Scratch

It should be 3, and not -3 as far as I know. Is that just a typo or did it tell you -3?

aweman wrote:

Today in school we worked on simple quadratic equation. I plugged (X+1)(X-3)=0, and it said false. I got rid of the =0 to make it (X+1)(X-3), and it said -3. This is correct, but now plug in -1. That's right too! Just something interesting to think about.

What were you plugging it in on? If you're using a standard calculator then you need to be in some sort of solver mode to get it to give you the answer, not just evaluate what you type in. If you just plugged in X without an equals sign (assuming you haven't assigned it a value, making it zero), then (0+1)(0-3) would of course equal -3!

By the way, I moved this to Misc. where it belongs. 

Kileymeister wrote:

On most calculators, not giving X a value makes the calculator give you the answer in terms of X.  In this case, X^2 - 2X - 3.

Really? What sort of calculator are you using? I gotta get me one of those 

aweman wrote:

I'm not sure we're talking about the same quadratic equations here XD. Mine are trinomials with an equation ax^2+bx+c. When you do (x+1)=0, x=-1, then 0(x-3) HAS to be one because 0 times anything equals 0. However, when you do (x-3)=0, you get 3. The same principle applies; 0(x+3)=0. Therefore x=-1 and 3, but Scratch says -3, which isn't correct! I just though it was interesting. Also, I didn't use a calculator, I did that part in my head.

I don't follow.  In the equation 0(3X+3)=0, X is equal to every number in existence, because anything multiplied by 0 is 0.  Therefore X is undefined.

It's the same principal as 0/0, that is equal to every number in existence and is therefore undefined (this is not true for 1/0 or any other number, that is undefined because no value can render it correct).

Last edited by Kileymeister (2011-03-14 21:08:21)

I think we've established that we know how quadratics work.    (By the way, ax^2 + bx + c is standard form; (dx + e)(fx + g) is factored form.)

What we really need to know is what you're "putting into" Scratch to get the output you are describing. As I've said before, Scratch evaluates it fine for me. It's likely that you made a simple error. Could we see a screenshot or an upload of this project?

wow, you actually are just starting quadratic equations?
i did that like 3 years ago
now it's ∭_y!^∑▒abc▒〖√x〗  if a2x = y3 /  b!c!

aweman wrote:

I see your point, but X has already been decided as -1 in that situation. And you meant -3, not +3.

Oh, wait, I only took a simplified version and mistook it to be a separate equation.

You do know that in most quadratics you get multiple answers, right?  Often up to four or sometimes more.  I assumed you did and this may be the cause of the confusion.  I thought you were trying to prove something different.

Last edited by Kileymeister (2011-03-15 18:57:00)

MrsCullen2222 wrote:

I'm doing the same thing! But we haven't learned how to factor them yet. We learned to solve them this way:

x= -b - (radical sign) b(squared) - 4ac
     ---------------------------------------- (its a fraction bar    )
              2a

but since all quadratic equations are porabolas, you should get 2 different answers for x

(sorry if this is wrong- we just started learning it)

That is correct, however sometimes you can get more than 2 answers (or sometimes one answer if one of the answers doesn't factor in correctly, which can happen), because sometimes you get imaginary answers in terms of i (the square root of -1) which still work when plugged in.

Last edited by Kileymeister (2011-03-16 16:33:19)

Kileymeister wrote:

MrsCullen2222 wrote:

I'm doing the same thing! But we haven't learned how to factor them yet. We learned to solve them this way:

x= -b - (radical sign) b(squared) - 4ac
     ---------------------------------------- (its a fraction bar    )
              2a

but since all quadratic equations are porabolas, you should get 2 different answers for x

(sorry if this is wrong- we just started learning it)

That is correct, however sometimes you can get more than 2 answers (or sometimes one answer if one of the answers doesn't factor in correctly, which can happen), because sometimes you get imaginary answers in terms of i (the square root of -1) which still work when plugged in.

oh ok! we havent learned to factor the equation yet... we skipped to polynomials