x negativity

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by Stockmoose16 » Tue Oct 28, 2008 4:52 pm
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?

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by Stacey Koprince » Tue Oct 28, 2008 5:38 pm
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.
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by linfongyu » Tue Oct 28, 2008 8:31 pm
I just have one question - how the world does anyone solve this in a span of around 2 minutes? :shock:

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by logitech » Tue Oct 28, 2008 9:27 pm
linfongyu wrote:I just have one question - how the world does anyone solve this in a span of around 2 minutes? :shock:
You will be surprised! 8)
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by Stockmoose16 » Tue Oct 28, 2008 10:45 pm
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,

Something still doesn't make sense here. For statement #2, when you factorize and come out with the values X<-1 or X>1, the X values don't work in the original equation (X^2-1<0)

For example, if you were to put -2 into the equation, you'd come out with 4-1<0, or 3<0. That's obviously wrong. Why is the algebra spitting out answers that make the equation untrue?

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by Stacey Koprince » Wed Oct 29, 2008 6:08 am
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.
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by Stockmoose16 » Wed Oct 29, 2008 9:16 am
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.
But just so I get the logic correct, when my factorization for stmt #2 comes out X<-1, X>1, should I be saying, "Oh wait, that doesn't make any sense since it doesn't work in the original equation." Or, are those values actually correct. I guess I'm confused as to why the factorization is pumping out values that don't work in the original equation.

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by Stacey Koprince » Wed Oct 29, 2008 9:29 am
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.
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by Stockmoose16 » Wed Oct 29, 2008 9:34 am
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.
Ah, very interesting. Thanks, Stacey.

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by simba12123 » Mon Nov 03, 2008 6:10 am
THe main problem was very bad rephrasing and making things harder than they really were. ONE FINAL QUESTION.

In picking numbers but following every step with Stacy:
statement one yields x= -2 and x= 1/2
statement two yields x = 1/2 and x =-1/2

combining statements 1 and 2, we must only bring in what both statements agree on. Cant we argue here that x=1/2 and hence positive?




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another method....

by jimmiejaz » Mon Nov 03, 2008 6:55 am
hi simba,

here's my approach if it helps...

From 1.) x3(1-x2)<0 two cases arise:
either (x3>0 and 1-x2<0)...I or (x3<0 and 1-x2>0)....II
From I,
x3>0 implies x>0, 1-x2<0 gives x2>1 or |x| > 1 which gives x>1 or x<-1.
x<-1 is not possible since x>0 from x3>0.
From II,
x3<0 implies x<0, 1-x2>0 gives x2<1 or |x| < 1 which gives -1<x<1.
But, 0<x<1 is not true since x<0 from x3<0.
so, from 1 we get x>1 or -1<x<0 hence not sufficient.

From 2.) x2-1<0
it gives x2<1 or |x| < 1which is equal to
-1<x<1. x can be both +ve or -ve hence 2 is also not sufficient.

Combining both, we get

-1<x<0 which is -ve. So, hence the ans is C.
Hope it helps.

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by EricLien9122 » Mon Nov 03, 2008 7:21 am
statement 1:

X^3-X^5<0

x=2, 8-64= neg answer, x is not neg.

x=-1/2--> -1/8 + 1/64=neg answer, x is neg.

Statement 1 is insufficient.

Statement 2:

x^2 < 1

-1<x<1

x=1/2, x is not neg.
x=-1/2, x is neg.

Statement 2 is insufficient.

1+2

we know that x has to be a fraction from statement 2, and x^3-x^5<0 from statement 1. Therefore, x is negative.

answer choice is c.

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by scoobydooby » Mon Nov 03, 2008 8:38 am
Stmnt 1: x^3(1-x^2)<0
or x^3-x^5<0
or x^3<x^5
this can happen if x is a positive integer or a negative fraction
say x=2, 8<32, x is not negative
say x=-1/2, -1/8 0r -4/32<-1/32, x is negative
we cant say definitely if x<0 not suff

Stmnt 2: x^2-1<0
or x^2<1. x must be a fraction, positive or negative, not sufficient

combining x must be negative fraction or C