Damn Math... 25k Credits For Help?

[25thhour]

I need to know how to get the answer... I have no idea how to get the actual answer with the inequality signs.

the question is this:

3 + X > 4/X
Where X cannot = 0
This is what I get:

Case 1:
Where x > 0

x > -4 , x > 1



Case 2:
Where x < 0

x < -4, x < 1

I have no clue how to arrange the answers and we are not allowed to use our graphing calculator. I would like help in trying to understand the process of rearranging it. I'm good with the math.

I hope I made it clear enough, if not just say what other clarification you may need.

Thanks

[Sephiroth28]

Re: Damn Math... 25k Credits For Help?

So you want to know how to graph the inequalities without the calculator?

Haha well iironiic kinda beat me to the explanation. lol

[25thhour]

Re: Damn Math... 25k Credits For Help?

no, I want to know how to rearrange the answers that I got to the cases to find the right answer. :P

All I know is that you have to eliminate some of them with the x > or < 0 but Im not sure which ones.

[iironiic]

Re: Damn Math... 25k Credits For Help?

3 + X > 4/X

Let X < 0
Therefore, 3X + X^2 < 4
Or, X^2+3X - 4 < 0
Or, (X+4)(X-1) < 0
So we know that one (X-1) or (X+4) is negative and the other is positive
a) Suppose (X-1)>0, or X>1
This implies that (X+4)<0 but for all X>1, that's not true.

b) Suppose (X+4) > 0, or X> -4
This implies that (X-1) < 0, which is true for -4<X<1
But remember X < 0, so -4<X<0 for the first case.

Now let X>0
Therefore, 3X +X^2 >4
Or, (X+4)(X-1) > 0
Which implies that they both have to share the same sign.
Let (X+4) > 0 and (X-1) > 0. Or, X > -4 and X > 1. This is only true when X>1.

Now let (X+4) < 0 and (X-1) < 0. Or, X < -4 and X < 1. This is only true when X<-4.

This implies that -4 < X < 1 but note that X > 0. So 0 < X < 1.

So the final solution is the union of the two, or -4 < X < 1.

I hope this helps. (more of a spoiler oops)

EDIT: Exclude 0 in the set of solutions. I forgot we're dividing by X lol.

[25thhour]

Re: Damn Math... 25k Credits For Help?

3 + X > 4/X

Let X < 0
Therefore, 3X + X^2 < 4
Or, X^2+3X - 4 < 0
Or, (X+4)(X-1) < 0
So we know that one (X-1) or (X+4) is negative and the other is positive
a) Suppose (X-1)>0, or X>1
This implies that (X+4)<0 but for all X>1, that's not true.

b) Suppose (X+4) > 0, or X> -4
This implies that (X-1) < 0, which is true for -4<X<1
But remember X < 0, so -4<X<0 for the first case.

Now let X>0
Therefore, 3X +X^2 >4
Or, (X+4)(X-1) > 0
Which implies that they both have to share the same sign.
Let (X+4) > 0 and (X-1) > 0. Or, X > -4 and X > 1. This is only true when X>1.

Now let (X+4) < 0 and (X-1) < 0. Or, X < -4 and X < 1. This is only true when X<-4.

This implies that -4 < X < 1 but note that X > 0. So 0 < X < 1.

So the final solution is the union of the two, or -4 < X < 1.

I hope this helps. (more of a spoiler oops)

EDIT: Exclude 0 in the set of solutions. I forgot we're dividing by X lol.

The back of the book gets
-4 < x < 0, X > 1.

:S

[Jtehanonymous]

Re: Damn Math... 25k Credits For Help?

You have to use discontinuities in your interval test as well.

So test, - infinity to -4, -4 to 0, 0 to 1, and 1 to infinity.

If you test those you get your answer.

[MarioNintendo]

Re: Damn Math... 25k Credits For Help?

The back of the book gets
-4 < x < 0, X > 1.

:S

That.. doesn't make sense!?

[iironiic]

Re: Damn Math... 25k Credits For Help?

Oh found my mistake rofl

Now let X>0
Therefore, 3X +X^2 >4
Or, (X+4)(X-1) > 0
Which implies that they both have to share the same sign.
Let (X+4) > 0 and (X-1) > 0. Or, X > -4 and X > 1. This is only true when X>1.

Now let (X+4) < 0 and (X-1) < 0. Or, X < -4 and X < 1. This is only true when X<-4.

This implies that X<-4 and X>1 but note that X > 0. So X > 1.

So the final solution is the union of the two, or -4 < X < 0 and X>1.

I hope this helps. (more of a spoiler oops)

EDIT: Exclude 0 in the set of solutions. I forgot we're dividing by X lol.

Rofl, stupid sign mistakes xD

EDIT:

You have to use discontinuities in your interval test as well.

So test, - infinity to -4, -4 to 0, 0 to 1, and 1 to infinity.

If you test those you get your answer.

Nice trollpost >=(

[leonid]

Re: Damn Math... 25k Credits For Help?

3 + X > 4/X

If X > 0,
X^2 + 3X - 4 > 0
=> (X+4)(X-1) > 0
=> X+4 and X-1 are both positive or both negative.
Since X>0, X+4 can't be negative, so they are both positive, which means X>1

If X<0,
X^2 + 3X - 4 < 0
=> (X+4)(X-1) < 0
=> X+4 and X-1 have different signs.
Since X<0, X-1 cannot be positive, so X+4 is positive and X-1 is negative
Therefore, X<1 and X>-4. Since X<0, -4<X<0

Thus, the final answer is -4<X<0 and X>1

[darkshark]

Re: Damn Math... 25k Credits For Help?

You know what **** you people that are good at math. Seriously I look at equations like they're written in Chinese or something. I hope I'm not alone in this haha.

[DossarLX ODI]

Re: Damn Math... 25k Credits For Help?

You know what **** you people that are good at math. Seriously I look at equations like they're written in Chinese or something. I hope I'm not alone in this haha.

Same for me.

[Rubin0]

Re: Damn Math... 25k Credits For Help?

I hate math but with all my GRE prep I actually get this question.

[leonid]

Re: Damn Math... 25k Credits For Help?

come on now are you calling this hard or something

[korny]

Re: Damn Math... 25k Credits For Help?

Last time I made a thread about getting help for math homework my thread was immediately deleted and I was given a warning. How this place has changed.

[DossarLX ODI]

Re: Damn Math... 25k Credits For Help?

The homework forum is dead now