到目前为止我有这个代码，但是每次我运行并将三个数字放在一起得到根源是NaN可以请一些人帮助或指出我哪里出错了。
import java.util.Scanner;
class Quadratic {
public static void main(String[] args) {
System.out.println("Enter three coefficients");
Scanner sc = new Scanner(System.in);
double a = sc.nextDouble();
double b = sc.nextDouble();
double c = sc.nextDouble();
double root1= (-b + Math.sqrt( b*b - 4*a*c ) )/ (2*a);
double root2= (-b - Math.sqrt( b*b - 4*a*c ) )/ (2*a);
System.out.println("The roots1 are: "+ root1);
System.out.println("The roots2 are: " + root2);
}
}
答案 0 :(得分：4)
你必须记住，并非每个二次方程都有根，可以用实数表示。更具体地说，如果b*b - 4*a*c < 0，则根将具有虚部，并且将返回NaN，因为负数的Math.sqrt返回NaN，如{ {3}}。这适用于b*b - 4*a*c >= 0的系数，但是：
Enter three coefficients 1 5 6 The roots1 are: -2.0 The roots2 are: -3.0
如果你想考虑非真实的根，你可以做类似
的事情double d = (b * b - 4 * a * c);
double re = -b / (2 * a);
if (d >= 0) { // i.e. "if roots are real"
System.out.println(Math.sqrt(d) / (2 * a) + re);
System.out.println(-Math.sqrt(d) / (2 * a) + re);
} else {
System.out.println(re + " + " + (Math.sqrt(-d) / (2 * a)) + "i");
System.out.println(re + " - " + (Math.sqrt(-d) / (2 * a)) + "i");
}
答案 1 :(得分：0)
希望这有帮助 -
import java.util.Scanner;
class QuadraticCalculator
{
public static void main(String args[])
{
Scanner s=new Scanner(System.in);
double a,b,c,quad_dis,quad_11,quad_1,quad_21,quad_2;
System.out.println("Enter the value of A");
a=s.nextDouble();
System.out.println("\nEnter the value of B");
b=s.nextDouble();
System.out.println("\nEnter the value of C");
c=s.nextDouble();
quad_dis=b*b-4*a*c;
quad_11=(-1*b)+(Math.sqrt(quad_dis));
quad_1=quad_11/(2*a);
quad_21=(-1*b)-(Math.sqrt(quad_dis));
quad_2=quad_21/(2*a);
int choice;
System.out.println("\n\nWhat do you want to do with the numbers you entered ?\n(1) Calculate Discriminant\n(2) Calculate the values\n(3) Find the nature of roots\n(4) All of the above");
choice=s.nextInt();
switch(choice)
{
case 1: System.out.println("\nDiscriminant: "+quad_dis);
break;
case 2: System.out.println("\nValues are: "+quad_1+", "+quad_2);
break;
case 3: if(quad_dis>0)
{
System.out.println("\nThe roots are REAL and DISTINCT");
}
else if(quad_dis==0)
{
System.out.println("\nThe roots are REAL and EQUAL");
}
else
{
System.out.println("\nThe roots are IMAGINARY");
}
break;
case 4: System.out.println("\nDiscriminant: "+quad_dis);
System.out.println("\nValues are: "+quad_1+", "+quad_2);
if(quad_dis>0)
{
System.out.println("\nThe roots are REAL and DISTINCT");
}
else if(quad_dis==0)
{
System.out.println("\nThe roots are REAL and EQUAL");
}
else
{
System.out.println("\nThe roots are IMAGINARY");
}
break;
}
System.out.println("\n\nThank You for using this Calculator");
}
}
答案 2 :(得分：0)
您可以使用以下代码。首先，它将检查输入方程是否是二次方程。如果输入方程是二次的，那么它将找到根。 这段代码也能找到复杂的根源。
public static void main（String [] args）{
// Declaration of variables
float a = 0, b = 0, c = 0, disc, sq_dis;
float[] root = new float[2];
StringBuffer number;
Scanner scan = new Scanner(System.in);
// Input equation from user
System.out.println("Enter Equation in form of ax2+bx+c");
String equation = scan.nextLine();
// Regex for quadratic equation
Pattern quadPattern = Pattern.compile("(([+-]?\\d*)[Xx]2)+((([+-]?\\d*)[Xx]2)*([+-]\\d*[Xx])*([+-]\\d+)*)*|((([+-]?\\d*)[Xx]2)*([+-]\\d*[Xx])*([+-]\\d+)*)*(([+-]?\\d*)[Xx]2)+|((([+-]?\\d*)[Xx]2)*([+-]\\d*[Xx])*([+-]\\d+)*)*(([+-]?\\d*)[Xx]2)+((([+-]?\\d*)[Xx]2)*([+-]\\d*[Xx])*([+-]\\d+)*)*");
Matcher quadMatcher = quadPattern.matcher(equation);
scan.close();
// Checking if given equation is quadratic or not
if (!(quadMatcher.matches())) {
System.out.println("Not a quadratic equation");
}
// If input equation is quadratic find roots
else {
// Splitting equation on basis of sign
String[] array = equation.split("(?=[+-])");
for (String term : array) {
int len = term.length();
StringBuffer newTerm = new StringBuffer(term);
// If term ends with x2, then delete x2 and convert remaining term into integer
if (term.endsWith("X2") || (term.endsWith("x2"))) {
number = newTerm.delete(len - 2, len);
a += Integer.parseInt(number.toString());
}
// If term ends with x, then delete x and convert remaining term into integer
else if (term.endsWith("X") || (term.endsWith("x"))) {
number = newTerm.deleteCharAt(len - 1);
b += Integer.parseInt(number.toString());
}
// If constant,then convert it into integer
else {
c += Integer.parseInt(term);
}
}
// Display value of a,b,c and complete equation
System.out.println("Coefficient of x2: " + a);
System.out.println("Coefficient of x: " + b);
System.out.println("Constent term: " + c);
System.out.println("The given equation is: " + a + "x2+(" + b + ")x+(" + c + ")=0");
// Calculate discriminant
disc = (b * b) - (4 * a * c);
System.out.println(" Discriminant= " + disc);
// square root of discriminant
sq_dis = (float) Math.sqrt(Math.abs(disc));
// conditions to find roots
if (disc > 0) {
root[0] = (-b + sq_dis) / (2 * a);
root[1] = (-b - sq_dis) / (2 * a);
System.out.println("Roots are real and unequal");
System.out.println("Root1= " + root[0]);
System.out.println("Root2= " + root[1]);
}
else if (disc == 0) {
root[0] = ((-b) / (2 * a));
System.out.println("Roots are real and equal");
System.out.println("Root1=Root2= " + root[0]);
}
else {
root[0] = -b / (2 * a);
root[1] = Math.abs((sq_dis) / (2 * a));
System.out.println("Roots are complex");
System.out.println("ROOT1= " + root[0] + "+" + root[1] + "+i");
System.out.println("ROOT2= " + root[0] + "-" + root[1] + "+i");
}
}
答案 3 :(得分：-1)
else {
if ((Math.sqrt(-d) / (2*a)) > 0) {
System.out.println(r + " + " + (Math.sqrt(-d) / (2*a)) + " i");
System.out.println(r + " - " + (Math.sqrt(-d) / (2*a)) + " i");
}
else if ((Math.sqrt(-d) / (2*a)) == 0){
System.out.println(r);
}
else {
System.out.println(r + " - " + (Math.sqrt(-d) / (2*a)) + " i");
System.out.println(r + " + " + (Math.sqrt(-d) / (2*a)) + " i");
}