二次方程 [英] quadratic equation
问题描述
大家好！
我是c编程的新生，我们得到了一个创建程序的任务，该程序吐出了二次方程的根源！
我试了一下，但似乎错了，这就是为什么我需要你的帮助！谢谢
公式为x =（ - 1/2）* p + - sqrt（1/4）* p ^ 2 -q）及其（p， q）是我的数据。如果等式没有真正的根printf（它没有真正的根）;
我的代码
通过调用来阻止它功能!!
//
//由Haidar Wahid提供
//
//
#pragma warning（禁用：4996）
#include< stdio.h>
#include< stdlib.h>
#include< math.h>
int squareRoot_of（int p，int q，double a）; //原型
int main（）{
int x，y;
double z;
z = 0.5;
x = 55;
y = 10;
printf（"％d \ n％d \ n"，squareRoot_of（x，y，z））;
系统（暂停）;
}
int squareRoot_of（int p，int q，double a）// def
{
return（（-a * p）+ - sqrt（a * a * p * p - q））;
}
Hi everyone !
i''m a freshmen in c programming, and we got an assignment to create a program that spit out the roots of an quadratic equation!
I''ve given it a try but it seems to be wrong somehow that''s why i need your help! thanks
the formula is x= (-1/2)*p +- sqrt(1/4)*p^2 -q) and its (p,q) is my in-data. if the equation does not have real root printf("itdoes not have real roots");
my code
obs its by calling a function !!
//
// by Haidar Wahid
//
//
#pragma warning (disable:4996)
#include <stdio.h>
#include<stdlib.h>
#include<math.h>
int squareRoot_of(int p, int q, double a); // prototype
int main() {
int x, y;
double z;
z = 0.5;
x = 55;
y = 10;
printf("%d\n%d\n", squareRoot_of(x,y,z));
system("pause");
}
int squareRoot_of(int p, int q, double a) // def
{
return ((-a*p) +- sqrt(a*a*p*p - q));
}
推荐答案
你想找到这类方程的真正根源吗？
a * x ^ 2 + b * x + c = 0.
如果是的话，我会尝试给你一个答案而不会泄露解决方案。
好的，你首先应该找到歧视，说D，这就是
D = b ^ 2 - 4 * a * c。
现在，您要测试D的案例。如果D> 0，那意味着有可能真正的根源。如果D = 0，那么只有一个真实的根。
如果D< 0，那么你可以简单地输出没有真正的根。
鉴于找到根的公式，比如说x1，x2，如果D> 0是：
x1 =（ - b + sqrt（D））/ 2 * a
x2 =（ - b - sqrt（D））/ 2 * a
如果D = 0（来自上面的公式）：
x =（ - b）/ 2 * a
我认为你可以做出很好的解决方案！
希望它有所帮助！
Do you want to find real roots for this type of equation?
a * x^2 + b * x + c = 0.
If yes, I will try to give you an answer without revealing the solution.
OK, you first should find the discriminent, say D, and this is
D = b^2 - 4 * a * c.
Now, you want to test cases for D. If D > 0, then that means that there are
2 possible real roots. If D = 0, then there is only one real root.
If D < 0, then you can just simply output that there are no real roots.
Given that the formula for finding the roots, say x1, x2, if D > 0 is:
x1 = (-b + sqrt(D)) / 2 * a
x2 = (-b - sqrt(D)) / 2 * a
and if D = 0 is(which is derived from the above formula):
x = (-b) / 2 * a
I think that you can make a good solution!
Hope it helped!
非常感谢你很多zuko32！我马上开始，希望它有效！
坦克再次
thank you very much zuko32! I will start with it right away and hope it works !
tanks again
#pragma warning（禁用：4996）
#include< stdio.h>
#include< stdlib.h>
#include< math.h>
int squareRoot_of（int p，int q，double a，int d，int x1，int x2）; //原型
int main（）{
int x，y，c，v，g，h;
double b;
x = 7;
y = 2;
b = 0.5;
g = 0;
h = 0;
v =（（b * b * x * x） - y）;
c = squareRoot_of（x，y，b，g，h ，v）;
printf（"％d％d \ n"，c）;
system（ 暂停）;
}
int squareRoot_of（int p，int q，double a ，int d，int x1，int x2）// def
{
d =（（a * a * p * p） - q）;
if（d> = 0）{
printf（"％d％d \ n，p ，q）;
}
其他
{
printf（没有root \ n ;）;
}
x1 =（（-a * p）+ sqrt（（a * a * p * p）） - q）;
x2 =（ - a * p - sqrt（（a * a * p * p） - q））;
返回x1 ，x2;
}
您认为zuko32是什么？
#pragma warning (disable:4996)
#include <stdio.h>
#include<stdlib.h>
#include<math.h>
int squareRoot_of(int p, int q, double a, int d ,int x1, int x2); // prototype
int main() {
int x, y, c, v, g, h;
double b;
x = 7;
y = 2;
b = 0.5;
g = 0;
h = 0;
v = ((b*b*x*x) - y);
c = squareRoot_of( x,y,b,g,h,v);
printf("%d%d\n", c);
system("pause");
}
int squareRoot_of(int p, int q, double a, int d, int x1, int x2) // def
{
d = ((a*a*p*p) - q);
if (d >= 0) {
printf("%d%d\n",p,q);
}
else
{
printf("there are no roots\n");
}
x1 = ((-a*p) + sqrt((a*a*p*p)) - q);
x2 = (-a*p - sqrt((a*a*p*p) - q));
return x1, x2;
}
what do u think zuko32?
这篇关于二次方程的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
