Question

JAVA 中的求一元二次方程的根， 帮我看看程序错在哪里了，为什么不能运行，谢谢了，哪位帮我搞定一下

public class Fcg
{
public static void main(String args[])

{
double a;
double b;
double c;
System.out.println("请输入三个数：");
System.out.println("此方程为"+a+"x3"+b+"b"+c+"的根");
fcg(a,b,c);

}
static void fcg(double a,double b,double c)
{
double x1;
double x2;
if((b*b-4*a*c)<0)
System.out.println("此方程无实根");
else
if((b*b-4*a*c)==0)
System.out.println("x1=x2="+b/2*a+Math.sqrt(b*b-4*a*c)/2*a);
else
{System.out.println("x1="+b/2*a+Math.sqrt(b*b-4*a*c)/2*a);
System.out.println("x2="+b/2*a-Math.sqrt(b*b-4*a*c)/2*a);
}
}
}

Answer 1

public class Fcg
{
public static void main(String args[])

{
double a=1;//需要初始化
double b=2;
double c=1;
System.out.println("请输入三个数：");
System.out.println("此方程为"+a+"x3"+b+"b"+c+"的根");
fcg(a, b, c);

}

static void fcg(double a, double b, double c)
{
double x1;
double x2;
if((b*b-4*a*c)<0)
System.out.println("此方程无实根");
else if((b*b-4*a*c)==0)
System.out.println("x1=x2="+b/2*a+Math.sqrt(b*b-4*a*c)/2*a);
else
{
x1=b/2*a+Math.sqrt(b*b-4*a*c)/2*a;
x2=b/2*a-Math.sqrt(b*b-4*a*c)/2*a;//你的那种是字符串运算，+相当于连字符，而此处出现"-"，所以出错
System.out.println("x1="+x1);
System.out.println("x2="+x2);
}
}