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>

int main()

{

    double a, b, c, x1=0, x2=0;

    scanf("%lf%lf%lf", &a, &b, &c);

    if ((b * b - 4 * a * c) < 0)//方程无实数解

    {

        x1 = -b / (2 * a);

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

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

        printf("x1=%.3lf+%.3lfi x2=%.3lf-%.3lfi\n", x1, x2, x1, x2);

    }

    else if ((b*b - 4 * a * c) == 0)//方程有两个相同的解

    {

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

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

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

    }

    else//方程有两个不同的解

    {

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

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

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

    }

    return 0;

}