Write a GUI interface for a calculator that finds the roots of a quadratic equat

Question

Write a GUI interface for a calculator that finds the roots of a quadratic equation. JAVA

Explanation / Answer

//import java.util.Scanner;//delete this part after public class findingRoots { public static void main(String[] args) { double temp = printRoots(Integer.parseInt(args[0]), Integer.parseInt(args[1]), Integer.parseInt(args[2])); } public static double printRoots (int a, int b, int c){ //should it be double here or int? //read in the coefficients a,b,and c //now compte the discrimintat d double discriminant = (Math.pow(b, 2.0)) - (4 * a * c); double d=Math.sqrt(discriminant); double X = 0,Y = 0; //root 1 & root 2, respectively // is the step double X,Y necessary? d = (b*b)-(4.0*a*c); if (discriminant > 0.0) { X = (-b + d)/(2.0 * a ); //X= root 1, which is larger //do i put int or double in front of X?** Y = (-b - d)/(2.0 *a); //Y= root 2, which is the smaller root String root1 = Double.toString(X); String root2 = Double.toString(Y); System.out.println("The two roots are X and Y:" + root1 + "and" + root2); } else if (discriminant==0.0){ X = (-b + 0.0)/(2.0 * a);//repeated root String root2 = Double.toString(X); System.out.println("The equation has one root X:" + root2);//where X is the only root } else if (discriminant < 0.0){ double X1 = -b/(2*a); double Y1 = (Math.sqrt(-c))/(2*a); double X2 = -b/(2*a); double Y2 = (-(Math.sqrt(-c)))/(2*a); String root1 = Double.toString(X1); String root2 = Double.toString(Y1); String root3 = Double.toString(X2); String root4 = Double.toString(Y2); System.out.println("The equation has two roots:" + root1 + root2 + "and" + root3 + root4); // where i represents the square root of negative 1 } return -1; } }

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