[编程入门]自定义函数求一元二次方程

解题思路:

注意事项:

参考代码:

#include<stdio.h>

#include<math.h>

float n1(float a,float b,float c);

float n2(float a,float b,float c);

float n3(float a,float b,float c);

int main()

{

float m,a,b,c;

scanf("%f%f%f",&a,&b,&c);

m=(b*b)-(4*a*c);

if(m>0)

n1(a,b,c);

if(m==0)

n2(a,b,c);

if(m<0)

  n3(a,b,c);

return 0;

}

float n1(float a,float b,float c)

{

float x1,x2;

x1=((-b+sqrt((b*b)-(4*a*c)))/(2*a));

x2=((-b-sqrt((b*b)-(4*a*c)))/(2*a));

printf("x1=%.3f,x2=%.3f",x1,x2);

}

float n2(float a,float b,float c)

{

float x1,x2;

x1=((-b+sqrt((b*b)-(4*a*c)))/(2*a));

x2=((-b-sqrt((b*b)-(4*a*c)))/(2*a));

printf("x1=x2=%.3f",x1);

}

float n3(float a,float b,float c)

{

float m,n;

    m=(-b)/(2*a); 

    n=sqrt(4*a*c-b*b)/(2*a);

    printf("x1=%.3f+%.3fi x2=%.3f-%.3fi\n",m,n,m,n);

}