x negativity

Stacey Koprince wrote:multiple answers to multiple questions.

Statement (2)
............x^2 - 1 < 0
............(x+1)(x-1) < 0
if x=+...(pos)(neg) < 0

x+1 > 0, therefore x > -1
x-1 < 0, therefore x < 1

We started with the assumption that x is pos, so 0 < x < 1. So if this is the case, then x is pos.

............x^2 - 1 < 0
............(x+1)(x-1) < 0
if x=-....(pos)(neg) < 0

x+1 > 0, therefore x > -1
x-1 < 0, therefore x < 1

We started with the assumption that x is neg, so -1 < x < 0. So if this is the case, then x is neg.

Stacey,

Here's my confusion. For Statement 2, why are you making assumptions about X being negative/positive, rather than making assumptions about an entire term?

For example:

Assuming (X+1) is positive, then (X-1) MUST be negative... SO:

X-1<0
X<1

&

X+1>0
X>-1

Then, you also have to try the other way around (X+1, negative; X-1, positive)

X+1<0
X<-1

X-1>0
X>1

So, for statement 2, you actually get 4 possible answers:

X<1, X>-1, X<-1, X>1

...Is this method completely wrong? Do you have to make assumptions about the individual variables as opposed to the terms?

linfongyu wrote:I just have one question - how the world does anyone solve this in a span of around 2 minutes?

Stacey Koprince wrote:I did it based on the variables because the form matches what I was ultimately asked about (the individual variable), so I find it easier to think about it that way.

For what you did above, the math is right, but here's how you would interpret it in a useful way:

when -1 < x < 1, (x+1) is pos and (x-1) is neg

when x < -1 or x > 1, (x+1) is neg and (x-1) is pos

Do I know which scenario is the right scenario? Nope. Either is possible. And even if I did know that one was the right scenario - each scenario leaves open the possibility that x could be neg or pos. So insufficient.

Stacey Koprince wrote:There's another reason why it's better to do it the way I did it (looking at the individual variables). The way you did it, you just split the number line into big categories, but you didn't actually narrow the range of values to things that would make your original statement true. You'd have to do more work to do it your way.

If you do it via the individual variables, the way I did, then you actually narrow the range down to just those numbers that would make the statements true.

Stacey Koprince wrote:You're assuming that both scenarios can work (that is, each term can be pos or neg). But that may or may not be true - you'd have to go check. If a particular scenario / assumption doesn't work, then you throw it out.

It's fairly common, with more complex equations and inequalities, for some apparent solutions not to work. That's why, in high school algebra, we were taught to plug the solutions back in for quadratic equations and the like - the only valid solutions are the ones that work when you plug them back into the original.

Go look at the work I did. I started with assumptions about just the individual variables, right? And some parts of the solutions didn't make sense - but I checked for that and threw out the parts that didn't make sense.

eg, I wrote:

............x^2 - 1 < 0
............(x+1)(x-1) < 0
if x=+...(pos)(neg) < 0

x+1 > 0, therefore x > -1
x-1 < 0, therefore x < 1

the full range I calculated was technically -1 < x < 1. But my original assumption was that x is positive. So I didn't write the solution as -1 < x < 1, because all of the negative solutions don't work (assuming x is positive). That's why I wrote the actual range as 0 < x < 1, because that's the only part of the original range in which all of the numbers are positive.