I tried to write an easy algorithm but it doesn't work

[rian bron]

Hello guys,  i tried to write the formula of Quadratic equation in  java, but for some reason it seems that i wrote it right but the answer is wrong.
the code i wrote :

public class RootFormula
{
    private int a;
    private int b;
    private int c;
    public RootFormula(int a,int b, int c)
    {
        this.a=a;
        this.b=b;
        this.c=c;
    }
    public int getA()
    { return a;}
    public int getB()
    {return b;}
    public int getC()
    {return c;}
    public String findRootFormula()
    {
     double firstX=(getB()+(Math.sqrt(Math.pow(getB(),2))-4*(getA()*getC())))/2*getA();
     double secondX=(getB()-(Math.sqrt(Math.pow(getB(),2))-4*(getA()*getC())))/2*getA();
     return "x1= "+firstX+"\nx2= "+secondX;
    }
  
    }
    

the algorithm is at lines 20,21
could you please tell me what i did wrong?
Thank you

[Paul Clapham]

Well, first of all your code is very hard to read because you use the accessor methods instead of the instance variables. If I rewrite your code to remove that idea (and to remove the complicated method of squaring a number, and to remove unnecessary brackets) then it looks like this:

double firstX=(b+(Math.sqrt(b*b)-4*a*c))/2*a;

When I compare that to the quadratic formula I can see at least two differences, probably three. And that's before considering the possibility of complex numbers as the solution.

[rian bron]

Paul Clapham wrote:Well, first of all your code is very hard to read because you use the accessor methods instead of the instance variables. If I rewrite your code to remove that idea (and to remove the complicated method of squaring a number, and to remove unnecessary brackets) then it looks like this:

double firstX=(b+(Math.sqrt(b*b)-4*a*c))/2*a;

When I compare that to the quadratic formula I can see at least two differences, probably three. And that's before considering the possibility of complex numbers as the solution.

well, your answer is indeed easier to read, but still if i insert a=1,b=2,c=-3 the answer i get according to this code is x1=8,x2=-6 versa the correct answer is x1=-3,x2=1
and i don't get why the answer is incorrect although the algorithm seems to be right

[fred rosenberger]

first of all, what is the quadratic formula?  from the get go, I think your very first term is wrong.

What I would suggest you do is break it apart.  you can't tell where the math is wrong, because you do it all in one line.  so, i'd suggest breaking it apart.

is "b" correct?

what does "(Math.sqrt(b*b)-4*a*c))" give you, and is it correct? if not, break it apart into smaller pieces.

if you are not getting the right answer, you need to figure out what part is wrong, and focus on that.

[Carey Brown]

This is suspect. Sqrt of 'b' squared is 'b'.

Math.sqrt(b*b)

[Paul Clapham]

Carey Brown wrote:This is suspect. Sqrt of 'b' squared is 'b'.

Math.sqrt(b*b)

Indeed. That's just one of the errors in the calculation -- the brackets are in the wrong place there.

[Campbell Ritchie]

Paul Clapham wrote:. . . just one of the errors in the calculation . . .

I can see two more errors, I think. Please remind yourself what the formula for a quadratic equation is.
You doubtless already know that you can't directly implement ± in Java®, so you will have to calculate the two solutions separately. Yes, you are calculating them separately. It would make sense mathematically to calculate the determiner first; if it is > 0 you have two real solutions, if it is 0 you have one real solution and if it is < 0 you have two solutions which are complex numbers.

[rian bron]

you were right guys, the problem was with the brackets,
the right solution is :

 double firstX=(b+(Math.sqrt(b*b-4*a*c)))/(2*a);
     double secondX=(b-(Math.sqrt(b*b-4*a*c)))/(2*a);

thank you all for help

[Piet Souris]

rian bron wrote:
the right solution is :

Hmmm, are you sure?    

[Knute Snortum]

The first b needs to be negative, correct?

https://www.calculatorsoup.com/calculators/algebra/quadratic-formula-calculator.php