用C ++求解二次方程

问题描述

我正在尝试用C ++写一个函数，该函数使用二次方程式求解X.这是我最初写的，只要答案没有复数，它似乎就可以工作:

I am trying to write a function in C++ that solves for X using the quadratic equation. This is what I have written initially, which seems to work as long as there are no complex numbers for an answer:

float solution1 = (float)(-1.0 * b) + (sqrt((b * b) - (4 * a * c)));
solution1 = solution1 / (2*a);

cout << "Solution 1: " << solution1 << endl;

float solution2 = (float)(-b) - (sqrt((b*b) - (4 * a * c)));
solution2 = solution2 / (2*a);
cout << "Solution 2: " << solution2;

例如，如果我使用等式:x ^ 2-x-6，我将正确地获得解3，-2.

If, for example, I use the equation: x^2 - x - 6, I get the solution 3, -2 correctly.

我的问题是如何计算复数....例如，给定等式:

My question is how would I account for complex numbers....for example, given the equation:

x ^ 2 + 2x + 5

x^2 + 2x + 5

用手解决，我会得到-1 + 2i，-1-2i.

Solving by hand, I would get -1 + 2i, -1 - 2i.

好吧，我猜有两个问题，我可以更好地写上上面的内容，并考虑到复数吗?

Well, I guess two question, can I write the above better and also make it account for the complex number?

感谢您的帮助！

推荐答案

类似的方法会起作用:

struct complex { double r,i; }
struct pair<T> { T p1, p2; }

pair<complex> GetResults(double a, double b, double c)
{
  pair<complex> result={0};

  if(a<0.000001)    // ==0
  {
    if(b>0.000001)  // !=0
      result.p1.r=result.p2.r=-c/b;
    else
      if(c>0.00001) throw exception("no solutions");
    return result;
  }

  double delta=b*b-4*a*c;
  if(delta>=0)
  {
    result.p1.r=(-b-sqrt(delta))/2/a;
    result.p2.r=(-b+sqrt(delta))/2/a;
  }
  else
  {
    result.p1.r=result.p2.r=-b/2/a;
    result.p1.i=sqrt(-delta)/2/a;
    result.p2.i=-sqrt(-delta)/2/a;
  }

  return result;
}

这样，您就可以以相似的方式获得真实和复杂结果的结果(实际结果只是将虚部设置为0).看起来会更漂亮！

That way you get the results in a similar way for both real and complex results (the real results just have the imaginary part set to 0). Would look even prettier with boost!

edit:修复了delta的问题，并添加了对退化案例(如a = 0)的检查.失眠的夜晚！

edit: fixed for the delta thing and added a check for degenerate cases like a=0. Sleepless night ftl!

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