二次方程因子计算

Question

Regarding quadratic equations (learn more here ), I've taken the a , b , and c of the equation as input.

A sample equation will be this: 21x^2 - 8x - 4 Here, a = 21, b = -8, c = -4. So, on solving (without formula), => 21x^2 - 14x + 6x - 4 = 0.

I need the two middle numbers, that is, in this case 14 and 6 (read factors). I think I have done all correct, but the input seems to be infinite and does not stop at all. Can you please rectify the mistake? I am also curious to learn why that happened.

import java.util.Scanner;
public class QuadFact {
    static Scanner sc = new Scanner(System.in); 
    static int a,b,c; 
    static int P, diff, p; 
    static int i;
    static boolean found = false;

    void accept(){
        System.out.println("Enter the a, b, c");
        a = sc.nextInt(); b = sc.nextInt(); c = sc.nextInt();
    }

    void compute(){
        P = a * c;
        diff = 0;
        while(!found){
           for (i = b + 1;;i++){
                diff = i - b;
                p = i * diff;
                if (p==P) {
                    found = true;
                    break; 
                }
            }
        }
    }

    void display(){
        System.out.print("These are the raw numbers, should be   correct.  
        Still,\n it is advisable you verify it.");
        System.out.println("One factor: " + i);
        System.out.println("Other factor: " + diff);
    }

    public static void main(String[] args){
        QuadFact a = new QuadFact();
        a.accept();
        a.compute();
        a.display();
    }
}

Answer 1

I think you have to look "on both sides" of b for a factor pair that adds up to b and produces the product a*c.

void compute(){
    P = a * c;
    while(!found){
    for( i = 1; ; i++ ){
            diff = b - i;
            if (i * diff == P) {
                found = true;
                break; 
            }
            diff = b + i;
            if (-i * diff == P) {
                found = true;
                break; 
            }
        }
    }
}

Answer 2

ok, I wrote a code for that.

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        //Declare and get the variables 
        int a, b,c;

        Scanner s = new Scanner(System.in);

        System.out.println("Enter A");

        a = s.nextInt();

        System.out.println("Enter B");

        b = s.nextInt(); 

        System.out.println("Enter c");

        c = s.nextInt(); 

        //A should be 1 if not divide a, b and c by a
        if (a>1) {

            b= b/a;
            c=c/a;
            a= a/a;
        }
        //Just printing what the values of ABC IS AGAIN
        System.out.println("A = "+a+" B = "+b+" C = "+c);
        //Just printing what the values of ABC IS AGAIN but in reverse, that is if c was 5 it becomes -5 
        //another way to reverse positive to negative and vice versa
        System.out.println("A = "+(0-a)+" B = "+(0-b)+" C = "+(0-c));

        //Set i as c and start the loop from highest to lowest. 
        for (int i = Math.abs(c); i>0 ;i-- ) {

            //if a multiple is found it proceeds and checks for the 
            //multiple combination that when multiplied you get C and the addition or subtraction gives you B
            if (c%i==0) { 
                int fac1 = c/i;
                //Displays the multiples found

                System.out.println(i+" x "+fac1+" = "+c);

                //There are 4 possible outcomes or cases
                //case 1 multiple 1 - multiple 2 = b
                //case 2 multiple 2 - multiple 1 = b
                //case 3 multiple 1 + multiple 2 = b
                //case 4 -multiple 1 + -multiple 2 = b

                if (i-fac1 == b) {
                    //System.out.println("case 1: " + i+"-"+fac1+"="+b);
                    answer(i,fac1);
                    break;
                }
                else if(fac1 - i == b){

                    //System.out.println("case2: "  + fac1+"-"+i+"="+b);
                    answer(fac1,i);
                    break;
                }

                else if (i +fac1 == b ) {

                    //System.out.println("case3: "  + i+"+"+fac1+"="+b);
                    answer(i,fac1);
                    break;
                }
                else if((0-Math.abs(i)) + (0- Math.abs(fac1))==b) {

                    //System.out.println("case4: "+"-"+i+ " + "+ "-"+ fac1 +"="+b);

                    answer((0-Math.abs(i)),(0- Math.abs(fac1)));
                    break;
                }


                else {
                    System.out.println("Probably not a factorizable Equation");
                }

            }

        }

    }
    //Use this method to show the final answer
    private static void answer(int f1, int f2) {

        System.out.println("x = "+(0-f1) +" or x = "+(0-f2));

    }

}