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Thread: quadratic equation using functions

  1. #1

    Thread Starter
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    Question quadratic equation using functions

    How can i put this Vb code in Vb.net Code, and make the same thing but using functions.

    Thank's!


    Dim A, B, C As Single

    Private Sub Command1_Click()
    A = Val(A1.Text) '''1
    B = Val(B1.Text) '''6
    C = Val(C1.Text) '''8

    Dim h1, h2, x1, x2


    '''h1 = B * B
    h2 = B * B - 4 * A * C '''4
    '''h2 = h1 + h2
    h1 = Sqr(h2) '''2

    '''**** h1 ****

    '''****** X1 ******
    x1 = -1 * B + h1
    x1 = x1 / (2 * A)

    x2 = -1 * B - h1
    x2 = x2 / (2 * A)

    x11.Caption = "X1 = " + Str(x1)
    x22.Caption = "X2 = " + Str(x2)

    End Sub

  2. #2
    Frenzied Member
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    Orlando, FL
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    1,618
    Ugh, I just had a horrible Calculus flashback....

    What is the whole equation again, I forget offhand, that'll help come up with an answer for you.
    Sean

    Some days when I think about the next 30 years or so of my life I am going to spend writing code, I happily contemplate stepping off a curb in front of a fast moving bus.

  3. #3
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    Oct 2000
    Location
    Philadelphia
    Posts
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    try this

    create three textbox for input
    create two textbox for output

    Code:
    dim a, b, c, h, tx1,tx2 as single
    dim x1,x2 as string
    if isnumeric(textbox1.text) and  isnumeric(textbox2.text)  and isnumeric(textbox2.text) 
    a = ctype(textbox1.text,single)
    b = ctype(textbox2.text,single)
    c = ctype(textbox3.text,single)
    
    h=b^2-4*a*c
    if h > 0
       x1 = (-b+math.sqrt(h))/(2*a)
       x2 = (-b-math.sqrt(h))/(2*a)
    else
       tx1 = -b/(2*a)
       tx2 = math.sqrt(math.abs(h)))/(2*a)
       x1 = tx1 & " + i" & tx2
       x2 = tx1 & " - i" & tx2
    end if
    
    textBox4.text = x1
    textbox5.text = x2
    U S A
    Visual Studio .NET
    Windows XP

  4. #4

    Thread Starter
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    May 2004
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    Procedures to make quadratic equation

    thank's, but i can´t do it like that, i have to use procedures, something like this

    Public Function ...(Byval ... as integer)

    Code...

    End Function

    I have to use Functions like that to made the program, and call them from one to others, the command1_click will have very few code.

    sorry about my english is because im Portuguese.

  5. #5
    Member
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    Oct 2000
    Location
    Philadelphia
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    Using Structures and a Function

    Here you go...

    Code:
        Private Structure quadratic
            Dim x1 As String
            Dim x2 As String
        End Structure
    
        Private Function retQuadratic(ByVal a As Single, ByVal b As Single, ByVal c As Single) As quadratic
            Dim h, tx1, tx2 As Single
            h = b ^ 2 - 4 * a * c
            If h > 0 Then
                retQuadratic.x1 = (-b + Math.Sqrt(h)) / (2 * a)
                retQuadratic.x2 = (-b - Math.Sqrt(h)) / (2 * a)
            Else
                tx1 = -b / (2 * a)
                tx2 = Math.Sqrt(Math.Abs(h)) / (2 * a)
                retQuadratic.x1 = tx1 & " + i" & tx2
                retQuadratic.x2 = tx1 & " - i" & tx2
            End If
        End Function
    
        Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
            Dim vals As quadratic
            vals = retQuadratic(1, 5, 3)
            TextBox1.Text = vals.x1
            TextBox2.Text = vals.x2
        End Sub
    Good Luck
    U S A
    Visual Studio .NET
    Windows XP

  6. #6

    Thread Starter
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    Thank's a lot, really... Thank you!!!

  7. #7

    Thread Starter
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    May 2004
    Posts
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    create a class for quadratic equation

    Hi again!

    Your code where very helpfull for me thank's a lot.

    But now i need to create a class to resolve the equation(quadratic equation), could you help me doing that!?

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