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Math Help - Completing the square

  1. #1
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    Completing the square

    I need to complete the square on the following equation, can anyone help?

    {3(x')}^{2 } + {2(y')}^{2 } + {(z')}^{2 } - \sqrt{2}  x'+ \sqrt{2} z'=0
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  2. #2
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    Quote Originally Posted by Arron View Post
    I need to complete the square on the following equation, can anyone help?

    {3(x')}^{2 } + {2(y')}^{2 } + {(z')}^{2 } - \sqrt{2}  x'+ \sqrt{2} z'=0
    I will do the first third (the x part)

    1: gather all of the x,y and z stuff together this gives

    3x^2+\sqrt{2}x+2y^2+z^2+\sqrt{2}z=0

    First we factor a 3 (the lead coefficient of the x squared term) out of only the x terms this gives

    3\left(x^2+\frac{\sqrt{2}}{3}x \quad \right)+2y^2+z^2+\sqrt{2}z=0

    Now we take half of the coefficient of the x term and square it.

    \left( \frac{\sqrt{2}}{6}\right)^2=\frac{1}{18}

    Now we add this inside the parentheses on the left to get

    3\left(x^2+\frac{\sqrt{2}}{3}x +\frac{1}{18} \right)+2y^2+z^2+\sqrt{2}z=0

    but we also have to add this to the other side so we don't change the equation. Also don't forget to multiply it by the 3 out side the parenthesis to get

    3\left(x^2+\frac{\sqrt{2}}{3}x +\frac{1}{18} \right)+2y^2+z^2+\sqrt{2}z=3\cdot \frac{1}{18}=\frac{1}{6}

    Now factor the x's to get

    3\left(x+\frac{\sqrt{2}}{6}\right)^2+2y^2+z^2+\sqr  t{2}z=3\cdot \frac{1}{18}=\frac{1}{6}

    Now you just need to do the z's the y is already a perfect square.
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  3. #3
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    Thanks, I have the complete answer as follows

    3{(x+\frac{\sqrt{2} }{ 6}) }^{2 }+{y}^{2 } +{(z+\frac{1}{2 }) }^{ 2}  = \frac{2}{3 }

    Is this correct?
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  4. #4
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    Quote Originally Posted by Arron View Post
    Thanks, I have the complete answer as follows

    3{(x+\frac{\sqrt{2} }{ 6}) }^{2 }+{y}^{2 } +{(z+\frac{1}{2 }) }^{ 2}  = \frac{2}{3 }

    Is this correct?
    No if you multiply out the z term you get

    (z+\frac{1}{2})^2=z^2+z+\frac{1}{4}

    The term inside the square is always half of the coefficient on the linear term is should be

    \frac{\sqrt{2}}{2}
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    Thanks,

    So the term outside will be \frac{1}{4 } +\frac{1}{6 }=\frac{5}{12 }
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    no it should be

    \frac{1}{4}
    Last edited by TheEmptySet; May 14th 2011 at 10:40 AM. Reason: I was wrong.
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  7. #7
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    Thanks

    {3(x'+\frac{\sqrt{2} }{6 }) }^{2 }+2{y'}^{2 }+{(z'+\frac{\sqrt{2} }{2 }) }^{2 }-\frac{5}{12 } = 0

    I need to simplify this equation and substitute
    x" = x'+ \frac{\sqrt{2} }{6 } , y" = y' and z" = z+ \frac{\sqrt{2} }{2 }
    Last edited by Arron; May 14th 2011 at 11:21 AM.
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  8. #8
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    Quote Originally Posted by Arron View Post
    Thanks

    {3(x'+\frac{\sqrt{2} }{6 }) }^{2 }+2{y'}^{2 }+{(z'+\frac{1}{2 }) }^{2 }-\frac{5}{12 } = 0

    I need to simplify this equation and substitute
    x" = x'+ \frac{\sqrt{2} }{6 } , y" = y' and z" = z+ \frac{1}{2 }
    as I said above the z term is still incorrect is should be

    (z+\frac{\sqrt{2}}{2})^2=z^2+\sqrt{2}z+\frac{1}{4}

    Note: I was wrong in post number #6 and have corrected the post.
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  9. #9
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    So am I correct with the following?

    {3(x+\frac{\sqrt{2} }{6 } )}^{2 } + {2y}^{2 } +{(z+\frac{\sqrt{2} }{2 }) }^{2 }=\frac{1}{6 }+\frac{1}{4 }=\frac{5}{12 }
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  10. #10
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    Yes you can ( and should ) multiply it out to verify it is correct.
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  11. #11
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    Quote Originally Posted by Arron View Post
    Thanks

    {3(x'+\frac{\sqrt{2} }{6 }) }^{2 }+2{y'}^{2 }+{(z'+\frac{\sqrt{2} }{2 }) }^{2 }-\frac{5}{12 } = 0

    I need to simplify this equation and substitute
    x" = x'+ \frac{\sqrt{2} }{6 } , y" = y' and z" = z+ \frac{1}{2 }
    Quote Originally Posted by Arron View Post
    So am I correct with the following?

    {3(x+\frac{\sqrt{2} }{6 } )}^{2 } + {2y}^{2 } +{(z+\frac{\sqrt{2} }{2 }) }^{2 }=\frac{1}{6 }+\frac{1}{4 }=\frac{5}{12 }
    With the exception that you need to correct the z term in your original question, then yes your question about post #7 is correct.

    -Dan
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    With regard post 7, am I right with the following, we obtain

    {3(x")}^{2 }+{2(y")}^{2 }+{(z")}^{2 }-\frac{5}{12 }=0

    Finally I need to put this quartic in standard form, and decide what type of non-denegenerate quartic it represents.
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  13. #13
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    Quote Originally Posted by Arron View Post
    Thanks

    {3(x'+\frac{\sqrt{2} }{6 }) }^{2 }+2{y'}^{2 }+{(z'+\frac{\sqrt{2} }{2 }) }^{2 }-\frac{5}{12 } = 0

    I need to simplify this equation and substitute
    x" = x'+ \frac{\sqrt{2} }{6 } , y" = y' and z" = z+ \frac{\sqrt{2} }{2 }
    Quote Originally Posted by Arron View Post
    With regard post 7, am I right with the following, we obtain

    {3(x")}^{2 }+{2(y")}^{2 }+{(z")}^{2 }-\frac{5}{12 }=0
    As long as you are using the correct expression for z'' then yes. Your post 7 definition of z'' does not have the correct expression. (Note that your discussion with TheEmptySet has changed what z'' should be.)

    -Dan

    Edit: Okay, you edited post 7. I didn't see that. (Please don't do that or mention that you have done so, else it causes confusion, like in this instance.)

    Yes, your substitutions are correct.
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  14. #14
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    Sorry and thank you.

    I just need to put this quartic in standard form, and decide what type of non-denegenerate quartic it represents.

    Can anyone help?
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  15. #15
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    Quote Originally Posted by Arron View Post
    Sorry and thank you.

    I just need to put this quartic in standard form, and decide what type of non-denegenerate quartic it represents.

    Can anyone help?
    {3(x")}^{2 }+{2(y")}^{2 }+{(z")}^{2 }-\frac{5}{12 }=0

    You have + signs between all the terms. That means that you either have an ellipsoid or a sphere. To have a sphere the equation must have the coefficients of all terms be equal. This is not the case. So your equation is for an ellipsoid.

    -Dan
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