Question

用C语言编写求一元二次方程根的程序

条件要判断的充分

Answer 1

用C语言编写求一元二次方程根的程序，条件判断的充分，步骤如下：

void main()
{
float a,b,c,delta;
scanf("%f%f%f",&a,&b,&c);
if(a!=0)
{
delta=b*b-4*a*c;
if(delta==0)
printf("x1=x2=%7.2f",-b/(2*a));
else if(delta>0)
{
printf("x1=%7.2f",(-b+sqrt(delta))/(2*a));
printf("x2=%7.2f",(-b-sqrt(delta))/(2*a));
}
else
{
printf("x1=%7.2f+i%7.2f",-b/(2*a),sqrt(-delta)/(2*a));
printf("x2=%7.2f-i%7.2f",-b/(2*a),sqrt(-delta)/(2*a));
}
}
else if(b!=0)
printf("x=%7.2f",-c/b);
else if(c==0)
printf("0=0,x为任意解
");
else
printf("%f=0,error!",c);
}

Answer 2

#include <stdio.h>
#include <math.h>
int main()
{
	double a,b,c,disc,x1,x2,realpart,imagpart;
	scanf("%lf%lf%lf",&a,&b,&c);
	printf("The equation");
	if (fabs(a)<=1e-6)
		printf(" is not a quadratic\n");
	else
	{
		disc=b*b-4*a*c;
		if (fabs(disc)<=1e-6)
			printf(" has two equal roots:%8.4f\n",-b/(2*a));
		else
			if (disc>0)
			{
				x1=(-b+sqrt(disc))/(2*a);
				x2=(-b-sqrt(disc))/(2*a);
				printf(" has distinct real roots:%8.4f and %8.4f\n",x1,x2);
			}
			else
			{
				realpart=-b/(2*a);
				imagpart=sqrt(-disc)/(2*a);
				printf(" has complex roots:\n");
				printf("%8.4f+%8.4fi\n",realpart,imagpart);
				printf("%8.4f-%8.4fi\n",realpart,imagpart);
			}
	}
	return 0;
}

Answer 3

#include "stdio.h"
#include "math.h"
double x1,x2,p;

float file1(float a,float b)
{
x1=(-b+sqrt(p))/2*a;
x2=(-b-sqrt(p))/2*a;
return 0;
}

float file2(float a,float b)
{
x1=x2=(-b+sqrt(p))/2*a;
return 0;
}

void main()
{
float a,b,c;
scanf("%f%f%f",&a,&b,&c);
p=b*b-4*a*c;
printf("方程是：%.3f*x*x+%.3f*x+%.3f=0\n",a,b,c);
if(p>0)
{
file1(a,b);
printf("X1=%f\tX2=%f\n",x1,x2);
}
else if(p==0)
{
file2(a,b);
printf("X1=%f\tX2=%f\n",x1,x2);
}
else printf("方程无解");
}

运行正确 输入： 2 3.7 1.2
输出 X1=-1.677625 X2=-5.722375

Answer 4

#include<stdio.h>
#include<math.h>
int main(void)
{
float a,b,c;
double x1,x2;
double delta;
printf("输入a,b,c数值：\n");
scanf("%f%f%f",&a,&b,&c);

if (a!=0)
{delta=pow(b,2)-4*a*c;
if(delta==0){

printf("x1=x2=%.2f\n",-b/(2*a));}

if(delta>0) {

x1=(-b+sqrt(delta))/(2*a);
x2=(-b-sqrt(delta))/(2*a);
printf("x1=%.2f x2=%.2f\n",x1,x2);}

if(delta<0) {

printf("无解！\n");}

}
return 0 ;
}