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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 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 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

arg1 args [1] 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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08-06 08:59