使用二次公式的简单二次方程计算器

Question

我想知道是否有人可以帮助我解决这个（初学者）问题。 我正在创建一个非常基本的二次方程计算器。我在下面列出了我的代码以及 cmets（代码顶部的那些 sn-p 解释了我必须做什么）。我在网上查看了多种解决方案并尝试了自己，但似乎我一直得到不正确的 x1 和 x2 值。如果有人可以指导我，我会非常高兴。干杯。

/* 
create program to calculate x for quadratic equation. (a, b and c)
1) create function which prints out roots of quad equation.
2) throw exception if b2 - 4ac is less than 0.
3) call the function from main.
*/

#include <iostream>
#include <cmath>
#include <string>

using namespace std;

double roots() { //return type, function name, parameters and body of function.
    double a = 0, b = 0, c = 0, x1, x2;
    double discriminant = (b * b) - 4 * a * c;
    cout << "Enter roots of quad equation (in order: a, b, c) " << endl;
    cin >> a >> b >> c;

        if (discriminant >= 0) {

            cout << "Your quadratic equation is: " << a << "x squared + " << b << "x + " << c << " = 0 " << endl;
            x1 = -b + sqrt(discriminant) / (2 * a);
            x2 = -b - sqrt(discriminant) / (2 * a);
            cout << "x1 = " << x1 << endl;
            cout << "x2 = " << x2 << endl;
        } 
        else {
            cout << "Negative value returned from (b2 - 4ac), please try again! " << endl;
            exit(1);
        }   
}

int main() {

    cout << "Quadratic Equation Calculator " << endl;
    roots();




    return 0;
}

Answer 1

这是一个简单的错误。您只是以错误的顺序放置了代码行。这应该可以解决它：

/* 
create program to calculate x for quadratic equation. (a, b and c)
1) create function which prints out roots of quad equation.
2) throw exception if b2 - 4ac is less than 0.
3) call the function from main.
*/

#include <iostream>
#include <cmath>
#include <string>

using namespace std;

double roots() { //return type, function name, parameters and body of function.
    double a = 0, b = 0, c = 0, x1, x2;
    cout << "Enter roots of quad equation (in order: a, b, c) " << endl;
    cin >> a >> b >> c;
    double discriminant = (b * b) - 4 * a * c; // restructed this line

        if (discriminant >= 0) {

            cout << "Your quadratic equation is: " << a << "x squared + " << b << "x + " << c << " = 0 " << endl;
            x1 = -b + sqrt(discriminant) / (2 * a);
            x2 = -b - sqrt(discriminant) / (2 * a);
            cout << "x1 = " << x1 << endl;
            cout << "x2 = " << x2 << endl;
        } 
        else {
            cout << "Negative value returned from (b2 - 4ac), please try again! " << endl;
            exit(1);
        }   
}

int main() {

    cout << "Quadratic Equation Calculator " << endl;
    roots();




    return 0;
}

Answer 2

您遇到的判别式问题可能会教您编写一个函数来计算判别式，例如：

double f_discriminant(double a,b,c){
  return b*b-4*a*c;
}

像这样，你的代码变成（基于 PranavaGande 的改进）：

...
double roots() { //return type, function name, parameters and body of function.
    double a = 0, b = 0, c = 0, x1, x2;
    cout << "Enter roots of quad equation (in order: a, b, c) " << endl;
    cin >> a >> b >> c;
    double discriminant = f_discriminant(a, b, c); // restructed this line
...

这里的好处是，如果你需要再次计算判别式，你可以只使用你为它编写的函数。 （这个例子很明显，但是你会遇到更难开发的功能）

Answer 3

问题在于计算判别式的那一行：

double discriminant = (b * b) - 4 * a * c;

这一行出现在您为 a、b 和 c 分配非零值之前。在执行此操作时，所有这些变量都等于零，因此判别式也为零。要使用用户的值计算判别式，该行需要在该行之后：

cin >> a >> b >> c;

原因是算术计算只执行一次，就在它写入的那一行；之后， = 运算符（赋值）将该结果放入变量中。之后，该变量不记得该值是如何到达那里的，也无法使用旧表达式中的新值“刷新”自身。您可以通过将判别式设为 a、b 和 c 的函数来近似。