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How do I make a quadratic function program on my ti 84 calculator?

I want a program that will run just like this:

Having a trinomial such as ax^2+bx+c

Program asks for "a".

User inputs "a".

Program asks for "b".

User inputs "b".

Program asks for "c".

User inputs "c".

Program calculates the roots using the quadratic equation:

variable1 = (-b + sqrt(b^2-4ac))/(2a)

varible2 = (-b - sqrt(b^2-4ac))/(2a)

Program displays variable1 and variable2.

This is what i have so far:

PROGRAM:QUAD

:ClrHome

:Input "A VALUE?",A

:Input "B VALUE?",B

:Input "C VALUE?",C

:Q=((-1*B)+(sqrt((B*B)-(4*A*C))))/(2A)

:O=((-1*B)-(sqrt((B*B)-(4*A*C))))/(2A)

:Disp "ROOTS ARE "

:Disp Q

:Disp " AND "

:Disp O

:Pause

The program isn't working because my calculator has Q and O stored as zero so whenever i run it, i always get the output to say:

ROOTS ARE

0

AND

0

How do i create a "fresh" variable for my program? (meaning one that isn't already defined)

Otherwise, the program works flawlessly. I get no errors unless I put in number that give an imaginary number.

If you're trying to figure this one out, here's a trinomial that does have real roots:

x^2 - 2x - 3

Roots should be, -1 and +3.

Thank you so much for helping me out.

And on a side note, I need this for my chem class because we have trinomials that are like this:

5.56*10^(-11) - 5.56*10^(-10)x - x^2

and i can't just graph it and find the y-intercept because the value of x is so small it doesn't detect a change in sign.

2 Answers

Relevance
  • 7 years ago
    Favorite Answer

    You're storing variables incorrectly. In TI-Basic, "=" is an equality check (equivalent to == in other programming languages). If you want to store something into a variable, you should use the [Sto►] button (above [On]).

    Example:

    :5→A

    :3→B

    :A+B→C // the C variable now contains the value 8

  • 7 years ago

    In TI-84 BASIC to assign a value to a variable you need to use the STO command. For example, assign 3 to x you would enter

    3 STO x

    So in your case to assign Q and O the syntax would be:

    ((-1*B)+(sqrt((B*B)-(4*A*C))))/(2A) STO Q

    ((-1*B)-(sqrt((B*B)-(4*A*C))))/(2A) STO O

    STO should come up as an arrow so it would actually look like

    ((-1*B)+(sqrt((B*B)-(4*A*C))))/(2A)->Q

    ((-1*B)-(sqrt((B*B)-(4*A*C))))/(2A)->O

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