用逼近法求一元二次方程近似解

/**

* 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();

}

}