quadratic dillema

[Imparcticle]

We were recently reviewing quadratic equations in my algebra 1 class. As my teacher simplified equation after equation on the board, I began to get this nagging feeling there was something incorrect.
I have pin pointed where I believe an error was made.

At this point in solving a quadratic equation, (6 +/- 2 root24)/2, my teacher simply cancels out the 6 and the 2 at once. I disagree here. It is a rule that you cannot cancel each component of an equation where a term is seperated by a + or - sign (of course, one can cancel the two since it is being multiplied with the "root24"). Instead, it I believe one must factor out a 2 from the numerator, then cancel out the 2 in the denominator.
my way:

1.)
6 +/- 2 root24 2(3 +/- root24)
-------------- = ---------------- = 3 +/- root24
2 2

2.)
The way my teacher does it:

6 +/- 2 root24 6/2 +/- 2/2 root24 = 3 +/- root24
--------------=
2


I realize that essentially, when you factor (as I did) , you are dividing each term, seperately by 2. However on the second example, one is dividing each term by the exact same integer.


is my analysis correct or incorrect?

[Integral]

It is very difficult to figure out your equations. This site has very nice equation capabilities. I suggest that you read this thread (http://www.physicsforums.com/showthread.php?t=8997)

[cookiemonster]

Factoring the 2 first or distributing the 1/2 first doesn't change things. Multiplication is distributive.

cookiemonster

[ahrkron]

I wonder if using [ code ][ /code ] will help:


1.)
6 +/- 2 root24 2(3 +/- root24)
-------------- = ---------------- = 3 +/- root24
2 2

2.)
The way my teacher does it:

6 +/- 2 root24 6/2 +/- 2/2 root24 = 3 +/- root24
--------------=
2

[cookiemonster]

How about

\frac{6 \pm 4\sqrt{6}}{2} = \frac{2(3 \pm 2\sqrt{6})}{2} = 3 \pm 2\sqrt{6}

and

\frac{6 \pm 4\sqrt{6}}{2} = \frac{6}{2} \pm \frac{4\sqrt{6}}{2} = 3 \pm 2\sqrt{6}

cookiemonster

[Imparcticle]

thank you for your clarification, cookie. I totally understand now.

[JonF]

You know I’ve never seen how the quadratic equation is derived, or a proof for it. Would some one post (or link) one please?

[Zurtex]

You know I’ve never seen how the quadratic equation is derived, or a proof for it. Would some one post (or link) one please?
It is just a generalisation of completing the square method: http://mathworld.wolfram.com/QuadraticEquation.html

[JonF]

I feel silly for asking now…

[arildno]

I feel silly for asking now…
Really, why?
It's not your fault that too many math teachers say "that's just the way it is" (usually to cover up their own ignorance/lack of understanding.)

[FlatlineLemon]

Really, why?
It's not your fault that too many math teachers say "that's just the way it is" (usually to cover up their own ignorance/lack of understanding.)

BRAVO!!!, right on the money there. lol

[TALewis]

My favorite: "The proof is left as an exercise to the reader." :)