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

1. lgbasallote Group Title

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

2. Tombrokaw Group Title

yes that is the problem

3. lgbasallote Group Title

cross multiply..what do you get?

4. Tombrokaw Group Title

1x=0

5. Tombrokaw Group Title

?

6. lgbasallote Group Title

= ?

7. lgbasallote Group Title

where did you get =?

8. Tombrokaw Group Title

1x = 0

9. lgbasallote Group Title

should be $$\ge$$ not =

10. lgbasallote Group Title

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

11. Tombrokaw Group Title

right

12. Tombrokaw Group Title

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

13. lgbasallote Group Title

nothing more

14. lgbasallote Group Title

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

15. LolWolf Group Title

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

16. lgbasallote Group Title

you sure?

17. Algebraic! Group Title

yep:)

18. Tombrokaw Group Title

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

19. lgbasallote Group Title

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

20. LolWolf Group Title

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

21. lgbasallote Group Title

i did say x cannot be 2 or -2

22. LolWolf Group Title

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.

23. lgbasallote Group Title

you did read what i was saying right?

24. LolWolf Group Title

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

25. Algebraic! Group Title

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

26. Algebraic! Group Title

simplest way I think...

27. lgbasallote Group Title

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

28. Tombrokaw Group Title

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

29. lgbasallote Group Title

and i was also insulting the mathematicians

30. Algebraic! Group Title

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

31. Algebraic! Group Title

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.

32. satellite73 Group Title

$\frac x{x^2 - 4} \ge 0$ $\frac x{(x-2)(x+2)} \ge 0$ 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 $(-\infty,-2), (-2,0), (0,2), (2,\infty)$

33. LolWolf Group Title

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

34. satellite73 Group Title

counter example to what?

35. lgbasallote Group Title

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

36. LolWolf Group Title

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

37. satellite73 Group Title

oh, ok

38. lgbasallote Group Title

oh

39. lgbasallote Group Title

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

40. lgbasallote Group Title

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

41. Algebraic! Group Title
42. Tombrokaw Group Title

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 .

43. Algebraic! Group Title

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

44. Tombrokaw Group Title

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

45. Algebraic! Group Title

46. Tombrokaw Group Title

Nope!

47. Tombrokaw Group Title

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

48. Tombrokaw Group Title

would this be correct?