Quadratic Formula

[l270ucas]

Just trying to get a little more experience with this whole programing thing out. Im trying to make it easier to solve the quadratic equation.

msgbox = MsgBox(4,"Quadratic Formula", "Solve Equation?")

If $msgbox = 6 Then
   $a = InputBox("Quadratic Formula", "What is the value of a?")
   $b = InputBox("Quadratic Formula", "What is the value of b?")
   $c = InputBox("Quadratic Formula", "What is the value of c?")

$quadraticformula = (-$b + ($b^2-4*$a*$c)^1/2)/(2*$a)
$quadraticforumla1 = (-$b + ($b^2+4*$a*$c)^1/2)/(2*$a)



EndIf

MsgBox(0, "Quadratic Formula", "X = , , " & $quadraticformula & $quadraticforumla1)

This is what i  have so far and it runs. However When the final msg box comes up it says my value for x but then it also says #IND. Just that mean i need to tell it to round to a certain digit? Would just like a little help cleaning this up if its possible.

 

[Melba23]

l270ucas,

You were not correctly evaluating the formula - you needed a whole lot more parentheses to get the various calculations done in the correct order:

 

$a = InputBox("Quadratic Formula", "What is the value of a?")
$b = InputBox("Quadratic Formula", "What is the value of b?")
$c = InputBox("Quadratic Formula", "What is the value of c?")

$quadraticformula = (($b * -1) + ((($b ^ 2) - (4 * $a * $c)) ^ (1 / 2))) / (2 * $a)
$quadraticformula1 = (($b * -1) - ((($b ^ 2) - (4 * $a * $c)) ^ (1 / 2))) / (2 * $a)

MsgBox(0, "Quadratic Formula", "X = " & $quadraticformula & " and " & $quadraticformula1) ; And it helps if you use the correct variable here

That runs for me - and gives the correct answers when I test.

M23

[jchd]

Guys,

You also have to check that the determinant Δ = b² - 4ac is not negative, else your solutions {x1, x2} escape from ℝ (reals) and "fall" into ℂ (complex).

Finally Δ = 0 leads to a double solution, with x₁ = x₂

[l270ucas]

That does help with the right parentheses. And I wasn't sure how to display two variables in a msg box. So thanks for that!

Jchd But I'm confused about your statement of - (b^2 - 4ac)^(1/2). Because the quadratic formula uses plus or minus, but I don't know how to do that so I wrote two variables, one positive, one negative.

[jchd]

• Δ < 0 : no real solution

{x_1, x_2} = ∅

• Δ = 0 : one (double) real solution

{x_1, x_2} = {-b/(2 a)}

• Δ = 0 : two real solutions

{x₁, x₂} = {(-b - Sqrt(Δ)) / (2 a), (-b + Sqrt(Δ)) / (2 a)}

Edited February 3, 2014 by jchd