Option Explicit Dim A As Double Dim B As Double Dim C As Double Dim BSqrd As Double Dim FourAC As Double Dim TwoA As Double Dim OutputBoxBA As String Dim OutputBoxAB As String ' Private Sub CalculateButton_Click() 'My values A = Val(InputBox.Text) B = Val(InputBox.Text) C = Val(InputBox.Text) 'The squares BSqrd = (B * B) FourAC = (4 * (A * C)) TwoA = (A * A) 'The if/then statement x2 Containing an error trap If A < 0 Then MsgBox ("Runtime Error 8675309") Else OutputBoxA.Text = -B + (Sqr(BSqrd) - (FourAC)) / (TwoA) OutputBoxB.Text = -B - (Sqr(BSqrd) - (FourAC)) / (TwoA) End If If OutputBoxA.Text < 0 Or OutputBoxB.Text < 0 Then MsgBox ("Runtime Error 8675309") MsgBox ("Square Root Of A Negitive NumBer!!!") OutputBoxA.Text = "" OutputBoxB.Text = "" End If End Sub Private Sub ExitButton_Click() Unload Me End Sub Private Sub ResetButton_Click() InputBox.Text = "" OutputBoxA.Text = "" OutputBoxB.Text = "" End Sub

# Quadratic Equation factoring

Page 1 of 1## 7 Replies - 3729 Views - Last Post: 02 December 2008 - 04:20 PM

### #1

# Quadratic Equation factoring

Posted 22 November 2008 - 03:43 PM

There is a Bug in here some where but I can't find it

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**Replies To:** Quadratic Equation factoring

### #2

## Re: Quadratic Equation factoring

Posted 22 November 2008 - 03:56 PM

Well, if you run your app from within VB6 it will halt at the erraneous code. Where does it halt?

Where is OutputBoxA declared?

The statement OutputBoxA.Text < 0 seems a little strange. "0" is a number and you are comparing it to a string...

Why such nasty error boxes? Better to tell the user what they did wrong.

You should test if there's a negative number before you try to calculate the square root of it.

To calculate the square root I think you should use a function sqrt() not sqr, but I'm a little unsure.

/Jens

Where is OutputBoxA declared?

The statement OutputBoxA.Text < 0 seems a little strange. "0" is a number and you are comparing it to a string...

Why such nasty error boxes? Better to tell the user what they did wrong.

You should test if there's a negative number before you try to calculate the square root of it.

To calculate the square root I think you should use a function sqrt() not sqr, but I'm a little unsure.

/Jens

### #3

## Re: Quadratic Equation factoring

Posted 01 December 2008 - 04:48 PM

OutputBoxA Is Declared as OutputBoxAB in the Dim statement

Very last Dim statement

I guess my error messages are harsh

also my teacher told us not to worry about negitive numbers

Hence the "Square Root Of A Negitive Number!!" error message

My Teacher said The problem is in The math section of my code

Very last Dim statement

I guess my error messages are harsh

also my teacher told us not to worry about negitive numbers

Hence the "Square Root Of A Negitive Number!!" error message

My Teacher said The problem is in The math section of my code

OutputBoxA.Text = -B + (Sqr(BSqrd) - (FourAC)) / (TwoA) OutputBoxB.Text = -B - (Sqr(BSqrd) - (FourAC)) / (TwoA)

### #4

## Re: Quadratic Equation factoring

Posted 01 December 2008 - 06:55 PM

I did not understand the math but any way I think this is the problem

Vace

'My values A = Val(InputBox.Text) B = Val(InputBox.Text) C = Val(InputBox.Text) 'The squares BSqrd = (B * B ) FourAC = (4 * (A * C ) ) TwoA = (A * A) 'If is two A it should be => TwoA = (2*a)

Vace

This post has been edited by **VB Crash**: 01 December 2008 - 06:56 PM

### #5

## Re: Quadratic Equation factoring

Posted 02 December 2008 - 12:23 PM

VB Crash, on 1 Dec, 2008 - 05:55 PM, said:

I did not understand the math but any way I think this is the problem

Vace

'My values A = Val(InputBox.Text) B = Val(InputBox.Text) C = Val(InputBox.Text) 'The squares BSqrd = (B * B ) FourAC = (4 * (A * C ) ) TwoA = (A * A) 'If is two A it should be => TwoA = (2*a)

Vace

Spot on Vace, first thing i noticed!

a * a = a squared

a + a = 2a

Very common problem with math code, people just forget and the simple things slide into their code

DempZ

### #6

## Re: Quadratic Equation factoring

Posted 02 December 2008 - 01:09 PM

Yeah, it can be A*2 or A+A .... I did't understand the name of the varible. Anyway, it's ok now?

Vace

Vace

### #7

## Re: Quadratic Equation factoring

Posted 02 December 2008 - 02:47 PM

My guess is that you are trying to solve somthing like this:

One way to do it:

1 ) Divide all of it with A =>

2 ) The equation

3 )

4 ) Apply this on your problem...

5a) p = B/A

5b) q = C/A

6 ) underTheSquareRoot = (p/2) * (p/2) - q = (p*p/4) - q

7 ) IF underTheSquareRoot < 0 THEN msgbox("No solutions") 'More stuff here that you'll have to think about

8a) X1 = -p/2 + sqrt(underTheSquareRoot)

8b) X2 = -p/2....

9 ) Present your solutions in e.g. two textboxes.

10) Done

Regards

/Jens

**Ax2 + Bx + C = 0**(Where x2 indicates x*x)One way to do it:

1 ) Divide all of it with A =>

**x2 + (B/A)x + C/A = 0**2 ) The equation

**x2 + px + q = 0**has (usually) two solutions, one of them is...3 )

**X = -p/2 + sqrt((p/2)*(p/2)-q)**'You probably recognize this one, the other solution is just "a sign" away.4 ) Apply this on your problem...

5a) p = B/A

5b) q = C/A

6 ) underTheSquareRoot = (p/2) * (p/2) - q = (p*p/4) - q

7 ) IF underTheSquareRoot < 0 THEN msgbox("No solutions") 'More stuff here that you'll have to think about

8a) X1 = -p/2 + sqrt(underTheSquareRoot)

8b) X2 = -p/2....

9 ) Present your solutions in e.g. two textboxes.

10) Done

Regards

/Jens

### #8

## Re: Quadratic Equation factoring

Posted 02 December 2008 - 04:20 PM

according to you

delta = b^2 - 4*a*c 'where delta determine the result of the equation if delta < 0 then msgbox "roots are imaginary" elseif delta>=0 then x= -b +((delta)^(0.5))/2*a y= -b -((delta)^(0.5))/2*a end if

This post has been edited by **thava**: 02 December 2008 - 04:20 PM

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