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Quadratic solver with complex roots

SomeCodingNoob (29)
Another quadratic solver but this time with imaginary roots included. Ignoring the fact that the imaginary part gives the wrong answer (although advice would be welcome) when I build and run the program, it just prints two numbers (e.g. 16i and -12i) because the return function adds the two parts together (instead of printing one then another). It doesn't print a complex root (e.g. 2 + 3i and 2 - 3i). Would I have to split up the functions that calculate the complex roots into two separate functions that calculate the real and imaginary bits separately? Please advise.

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#include <iostream>
#include <cmath>

using namespace std;

double RealRootOne (double a, double b, double c)
{
    double RealRootOne = ((-1*b) + sqrt((b*b) - (4*a*c))) / (2*a);
    return RealRootOne;
}

double RealRootTwo (double a, double b, double c)
{
    double RealRootTwo = ((-1*b) - sqrt((b*b) - (4*a*c))) / (2*a);
    return RealRootTwo;
}

void printRealResult (double a, double b, double c)
{
    cout << "Your results are " << RealRootOne(a, b, c) << " and " << RealRootTwo(a, b, c) << endl;
}

double ComplexRootOne (double a, double b, double c)
{
    double ComplexRealOne = (-1*b) / (2*a);
    double ComplexImOne = (-1*((b*b) - (4*a*c))) / (2*a);
    return ComplexRealOne + ComplexImOne;
}

double ComplexRootTwo (double a, double b, double c)
{
    double ComplexRealTwo = (-1*b) / (2*a);
    double ComplexImTwo = (-1*((b*b) - (4*a*c))) / (2*a);
    return ComplexRealTwo - ComplexImTwo;
}

void printComplexResult (double a, double b, double c)
{
    cout << "You're results are " << ComplexRootOne(a, b, c) << "i and " << ComplexRootTwo(a, b, c) << "i" << endl;
}

int main()
{
    cout << "Enter a, b and c for quadratic in the form ax^2 + bx + c = 0" << endl;
    double a, b, c;
    cin >> a >> b >> c;

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

    if (disc < 0)
    {
        printComplexResult(a, b, c);
    }

    if (disc == 0)
    {
        cout << "One real and repeated root " << ((-1*b) / (2*a));
    }

    else if (disc > 0)
    {
        printRealResult(a, b, c);
    }

    return 0;
}
TwilightSpectre (1392)
The easiest way would be to use complex numbers to return a complex number rather than a plain double. For examples, look at http://www.cplusplus.com/reference/complex/complex/ and follow the links for more details about the class.

You'll probably just want to use the 'real' root functions you have, and change it so it takes a complex value instead. After all, even when using complex numbers, the formula never changes.
Last edited on
SomeCodingNoob (29)
Ahh I didn't know about that! Thank you very much, more stuff to learn *rubs hands eagerly* :)
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