# Thread: 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.)

Yes, your substitutions are correct.

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

Page 1 of 2 12 Last