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Identify Quadric

  1. Jun 25, 2007 #1
    1. The problem statement, all variables and given/known data
    Identify the quadric:


    2. Relevant equations
    [tex]4x^2+y^2+4z^2-4y-24z+36=0[/tex]


    3. The attempt at a solution

    [tex]4x^2+(y-2)^2+4(z-3)^2=-36+9+4[/tex]
    Right?
    Then
    [tex]4x^2+(y-2)^2+4(z-3)^2=-23[/tex]
    But this is wrong. How?
     
  2. jcsd
  3. Jun 25, 2007 #2
    Check your algebra in step 3.

    More specifically, check the terms associated with Z.
     
  4. Jun 25, 2007 #3
    Mmm. I am not quite sure what you mean. If your talking about the 4 I just factored that out.
     
  5. Jun 25, 2007 #4

    VietDao29

    User Avatar
    Homework Helper

    Yes, that's what he means. Since you factor the 4 out, and you complete the square, like this:
    [tex]4z ^ 2 - 24z = 4 (z ^ 2 - 6z) = 4 (z ^ 2 - 6z) + 36 - 36 = 4 (z ^ 2 - 6z + 9) - 36 = 4 (z - 3) ^ 2 \textcolor{red}{- 36}[/tex]

    In fact, you should add and subtract 36, instead of 4, as you did. :)
     
  6. Jun 25, 2007 #5
    Shouldn't it be:

    [tex] 4x^2+y^2+4z^2-4y-24z+36=0[/tex]
    [tex] 4x^2+(y^2-4y+4)+(4z^2-24z+144)=-36+144+4[/tex]?
     
  7. Jun 25, 2007 #6
    Actually I think I see it:
    I have to complete the square first, then I can factor, then subtract.
     
  8. Jun 25, 2007 #7

    VietDao29

    User Avatar
    Homework Helper

    Nope, note that 4z2 = (2z)2

    So, it should be:

    [tex]4x^2+(y^2-4y+4)+((2z)^2- 2 \times (2z) \times 6 + 6 ^ 2)=-36 \textcolor{red}{+ 36} + 4 \Rightarrow ...[/tex]

    Can you take it from here? :)


    -----------------

    Edit:
    No, you can factor first, like what I've done in the previous post. That's okay, but you should be extremely careful when doing this.
     
    Last edited: Jun 25, 2007
  9. Jun 25, 2007 #8
    lol. I feel like a moron.
     
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