pq method math to the function of axx+b*x+c=0

Question

Question : I thought the source code is right and it worked well a lot of times for example at functions like : x^2 + 4x +4 or 3x^2 +5x+1 but in some functions the output is NaN and i don't understand why (function that outputs NaN : 4x^2 - 2x +8

import java.util.*; 
public class QuadGl {

    /**      * @param args   */
    public static void main(String[] args) {

    double a,b,c,rechnung1,rechnung11,rechnung2,rechnung22,rechnung23,ergebnis1,ergebnis2;//BENÖTIGTE DATENTYPEN

    Scanner input = new Scanner(System.in);             
    System.out.println("Bitte geben sie einen Wert für a ein : "); a = input.nextDouble();
    System.out.println("Bitte geben sie einen Wert für b ein : "); b = input.nextDouble(); 
    System.out.println("Bitte geben sie einen Wert für c ein : "); c = input.nextDouble();

    //ERSTER RECHNUNGSSCHRITT DER P.Q.FORMEL 
    rechnung1 = b/a;//b=p
    rechnung11 = c/a; //c=q 
    System.out.println(rechnung1);
    System.out.println(rechnung11);


    //ZWEITER RECHNUNGSSCHRITT 
    rechnung2 = (-rechnung1/2);// -b/2
    rechnung22 = (rechnung1/2); // b/2 
    rechnung23 = Math.sqrt(( rechnung22 * rechnung22 -rechnung11));//

    System.out.println(rechnung2);
    System.out.println(rechnung22);
    System.out.println(rechnung23);

    ergebnis1 = rechnung2 + rechnung23;//-b/2 + b/2 ^2 - c 
    ergebnis2 = rechnung2 - rechnung23;//-b/2 - b/2 ^2 - c 

    System.out.println("x(1)= "+ ergebnis1);
    System.out.println("x(2)= " + ergebnis2);
    }

}

@alk As far as I know, no. But you can create one! I know that in Ruby, you can do it (calculation with complex numbers) — dryairship, Jun 25 '16 at 12:06
@Hackerdarshi: In C there is one; so perhaps retag this C again? — alk, Jun 25 '16 at 12:11
@alk Ahh... No. Please. The question has been answered. There is no need for anything now. Moreover, the code is in Java. So tagging C would be like ... what? — dryairship, Jun 25 '16 at 12:16
It had been tagged C in the beginning as well, perhaps with the intention in mind to received a C based answer using complex numbers? ;-) @Hackerdarshi — alk, Jun 25 '16 at 12:17
@alk Well, but if he asked this question, then that means that he was probably unaware that it shows NaN because it is a complex number... — dryairship, Jun 25 '16 at 12:22

score 4 · Answer 1 · edited Jun 25 '16 at 12:10

4

4x² - 2x +8 shows NaN because, the formula to solve a quadratic equation is :

(-b ± (b² - 4ac)^1/2) / 2a

Here, b² - 4ac is -ve (4 - 4*16*8), and hence, its square root is NaN

b² - 4ac is called Discriminant.

When it is equal to 0, the roots are equal.
When is is less than 0, you will get NaN, as the roots are imaginary.
When it is greater than 0, you get two different roots.

edited Jun 25 '16 at 12:10

Boann

48,794
16
117
146

answered Jun 25 '16 at 11:53

dryairship

6,022
4
28
54

Thanks @Boann! I didn't see that! – dryairship Jun 25 '16 at 12:18
Note: you could use complex numbers to get an imaginary solution. – Peter Lawrey Jun 25 '16 at 13:06
1

@PeterLawrey Is there any class like `ComplexNumbers` in java? – dryairship Jun 25 '16 at 13:10
Not built in but you could add this handling to the code by checking for a negative value before using Math.sqrt. There are libraries for complex numbers on the web. – Peter Lawrey Jun 25 '16 at 14:55
You could also just output the result as a complex number in that case, this is much much shorter than implementing a complex data type. – Lutz Lehmann Jun 25 '16 at 19:01

score 0 · Answer 2 · answered Jun 25 '16 at 19:08

Just format the output according to the discriminant

disc = b*b-4*a*c
if( disc > 0 ) {
    rdisc = sqrt( disc )/(2*abs(a));
    print("1st root = " + ( -b / (2*a) + rdisc ) );
    print("2nd root = " + ( -b / (2*a) - rdisc ) );
} else {
    rdisc = sqrt( -disc )/(2*abs(a));
    print("1st root = " + ( -b / (2*a) ) + " + i*" + rdisc );
    print("2nd root = " + ( -b / (2*a) ) + " - i*" + rdisc );

}

pq method math to the function of a*x*x+b*x+c=0

2 Answers2

pq method math to the function of axx+b*x+c=0