n", X[0], X[1]);
       }
       else if ( D == 0 )                        /* 重根 */
       {
               X[0] = ( - B ) / ( 2 * A );
               printf("X = %lf\n", X[0]);
       }
       else                                     /* 虛根 */
       {
               tmp = sqrt(-D);

               X[0] = ( -B ) / ( 2 * A );
               tmp /= 2 * A;
               printf("X = %lf + %lfi, %lf - %lfi\n", X[0], tmp, X[0], tmp);
       }
}

在此感謝當年在數位高手(NBP)分享的朋友，本人將資料彙整於此並盡量將當初發表者名稱註記。