C++ Program with quadratic equation, for unreal and real solutions

[oN3k92yv]

I made a quadratic formula code that works - it accepts the numbers for A B and C, but it doesn't tell you when a solution is real, or unreal. How do I make it do that?

  #include <iostream>
#include <cmath>
#include <iomanip>

using namespace std;


int main ()
{
	double a = 0;
	double b = 0;
	double c = 0;
	float x1, x2 = 0;
	double disc = 0;

	cout << "===============================================================================" << endl;
	cout << "This program will provide solutions for an equation of the form A*x^2 + B*x + C where A, B, an C are intergers, and A is not equal to zero." << endl;
	cout << "Please enter value of a: " << endl;
	cin >> a;
	cout << "Please enter value of b: " << endl;
	cin >> b;
	cout << "Please enter value of c: " << endl;
	cin >> c;
	disc = ((b * b) - (4 * a* c)); 

	x1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
	x2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);

	if (disc>0)
	{
		x1 = (-1 * b + sqrt(disc)) / (2 * a);
		x2 = (-1 * b - sqrt(disc)) / (2 * a);
	}
	else if (disc == 0)
	{
		x1 = x2 = (-1 * b) / (2 * a);
	}
	else
	{
		disc = -1 * disc;
		cout << "x1=" << "(" << (-1 * b) << "+ i" << sqrt(disc) << ")/" << 2 * a;
		cout << "x1=" << "(" << (-1 * b) << "+ i" << sqrt(disc) << ")/" << 2 * a;
		exit(0);
	}
	cout << "x1=" << x1 << "\n";
	cout << "x2=" << x2 << "\n";



	return 0;
}

[DyslexicChciken]

The solution is unreal when it includes imaginary number(s), or when the radicand is negative. This is the case in your else statement.

[48T7M4Gy]

the solution is unreal when the dicriminant is negative

[oN3k92yv]

Okay, how does this look?

#include <iostream>
#include <cmath>
#include <iomanip>

using namespace std;


int main ()
{
	double a = 0;
	double b = 0;
	double c = 0;
	double x1, x2, x3 = 0;
	double disc = 0;

	cout << "===============================================================================" << endl;
	cout << "This program will provide solutions for an equation of the form A*x^2 + B*x + C where A, B, an C are intergers, and A is not equal to zero." << endl;
	cout << "Please enter value of a: " << endl;
	cin >> a;
	cout << "Please enter value of b: " << endl;
	cin >> b;
	cout << "Please enter value of c: " << endl;
	cin >> c;
	disc = ((b * b) - (4 * a* c)); 

	x1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
	x2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);

	if (disc>0)
	{
		cout << "This equation has 2 solutions." << endl;
		x1 = (0 - b + sqrt(disc)) / (2 * a);
		cout << "The first solution is: " << x1 << endl;
		x2 = (0 - b - sqrt(disc)) / (2 * a);
		cout << "The second solution is: " << x2 << endl;
	
	}
	else if (disc == 0)
	{
		cout << "This equation has one real solution." << endl;
		x1 = x2 = (-1 * b) / (2 * a);
	}
	else if (disc<0)
	{
		cout << "This equation has 2 unreal solutions." << endl;
		disc = -1 * disc;
		cout << "x1=" << "(" << (-1 * b) << "+ i" << sqrt(disc) << ")/" << 2 * a;
		cout << "x1=" << "(" << (-1 * b) << "+ i" << sqrt(disc) << ")/" << 2 * a;
		exit(0);
	}

	


	return 0;
}

[oN3k92yv]

When I ran it, instead of only saying it won't calculate a value that has A = 0, it still displayed the other things. Look

http://i.imgur.com/OwxddKo.png

How do I stop that?

[Ispil]

Stick a return 0; in the if statement for if (a == 0). That way, once it tells you that A can't be zero, it exits main.

[oN3k92yv]

You're fucking great, that fixed that. Which data type will make the answers be something like x= 3.0000+E00? Is it float? Should I add setprecision in there? Alright, I'm going to try to make it output if there is only one real solution as well. BRB

[oN3k92yv]

updated below

[48T7M4Gy]

@aixth Sounds like your code supplier has let you down. If you wrote the original I doubt whether you'd be asking these sort of questions. Ask for your money back. :)

[oN3k92yv]

I fixed it, however it the program doesn't work if there is only one solution or two imaginary ones. this is what I got

#include <iostream>
#include <cmath>
#include <iomanip>

using namespace std;


int main ()
{
	float a = 0;
	float b = 0;
	float c = 0;
	float x1, x2, x3 = 0;
	int disc = 0;

	cout << "===============================================================================" << endl;
	cout << "This program will provide solutions for an equation of the form A*x^2 + B*x + C where A, B, an C are intergers, and A is not equal to zero." << endl;
	cout << "Please enter value of a: " << endl;
	cin >> a;
	cout << "Please enter value of b: " << endl;
	cin >> b;
	cout << "Please enter value of c: " << endl;
	cin >> c;
	disc = ((b * b) - (4 * a* c)); 
	cout.precision(4);
	cout << std::scientific;

	x1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
	x2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);

	if (a == 0)
	{
		cout << "No solutions will be calculated for the leading coefficient of 0!" << endl;
		return 0;
	}

	if (disc>0)
	{
		cout << "This equation has 2 solutions." << endl;
		x1 = (0 - b + sqrt(disc)) / (2 * a);
		cout << "The first solution is: " << x1 << endl;
		x2 = (0 - b - sqrt(disc)) / (2 * a);
		cout << "The second solution is: " << x2 << endl;
	
	}
	else if (disc == 0)
	{
		cout << "This equation has one real solution." << endl;
		x1 = x2 = (0 - b - sqrt(disc)) / (2 * a);
	}

	else if (disc<0)
	{
		cout << "This equation has 2 unreal solutions." << endl;
		disc = -1 * disc;
		cout << "x1=" << "(" << (-1 * b) << "+ i" << sqrt(disc) << ")/" << 2 * a;
		cout << "x1=" << "(" << (-1 * b) << "+ i" << sqrt(disc) << ")/" << 2 * a;
		exit(0);
	}

	


	return 0;
}

[48T7M4Gy]

Well done!

[oN3k92yv]

I'm a student currently learning programming. Should I change the way I made the (disc<0) to match the way if disc was >0?

[48T7M4Gy]

There's no reason why not. Complex numbers can have a 0i component

[oN3k92yv]

I tried it, that did not work.

http://i.imgur.com/DpfhXbW.png

[48T7M4Gy]

Hint 1: if the discriminant is negative then there is only one solution to it's square root.

Hint 2: There are always two solutions to a quadratic equation whether they are real or complex. And a real number is just a special case anyway.

if discriminant is negative
x1 = ...
x2 = ...

if discriminant is zero
x1 = ...
x2 = ... (x1 and x2 could be the same but who cares?)

if discriminant is positive (because zero is neither positive or negative)
x1 = ...
x2 = ...

display x1 and x2

http://www.purplemath.com/modules/quadform2.htm

[oN3k92yv]

the discriminant is negative, but I'm trying to find the two imaginary solutions. that's the only part left

[48T7M4Gy]

http://www.purplemath.com/modules/quadform2.htm

[oN3k92yv]

That's not what I'm having a problem with, I know how the quadratic formula works. Nevermind.

[48T7M4Gy]

#include <iostream>
#include <cmath>
#include <iomanip>

using namespace std;


int main ()
{
    float a = 0;
    float b = 0;
    float c = 0;
    float x1 = 0, x2 = 0;
    int disc = 0;
    double Im = 0;
    
    cout << "===============================================================================" << endl;
    cout << "This program will provide solutions for an equation of the form A*x^2 + B*x + C where A, B, an C are intergers, and A is not equal to zero." << endl;
    cout << "Please enter value of a: " << endl;
    cin >> a;
    cout << "Please enter value of b: " << endl;
    cin >> b;
    cout << "Please enter value of c: " << endl;
    cin >> c;
    
    disc = ((b * b) - (4 * a* c));
    
    if (disc > 0)
    {
        cout << "This equation has 2 solutions." << endl;
        
        x1 = (- b + sqrt(disc)) / (2 * a);
        x2 = (- b - sqrt(disc)) / (2 * a);
    }
    
    if (disc == 0)
    {
        cout << "This equation still has two real solutions, both are the same :)." << endl;
        x1 = x2 = -b / (2 * a);
    }
    
    if (disc < 0)
    {
        cout << "This equation has 2 complex solutions." << endl;
        disc = -1 * disc;
        x1 = x2 = b / (2 * a);
        Im = sqrt(disc)/(2 * a);
    }
    
    cout << " First: x2 = " << x1 << " + " << Im << "i" << endl;
    cout << "Second: x1 = " << x2 << " - " << Im << "i" << endl;
    
    return 0;
}