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

- anonymous

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

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- schrodinger

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- lgbasallote

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

- anonymous

yes
that is the problem

- lgbasallote

cross multiply..what do you get?

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## More answers

- anonymous

1x=0

- anonymous

?

- lgbasallote

= ?

- lgbasallote

where did you get =?

- anonymous

1x = 0

- lgbasallote

should be \(\ge\) not =

- lgbasallote

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

- anonymous

right

- anonymous

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

- lgbasallote

nothing more

- lgbasallote

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

- anonymous

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

- lgbasallote

you sure?

- anonymous

yep:)

- 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

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

- anonymous

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

- lgbasallote

i did say x cannot be 2 or -2

- 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

you did read what i was saying right?

- anonymous

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

- anonymous

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

- anonymous

simplest way I think...

- lgbasallote

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

- anonymous

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

- lgbasallote

and i was also insulting the mathematicians

- anonymous

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

- 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

\[\frac x{x^2 - 4} \ge 0\]
\[\frac x{(x-2)(x+2)} \ge 0\]
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

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

- anonymous

counter example to what?

- 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

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

- anonymous

oh, ok

- lgbasallote

oh

- lgbasallote

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

- lgbasallote

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

- anonymous

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

- 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

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

- anonymous

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

- anonymous

Hope you're not asking about adding fractions :P

- anonymous

Nope!

- anonymous

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

- anonymous

would this be correct?

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