anonymous
  • anonymous
This one is a doozy.. X/x^2-4 (greater than or equal to sign) 0. How would I solve ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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lgbasallote
  • lgbasallote
\[\frac x{x^2 - 4} \ge 0\] ???
anonymous
  • anonymous
yes that is the problem
lgbasallote
  • lgbasallote
cross multiply..what do you get?

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More answers

anonymous
  • anonymous
1x=0
anonymous
  • anonymous
?
lgbasallote
  • lgbasallote
= ?
lgbasallote
  • lgbasallote
where did you get =?
anonymous
  • anonymous
1x = 0
lgbasallote
  • lgbasallote
should be \(\ge\) not =
lgbasallote
  • lgbasallote
your initial sign was \(\ge\) so you can't change it
anonymous
  • anonymous
right
anonymous
  • anonymous
so my answer is correct ? or is there more to do ?
lgbasallote
  • lgbasallote
nothing more
lgbasallote
  • lgbasallote
or if you want to be meticulous like other "mathematicians" note that x cannot be 2 or -2
anonymous
  • anonymous
Uh... @lgbasallote , that answer isn't right.
lgbasallote
  • lgbasallote
you sure?
anonymous
  • anonymous
yep:)
anonymous
  • anonymous
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
lgbasallote
  • lgbasallote
then why don't you show how it's done, oh great one?
anonymous
  • anonymous
It's \(x\in (-2, 0]\cup (2,\infty)\) Well, that's what I'm getting... let me check.
lgbasallote
  • lgbasallote
i did say x cannot be 2 or -2
anonymous
  • anonymous
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.
lgbasallote
  • lgbasallote
you did read what i was saying right?
anonymous
  • anonymous
Yes, but, it's supposed to be: \[ x>2\\ -2
anonymous
  • anonymous
I'd look at where x^3-4x is positive and negative...
anonymous
  • anonymous
simplest way I think...
lgbasallote
  • lgbasallote
i didn't say any signs did i? i said x cannot be 2 or -2
anonymous
  • anonymous
So how does one get to this conclusion? If you dont mind me asking.
lgbasallote
  • lgbasallote
and i was also insulting the mathematicians
anonymous
  • anonymous
multiply both sides by (x^2-4 )^2
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[\frac x{x^2 - 4} \ge 0\] \[\frac x{(x-2)(x+2)} \ge 0\] 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 \[(-\infty,-2), (-2,0), (0,2), (2,\infty)\]
anonymous
  • anonymous
Here's a counter example to your solution: \[ x=1\ge0\\ \frac{1}{1-4}=-\frac{1}{3}\not \ge\; 0 \]
anonymous
  • anonymous
counter example to what?
lgbasallote
  • lgbasallote
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
anonymous
  • anonymous
(@lgbasallote 's not yours, @satellite73 , haha)
anonymous
  • anonymous
oh, ok
lgbasallote
  • lgbasallote
oh
lgbasallote
  • lgbasallote
anyway..if you don't need the domain of this then plain \(x \ge 0\) is fine
lgbasallote
  • lgbasallote
if it's a computer you're trying to please
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=plot++x%2F%28x%5E2-4%29++&dataset=&asynchronous=false&equal=Submit
anonymous
  • anonymous
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 .
anonymous
  • anonymous
add/subtract the exponents on like bases eg. : a^(2/3) * a^(5/3) = a^(7/3)
anonymous
  • anonymous
for example on (a) would leave denominator 3 alone and just add the numnerators?
anonymous
  • anonymous
Hope you're not asking about adding fractions :P
anonymous
  • anonymous
Nope!
anonymous
  • anonymous
My answer is A^1 * b^6/3 * c^5
anonymous
  • anonymous
would this be correct?

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