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Destinyyyy

  • one year ago

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  1. Destinyyyy
    • one year ago
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    Solve by completing the square. x^2+11x-9=0

  2. Destinyyyy
    • one year ago
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    @Nnesha

  3. Michele_Laino
    • one year ago
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    we can write this: \[\Large {x^2} + 11x - 9 = {x^2} + 11x + \frac{{121}}{4} - \frac{{121}}{4} - 9\]

  4. Destinyyyy
    • one year ago
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    Um where did you get 121 and 4 from???

  5. Michele_Laino
    • one year ago
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    since we have this: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} = {x^2} + 11x + \frac{{121}}{4}\]

  6. Destinyyyy
    • one year ago
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    Um okay.. Can you start at the beginning on how to solve this?

  7. Michele_Laino
    • one year ago
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    ok!

  8. Michele_Laino
    • one year ago
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    I rewrite the left side of your equation, adding and subtracting the same quantity, namely 121/4, so I get this equation: \[\Large {x^2} + 11x + \frac{{121}}{4} - \frac{{121}}{4} - 9 = 0\]

  9. Michele_Laino
    • one year ago
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    then, I use this identity: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} = {x^2} + 11x + \frac{{121}}{4}\]

  10. Michele_Laino
    • one year ago
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    so I can rewrite your starting equation, as below: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{121}}{4} - 9 = 0\]

  11. Michele_Laino
    • one year ago
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    next, I nothe that: \[ \Large - \frac{{121}}{4} - 9 = - \frac{{157}}{4}\]

  12. Michele_Laino
    • one year ago
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    after a substitution, I get this: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{157}}{4} = 0\]

  13. Michele_Laino
    • one year ago
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    or: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} = \frac{{157}}{4}\]

  14. Destinyyyy
    • one year ago
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    Uh? Im here- x^2+11x-9=0 (x+11/2)^2 -9=0 x^2+ 121/4 -9=0 I dont understand past that

  15. Destinyyyy
    • one year ago
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    Ive been using this method or factoring https://www.youtube.com/watch?v=bclm1tJB-3g

  16. Michele_Laino
    • one year ago
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    In order to get an equivalent equation, I have to add and contemporarily I have to subtract 121/4, namely the same quantity

  17. Destinyyyy
    • one year ago
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    Um okay

  18. Michele_Laino
    • one year ago
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    so I get this equation: \[\Large {x^2} + 11x + \frac{{121}}{4} - \frac{{121}}{4} - 9 = 0\]

  19. Michele_Laino
    • one year ago
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    the first three terms at the left side are the square of the subsequent binomial: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2}\]

  20. Destinyyyy
    • one year ago
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    Okay..

  21. Michele_Laino
    • one year ago
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    so, substituting, I can write this: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{121}}{4} - 9 = 0\]

  22. Destinyyyy
    • one year ago
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    Okay

  23. Michele_Laino
    • one year ago
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    next I do this computation: \[\Large - \frac{{121}}{4} - 9 = - \frac{{157}}{4}\]

  24. Michele_Laino
    • one year ago
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    and, again I substitute: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{157}}{4} = 0\]

  25. Destinyyyy
    • one year ago
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    What? I seriously do not understand. How in the world do you get 157? I understand that (x+ 11/2)^2 equals x^2 +121/4 .. I assume the 11 over 2 is 11x divided by 2 which cant happen so it stays as a fraction. You said something about subtracting 121/4 and I assumed you meant to the other side of the equal sign but wasn't entirely sure. Im really trying to follow what your saying and I'm good in math but all this makes absolutely no sense so far.

  26. Destinyyyy
    • one year ago
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    Im still where I was earlier.

  27. Michele_Laino
    • one year ago
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    here is more details of my computation: |dw:1440704847919:dw|

  28. Michele_Laino
    • one year ago
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    here are*

  29. Destinyyyy
    • one year ago
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    Where are you getting the negative in front of 121/4 from?

  30. Michele_Laino
    • one year ago
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    since I have used +121/4 in order to have the square of binomial, then the remaining -121/4 has to be summed to -9

  31. Destinyyyy
    • one year ago
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    "then the remaining -121/4 has to be summed to -9" ??

  32. Michele_Laino
    • one year ago
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    yes! correct!

  33. Nnesha
    • one year ago
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    according to the video u watched 1st) move the constant term to the right side 2nd) divide `b` by 2 and then take square of the result in that case you you would `add` \[\rm (\frac{ b }{ 2})^2\]to the right side but if you keep the constant term at left side then you should subtract

  34. Destinyyyy
    • one year ago
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    @Michele_Laino I put question marks at the end saying what to it... @Nnesha yeah I tried that and 11 divided by 2 is 5.5 so I stopped

  35. Nnesha
    • one year ago
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    http://prntscr.com/89mc1g http://prntscr.com/89mbs3 so you need to be careful about this

  36. Nnesha
    • one year ago
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    that's right 5.5 which is same as 11/2 u got the decimal but he/she is keptt the fraction form but same thing! :=)

  37. Destinyyyy
    • one year ago
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    Yes I understood that part. After that I get lost

  38. Nnesha
    • one year ago
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    so do you want to move the constant to the right side or do you want to keep it at left side ?

  39. Destinyyyy
    • one year ago
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    The video says to move it to the right.. But I honestly dont know.

  40. Nnesha
    • one year ago
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    you will get the same answer

  41. Michele_Laino
    • one year ago
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    we have to add to both sides +157/4, so we get this: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{157}}{4} + \frac{{157}}{4} = \frac{{157}}{4}\]

  42. Destinyyyy
    • one year ago
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    I understand the adding to both sides but I dont get where you got 157 from

  43. Nnesha
    • one year ago
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    ye so doesn't matter you can move the constant term to right side at the beginning or after completing the square of (x^2+11x)

  44. Nnesha
    • one year ago
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    when you add -9 and -121/4 you will get that -157/4

  45. Nnesha
    • one year ago
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    \[\huge\rm x^2+11x=9\] divide b by 2 and then take square of that result

  46. Destinyyyy
    • one year ago
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    Do you mean (x^2+121/4) or 11/2= 5.5^2= 30.25

  47. Destinyyyy
    • one year ago
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    Can you explain this more " when you add -9 and -121/4 you will get that -157/4 " because im not getting -157... -9+-121= -130 .. I dont see where the plus sign came from.

  48. Nnesha
    • one year ago
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    yes right \[\huge\rm (x+5.5)^2=9+(5.5)^2\] okay i will

  49. Nnesha
    • one year ago
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    |dw:1440706398196:dw| there is one under 9 so 4 is the common denominator right ?

  50. Destinyyyy
    • one year ago
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    Okay now I understand where the -157 come from.

  51. Destinyyyy
    • one year ago
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    5.5*5.5= 30.25 +9= 39.25

  52. Nnesha
    • one year ago
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    yes right common denominator multiply the numerator of first fraction by the denominator of 2nd fraction multiply the numerator of 2nd fraction by the denominator of first fraction |dw:1440706515919:dw|

  53. Destinyyyy
    • one year ago
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    Can you write it down how to solve this on paper and take a picture and post it on here? Maybe I can actually understand then

  54. Nnesha
    • one year ago
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    yes right 39.25 is same as 157/4

  55. Nnesha
    • one year ago
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    same thing basically

  56. Nnesha
    • one year ago
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    and after that i'm pretty sure you know how to solve for x right :D

  57. Destinyyyy
    • one year ago
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    Yeah if I knew what the equation was.. All I have written down in front of me is x^2+11x+ ____=-9+____ x^2+11x-9=0 (x+11/2)^2 -9=0 x^2+ 121/4 -9=0

  58. Nnesha
    • one year ago
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    uh-oh why it's negative 9 at right sdie ?

  59. Nnesha
    • one year ago
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    side*

  60. Destinyyyy
    • one year ago
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    Its suppose to be a positive mistype

  61. Nnesha
    • one year ago
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    yes right

  62. Destinyyyy
    • one year ago
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    I have x^2- 157/4 =0

  63. Nnesha
    • one year ago
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    it's not x^2

  64. Nnesha
    • one year ago
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    cuz here |dw:1440706885310:dw| like this

  65. Nnesha
    • one year ago
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    or you can skip that factor step shortcut!|dw:1440707138935:dw|

  66. Nnesha
    • one year ago
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    divide `b` by 2 write that in the parentheses square would be to (x+b/2)^2 but you would add square of (b/2)^2 to the right side

  67. Nnesha
    • one year ago
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    that's another of saying x^2+!1x+ 30.25 = (x+5.5)^2 x+5.5 is a factor of x^2+!1x+30.25

  68. Destinyyyy
    • one year ago
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    Yeah

  69. Nnesha
    • one year ago
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    ask question if you didn't get what i mean ......

  70. Destinyyyy
    • one year ago
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    I understand that

  71. Nnesha
    • one year ago
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    this is good example here \[\huge\rm x^2+bx+C=0\]\[\huge\rm x^\color{Red}{2}+Bx=-C\]\[\huge\rm (x+\frac{ b }{ 2})^\color{red}{2}=-C +(\frac{ b }{ 2 })^2\]

  72. Destinyyyy
    • one year ago
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    Okay.. I have (x+5.5)^2=+- 39.25 square root x+5.5= +- square root 39.25 x+5.5=+-

  73. Nnesha
    • one year ago
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    yes right \[x+\frac{ 11 }{ 2 } = \pm \sqrt{ \frac{ {157} }{ 4 }}\] or you can keep it in fraction form

  74. Nnesha
    • one year ago
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    add you can see 4 is perfect square root so can take square of 4

  75. Destinyyyy
    • one year ago
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    Okay I get it.. I think

  76. Nnesha
    • one year ago
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    if keep the decimal we will not be able to know if there is any perfect square root or not

  77. Destinyyyy
    • one year ago
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    The final answer is x= 11/2 +- square root 157/ 4

  78. Nnesha
    • one year ago
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    okay practice 2 or 3 time you will definitely understand

  79. Nnesha
    • one year ago
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    hmmmm

  80. Destinyyyy
    • one year ago
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    I hope so

  81. Destinyyyy
    • one year ago
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    Opp negative in front of the 11

  82. Nnesha
    • one year ago
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    \[x+\frac{ 11 }{ 2 } = \pm \sqrt{ \frac{ {157} }{ 4 }}\] \[x+\frac{ 11 }{ 2 } =\frac{ \sqrt{157} }{ \sqrt{4}}\] square root of 4 is 2 right

  83. Nnesha
    • one year ago
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    \[\huge\rm x=\frac{ -11 }{ 2 } \pm \frac{ \sqrt{157} }{ 2}\] so same denominator

  84. Destinyyyy
    • one year ago
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    Ohh yeah

  85. Nnesha
    • one year ago
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    yes good luck!

  86. Destinyyyy
    • one year ago
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    Thanks

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