Block of Code

Practical Examples for Programmers

Quadratic Equation

In Visual Basic, write a program to determine the real roots of the quadratic equation after requesting the values of a, b, and c.  Before finding the roots, ensure that a is nonzero.  Note:  The equation has 2,1, or 0 solutions depending on whether the value of b^2 – 4*a*c is positive, zero, or negative.  In the first two cases, the solutions are given by the quadratic formula:
Quadratic FormulaQuadratic Formula 2

Below is an example of what the application should look like:

Quadratic Equation

Hints:

  • There will either be 0, 1 or 2 solutions to the equation
  • If the discriminant of the equation (the part of the numerator under the square root) is a negative number, there are no real solutions.
  • If the discriminant is zero, there will only be one solution.
  • If the discriminant is positive, there will be two solutions.

Suggested Control Names and Attributes:

Name Property Text Property Control Type Notes
 frmQuadratic  Quadratic Form Holds Controls
 txtA TextBox Captures value of “A” variable
 txtB TextBox Captures value of “B” variable
 txtC TextBox Captures value of “C” variable
 btnSolution Find Solutions Button Triggers event to display solutions
 txtSolutions TextBox Displays solutions

Write the Code:

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' Project: Quadratic Equation
' Programmer: Janice Wallace
' Date:  July 26, 2014
' Description:  Solves the quadratic equation
 
Public Class frmQuadratic
 
    Private Sub btnSolution_Click(sender As Object, e As EventArgs) Handles btnSolution.Click
        Dim varA As Integer = CInt(txtA.Text)
        Dim varB As Integer = CInt(txtB.Text)
        Dim varC As Integer = CInt(txtC.Text)
        Dim discriminant As Integer = (varB * varB) - (4 * varA * varC)
        Dim solution1 As Double = 0
        Dim solution2 As Double = 0
 
        ' Check to make sure A is a positive number
        If (varA = 0) Then
            MessageBox.Show("The value of A cannot be zero.")
        End If
 
        ' If the disciminant is negative there is no real solution
        If (discriminant < 0) Then
            txtSolutions.Text = "There is no solution"
        End If
 
        ' If the discriminant is zero there is one solution
        If (discriminant = 0) Then
            solution1 = (-varB / 2 * varA)
            txtSolutions.Text = "One solution: " & solution1
        End If
 
        ' Otherwise there are two solutions
        If (discriminant > 0) Then
            solution1 = ((-varB + Math.Sqrt(discriminant)) / (2 * varA))
            solution2 = ((-varB - Math.Sqrt(discriminant)) / (2 * varA))
            txtSolutions.Text = "Two solutions: " & solution1 & " and " & solution2
        End If
 
    End Sub
End Class