class FindRoot{ //计算一元二次方程的根
	static double[] root(double a, double b, double c) throws Exception
	{
		double x[]=new double[2];
		if (a== 0)
		{
			throw new Exception("a 不能为零.");
		}
		else{
			double disc = (b*b) - 4*a*c;
			if (disc < 0) 
				throw new Exception("b*b-4ac值不能小于零.");
			
			x[0]=(-b + Math.sqrt(disc)) / (2*a);
			x[1]=(-b - Math.sqrt(disc)) / (2*a) ;
			return x;
			}
	}
	public static void main(String arg[]) {
		try {
			double x[] = root(2.0,-3.0,1);	//a,b,c
			System.out.println("方程2*x*x-3*x+1=0的根为：" +x[0]+"，"+x[1]);
		}
		catch(Exception e) { System.out.println(e); }
	}
}

一元二次方程ax2+bx+c=0，输入a,b,c三个系数，求解方程，结果有三种情况：两个实数根、一个实数根、两个复数根。
/**
* Equation.java
*/
package cn.edu.uibe.oop;
/**
* 计算一元二次方程的根
* @author TongQiang
*
*/
public class Equation {
double a,b,c; //方程的系数
double x1,x2; //两个根
double r; //实部
double v; //虚部
int type; //0表示一个根，1表示两个实根，2表示两个复数的根
public Equation(double a,double b,double c){
this.a = a;
this.b = b;
this.c = c;
}
private double delta(){
return b*b-4*a*c;
}
public void calculate(){
double d = delta();
if(Math.abs(d) < 1E-5){
type = 0;
x1 = -b/(2*a);
x2 = x1;
}else if(d > 0){
type = 1;
x1 = (-b+Math.sqrt(d))/(2*a);
x2 = (-b-Math.sqrt(d))/(2*a);
}else{
type = 2;
r = -b/(2*a);
v = Math.sqrt(-d)/(2*a);
}
}
public void showResult(){
switch(type){
case 0:
System.out.println("方程只有一个实根，x1=x2="+x1);
break;
case 1:
System.out.println("方程有两个实根，分别是：x1="+x1+"\tx2="+x2);
break;
case 2:
System.out.println("方程有两个复数根，分别是："+r+"+"+v+"i,\t"+r+"-"+v+"i");
break;
}
}
public static void main(String[] args) {
Equation eq1 = new Equation(1,2,2);
eq1.calculate();
eq1.showResult();
Equation eq2 = new Equation(1,-2,1);
eq2.calculate();
eq2.showResult();
Equation eq3 = new Equation(1,6,5);
eq3.calculate();
eq3.showResult();
}
}
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2009-09-25 23:46
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计算一元二次方程的两个实数根
import java.util.Scanner;

public class Psg {
    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);
        System.out.println("请输入方程的系数a、b、c：");
        int a, b, c;
        a = reader.nextInt();
        b = reader.nextInt();
        c = reader.nextInt();
        int delta;
        double x1, x2;
        delta = b*b-4*a*c;
        if (delta>0){
            x1=(-b+Math.sqrt(delta));
            x2=(-b-Math.sqrt(delta));
            System.out.println("方程有两个实数根。");
            System.out.println(x1/2*a);
            System.out.println(x2/2*a);
            return;
        }
        if (delta==0){
            System.out.println("方程有一个实数根。");
            return;
        }
        if (delta<0){
            System.out.println("方程无实数根。");
        }
    }
}

求一元二次方程的根。
【问题描述】编写程序，从键盘输入三个系数ax2+bx+c=0，计算方程的解并输出。需要考虑方程有虚根、方程有实根、方程是一元一次方程、没有根等。输出方程的解并保留6位小数。
import java.text.DecimalFormat;
import java.util.*;
public class  Equation{
      public static void main(String[] args) {
            Scanner reader = new Scanner(System.in);
            System.out.println("请输入方程的系数a、b、c：");
            int a, b, c;
            a = reader.nextInt();
            b = reader.nextInt();
            c = reader.nextInt();
            DecimalFormat df = new DecimalFormat("0.000000");
            int delta;
            double x1, x2;
            delta = b * b - 4 * a * c;
            if (a == 0 && b == 0 && c == 0) {
                System.out.println("方程有无限根。");
                return;
                }
                if (a == 0 && b == 0 & c != 0) {
                    System.out.println("方程无解。");
                    return;
                    }
                    if (delta == 0) {
                        x1 = b / -2.0 / a;
                        x2 = b / -2.0 / a;
                        System.out.println("方程有两个相同实根, x1 =  "
                        + df.format(x1) + ",  x2 = " + df.format(x2));
                        }
                     if (delta > 0) {
                          x1 = (-b + Math.sqrt(delta)) / 2.0 / a;
                          x2 = (-b - Math.sqrt(delta)) / 2.0 / a;
                          System.out.println("方程有两个不同实根, x1 = "
                          + df.format(x1) + ",  x2 = " + df.format(x2));
                          }
                     if (delta < 0) {
                          double s, x;
                          s = -b / 2.0 / a;
                          x = Math.sqrt(-delta) / 2.0 / a;
                          System.out.println("方程有虚根, x1 = "
                          + df.format(s) + " + " + df.format(x) + "i,  x2 = "
                          + df.format(s) + " - " + df.format(x) + "i");
                          }
            }
 }

运行结果：