Where is the mistake?

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
kukukachoo
Posts: 1
Joined: Fri Mar 15, 2013 3:13 am
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Where is the mistake?

My fiance cooked up a ridiculous off-the-cuff equation to simply illustrate the difference in complexity between a high school algebra I textbook (which I've seen) and a college level algebra textbook that covers the same material (which he's found through experience is much more complex than the high school textbook.)

That equation is:

0 = ( (x - x2) / ( - (1 / x2) ) ) / (x - x2)

I didn't just accept the point like I should have, I had to try to solve for x. After much deliberation, frustration, and necessary algebra review, I arrived at the following:

0 = -x2

Therefore, x = 0. However...

When you plug 0 into the original equation, the solution is undefined and x could not possibly equal 0.

I then substituted 0 for y to graph the equation (y = -x2) to see where the line crosses the x axis. When I graphed the equation in Google, the line appears to intersect the x-axis at 0. I then had graphed the original equation (with y instead of 0,) and Google gave me back (from what I could tell by comparing two corresponding points on each graph) the same exact line.

Therefore, I don't believe I simplified the equation incorrectly, and both graphs appear to give the answer x = 0. Why then, can x = 0 on the graph (for both the original equation and its simplification) and when checked against the simplification, but zero cannot be substituted for x in the original equation?

Please help me understand where I've made an error in my logic, or if my question deals with a mathematical principle I have yet to learn. It'd be great if someone could point me in the right direction.

I got halfway through high school algebra II 7 years ago; haven't touched math since. It's taken me literally all day to refresh my memory and figure out how to get this far, plus another couple hours stumbling around my conundrum. I've spent all this time scouring the internet and I'm more confused now than I was when I started out. I really want to understand how x can equal 0, and at the same time cannot equal 0.

Ultimately the answer doesn't matter, the question was just made-up to begin with and no one was supposed to try to solve for x. But I just had to try, and here I am seven hours later. Please help.

buddy
Posts: 191
Joined: Sun Feb 22, 2009 10:05 pm
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Re: Where is the mistake?

kukukachoo wrote:That equation is:

0 = ( (x - x2) / ( - (1 / x2) ) ) / (x - x2)

I didn't just accept the point like I should have, I had to try to solve for x. After much deliberation, frustration, and necessary algebra review, I arrived at the following:

0 = -x2

Therefore, x = 0. However...

When you plug 0 into the original equation, the solution is undefined and x could not possibly equal 0.

Does it look like this:

$0\, =\, \frac{\left(\frac{x\, -\, x^2}{-\frac{1}{x^2}}\right)}{x\, -\, x^2}$

YOu didn't show what you did so I can't help with that. Try simplifying first, like they show here.

$x\, -\, x^2\, =\, x(1\, -\, x)$

$\frac{x(1\, -\, x)}{-\frac{1}{x^2}}\, =\, x(1\, -\, x)(-x^2)\, =\, -x^3 (1\, -\, x)$

$\frac{-x^3 (1\, -\, x)}{x(1\, -\, x)}\, =\, (-1) \left(\frac{x^3}{x}\right) \left(\frac{1\, -\, x}{1\, -\, x}\right)\, =\, y$

If you "simplify" this you have to keep track of the change you make to the domain. In the original you can't have x = 0 or x = 1 in the denominator, so you have to put those rules in the simplified version too:

$y\, =\, -x^2,\, \mbox{ for }\, x\, \neq\, 0,\, 1$

This is "the case with the hole".