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  1. anonymous
    • one year ago
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    \[2x^3+ 5x^2 -1 \le 0\]

  2. anonymous
    • one year ago
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    solve the inequality

  3. anonymous
    • one year ago
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    x<=-1-√2

  4. anonymous
    • one year ago
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    how?

  5. xapproachesinfinity
    • one year ago
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    what course are you taking! now of some factoring thoerems

  6. xapproachesinfinity
    • one year ago
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    i meant do you know of some factoring theorems or not

  7. xapproachesinfinity
    • one year ago
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    \(2x^3+5x^2-1=0\) has one solution we start by factoring this equation first so we can draw conclusion about the inequality

  8. xapproachesinfinity
    • one year ago
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    @Deeezzzz

  9. anonymous
    • one year ago
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    i found a factor but don't know how to continue.

  10. anonymous
    • one year ago
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    |dw:1436640267462:dw|

  11. anonymous
    • one year ago
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    The course I am currently taking is precalc @xapproachesinfinity

  12. xapproachesinfinity
    • one year ago
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    oh good then what does it mean for ab<0

  13. xapproachesinfinity
    • one year ago
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    one has to be negative right? either a<0, b>0 or a>0, b<0

  14. anonymous
    • one year ago
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    No idea??

  15. anonymous
    • one year ago
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    When ab multiply they must be less than zero?

  16. xapproachesinfinity
    • one year ago
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    if it was numbers case if we multiply two negative give positive multiplying two positive return a positive so the remaining case is one negative the other positive

  17. anonymous
    • one year ago
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    Yes that is true so one of the variables has to be a negative

  18. xapproachesinfinity
    • one year ago
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    now we ask which should be positive and which negative so (x+1/2)<0 and the other>0 or (x+1/2)>0 and parabola<0

  19. anonymous
    • one year ago
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    x<-1/2

  20. xapproachesinfinity
    • one year ago
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    let's simplify first we can write it as (2x+1)(x^2+2x-1)<=0

  21. xapproachesinfinity
    • one year ago
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    just hold on before that

  22. xapproachesinfinity
    • one year ago
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    can we look at that parabola first

  23. anonymous
    • one year ago
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    yes

  24. xapproachesinfinity
    • one year ago
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    x^2+2x-1 can this be <0?

  25. xapproachesinfinity
    • one year ago
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    if so when does it be <0 if not it must be >0 just the kind of thinking you need here okay

  26. xapproachesinfinity
    • one year ago
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    first that parabola opens up yes

  27. anonymous
    • one year ago
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    Yes it does with minimum (-1,-2)

  28. xapproachesinfinity
    • one year ago
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    here is a picture of that parabola http://prntscr.com/7rixyx

  29. xapproachesinfinity
    • one year ago
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    so it has part where it is <0 and as well as part where it is >0 good enough

  30. anonymous
    • one year ago
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    -1<x<0 F(x)<0

  31. xapproachesinfinity
    • one year ago
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    not really f(x)<0 when x1<x<x2 x1,x2 are root of x^2+2x-1

  32. xapproachesinfinity
    • one year ago
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    there are irrational roots

  33. xapproachesinfinity
    • one year ago
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    and f(x)>0 for x<x1 and x>x2

  34. xapproachesinfinity
    • one year ago
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    you got that part?

  35. anonymous
    • one year ago
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    Yes

  36. xapproachesinfinity
    • one year ago
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    let's go back with the first step (2x+1)(x^2+2x-1)<=0 implies 2x+1> 0 and x^2+2x-1<0 first case

  37. xapproachesinfinity
    • one year ago
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    we solve for that case by itself!

  38. xapproachesinfinity
    • one year ago
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    second case (2x+1)<0 and (x^2+2x-1)>0

  39. xapproachesinfinity
    • one year ago
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    we already decided when x^2+2x+1<=0 and when x^2+2x-1>=0 just a matter of solving 2x+1<0 2x+1>0 then combine

  40. xapproachesinfinity
    • one year ago
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    or means union

  41. xapproachesinfinity
    • one year ago
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    you got it now?

  42. xapproachesinfinity
    • one year ago
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    let see the first case 2x+1>0 and x^2+2x-1<0 x>-1/2 and x in [x1,x2] well you still need to decide what is x1 and x2 though

  43. anonymous
    • one year ago
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    so the final answer would be |dw:1436641853588:dw|

  44. xapproachesinfinity
    • one year ago
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    hm where did you [-1.0] i said x1 and x2 are not -1 and 0

  45. xapproachesinfinity
    • one year ago
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    and what -3 there? you have to explain your work

  46. xapproachesinfinity
    • one year ago
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    why*

  47. anonymous
    • one year ago
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    |dw:1436641991554:dw|

  48. xapproachesinfinity
    • one year ago
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    roots are x1=-1-root2 , x2=root2-1

  49. anonymous
    • one year ago
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    |dw:1436642079744:dw|

  50. xapproachesinfinity
    • one year ago
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    so first case is x>=-1/2 and -1-root2<=x<=root2-1

  51. xapproachesinfinity
    • one year ago
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    coming back i have a call

  52. anonymous
    • one year ago
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    yes that is correct so i just need the roots of the second quadratic

  53. anonymous
    • one year ago
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    |dw:1436642261054:dw|

  54. xapproachesinfinity
    • one year ago
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    intersection here not union

  55. xapproachesinfinity
    • one year ago
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    root2 not root6

  56. xapproachesinfinity
    • one year ago
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    union is the last step

  57. anonymous
    • one year ago
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    so how would the final answer look like?

  58. xapproachesinfinity
    • one year ago
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    i don;t know yet, i'm doing with you lol

  59. xapproachesinfinity
    • one year ago
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    so we have first case (-oo, -1/2] with [x1, x2] can you find that interval what is ?

  60. anonymous
    • one year ago
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    x1= -3.44 x2= 1.44

  61. xapproachesinfinity
    • one year ago
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    do number line

  62. anonymous
    • one year ago
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    (-oo, -3.44) U (-1/2, 1.44]

  63. xapproachesinfinity
    • one year ago
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    oh wait a minute error you would have (-1/2, oo) not -oo, -1/2 first case! look above

  64. xapproachesinfinity
    • one year ago
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    (-1/2,oo) intersection with [-3.4, 1.4]

  65. xapproachesinfinity
    • one year ago
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    so that would be [-1/2, 1.44]

  66. xapproachesinfinity
    • one year ago
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    that's what they share

  67. xapproachesinfinity
    • one year ago
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    now second case x<-1/2 and x<-3.44 and x>1.44

  68. xapproachesinfinity
    • one year ago
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    (-oo, -0.5) intersect (-oo,-3.44) intersect (1.44, oo)

  69. xapproachesinfinity
    • one year ago
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    you there?

  70. xapproachesinfinity
    • one year ago
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    i think we don't need [1.4, oo) there not intersection with that part

  71. xapproachesinfinity
    • one year ago
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    just (-oo, -0.5] and (-oo, -3.4] find the interval of intersection

  72. xapproachesinfinity
    • one year ago
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    you lost ? lol

  73. anonymous
    • one year ago
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    Just on whether its a union or a intersection

  74. anonymous
    • one year ago
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