Question

有人可以查看我的代码中的二次方程吗？我总是在root2中遇到错误（“root1的原始类型double没有字段root2）。我只需打印出两个根。谢谢。

public class QuadraticEqn {
public static void main(String[] args) {
    System.out.print(quadratic(-7, 4, 3));
}
    public static double quadratic(int a, int b, int c){
        double discriminant = (b*b)-4*a*c;
        double root1 = -1*b + Math.sqrt(discriminant);
        double root2 = -1*b - Math.sqrt(discriminant);
        return (root1, root2);
}
}

Answer 1

您仍然需要除以2a。

Answer 2

Java不能像这样返回元组，可能最容易返回一个数组，因为它们属于同一类型。

    public static double[] quadratic(int a, int b, int c){
        double discriminant = (b*b)-4*a*c;
        double root1 = -1*b + Math.sqrt(discriminant);
        double root2 = -1*b - Math.sqrt(discriminant);
        double[] array = {root1, root2};
        return array;
    }

Answer 3

这种方式无法返回两个double值。尝试返回一个duble值数组。

    public class QuadraticEqn {
        public static void main(String[] args) {
            double[] root = quadratic(-7, 4, 3);
            System.out.print(root[0] + " " + root[1]);
        }

        public static double[] quadratic(int a, int b, int c) {
            double discriminant = (b * b) - 4 * a * c;
            double[] root = new double[2];
            root[0] = -1 * b + Math.sqrt(discriminant);
            root[1] = -1 * b - Math.sqrt(discriminant);
            return root;
        }
    }

Answer 4

不确定你想要完成什么，我使用javascript制作了一个类似的程序，并且继承了代码，它非常简单。顺便说一句，二次方程是-b±√D/ 2a，我不认为你把它完全放入。继承人我写的代码：此外，我认为问题出在您返回的部分：（root1，root2），将其变为数组或只执行2次返回命令。即返回（root1） return（root2）不确定你是否有2个返回函数，而不是java程序员，对不起，但我试过了：D

else if(choice == 2){//quadratic start
     alert("the equation is of the form : ax^2 + bx + c == 0 , only input the coefficients i.e - the value of ax^2 is a, or the value of bx is b, not bx. The value of b for the equation 5x^2 + 7x +3 is 7, not 7x");
     var a = prompt("Put in the value of a");//declaring variables
     var b = prompt("Put in the value of b, if the bx part of the equation doesn't exist, input 0. Ex for equation 2x^+6==0 , b ==0, since its technically 2x^2 + 0b + 6 == 0");
     var c = prompt("Put in the value of c, if the c part of the equation doesn't exist, input 0. Ex for equation 2x^+6x==0 , c ==0, since its technically 2x^2 + 6b + 0 == 0");
     var D = ((b*b)-(4*a*c));//computing discriminant
     if(D < 0){
     alert("The quadratic equation doesn't have real roots; the closest value is : " + (-b/2) +" + i/2");
     }
     else{
     root1 = (- b + Math.sqrt(D))/(2*a);
     root2 = (- b - Math.sqrt(D))/(2*a);
     }
     if(D===0){
     console.log("Both roots are equal, their value is " + root1);
     alert("Both roots are equal, their value is " + root1);
     }
     else if ( D > 0){
     console.log("The roots of the equation are: " + root1 + " and " + root2);
     alert("The roots of the equation are: " + root1 + " and " + root2 );
     }
     }//quadratic end

并且是完整的程序：

<!DOCTYPE html>
<head>
 <script type="text/javascript" src ="code.js"></script>
 <script type="text/javascript">
 var main = function(){//Linear in 2 start
 var choice = prompt("Choose your type of equation : Type 1 for linear in 2 variables, 2 for quadratic in one variable ");
 if(choice ==1){
 alert("The two equations are of the forms a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 respectively")
 var a1 = prompt("Enter value of a (a1) for equation 1");
 var b1 = prompt("Enter value of b (b1) for equation 1");
 var c1 = prompt("Enter value of c (c1) for equation 1");
 var a2 = prompt("Enter value of a (a2) for equation 2");
 var b2 = prompt("Enter value of b (b2) for equation 2");
 var c2 = prompt("Enter value of c (c2) for equation 2");
 if(a1/a2===b1/b2===c1/c2){
 alert("No solution is possible for the equation, i.e. the lines are parallel")
 }
 else if(a1/a2===b1/b2 && b1/b2!=c1/c2){
 alert("The lines are collinear and the values of x and y are infinite")
 }
 else{
 var ansy = (c2*a1-c1*a2)/(b1*a2-b2*a1);
 var ansx = (c1-b1*ansy/a1);
 alert("x is equal to : " + ansx);
 alert("y is equal to : " + ansy);
 }
 }
 else if(choice == 2){//quadratic start
 alert("the equation is of the form : ax^2 + bx + c == 0 , only input the coefficients i.e - the value of ax^2 is a, or the value of bx is b, not bx. The value of b for the equation 5x^2 + 7x +3 is 7, not 7x");
 var a = prompt("Put in the value of a");//declaring variables
 var b = prompt("Put in the value of b, if the bx part of the equation doesn't exist, input 0. Ex for equation 2x^+6==0 , b ==0, since its technically 2x^2 + 0b + 6 == 0");
 var c = prompt("Put in the value of c, if the c part of the equation doesn't exist, input 0. Ex for equation 2x^+6x==0 , c ==0, since its technically 2x^2 + 6b + 0 == 0");
 var D = ((b*b)-(4*a*c));//computing discriminant
 if(D < 0){
 alert("The quadratic equation doesn't have real roots; the closest value is : " + (-b/2) +" + i/2");
 }
 else{
 root1 = (- b + Math.sqrt(D))/(2*a);
 root2 = (- b - Math.sqrt(D))/(2*a);
 }
 if(D===0){
 console.log("Both roots are equal, their value is " + root1);
 alert("Both roots are equal, their value is " + root1);
 }
 else if ( D > 0){
 console.log("The roots of the equation are: " + root1 + " and " + root2);
 alert("The roots of the equation are: " + root1 + " and " + root2 );
 }
 }//quadratic end
}
var again = confirm("Proceed to equation selection menu?");
if(again === true){
 main();
}
 </script>
 <title>Equation Solver</title>
</head>
<body>
</body>
</html>

有人可以查看我的代码中的二次方程（在java中）吗？

4 个答案: