#include <stdio.h>
#include <math.h>
float a,b,c,x1,x2;
void deata_MoreThan_0(float a,float b,float c,float deata);
void deata_equal_0(float a,float b,float c,float deata);
void deata_less_0(float a,float b,float c,float deata);
void main()
{
float deata;
printf("input a,b and c:");
scanf("%f %f %f",&a,&b,&c);
deata = b*b - 4*a*c;
if(deata > 0)
{
deata_MoreThan_0(a,b,c,deata);
}
else if(deata == 0)
{
deata_equal_0(a,b,c,deata);
}
else
deata_less_0(a,b,c,deata);
}
void deata_MoreThan_0(float a,float b,float c,float deata)
{
printf("x1=%f x2=%f",(-b+sqrt(deata))/2*a,(-b-sqrt(deata))/2*a);
}
void deata_equal_0(float a,float b,float c,float deata)
{
printf("x1=%f x2=%f",(-b+sqrt(deata))/2*a,(-b-sqrt(deata))/2*a);
}
void deata_less_0(float a,float b,float c,float deata)
{
printf("x1=%f+%f i x2=%f-%f i",-b/2*a,sqrt(-deata)/2*a,-b/2*a,sqrt(-deata)/2*a);
}
结果:
input a,b and c:2 4 6
x1=-4.000000+5.656854 i x2=-4.000000-5.656854 i
input a,b and c:1 2 1
x1=-1.000000 x2=-1.000000
input a,b and c:1 0 4
x1=-0.000000+2.000000i x2=-0.000000-2.000000i
C语言刷题(8):求方程ax^2 + bx +c =0的根,用三个函数分别求当:b^2 -4ac大于0,小于 0,等于 0的根,并输出结果
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转载自blog.csdn.net/qq_38173631/article/details/103953953
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