为什么二次方程式的根结果是NaN? (Java)

问题描述

为什么要写出来根源是控制台中的NaN和NaN 吗?我已经读过有关NaN的文章，但是我找不到合适的解决方案，该如何解决错误...我已经尝试过广播，将判别和根加倍，但是不起作用.有人可以帮助我，在什么地方以及我需要重写什么?

Why write out The roots are NaN and NaN in console?I've read about NaN, but I don't find the right solution how can I repair the mistakes...I've tried casting to double the discriminant and the roots but doesn't work.Can somebody help me, where and what I need to rewrite?

public static void main(String args[]) {
    Scanner sc = new Scanner(System.in);
    Pattern newlineOrSpace = Pattern.compile(System
            .getProperty("line.separator")
            + "|\\s");
    sc.useDelimiter(newlineOrSpace);
    System.out.print("Enter a, b, c: ");
    double a = sc.nextDouble();
    double b = sc.nextDouble();
    double c = sc.nextDouble();
    // System.out.format("a = %f, b = %f, c = %f", a, b, c);

    double root1;
    double root2;
    double discriminant;
    discriminant = Math.sqrt(b * b - 4 * a * c);
    if (discriminant > 0) {
        System.out.println("There are no real roots ");
    } else {
        root1 = (-b + discriminant) / (2 * a);
        root2 = (-b - discriminant) / (2 * a);
        System.out.println("The roots are " + root1 + " and " + root2);
    }

推荐答案

Math.sqrt(x)当x为负数时返回NaN，然后在其余代码中传播.您需要测试之前取平方根的负数:

Math.sqrt(x) returns NaN when x is negative, which then propagates through the rest of your code. You'll want to test for negative numbers before taking the square root:

discriminant = b * b - 4 * a * c;
if (discriminant < 0) {
    System.out.println("There are no real roots ");
} else {
    root1 = (-b + Math.sqrt(discriminant)) / (2 * a);
    root2 = (-b - Math.sqrt(discriminant)) / (2 * a);
    System.out.println("The roots are " + root1 + " and " + root2);
}

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