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Math Help - Noob having problems with standard form

  1. #1
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    Noob having problems with standard form

    Hi all



    The below equation is a quadric. What is it in standard form?

    5x^2 + 3y^2 + 3z^2 - 2xy + 2yz - 2xz - 10x + 6y - 2z - 9 = 0


    Ok so in my book I understand the majority of how to do it. I get to simplifying the 'completing the square part' and this leaves

    6(x'')^2 + 3(y'')^2 + 2(z'') - 18 = 0

    This next part is the bit I do not understand what is going on.

    The book says

    "the equation of the quadric in standard form is:

    (((x'')^2)/3) + (((y'')^2)/6) + (((z'')^2)/9) = 1

    this is the equation of an ellipsoid"

    Could some explain the rule how it goes from

    6(x'')^2 + 3(y'')^2 + 2(z'') - 18 = 0

    to get

    (((x'')^2)/3) + (((y'')^2)/6) + (((z'')^2)/9) = 1
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Noob having problems with standard form

    Hi kirstyaa9!

    6(x'')^2 + 3(y'')^2 + 2(z'') - 18 = 0

    Divide by 18:
    (6/18)(x'')^2 + (3/18)(y'')^2 + (2/18)(z'') - (18/18) = 0

    Rewrite:
    (((x'')^2)/3) + (((y'')^2)/6) + (((z'')^2)/9) - 1 = 0

    (((x'')^2)/3) + (((y'')^2)/6) + (((z'')^2)/9) = 1
    Thanks from kirstyaa9
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  3. #3
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    Re: Noob having problems with standard form

    Quote Originally Posted by ILikeSerena View Post
    Hi kirstyaa9!

    6(x'')^2 + 3(y'')^2 + 2(z'') - 18 = 0

    Divide by 18:
    (6/18)(x'')^2 + (3/18)(y'')^2 + (2/18)(z'') - (18/18) = 0

    Rewrite:
    (((x'')^2)/3) + (((y'')^2)/6) + (((z'')^2)/9) - 1 = 0

    (((x'')^2)/3) + (((y'')^2)/6) + (((z'')^2)/9) = 1

    Thanks, didn't think it was that simple .
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