创建一个java程序，按顺序给出3个术语作为输入（a，b，c）打印它们的根 [英] Creating a java program that given 3 terms as input(a,b,c) in that order prints their roots

问题描述

Wrie方法printRoots给出3个术语作为输入（a，b，c）按顺序打印它们的根

我们有以下给定的信息

1. 如果b²-4ac 是正数，您的程序应该打印两个根是X和Y其中X是较大的根，Y是较小的根




2. 如果b²-4ac * 等于0 * ，程序应该打印。 等式有一个X，其中X是唯一的根




3. 如果b²-4ac 是负数，该程序应打印。该等式有两个根（-X1 + Y1i）和（-X2和Y2i）

   
   
   

   该术语可以根据：

   
   
   
   
   
   • 如果b ^ 2 - 4ac是负数，则二次方程变为：（ - b +/-√C)/2a
     - 这意味着公式可以简化为（-b +/-√Ci）/ 2a，其中平方根不是正数
   
   

计算系数并打印出来（即X1是-b / 2a，Y1是sqrt（-C）/ 2i

注意：此问题不允许使用扫描仪

有人可以查看我的程序，并告诉我我哪里出错了，我只是删除我的扫描仪，使其成为没有扫描仪的程序？

---

  import java.util.Scanner; // delet e $ 
公共类查找之后的这部分语句{
 public static void main（String [] args）
 {
} 
 public static double printRoots（）{// should它在这里加倍还是int？ 
 //读入系数a，b和c 
扫描仪读卡器=新扫描仪（System.in）; 
 int a = reader.nextInt（）; 
 System.out.println（输入值）; 
 int b = reader.nextInt（）; 
 System.out.println（输入b的值）; 
 int c = reader.nextInt（）; 
 System.out.println（输入c的值）; 
 //现在compte the discriminimintat d 
 double discrimintant = d; 
倍X，Y; //根1& root 2，分别是
 //是步骤双X，Y必要吗？ 
 double d =（b * b） - （4.0 * a * c）; 
 if（d> 0.0）{
 d = Math.sqrt（d）; 
 System.out.println（两个根是X和Y）; 
 double X =（ -  b + d）/（2.0 * a）; // X =根1，更大
 double Y =（ -  b  -  d）/（2.0 * a）; // Y = root 2，它是较小的根
 System.out.println（Root 1= X和Root 2=Y）; 
} 
 else {
 if（d == 0.0）//then ...怎么写？ 
 System.out.println（公式有一个根X）//其中X是唯一的根
 double X =（ -  b + 0.0）/（2.0 * a）; //重复root 
 System.out.println（Root=X）; 
} 
 else {
 if（d< 0.0）
 System.out.println（等式有两个根（-X1 + Y1i）和（-X2 + Y2i） ））; 
 //其中i代表负1的平方根
 double X1 = -b /（2 * a）; 
 double Y1 =（Math.sqrt（-C））/（2 * a）; 
 double X2 = -b /（2 * a）; 
 double Y2 =（ - （Math.sqrt（-C）））/（2 * a）; 
 double Y2 =（ - （Math.sqrt（-C）））/（2 * a）; 
 System.out.println（Root 1=（ -  X1 + Y1i）和Root 2=（ -  X2 + Y2i））; 
} 
} 
} 
  

解决方案

你可以传递输入从命令行。你将获得args数组中的数据

in public static void main（String [] args）这里args是指使用

 <$运行java程序c $ c> java MyApp arg1 arg2 
  

主要 args [0]中的



是 arg1 和 args [1] 是 arg2

所以在您的情况下运行应用程序，如下面的命令

  java findingRoots 1 2 3 
  

and main

  int a = Integer.parseInt（args [0]）
  

NB 我想你想验证命令行参数检查args.length以及它们是否为int

Wrie a method printRoots that given 3 terms as input(a,b,c) in that order prints their roots

We have the following given information

1. If b²-4ac is a positive number, your program should print "The two roots are X and Y" where X is the larger root and Y is the smaller root

2. If b²-4ac *equals 0*, the program should print. "The equation has one X" where X is the only root

3. If b²-4ac is a negative number, the program should print." The equation has two roots(-X1 + Y1i) and (-X2 and Y2i)

   The term can be determined based on:

   • If b^2 - 4ac is a negative number, then the quadratic equation becomes: (-b+/- √C)/2a -This means the equation can be simplified to become (-b+/- √Ci)/2a where the square root is not a positive number

Calculate the coefficient and print that(i.e X1 is -b/2a and Y1 is sqrt(-C)/2i

Note: Not allowed to use Scanners for this question

Is it possible for someone to review my program and tell me where I have gone wrong and do i just remove my scanners to make it a program without scanners?

---

import java.util.Scanner;//delete this part after 
    public class findingRoots {
    public static void main(String[] args)
        {
        }
          public static double printRoots (){ //should it be double here or int? 
           //read in the coefficients a,b,and c 
          Scanner reader = new Scanner(System.in);
          int a=reader.nextInt();
         System.out.println("Enter the value of a");
          int b=reader.nextInt();
          System.out.println("Enter the value of b");
          int c=reader.nextInt();
          System.out.println("Enter the value of c");
          //now compte the discrimintat d 
           double discrimintant = d; 
          double X,Y; //root 1 & root 2, respectively
           // is the step double X,Y necessary? 
           double d = (b*b)-(4.0*a*c);
             if (d > 0.0){ 
             d = Math.sqrt(d);
             System.out.println("The two roots are X and Y");
             double X = (-b + d)/(2.0 * a ); //X= root 1, which is larger 
             double Y = (-b - d)/(2.0 *a); //Y= root 2, which is the smaller root 
             System.out.println("Root 1" = X "and" "Root 2" "=" Y);
           }
           else{
             if (d==0.0) //then...how to write?
               System.out.println("The equation has one root X")//where X is the only root 
            double X = (-b + 0.0)/(2.0 * a);//repeated root 
             System.out.println("Root" "=" X);
           }
           else{
             if(d < 0.0)
               System.out.println("The equation has two roots (-X1 + Y1i) and (-X2 +Y2i)");
             // where i represents the square root of negative 1 
             double X1 = -b/(2*a);
             double Y1 = (Math.sqrt(-C))/(2*a);
             double X2 = -b/(2*a);
             double Y2 = (-(Math.sqrt(-C)))/(2*a);
             double Y2 = (-(Math.sqrt(-C)))/(2*a);   
             System.out.println("Root 1" "=" (-X1 + Y1i) "and" "Root 2" "=" (-X2 +Y2i)");
           }
          }
      }

解决方案

you can pass input from command lines. You will get the data at args array

in public static void main(String[] args) here args refers to command line arguements

when you run a java program using

java MyApp arg1 arg2

in your main args[0] is arg1 and args[1] is arg2

So in your case run the app like following command

java findingRoots 1 2 3

and in main

int a= Integer.parseInt(args[0])

N.B I think you would like to validate the command line parameters. check both the args.length and if they are int or not

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