用c语言求ax²+bx+c=0方程的根,a,b,c由键盘输入,设b²-4ac>0
时间: 2024-09-06 12:03:58 浏览: 50
在C语言中,求解一元二次方程`ax^2 + bx + c = 0`的根,通常需要使用公式:
```c
x1 = (-b + sqrt(b*b - 4*a*c)) / (2*a)
x2 = (-b - sqrt(b*b - 4*a*c)) / (2*a)
其中`sqrt()`表示平方根函数,当判别式`b*b - 4*a*c > 0`(即一元二次方程有两个实数根)时有效。
以下是一个简单的程序示例,展示了如何获取用户输入的系数并计算根:
```c
#include <stdio.h>
#include <math.h>
int main() {
double a, b, c, x1, x2;
printf("请输入一元二次方程的系数a、b和c:\n");
scanf("%lf %lf %lf", &a, &b, &c);
// 检查判别式是否大于0
if (b * b - 4 * a * c > 0) {
x1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
x2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);
printf("方程的两个根分别为:%.2lf 和 %.2lf\n", x1, x2);
} else {
printf("方程无实数根。\n");
}
return 0;
}
相关问题
编写c语言程序,求一元二次方程ax²+bx+c=0的根
```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;
}
```
c语言求方程ax²+bx+c=0的根,用3个函数分别求当:b²-4ac大于0,等于0,小于0时的根并输出结果。从主函数输入a,b,c的值
以下是C语言求解一元二次方程的程序,其中包括三个函数分别求当b²-4ac大于0,等于0,小于0时的根并输出结果。从主函数输入a,b,c的值:
```c
#include <stdio.h>
#include <math.h>
void roots_greater_than_zero(float a, float b, float c);
void roots_equal_to_zero(float a, float b, float c);
void roots_less_than_zero(float a, float b, float c);
int main() {
float a, b, c;
printf("Please enter the value of a, b, c:\n");
scanf("%f%f%f", &a, &b, &c);
float discriminant = b * b - 4 * a * c;
if (discriminant > 0) {
roots_greater_than_zero(a, b, discriminant);
} else if (discriminant == 0) {
roots_equal_to_zero(a, b);
} else {
roots_less_than_zero(a, b, discriminant);
}
return 0;
}
void roots_greater_than_zero(float a, float b, float discriminant) {
float root1 = (-b + sqrt(discriminant)) / (2 * a);
float root2 = (-b - sqrt(discriminant)) / (2 * a); printf("The roots are real and different.\n");
printf("root1 = %.2f, root2 = %.2f\n", root1, root2);
}
void roots_equal_to_zero(float a, float b) {
float root = -b / (2 * a);
printf("The roots are real and equal.\n");
printf("root1 = root2 = %.2f\n", root);
}
void roots_less_than_zero(float a, float b, float discriminant) {
float realPart = -b / (2 * a);
float imaginaryPart = sqrt(-discriminant) / (2 * a);
printf("The roots are complex and different.\n");
printf("root1 = %.2f + %.2fi, root2 = %.2f - %.2fi\n", realPart, imaginaryPart, realPart, imaginaryPart);
}
```