[英]I can't figure out how to find the two imaginary numbers
所以,我寫了一個 Java 程序,它找到了一個二次方程的解,我的問題是我似乎無法編寫正確的代碼來找到“虛數”,當它打印出來時,我只得到“NaN”。 有什么解決辦法嗎?
import java.util.Scanner;
class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print("Enter the value for a: ");
double a = scan.nextDouble();
System.out.print("Enter the value for b: ");
double b = scan.nextDouble();
System.out.print("Enter the value for c: ");
double c = scan.nextDouble();
double result = b * b - 4.0 * a * c;
if(result > 0.0){
//to find two real solutions
double x1 = (-b + Math.pow(result, 0.5)) / (2.0 * a);
double x2 = (-b - Math.pow(result, 0.5)) / (2.0 * a);
System.out.println("There are two real solutions.");
System.out.println("x1 = " + x1);
System.out.println("x2 = " + x2);
//to find one real solution
} else if(result == 0.0){
double x1 = (-b / (2.0 * a));
System.out.println("There is one real solution");
System.out.println("x = " + x1);
//to find the imaginary numbers
} else if(result < 0.0){
double x1 = (-b + Math.pow(result, 0.5)) / (2.0 * a);
double x2 = (-b - Math.pow(result, 0.5)) / (2.0 * a);
System.out.println("There are two imaginary solutions.");
System.out.println("x1 = " + x1 + " + " + x2);
System.out.println("x2 = " + x1 + " - " + x2);
}
}
}
在處理復雜的根時(當結果 < 0 時),您的代碼中有幾個不正確的地方:
result
的平方根是否為負。 這將導致 NaN。 正確的方法是得到-result
的平方根來得到你的答案。-b/(2*a)
和相同的虛部值,僅符號不同。我在下面修復了您的代碼,以提供正確的 output。 計算實部,然后計算虛部。 然后用后綴“i”的虛部打印根來表示虛部。
double real = -b / (2*a);
double imag = Math.pow(-result, 0.5) / (2.0 * a);
System.out.println("There are two imaginary solutions.");
System.out.println("x1 = " + real + " + " + imag + "i");
System.out.println("x2 = " + real + " - " + imag + "i");
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