[编程入门]自定义函数求一元二次方程-题解（C语言代码）

解题思路:

注意事项:

参考代码:

#include<stdio.h>

#include<math.h>

void qiugen(double a,double b,double c)

{

double d,x1,x2;

d=b*b-4*a*c;

if(d==0)

{

x1=-b/(2*a);

x2=x1;

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

}

else if(d>0)

{

x1=(-b+pow(d,0.5))/(2*a);

        x2=(-b-pow(d,0.5))/(2*a);

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

}

else

{

printf("x1=%.3f+%.3fi ",-b/(2*a),sqrt(4*a*c-b*b)/(2*a));

printf("x2=%.3f-%.3fi",-b/(2*a),sqrt(4*a*c-b*b)/(2*a));

}

printf("\n");

}

int main()

{

double a,b,c;

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

qiugen(a,b,c);

return 0;

}