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

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\[\frac x{x^2 - 4} \ge 0\] ???
yes that is the problem
cross multiply..what do you get?

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Other answers:

= ?
where did you get =?
1x = 0
should be \(\ge\) not =
your initial sign was \(\ge\) so you can't change it
so my answer is correct ? or is there more to do ?
nothing more
or if you want to be meticulous like other "mathematicians" note that x cannot be 2 or -2
Uh... @lgbasallote , that answer isn't right.
you sure?
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
then why don't you show how it's done, oh great one?
It's \(x\in (-2, 0]\cup (2,\infty)\) Well, that's what I'm getting... let me check.
i did say x cannot be 2 or -2
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.
you did read what i was saying right?
Yes, but, it's supposed to be: \[ x>2\\ -2
I'd look at where x^3-4x is positive and negative...
simplest way I think...
i didn't say any signs did i? i said x cannot be 2 or -2
So how does one get to this conclusion? If you dont mind me asking.
and i was also insulting the mathematicians
multiply both sides by (x^2-4 )^2
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.
\[\frac x{x^2 - 4} \ge 0\] \[\frac x{(x-2)(x+2)} \ge 0\] 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 \[(-\infty,-2), (-2,0), (0,2), (2,\infty)\]
Here's a counter example to your solution: \[ x=1\ge0\\ \frac{1}{1-4}=-\frac{1}{3}\not \ge\; 0 \]
counter example to what?
you're a mathematician @LolWolf know that greater than and less than are under "inequalities" => not equal to and again i repeat...i was insulting the meticulousness of mathemeticians
(@lgbasallote 's not yours, @satellite73 , haha)
oh, ok
anyway..if you don't need the domain of this then plain \(x \ge 0\) is fine
if it's a computer you're trying to please
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 .
add/subtract the exponents on like bases eg. : a^(2/3) * a^(5/3) = a^(7/3)
for example on (a) would leave denominator 3 alone and just add the numnerators?
Hope you're not asking about adding fractions :P
My answer is A^1 * b^6/3 * c^5
would this be correct?

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