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wolfgirl Group Title

Solve the equation. Express the solution as a radical in the simplest form.

  • 2 years ago
  • 2 years ago

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  1. jhonyy9 Group Title
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    where is ?

    • 2 years ago
  2. wolfgirl Group Title
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    |dw:1314895631260:dw|

    • 2 years ago
  3. jhonyy9 Group Title
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    16(x+3)=x2 x2 -16x -48=0

    • 2 years ago
  4. wolfgirl Group Title
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    do you mean x^2?

    • 2 years ago
  5. fiddlearound Group Title
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    square both sides.

    • 2 years ago
  6. jhonyy9 Group Title
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    yes

    • 2 years ago
  7. wolfgirl Group Title
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    oh ok

    • 2 years ago
  8. fiddlearound Group Title
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    Just make sure that whatever solutions you find are valid for the original question.

    • 2 years ago
  9. wolfgirl Group Title
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    ok

    • 2 years ago
  10. wolfgirl Group Title
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    (x-12)(x-4)

    • 2 years ago
  11. fiddlearound Group Title
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    how did you get that ?

    • 2 years ago
  12. wolfgirl Group Title
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    I have no idea, let me try it again

    • 2 years ago
  13. fiddlearound Group Title
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    It's : x^2-16x-48=0 not: x^2-16x+48=0

    • 2 years ago
  14. wolfgirl Group Title
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    ok

    • 2 years ago
  15. wolfgirl Group Title
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    I got tht

    • 2 years ago
  16. wolfgirl Group Title
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    did I get the answer right though?

    • 2 years ago
  17. fiddlearound Group Title
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    You got this right: x^2-16x+48 = (x-12)(x-4) But we need to solve this: x^2-16x-48=0 Probably easiest to just use the quadratic formula.

    • 2 years ago
  18. wolfgirl Group Title
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    ok i'll try that

    • 2 years ago
  19. wolfgirl Group Title
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    24,-40

    • 2 years ago
  20. fiddlearound Group Title
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    can you show how you got that ?

    • 2 years ago
  21. fiddlearound Group Title
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    \[ax^2+bx+c=0\] \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\] What values did you use for: a=? b=? c=?

    • 2 years ago
  22. wolfgirl Group Title
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    sorry my comp was dissconnected again

    • 2 years ago
  23. fiddlearound Group Title
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    ok - so are you ready to continue solving the above ?

    • 2 years ago
  24. wolfgirl Group Title
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    yeah, I got -4,-10

    • 2 years ago
  25. fiddlearound Group Title
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    what values did you use for a=? b=? c=? when using the quadratic formula ?

    • 2 years ago
  26. wolfgirl Group Title
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    a=1 b=-16 c=-48

    • 2 years ago
  27. fiddlearound Group Title
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    right. can you show your full calculation ?

    • 2 years ago
  28. fiddlearound Group Title
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    first, can you calculate this part ? b^2-4*a*c That's the part inside the square root.

    • 2 years ago
  29. wolfgirl Group Title
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    |dw:1314900092634:dw||dw:1314899842897:dw|

    • 2 years ago
  30. wolfgirl Group Title
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    the top drawing is supposed to be on bottom

    • 2 years ago
  31. fiddlearound Group Title
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    I'll start with this guy under the square root. He is a special guy and even has a name: The discriminant. \[\Delta=b^2-4ac\] \[\Delta=b^2-4ac=(-16)^2-4(1)(-48)=256+192=448\] so, \[x=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-(-16) \pm \sqrt{448}}{2(1)}=\frac{16 \pm \sqrt{7*8^2}}{2}\] \[=\frac{16 \pm 8\sqrt{7}}{2}=8 \pm 4\sqrt{7}\]

    • 2 years ago
  32. fiddlearound Group Title
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    let me know if the above is clear.

    • 2 years ago
  33. wolfgirl Group Title
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    i'm trying to understand it

    • 2 years ago
  34. wolfgirl Group Title
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    I don't get it

    • 2 years ago
  35. fiddlearound Group Title
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    can you tell me where it becomes unclear ?

    • 2 years ago
  36. fiddlearound Group Title
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    you basically did the same thing in the drawings you posted - just the negative/positive signs got messed up a bit.

    • 2 years ago
  37. wolfgirl Group Title
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    oh I get it, thanks

    • 2 years ago
  38. fiddlearound Group Title
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    we're not done yet though ... so if anything above is unclear, let me know before we go on. :)

    • 2 years ago
  39. wolfgirl Group Title
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    ok, explain the third line

    • 2 years ago
  40. fiddlearound Group Title
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    do you mean the part where the number 8 suddenly pops out from under the square root ?

    • 2 years ago
  41. wolfgirl Group Title
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    7*8^2

    • 2 years ago
  42. fiddlearound Group Title
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    oh - this ? 448 = 7*64 = 7*8*8 = 7*8^2

    • 2 years ago
  43. wolfgirl Group Title
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    yeah

    • 2 years ago
  44. fiddlearound Group Title
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    the reason I wrote it that way, is because the question wants us to: "Express the solution as a radical in the simplest form" Which means that we need to take everything we can out from under the square root. \[\sqrt{448}=\sqrt{7*64}=\sqrt{7*8^2}=\sqrt{7}\sqrt{8^2}=\sqrt{7}*8=8\sqrt{7}\] Is that cleat now ?

    • 2 years ago
  45. wolfgirl Group Title
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    yeah

    • 2 years ago
  46. fiddlearound Group Title
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    ok but we're still not done ...

    • 2 years ago
  47. fiddlearound Group Title
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    The quadratic equation gave us two solutions for x: \[x=8+4\sqrt{7}\] or \[x=8-4\sqrt{7}\] Right ?

    • 2 years ago
  48. wolfgirl Group Title
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    yeah

    • 2 years ago
  49. fiddlearound Group Title
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    but if you look at the original question, which was: \[4\sqrt{x+3}=x\] Notice the there is a square root there, and (x+3) is inside the square root. Right ?

    • 2 years ago
  50. wolfgirl Group Title
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    yeah

    • 2 years ago
  51. fiddlearound Group Title
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    Can a negative number have a real square root ?

    • 2 years ago
  52. wolfgirl Group Title
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    no

    • 2 years ago
  53. fiddlearound Group Title
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    so that means that since (x+3) is inside the square root, we cannot allow (x+3) to be negative - right ?

    • 2 years ago
  54. wolfgirl Group Title
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    right

    • 2 years ago
  55. fiddlearound Group Title
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    so we need to check the two possible values we got for x from the quadratic equation, and check that they don't make the number (x+3) a negative number. One of the numbers is easy to check, try checking this one first: \[x=8+4\sqrt{7}\] When you plug the above value into: \[(x+3)\] Do you get a negative value, or not ?

    • 2 years ago
  56. wolfgirl Group Title
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    positive

    • 2 years ago
  57. wolfgirl Group Title
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    is tht the answer?

    • 2 years ago
  58. fiddlearound Group Title
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    yes it is.

    • 2 years ago
  59. wolfgirl Group Title
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    ok thanks :)

    • 2 years ago
  60. Ishaan94 Group Title
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    \[16(x + 3) = x^2\] \[x^2 - 16x - 48= 0 \]

    • 2 years ago
  61. Ishaan94 Group Title
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    I did squaring ....

    • 2 years ago
  62. Ishaan94 Group Title
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    Now I will find Solutions

    • 2 years ago
  63. Ishaan94 Group Title
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    \[x = 4(2 +\sqrt7)\] \[x = 4(2 - \sqrt7)\] http://www.wolframalpha.com/input/?i=x^2+-+16x+-+48+%3D+0

    • 2 years ago
  64. Ishaan94 Group Title
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    Now when we put the solution in Original function

    • 2 years ago
  65. Ishaan94 Group Title
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    \[4(\sqrt{x+3}) = x \]I mean try putting \(4(2 -\sqrt7)\)

    • 2 years ago
  66. Ishaan94 Group Title
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    You will get - in root which is not possible that is why only one solution is possible

    • 2 years ago
  67. fiddlearound Group Title
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    yes, but plugging in that value into (x+3), it is not negative: (x+3) = (4*(2-sqrt(7)))+3 = 0.41...

    • 2 years ago
  68. Ishaan94 Group Title
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    Yeah You are right ...I did -3 in 8 my mistake..hmm wolfram is wrong

    • 2 years ago
  69. Ishaan94 Group Title
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    Let me check wolfram then

    • 2 years ago
  70. Ishaan94 Group Title
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    http://www.wolframalpha.com/input/?i=4%28sqrt%283%2B8+-+4sqrt7%29%29%3D8+-+4sqrt7

    • 2 years ago
  71. Ishaan94 Group Title
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    Check this out

    • 2 years ago
  72. fiddlearound Group Title
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    Checking: 8+4*sqrt(7) - Wolfram says true: http://www.wolframalpha.com/input/?i=4*sqrt%28%288%2B4*sqrt%287%29%29%2B3%29%3D%288%2B4*sqrt%287%29%29 Checking: 8-4*sqrt(7) - Wolfram says false: http://www.wolframalpha.com/input/?i=4*sqrt%28%288-4*sqrt%287%29%29%2B3%29%3D%288-4*sqrt%287%29%29

    • 2 years ago
  73. Ishaan94 Group Title
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    Yup

    • 2 years ago
  74. Ishaan94 Group Title
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    Something wrong with Wolfram I guess

    • 2 years ago
  75. fiddlearound Group Title
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    hhhmmmm... http://www.wolframalpha.com/input/?i=-4*sqrt%28%288-4*sqrt%287%29%29%2B3%29%3D%288-4*sqrt%287%29%29

    • 2 years ago
  76. fiddlearound Group Title
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    A result of squaring both sides ...

    • 2 years ago
  77. fiddlearound Group Title
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    wolfram is right.

    • 2 years ago
  78. Ishaan94 Group Title
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    http://www.wolframalpha.com/input/?i=sqrt%2811+-4sqrt7%29*4 yeah you got it squaring removes the minus sign

    • 2 years ago
  79. Ishaan94 Group Title
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    Lol I tried to prove that wolfram can be wrong time to time ...but it's always right..

    • 2 years ago
  80. fiddlearound Group Title
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    yeah, question is - does the teacher expect wolfgirl to take the two values from the quadratic equation and validate them in the original question.

    • 2 years ago
  81. fiddlearound Group Title
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    coz only one of them is valid.

    • 2 years ago
  82. Ishaan94 Group Title
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    well I don't think teacher would expect that ...hmm maybe they allow calculators

    • 2 years ago
  83. fiddlearound Group Title
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    ok -- we can ask wolfgirl what they usually do in class. Maybe she's related to wolfram.

    • 2 years ago
  84. Ishaan94 Group Title
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    Lol ... Good Job though ...really nice

    • 2 years ago
  85. wolfgirl Group Title
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    I'm not related to wolfram lol. But so far I am clueless as to what they do in class. I do Ashworth online school if that helps any

    • 2 years ago
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