Java Math Completing the square calculator

Question

Hey im new to the whole programming scene im just a high schooler taking a java computer science course. Im trying to test out my abilities with what I know so far and to try understand stuff and practice. So im trying to make a couple math calculators for certain things. Im working on one to complete the square. So like (2 + 4x^2)^2 will come out to be (4 + 16x + 4x^2). My problem is that I cant get it to fully work, and the code is kinda klanky.

import java.util.Scanner;

public class SquaringDoubles {

    public static void main(String[] args) {

        //declaring
        Scanner input = new Scanner(System.in);

        //inputs
        System.out.println("Enter in the double with this format ( A + B )^2");

        System.out.print("A --> ");
        double A = input.nextInt();

        System.out.print("B --> ");
        double B = input.nextInt();
        input.close();

        System.out.println("You're Equation: " + A + " + " + B + "x");

        //Math
        //A + C + B
        double A2 = Math.pow(A, 2);
        double B2 = Math.pow(B, 2);
        double C = 2 * (A * B);

        //final
        System.out.print("You answer: ");
        System.out.println(A2 + " + "+ C + "x" + " + "+ B2+ "x^2");
    }
}

Answer 1

Inserting is the simplest general approach to get from one formula to an equivalent equation of form

Y(x) = A2 + B2x + C2(x)^2

As there are three variables(A2,B2,C2) we need three equations to solve the system. To get those equations we can simply put three x of our choice and their calculated Y(x) into the form we want. And solve the system of equations.

So in essence we calculate Y(x) for three x of our choice and just stick them into the formula. One can take any (defined) value but some make life easier.

So

X=0 is the first candidate as it eliminates everything with x and directly gives you A2.

Y(0)= C2*(0)^2+B2*(0)+A2 = A2
A2 = Y(0)

x=1: you get

 Y(1)= C2*(1)^2+B2*(1)+A2 = C2+B2+A2

x=-1: you get

Y(-1)= C2*(-1)^2+B2*(-1)+A2= C2-B2+A2

Eliminating C2:

Y(-1)+B2-A2 = Y(1) -B2 -A2
-> 2*B2=Y(1)-Y(-1)
B2=(Y(1)-Y(-1))/2

Finaly calculate C2 by inserting into C2+B2+A2=Y(1):

C2=Y(1) -B2 -A2

So in General - for any given(valid) equation to get to the form A2+B2·x+C2·x²:

• A2 = Y(0)
• B2=(Y(1)-Y(-1))/2
• C2=Y(1) -B2 -A2 = (Y(1)+Y(-1))/2 - Y(0)

In your example Y(1) = Y(-1) due to the square, so

Y(x) = (A + B * (x)^2)^2

Y(1) = (A + B * (1)^2)^2   =(A+B)^2  // x=1
Y(-1) = (A + B * (-1)^2)^2 =(A+B)^2  // x=-1
B2 =(Y(1)-Y(-1))/2 =0

And so

C2= Y(1) -A2 = (A+B)^2 - A2

So for (A + B*(x)^2)^2:

• A2 = Y(0) = A^2
• B2=0
• C2=Y(1) - Y(0) = (A+B)^2 - A2

Code:

public static void main(String[] args) {

        //declaring
        Scanner input = new Scanner(System.in);

        //inputs
        System.out.println("Enter in the double with this format (A + B(x)^2)^2");

        System.out.print("A --> ");
        double A = input.nextInt();

        System.out.print("B --> ");
        double B = input.nextInt();
        input.close();

        System.out.println("You're Equation: (" + A + " + " + B + "x^2)^2");

        //Math
        //A + C + B
        double A2 = Math.pow(A, 2);
        /**  old code:
         * 
        double B2 = Math.pow(B, 2);

        double C = 2 * (A * B);
        */     
        /** replacement : */
        //Y(1)=(A + B*(1)^2)^2 = (A+B)^2
        //Y(-1)=(A + B*(-1)^2)^2 = (A+B)^2
        //B2 = (Y(1)-Y(-1))/2 = ((A+B)^2 -(A+B)^2)/2 = 0
        double B2=0; // it is always 0 in this case
        //Y(1)=(A + B*(1)^2)^2 = (A+B)^2
        double C2=(A+B)*(A+B) - A2; //Y(1) -A2 
        //final
        System.out.print("You answer: ");
        System.out.println(A2 + " + " + B2 + "x" + " + " + C2 + "x^2");
    }

I added some comments, to show what is going on and how to do similar with other equation.