有人可以查看我的二次方程代码吗？root2 总是出错（“root1 的原始类型 double 没有字段 root2）。我只需要打印出两个根。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);
}
}
你仍然需要除以2a。
Java 不能返回这样的元组，可能最容易返回数组，因为它们是相同的类型。
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;
}
以这种方式返回两个双精度值是不可能的。尝试返回一个双重值数组。
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;
}
}
不知道你想完成什么，我用javascript做了一个类似的程序，代码如下，它真的很简单。顺便说一句，二次方程是 -b±√D/2a ，我不认为你把它完全放进去。这是我写的代码：另外，我认为问题出在你已经返回的部分：(root1,root2) ，将其放入一个数组或只执行 2 个返回命令。即return（root1）return（root2）不确定你是否有2个返回ins一个函数，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
这是完整的程序：
<!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>