设置一个程序，求ax2+bx+c=0的根，a，b，c由键盘输入。

以下是一个求解ax2+bx+c=0的根的程序，其中a、b、c由键盘输入：

#include <stdio.h>
#include <math.h>

int main()
{
double a, b, c, delta, x1, x2;

printf("请输入a、b、c的值：");
scanf("%lf%lf%lf", &a, &b, &c);

delta = b * b - 4 * a * c;

if (delta < 0)
{
printf("方程无实根！\n");
}
else if (delta == 0)
{
x1 = x2 = -b / (2 * a);
printf("方程有唯一实根：x1=x2=%.2lf\n", x1);
}
else
{
x1 = (-b + sqrt(delta)) / (2 * a);
x2 = (-b - sqrt(delta)) / (2 * a);
printf("方程有两个实根：x1=%.2lf，x2=%.2lf\n", x1, x2);
}

return 0;
}

希望这个程序能够帮到你！

相关问题

用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

以下是用C语言编写的求解ax2+bx+c=0方程根的程序：

#include <stdio.h>
#include <math.h>

int main()
{
    float a, b, c, delta, x1, x2;

    printf("请输入a, b, c的值：\n");
    scanf("%f%f%f", &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("方程的两个实根为：x1=%.2f, x2=%.2f\n", x1, x2);
    }
    else if (delta == 0)
    {
        x1 = x2 = -b / (2 * a);
        printf("方程的两个相等实根为：x1=x2=%.2f\n", x1);
    }
    else
    {
        printf("方程的两个虚根为：x1=%.2f+%.2fi, x2=%.2f-%.2fi\n", -b / (2 * a), sqrt(-delta) / (2 * a), -b / (2 * a), sqrt(-delta) / (2 * a));
    }

    return 0;
}

程序中先从键盘输入a、b、c的值，然后根据判别式delta的值来判断方程的根的情况，最后输出根的值。

用c语言编写程序求ax 2 +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, c: ");
    scanf("%f %f %f", &a, &b, &c);

    discriminant = b * b - 4 * a * c;

    if (discriminant > 0) {
        root1 = (-b + sqrt(discriminant)) / (2 * a);
        root2 = (-b - sqrt(discriminant)) / (2 * a);
        printf("Roots are real and different.\n");
        printf("Root1 = %.2f and Root2 = %.2f", root1, root2);
    } else if (discriminant == 0) {
        root1 = root2 = -b / (2 * a);
        printf("Roots are real and same.\n");
        printf("Root1 = Root2 = %.2f", root1);
    } else {
        float realPart = -b / (2 * a);
        float imaginaryPart = sqrt(-discriminant) / (2 * a);
        printf("Roots are complex and different.\n");
        printf("Root1 = %.2f + %.2fi and Root2 = %.2f - %.2fi", realPart, imaginaryPart, realPart, imaginaryPart);
    }

    return 0;
}

注：%.2f表示保留两位小数输出。