I don't know what doesn't work with my c++ equation's type of root test: It doesn't d

[AntoniaMouawad]

#include <iostream> 
#include <cmath>
using namespace std;
int 
main (int argc, char** argv)
{
	double a=0.0;
	double b=0.0;
	double c=0.0;
	double num=0.0;

	cout <<"Given this form of equation: aX2 + bX + c = 0"<<endl;
	cout <<"Input the value of a (the coefficient of X2)"<<endl;
	cin>>a;
	cout <<"Input the value of b (the coefficient of X)"<<endl;
	cin>>b;
	cout <<"Input the value of c (the constant value)"<<endl;
	cin>>c;
	cout<<"Your equation is the following"<<endl;
	cout<<a<<"X2 + "<<b<<"X + "<<c<<" = 0"<<endl;
	num= pow(b,2)-(4*a*c);
	if (num=0)
	{
		cout<<"This equation has a single (repeated) root."<<endl;
	}
	if (num<0)
	{
		cout<<"This equation has two complex roots."<<endl;
	}
	if (num>0)
	{
		cout<<"This equation has two real roots."<<endl;
	}
	return 0;
}

[Banfa]

Your first problem is that at line 22 you have a single = which is assignment and not a comparison ==.

This sets num to 0 which causes all the if statements to return false.

However correcting your comparison is unlikely to completely fix the problem because doubles hold only approximations to numbers and generally we say that you should never use == or != to compare a double value, == will always return false and != will always return true.

That said you code does appear to work in some cases for example a=1, b=2, c=1 but I would expect there to be cases where it fails to correctly determine that a given equation has a single (repeated) root, although I haven't found an actual case yet.


EDIT:

I have an example for the equation

X2 + 0.387368 X + 0.037513491856

Your program outputs "This equation has two complex roots." where it should output "This equation has a single (repeated) root." because that equation has a root at -0.193684.

This is due to rounding errors produce in the approximation of the numbers when stored as doubles.