求一元二次方程的解

求方程ax2+bx+c=0的根,分别用三个函数求当b2-4ac大于0,等于0,和小于0的根,并输出结果。从主函数输入a,b,c的值。

实现代码:

#include<iostream>
#include<math.h>
using namespace std;
int main(){
	double a,b,c,m,p,q;
	void dengyu(double,double);
	void dayu(double,double);
	void xiaoyu();
	cout<<"please input a,b,c:";
	cin>>a>>b>>c;
	cout<<"the result of  "<<a<<"x^2+"<<b<<"x+"<<c<<"\n";
	cout<<(m=b*b-4*a*c)<<endl;
	cout<<(p=-b/(2*a))<<endl;
	if(m>0){
	cout<<(q=sqrt(m)/(2*a))<<endl;
	dayu(p,q);
	}
	else if(m==0){
	q=sqrt(m)/(2*a);
	dengyu(p,q);
	}
	else if(m<0)xiaoyu();
	return 0;
}
void dayu(double p,double q){
    	cout<<"root1:"<<(p+q)<<endl;
    	cout<<"root2:"<<(p-q)<<endl;
}
void dengyu(double p,double q){
	cout<<"root1=root2="<<(p+q)<<endl;
}
void xiaoyu(){
	cout<<"there is no root"<<endl;
}


posted @ 2013-04-26 23:13  千手宇智波  阅读(173)  评论(0编辑  收藏  举报