一元二次方程的解

谁借莪１个温暖的怀抱￠ 2022-11-07 04:18 2阅读 0赞

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjA0ODQ2Mw_size_16_color_FFFFFF_t_70][]

##  ##

## 示例输出 ##

x1=-1
    x2=-5

#include<stdio.h>
    #include<math.h>
    int judge(double a, double b, double c)//判断一元二次方程是否有实数解
    {
        if(pow(b, 2)-4*a*c>0)
        {
            return 2;//有两个不同的实数解
        }
        else if(pow(b, 2)-4*a*c==0)
        {
            return 1;//有两个相同的实数解
        }
        else
        {
            return 0;//在实数范围内无解
        }
    }
    float Solution(double a, double b, double c)
    {
        if(judge(a, b, c)>=1)//有实数解
        {
            double delta=pow(b, 2)-4*a*c;
            double x1, x2;
            x1=(-b+sqrt(delta))/(2*a);
            x2=(-b-sqrt(delta))/(2*a);
            if(x1>x2)
            {
                printf("x1=%.2f\n", x1);
                printf("x2=%.2f\n", x2);
            }
            else
            {
                printf("x1=%.2f\n",x2);
                printf("x2=%.2f\n",x1);
            }
        }
        else//在实数范围内无解
        {
            printf("-1\n");
        }
    }
    int main()
    {
        double a, b, c;
        scanf("%lf%lf%lf", &a, &b, &c);
        Solution(a, b, c);
        return 0;
    }

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjA0ODQ2Mw_size_16_color_FFFFFF_t_70 1][]

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjA0ODQ2Mw_size_16_color_FFFFFF_t_70]: /images/20221023/3141a8e12d1140f6abd81cf0723a867f.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjA0ODQ2Mw_size_16_color_FFFFFF_t_70 1]: /images/20221023/e2fd3eeb2e1b4802b01f8d6ae327c2fd.png