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Tombrokaw

  • 2 years ago

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

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  1. lgbasallote
    • 2 years ago
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    \[\frac x{x^2 - 4} \ge 0\] ???

  2. Tombrokaw
    • 2 years ago
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    yes that is the problem

  3. lgbasallote
    • 2 years ago
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    cross multiply..what do you get?

  4. Tombrokaw
    • 2 years ago
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    1x=0

  5. Tombrokaw
    • 2 years ago
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    ?

  6. lgbasallote
    • 2 years ago
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    = ?

  7. lgbasallote
    • 2 years ago
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    where did you get =?

  8. Tombrokaw
    • 2 years ago
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    1x = 0

  9. lgbasallote
    • 2 years ago
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    should be \(\ge\) not =

  10. lgbasallote
    • 2 years ago
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    your initial sign was \(\ge\) so you can't change it

  11. Tombrokaw
    • 2 years ago
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    right

  12. Tombrokaw
    • 2 years ago
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    so my answer is correct ? or is there more to do ?

  13. lgbasallote
    • 2 years ago
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    nothing more

  14. lgbasallote
    • 2 years ago
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    or if you want to be meticulous like other "mathematicians" note that x cannot be 2 or -2

  15. LolWolf
    • 2 years ago
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    Uh... @lgbasallote , that answer isn't right.

  16. lgbasallote
    • 2 years ago
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    you sure?

  17. Algebraic!
    • 2 years ago
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    yep:)

  18. Tombrokaw
    • 2 years ago
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    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
    • 2 years ago
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    then why don't you show how it's done, oh great one?

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

  21. lgbasallote
    • 2 years ago
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    i did say x cannot be 2 or -2

  22. LolWolf
    • 2 years ago
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    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
    • 2 years ago
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    you did read what i was saying right?

  24. LolWolf
    • 2 years ago
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    Yes, but, it's supposed to be: \[ x>2\\ -2<x\le 0 \]

  25. Algebraic!
    • 2 years ago
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    I'd look at where x^3-4x is positive and negative...

  26. Algebraic!
    • 2 years ago
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    simplest way I think...

  27. lgbasallote
    • 2 years ago
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    i didn't say any signs did i? i said x cannot be 2 or -2

  28. Tombrokaw
    • 2 years ago
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    So how does one get to this conclusion? If you dont mind me asking.

  29. lgbasallote
    • 2 years ago
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    and i was also insulting the mathematicians

  30. Algebraic!
    • 2 years ago
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    multiply both sides by (x^2-4 )^2

  31. Algebraic!
    • 2 years ago
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    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
    • 2 years ago
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    \[\frac x{x^2 - 4} \ge 0\] \[\frac x{(x-2)(x+2)} \ge 0\] 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
    • 2 years ago
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    Here's a counter example to your solution: \[ x=1\ge0\\ \frac{1}{1-4}=-\frac{1}{3}\not \ge\; 0 \]

  34. satellite73
    • 2 years ago
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    counter example to what?

  35. lgbasallote
    • 2 years ago
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    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
    • 2 years ago
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    (@lgbasallote 's not yours, @satellite73 , haha)

  37. satellite73
    • 2 years ago
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    oh, ok

  38. lgbasallote
    • 2 years ago
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    oh

  39. lgbasallote
    • 2 years ago
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    anyway..if you don't need the domain of this then plain \(x \ge 0\) is fine

  40. lgbasallote
    • 2 years ago
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    if it's a computer you're trying to please

  41. Tombrokaw
    • 2 years ago
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    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 .

  42. Algebraic!
    • 2 years ago
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    add/subtract the exponents on like bases eg. : a^(2/3) * a^(5/3) = a^(7/3)

  43. Tombrokaw
    • 2 years ago
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    for example on (a) would leave denominator 3 alone and just add the numnerators?

  44. Algebraic!
    • 2 years ago
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    Hope you're not asking about adding fractions :P

  45. Tombrokaw
    • 2 years ago
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    Nope!

  46. Tombrokaw
    • 2 years ago
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    My answer is A^1 * b^6/3 * c^5

  47. Tombrokaw
    • 2 years ago
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    would this be correct?

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