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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} 2x^2+\dfrac{1}{x}\gt0\hspace{.33em}\\~\\ \end{align}}\)

  2. sepeario
    • one year ago
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    x>0 is a solution.

  3. MrNood
    • one year ago
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    multiply both side od inequality by x subtract 1 from both sides divide both sides by 2 take cube root

  4. freckles
    • one year ago
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    @MrNood if we multiply both sides by x. We have two cases. x is positive and x is negative.

  5. freckles
    • one year ago
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    if we assume x is positive the inequality sign's direction holds if we assume x is negative we flip the inequality sign's direction

  6. freckles
    • one year ago
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    if you want to avoid that can you try multiplying x^2 on both sides instead

  7. mathmath333
    • one year ago
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    how to solve this \(x^3+\dfrac12>0\)

  8. freckles
    • one year ago
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    subtracting 1/2 on both sides then taking cube root of both sides

  9. mathmath333
    • one year ago
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    u mean this ? \(\large \color{black}{\begin{align} x>\left(-\dfrac12\right)^{1/3}\hspace{.33em}\\~\\ \end{align}}\) x can only have real values

  10. freckles
    • one year ago
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    that is right

  11. freckles
    • one year ago
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    and since ()^(1/3) is an odd function you can actually bring the negative outside if you prefer

  12. freckles
    • one year ago
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    \[x>- \frac{1}{2^\frac{1}{3}}\]

  13. mathmath333
    • one year ago
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    but wolfram gives it as complex number http://www.wolframalpha.com/input/?i=%5Cleft%28-%5Cdfrac12%5Cright%29%5E%7B1%2F3%7D%3D

  14. freckles
    • one year ago
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    yes there are complex cube roots of -1/2 but you want the real cube root of -1/2

  15. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=real+cube+root+of+%28-1%2F2%29

  16. mathmath333
    • one year ago
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    can this be done in case of square root too ?

  17. freckles
    • one year ago
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    what do you mean exactly ?

  18. mathmath333
    • one year ago
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    this one http://www.wolframalpha.com/input/?i=real+square+root+of+%28-1%2F2%29

  19. freckles
    • one year ago
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    short answer: no there is no real square root of -1/2 here is a long answer involving demorive' theorem \[\frac{-1}{2}=\frac{1}{2}(\cos(\pi+2\pi n)+i \sin(\pi+2 \pi n) ) \\ \frac{-1}{2}=\frac{1}{2}e^{(\pi+2n \pi)i} \\ \text{ taking \square \root of both sides } \\ \sqrt{\frac{-1}{2}}=(\frac{1}{2})^\frac{1}{2}e^{\frac{1}{2}(\pi+2n \pi)i} \text{ note: just raised both sides \to } \frac{1}{2} \\ \sqrt{\frac{-1}{2}}=(\frac{1}{2})^\frac{1}{2}e^{\frac{1}{2} (\pi i)} = (\frac{1}{2})^\frac{1}{2}(\cos(\frac{\pi}{2})+i \sin(\frac{\pi}{2})) =(\frac{1}{2})^\frac{1}{2}(0+i)=(\frac{1}{2})^\frac{1}{2} i \\ \sqrt{\frac{-1}{2}}=(\frac{1}{2})^\frac{1}{2}e^{\frac{1}{2}(\pi+2 \pi)}=(\frac{1}{2})^\frac{1}{2}(\cos(\frac{3\pi}{2})+i \sin(\frac{3 \pi}{2}))=(\frac{1}{2})^\frac{1}{2}(-1i)\] in other words square root of a negative number is only going to lead to imaginary numbers

  20. freckles
    • one year ago
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    though that was an answer involving trig I think I can also give you a good algebraic reasoning that has nothing to do with trig the sqrt(x) function looks like this: |dw:1436970600020:dw| this doesn't exist to the left of the y-axis so sqrt(-1) is not going to have a real answer but cuberoot(x) function looks like this: |dw:1436970632089:dw|

  21. freckles
    • one year ago
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    the cube function actually does exist to the left of the y-axis

  22. mathmath333
    • one year ago
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    u mean even roots dont have real roots

  23. freckles
    • one year ago
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    I think I said square roots

  24. freckles
    • one year ago
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    yes i did that is what I meant

  25. freckles
    • one year ago
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    square root of negative numbers aren't real

  26. mathmath333
    • one year ago
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    i mean if \(\Large a^{1/b}\) if b is even then there is no real root ?

  27. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=4th+roots+of+%28-1%2F2%29 though if you get into higher roots we can also use demorive' theorem to show this but I actually think you are only suppose to consider real function and real roots

  28. MrNood
    • one year ago
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    all even roots are roots of the square root - so if the squre root is not real neither is any even root

  29. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=6th+roots+of+%28-1%2F2%29 yes even roots of negative numbers aren't going to have real and the word even means also two which we call the square root

  30. freckles
    • one year ago
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    the example I showed though was of the square root of a negative number since you asked about sqrt(-1/2)

  31. mathmath333
    • one year ago
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    ok next qustion \(\large \color{black}{\begin{align} \dfrac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}>1\hspace{.33em}\\~\\ \end{align}}\)

  32. freckles
    • one year ago
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    I would probably end up subtracting 1 on both sides and combining the fractions but that seems really long

  33. freckles
    • one year ago
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    only because there is so much multiplication we have to do in the numerator

  34. freckles
    • one year ago
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    \[\frac{(x-1)(x-2)(x-3)-(x+1)(x+2)(x+3)}{(x+1)(x+2)(x+3)} >0\] like you have to multiply (x-1)(x-2)(x-3) and multiply (x+1)(x+2)(x+3) combine like terms on top there might be a little trick to making a short cut like if you already know the top isn't going to have any 0's but I don't see how to do that without actually multiplying it all out and combing like terms anyways the bottom is easy to see where we have vertical asymptotes since (x+1)(x+2)(x+3)=0 is easy to solve

  35. freckles
    • one year ago
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    I would draw a number line and test intervals around those vertical asymptotes and any 0's if they do exist

  36. mathmath333
    • one year ago
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    there will come a cubic polynomial in numerator , how will i factor that

  37. freckles
    • one year ago
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    you get a cubic poly in the numerator ?

  38. freckles
    • one year ago
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    I think the x^3 will cancel in the top

  39. mathmath333
    • one year ago
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    oh i see its not cubic

  40. freckles
    • one year ago
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    \[(x-1)(x-2)(x-3) \\ =(x^2-3x+2)(x-3) \\ =x^3-3x^2+2x-3x^2-9x-6 \\ =x^3-6x^2-7x-6 \\ (x+1)(x+2)(x+3) \\ =(x^2+3x+2)(x+3) \\ =x^3+3x^2+2x+3x^2+9x+6 \\ =x^3+6x^2+11x+6\] might check my arithmetic

  41. mathmath333
    • one year ago
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    ok its easy nnow

  42. freckles
    • one year ago
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    darn it

  43. freckles
    • one year ago
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    I didn't multiply -3x(-3) correctly

  44. ikram002p
    • one year ago
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    |dw:1436971504692:dw|

  45. freckles
    • one year ago
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    \[(x-1)(x-2)(x-3) \\ =(x^2-3x+2)(x-3) \\ =x^3-3x^2+2x-3x^2+9x-6 \\ =x^3-6x^2+11x-6 \\ (x+1)(x+2)(x+3) \\ =(x^2+3x+2)(x+3) \\ =x^3+3x^2+2x+3x^2+9x+6 \\ =x^3+6x^2+11x+6\]

  46. ikram002p
    • one year ago
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    |dw:1436971702003:dw|

  47. ikram002p
    • one year ago
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    if you are not struggling with my graph then my solution would be x>0 and \(x <\frac{ -1 }{ \sqrt[3]{2} }\)

  48. freckles
    • one year ago
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    yep yep on that one I liked multiplying both sides by x^2 and drawing a number line

  49. ikram002p
    • one year ago
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    by x not x^2*

  50. freckles
    • one year ago
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    \[2x^4+x>0 \\ x(2x^3+1)>0 \\ x(2x^3+1)=0 \text{ when } x=0 \text{ or } x=\frac{-1}{\sqrt[3]{2}} \approx -.79 \\ \] |dw:1436972007081:dw|

  51. freckles
    • one year ago
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    no by x^2

  52. freckles
    • one year ago
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    you can also choose to multiply both sides by x but then you have cases

  53. freckles
    • one year ago
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    we see that the function is only positive on (-inf,-1/cubeoot(2)) union (0,infty) just as you got

  54. ikram002p
    • one year ago
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    oh im referring to the origin didnt see other stuff :)

  55. freckles
    • one year ago
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    I didn't actually solve it I just suggested multiplying both sides by x^2 and that was all I like to avoid those cases sometimes

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