Inequalities Help Fix Tex Error

[Trail_Builder]

stupid error in the "tex" thing, see first post.

 

[Trail_Builder]

damn latex keeps screwing up see next post, soz

 

[Trail_Builder]

just starting to learn some stuff on inequalities and need some help. Not sure where to put this topic, so move it if necessary. thnx

Heres one problem I think I've solved but need clarification from you it's right or where I've gone wrong. I also may have missed an obvious solution but I am going off a very short and "concise" introduction to inequalities so I havn't really got a full understanding of how to solve a given problem. Thnx

Find the set of real numbers x \neq 0 such that 2x + 1/x < 3:

I then manipulated 2x + 1/x < 3 to get to

\frac{(2x-1)(x-1)}{x} &lt; 0

Im pretty sure up till there is right, but it's the next stage that I think my working in dubious. I'm pretty sure I can't times both sides by x because it could be negative, could be positive.

I know \frac{(2x-1)(x-1)}{x} is negative. Because its less than O. This means that either 2x-1, x-1, or x is negative, or all 3. I kinda think all that makes sense.

(this is where is gets real dubious lol).

So, the following are the options.

2x-1>0, x-1<0, x>0
or
2x-1>0, x-1>0, x<0
or
2x-1<0, x-1<0, x<0
or
2x-1<0, x-1>0, x>0

which cancels too

x>1/2, x<1, x>0
or
x>1/2, x>1, x<0
or
x<1/2, x<1, x<0
or
x<1/2, x>1, x>0

the 2nd and 4th possiblilities are impossible so that leaves

1/2>x<1
or
x<0

the first doesn't work either because it says find real numbers and subbing in "1" give 3<3, which is false.

so then i conclude the answer has to be x \leq -1


is this right? i know i probably did it wrong, but need to know where I went wrong, thnx.

 

[Trail_Builder]

should be fine now, fixed the latex.

 

[uart]

Find the set of real numbers x \neq 0 such that 2x + 1/x &lt; 3:

I then manipulated 2x + 1/x < 3 to get to

\frac{(2x-1)(x-1)}{x} &lt; 0

Im pretty sure up till there is right, but it's the next stage that I think my working in dubious. I'm pretty sure I can't times both sides by x because it could be negative, could be positive.

The easiest thing to do is multiply both sides by x^2, which you know will be non-negative. This gets you to,

(2x-1)(x-1)(x) &lt; 0

This a cubic with three real roots at x=0, x=1/2 and x=1. Make a quick sketch of it if you like, it's easy to see that it is negative for x < 0 and for 1/2 < x < 1.

BTW: You don't really need to sketch it, that was just a suggestion in case you where not already familar with what a cubic looks like. If you are familar with a cubic you'll know that a positive cubic (x^3 term positive) will go from -ive to +ive through the first zero, from +ive to -ive through the second zero and so on. So it's not hard to figure out where it is negative.

 

[Trail_Builder]

o rite i see

was my answer correct?

 

[VietDao29]

...the 2nd and 4th possiblilities are impossible so that leaves

1/2>x<1
or
x<0

No, it's not 1/2 > x < 1, in fact, it should read:

1/2 < x < 1

the first doesn't work either because it says find real numbers and subbing in "1" give 3<3, which is false.

so then i conclude the answer has to be x \leq -1is this right? i know i probably did it wrong, but need to know where I went wrong, thnx.

No, you don't substitute "1" in. 1/2 < x < 1, x = 1 does not satisfy the requirement. The solution is 1/2 < x < 1, so if you want to check, well take any x that is greater than 1/2, and less than 1, you can choose .999, but definitely not 1. Can you see why? :)

Btw, how did you get x <= -1? You should note that x is a real number, not an integer.

There's a more common method, are you familiar with drawing tables, and testing the sign on each interval?

 

[Trail_Builder]

o crap haha, misread the question lol. stupid me.

yeah thnx for that, yeah i think the 1/2 > x < 1 was a typo.