Question

This one is a doozy.. X/x^2-4 (greater than or equal to sign) 0. How would I solve ?

Answer 1

\[\frac x{x^2 - 4} \ge 0\]
???

Answer 2

yes
that is the problem

Answer 3

cross multiply..what do you get?

Answer 4

1x=0

Answer 5

?

Answer 6

= ?

Answer 7

where did you get =?

Answer 8

1x = 0

Answer 9

should be \(\ge\) not =

Answer 10

your initial sign was \(\ge\) so you can't change it

Answer 11

right

Answer 12

so my answer is correct ?
or is there more to do ?

Answer 13

nothing more

Answer 14

or if you want to be meticulous like other "mathematicians" note that x cannot be 2 or -2

Answer 15

Uh... @lgbasallote , that answer isn't right.

Answer 16

you sure?

Answer 17

yep:)

Answer 18

Oh ok, just making sure because I am supposed to solve this as an inequality
and show a sign chart as well expressing in interval notation

Answer 19

then why don't you show how it's done, oh great one?

Answer 20

It's \(x\in (-2, 0]\cup (2,\infty)\) Well, that's what I'm getting... let me check.

Answer 21

i did say x cannot be 2 or -2

Answer 22

According to Mathematica, that seems to be the case. And, I'm not trying to make you look bad or anything, jeez, it's a mistake, and it happens, don't take it personally. Check the boundaries yourself.

Answer 23

you did read what i was saying right?

Answer 24

Yes, but, it's supposed to be:
\[
x>2\\
-2<x\le 0
\]

Answer 25

I'd look at where x^3-4x is positive and negative...

Answer 26

simplest way I think...

Answer 27

i didn't say any signs did i? i said x cannot be 2 or -2

Answer 28

So how does one get to this conclusion? If you dont mind me asking.

Answer 29

and i was also insulting the mathematicians

Answer 30

multiply both sides by (x^2-4 )^2

Answer 31

I'm actually not sure what the 'standard' way is to look at these sorts of inequalities... I've forgotten maybe. Someone else might have a more kosher method.

Answer 32

\[\frac x{x^2 - 4} \ge 0\]
\[\frac x{(x-2)(x+2)} \ge 0\]
changes sign at \(-2,0,2\) so only need to check on interval. say \(x<-2\) i pick \(x=-3\) we get \[\frac{-3}{(-3)^2-4}=\frac{-3}{5}\] which is negative
so it is negative, positive, negative, positive on the respective intervals
\[(-\infty,-2), (-2,0), (0,2), (2,\infty)\]

Answer 33

Here's a counter example to your solution:
\[
x=1\ge0\\
\frac{1}{1-4}=-\frac{1}{3}\not \ge\; 0
\]

Answer 34

counter example to what?

Answer 35

you're a mathematician @LolWolf ...you know that greater than and less than are under "inequalities" => not equal to

and again i repeat...i was insulting the meticulousness of mathemeticians

Answer 36

(@lgbasallote 's not yours, @satellite73 , haha)

Answer 37

oh, ok

Answer 38

oh

Answer 39

anyway..if you don't need the domain of this then plain \(x \ge 0\) is fine

Answer 40

if it's a computer you're trying to please

Answer 41

http://www.wolframalpha.com/input/?i=plot++x%2F%28x%5E2-4%29++&dataset=&asynchronous=false&equal=Submit

Answer 42

Thank you. The other question I have goes like this : a^5/3 * b^2/3 * c^3 / C^-2 * a^2/3 * b^-4/3 .

Answer 43

add/subtract the exponents on like bases eg. : a^(2/3) * a^(5/3) = a^(7/3)

Answer 44

for example on (a) would leave denominator 3 alone and just add the numnerators?

Answer 45

Hope you're not asking about adding fractions :P

Answer 46

Nope!

Answer 47

My answer is A^1 * b^6/3 * c^5

Answer 48

would this be correct?