## anonymous 3 years ago This one is a doozy.. X/x^2-4 (greater than or equal to sign) 0. How would I solve ?

1. anonymous

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

2. anonymous

yes that is the problem

3. anonymous

cross multiply..what do you get?

4. anonymous

1x=0

5. anonymous

?

6. anonymous

= ?

7. anonymous

where did you get =?

8. anonymous

1x = 0

9. anonymous

should be $$\ge$$ not =

10. anonymous

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

11. anonymous

right

12. anonymous

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

13. anonymous

nothing more

14. anonymous

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

15. anonymous

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

16. anonymous

you sure?

17. anonymous

yep:)

18. 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

19. anonymous

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

20. anonymous

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

21. anonymous

i did say x cannot be 2 or -2

22. 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.

23. anonymous

you did read what i was saying right?

24. anonymous

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

25. anonymous

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

26. anonymous

simplest way I think...

27. anonymous

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

28. anonymous

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

29. anonymous

and i was also insulting the mathematicians

30. anonymous

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

31. 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.

32. 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 $(-\infty,-2), (-2,0), (0,2), (2,\infty)$

33. anonymous

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

34. anonymous

counter example to what?

35. anonymous

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. anonymous

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

37. anonymous

oh, ok

38. anonymous

oh

39. anonymous

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

40. anonymous

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

41. anonymous
42. 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 .

43. anonymous

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

44. anonymous

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

45. anonymous

46. anonymous

Nope!

47. anonymous

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

48. anonymous

would this be correct?