Question

I have this code so far but every time i run and put the three numbers in a get the roots are NaN can some one please help or point me to where i went wrong.

 import java.util.Scanner;

 class Quadratic {
   public static void main(String[] args) {

   System.out.println("Enter three coefficients");
   Scanner sc = new Scanner(System.in);
   double a = sc.nextDouble();
   double b = sc.nextDouble();
   double c = sc.nextDouble();
   double root1= (-b + Math.sqrt( b*b - 4*a*c ) )/ (2*a);
   double root2= (-b - Math.sqrt( b*b - 4*a*c ) )/ (2*a);
   System.out.println("The roots1 are: "+ root1);
   System.out.println("The roots2 are: " + root2);


     } 
   }

Answer 1

You have to remember that not every quadratic equation has roots that can be expressed in terms of real numbers. More specifically, if b*b - 4*a*c < 0, then the roots will have an imaginary part and NaN will be returned, since Math.sqrt of a negative number returns NaN, as specified in the documentation. This works for coefficients such that b*b - 4*a*c >= 0, however:

Enter three coefficients
1
5
6
The roots1 are: -2.0
The roots2 are: -3.0

---

If you wanted to account for non-real roots as well, you could do something like

double d = (b * b - 4 * a * c);
double re = -b / (2 * a);

if (d >= 0) {  // i.e. "if roots are real"
    System.out.println(Math.sqrt(d) / (2 * a) + re);
    System.out.println(-Math.sqrt(d) / (2 * a) + re);
} else {
    System.out.println(re + " + " + (Math.sqrt(-d) / (2 * a)) + "i");
    System.out.println(re + " - " + (Math.sqrt(-d) / (2 * a)) + "i");
}