# Math Help - Completing the square

1. ## 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

2. Originally Posted by Arron
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.

3. 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?

4. Originally Posted by Arron
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

$\frac{\sqrt{2}}{2}$

5. Thanks,

So the term outside will be $\frac{1}{4 } +\frac{1}{6 }=\frac{5}{12 }$

6. no it should be

$\frac{1}{4}$

7. 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+ $\frac{\sqrt{2} }{2 }$

8. Originally Posted by Arron
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.

9. So am I correct with the following?

${3(x+\frac{\sqrt{2} }{6 } )}^{2 } + {2y}^{2 } +{(z+\frac{\sqrt{2} }{2 }) }^{2 }=\frac{1}{6 }+\frac{1}{4 }=\frac{5}{12 }$

10. Yes you can ( and should ) multiply it out to verify it is correct.

11. Originally Posted by Arron
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+ $\frac{1}{2 }$
Originally Posted by Arron
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

12. 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.

13. Originally Posted by Arron
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+ $\frac{\sqrt{2} }{2 }$
Originally Posted by Arron
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.)

14. 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?

15. Originally Posted by Arron
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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