C++ Quadratic equation outputs: -1.#IND

Question

My code works fine with: (1, -2, -8), It gives me the error above when I input a=1 b=0 c=1,

Here is my code:

    double x=0,a=0,b=0,c=0,d=0;
    complexType solu1;
    complexType solu2;


    cout << "\n\nEnter values of quadratic a,b,c:";
    cin >> a >> b >> c;

    double solution1 = (-1.0 * b) + (sqrt((b * b) - (4 * a * c)));
    solu1 = solution1 / (2*a);

    cout << setprecision(5) << solu1;

    double solution2 = (-b) - (sqrt((b*b) - (4 * a * c)));
    solu2 = solution2 / (2*a);
    cout << setw(5) << setprecision(5)  << solu2;

How can I remedy this?

This isn't an honest question: You claim to have runtime errors, but what you pasted isn't valid C++ code. Please post the real code that is giving you trouble. — Kerrek SB, Oct 13 '11 at 17:46
For `1 0 1`, the solution is imaginary. You'll be taking the square root of a negative number... — Mysticial, Oct 13 '11 at 17:55

score 3 · Accepted Answer · answered Oct 13 '11 at 18:05

You are trying to find a real solution to x^2 + 1 = 0, which only has imaginary solutions.

If you want to solve all quadratic equations, then you need to check whether the discriminant is positive to determine whether the solutions are real or complex:

double d = b*b - 4*a*c;
if (d >= 0) {
    double sol1 = (-b + sqrt(d))/(2*a);
    double sol2 = (-b - sqrt(d))/(2*a);
    std::cout << sol1 << ", " << sol2 << '\n';
} else {
    double real = -b/(2*a);
    double imag = sqrt(-d)/(2*a);
    std::cout << real << " +/- " << imag << "i\n";
}

You could do something neater with std::complex if you want.

Thank you, I used this in conjunction with my complexType class to get the correct answers. `if (d >= 0) { sol1 = (-b + sqrt(d))/(2*a); sol2 = (-b - sqrt(d))/(2*a); cout << x1 << ", " << x2 << '\n'; } else { sol3 = ( (-b/(2*a) )+( (sqrt(-d)/(2*a))*i)); sol4 = ( (-b/(2*a) )-( (sqrt(-d)/(2*a))*i)); cout << x3 << " " << x4 << "\n"; }` — Nick, Oct 13 '11 at 21:10

score 1 · Answer 2 · answered Oct 13 '11 at 18:07

sqrt((b*b) - (4 * a * c)) for your inputs is sqrt(-4). According to http://www.cplusplus.com/reference/clibrary/cmath/sqrt/, If the argument is negative, a domain error occurs, setting the global variable errno to the value EDOM. I don't see a definition for what it returns in that case. Either way, that's wrong.

I see that you have complexType in your code. If that's a typedef of std::complex<T>, then the code is easily fixable.

complexType top = b*b - 4*a*c;
solu1 = (-b + sqrt(top)) / (2*a);
solu2 = (-b - sqrt(top)) / (2*a);

Since std::sqrt has an overload for std::complex<T>.

The return value is implementation-defined if a domain error occurs. Most IEEE implementations will return `NaN`. — Mike Seymour, Oct 13 '11 at 18:59

C++ Quadratic equation outputs: -1.#IND

2 Answers2