In this tutorial, we will be discussing a program to find the roots of the Quadratic equation.
Given a quadratic equation of the form ax2 + bx + c. Our task is to find the roots x1 and x2 of the given equation.
For this, we are using the deterministic method, in this
D = √b2 - 4ac
UQ云商B2B2C系统
下载
UQCMS云商是一款B2B2C电子商务软件 ，非常适合初创的创业者，个人及中小型企业。程序采用PHP+MYSQL,模板采用smarty模板，二次开发，简单方便，无需学习其他框架就可以自行模板设计。永久免费使用，操作简单，安全稳定。支持PC+WAP+微信三种浏览方式，支持微信公众号。
then the roots of the equation will be
x1 = (-b + D)/2a ,and
x2 = (-b - D)/2a
Example
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
//calculating the roots of equation
void calc_roots(int a, int b, int c) {
if (a == 0) {
printf("Invalid Equation");
return;
}
int d = b*b - 4*a*c;
double sqrt_val = sqrt(abs(d));
if (d > 0) {
printf("Roots are both real and different </p><p>");
printf("%f</p><p>%f",(double)(-b + sqrt_val)/(2*a) , (double)(-b - sqrt_val)/(2*a));
}
else if (d == 0) {
printf("Roots are real and same </p><p>");
printf("%f",-(double)b / (2*a));
} else {
printf("Roots are complex </p><p>");
printf("%f + i%f</p><p>%f - i%f", -(double)b /(2*a),sqrt_val ,-(double)b / (2*a), sqrt_val);
}
}
int main() {
int a = 2, b = -5, c = 8;
calc_roots(a, b, c);
return 0;
}输出
Roots are complex 1.250000 + i6.244998 1.250000 - i6.244998
