Having trouble with the sqrt() function.

I'm getting strage output with the sqrt() function in a quadratic formula program. I'll also add that I had the same problem using pow(x,.5). It's definitely the function that's causing the problem, I've made sure of it.

Code:

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

using namespace std;

void quad(double, double, double, double&, double&);

int main() {
    double a, b, c, x1, x2;
    
    cout.setf(ios::fixed, ios::floatfield);
    cout.setf(ios::showpoint);

    cout << "Enter the values of A, B, and C for the quadratic equation."
         << "\n>";
    cin >> a >> b >> c;
    cin.ignore(3,'\n');
    quad(a,b,c,x1,x2);
    
    cout << setprecision(2) << "X = " << x1 << " or " << x2 << ".";
    cin.get();
    
    return 0;    
}

void quad(double a, double b, double c, double& xPos, double& xNeg) {       
     if (2*a != 0 && (b*b - 4*a*c) < 0) {
        xPos = (-b + sqrt(b*b - 4*a*c))/(2*a);
        xNeg = (-b - sqrt(b*b - 4*a*c))/(2*a);    
        } 
     else {
        xPos = 0;
        xNeg = 0;    
        }          
     return;
     }

Output:

Code:

Enter the values of A, B, and C for the quadratic equation.
>3 4 5
X = -1.#J or -1.#J.

b is 4. b*b is 16. a is 3, c is 5. 4*3*5 is 60. 16-60 is -44. You are taking the square root of a negative number.

The < in your if should probably be >.

A better test case might be 1 -5 6.

Hey, good call. A can't believe a genuine typo wasted 15 minutes of my life. What's the garbage that sqrt() returns if you attemp to take the sqrt of a negetive number. I know it's impossible in real life, but how does a computer come to the answer -.#J?

I know it's impossible in real life, but how does a computer come to the answer -.#J?

It is not impossible.
It is called ‘i’ which is short for imaginary.
‘i’ is just the square root of negative one.
Square root of -1 X square root of 4 is the same thing as the square root of -4.
Square root of 4 is two, square root of -1 is i, so the square root of -4 = 2i

You could just check what is under the root, if it is less than 0 then take the absolute value of it (meaning positive) evaluate the square root and tack on a ‘i’ to it.

I think the result is just a general error/problem message.

Yes, I know the value 'i', but I was refering of the spectrum of real numbers. The computer can't find a real number answer so it spits out garbage. ...but where does it find the garbage?

Since sqrt is only defined to work with non-negative numbers, you cannot expect valid output when you use it incorrectly. However, that isn't really garbage, that is probably just your implementation's way of expressing NaN (not a number).