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 x1, x2;

void dayu0(float a, float b, float c)

{

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);

}

void dengyu0(float a, float b, float c)

{

x1 = -(b/(2*a));

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

}

void xiaoyu0(float a, float b, float c)

{

float real = -(b / (2 * a));

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

printf("x1=%.3f+%.3fi x2=%.3f-%.3fi", real, imag, real, imag);

}

int main()

{

int n;

float a, b, c;

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

float delta = (b*b) - (4 * a * c);

if (delta > 0) n = 1;

else if (delta == 0) n = 2;

else n = 3;

switch (n)

{

case 1: dayu0(a, b, c); break;

case 2: dengyu0(a, b, c); break;

case 3: xiaoyu0(a, b, c); break;

default:break;

}

}