编写c语言程序,求一元二次方程ax²+bx+c=0的根
时间: 2024-01-16 22:02:26 浏览: 32
```c
#include <stdio.h>
#include <math.h>
int main()
{
double a, b, c, delta, x1, x2;
printf("请输入一元二次方程的三个系数:\n");
scanf("%lf%lf%lf", &a, &b, &c);
delta = b * b - 4 * a * c;
if (delta > 0) {
x1 = (-b + sqrt(delta)) / (2 * a);
x2 = (-b - sqrt(delta)) / (2 * a);
printf("方程有两个不相等的实根:x1=%.2f, x2=%.2f\n", x1, x2);
} else if (delta == 0) {
x1 = x2 = -b / (2 * a);
printf("方程有两个相等的实根:x1=x2=%.2f\n", x1);
} else {
printf("方程没有实根\n");
}
return 0;
}
```
相关问题
求解二次方程ax²+bx+c=0的c语言
可以使用以下代码来求解二次方程ax²+bx+c=:
#include <stdio.h>
#include <math.h>
int main()
{
float a, b, c, delta, x1, x2;
printf("请输入二次方程的系数a, b, c:\n");
scanf("%f%f%f", &a, &b, &c);
delta = b * b - 4 * a * c;
if (delta < )
{
printf("该方程无实数解!\n");
}
else if (delta == )
{
x1 = x2 = -b / (2 * a);
printf("该方程有唯一解:x1 = x2 = %.2f\n", x1);
}
else
{
x1 = (-b + sqrt(delta)) / (2 * a);
x2 = (-b - sqrt(delta)) / (2 * a);
printf("该方程有两个实数解:x1 = %.2f, x2 = %.2f\n", x1, x2);
}
return ;
}
c语言编写程序求ax²+bx+c=0方程的根,其中a,b,c从键盘输入。
以下是c语言编写的程序,用于求ax²+bx+c=0方程的根,其中a,b,c从键盘输入:
```
#include <stdio.h>
#include <math.h>
int main()
{
float a, b, c, discriminant, root1, root2;
printf("Enter coefficients a, b and c: ");
scanf("%f %f %f", &a, &b, &c);
discriminant = b*b - 4*a*c;
// 如果判别式小于0,则方程无实根
if (discriminant < 0)
{
printf("Roots are complex numbers.\n");
printf("Root1 = %.2f + %.2fi\n", -b/(2*a), sqrt(-discriminant)/(2*a));
printf("Root2 = %.2f - %.2fi\n", -b/(2*a), sqrt(-discriminant)/(2*a));
}
// 如果判别式等于0,则方程有一个实根
else if (discriminant == 0)
{
printf("Both roots are equal.\n");
root1 = -b/(2*a);
printf("Root1 = Root2 = %.2f", root1);
}
// 如果判别式大于0,则方程有两个实根
else
{
printf("Roots are real numbers.\n");
root1 = (-b + sqrt(discriminant))/(2*a);
root2 = (-b - sqrt(discriminant))/(2*a);
printf("Root1 = %.2f\n", root1);
printf("Root2 = %.2f\n", root2);
}
return 0;
}
```