I need to complete the square on the following equation, can anyone help?
$\displaystyle {3(x')}^{2 } $+$\displaystyle {2(y')}^{2 } $+$\displaystyle {(z')}^{2 } $-$\displaystyle \sqrt{2} $x'+$\displaystyle \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
$\displaystyle 3x^2+\sqrt{2}x+2y^2+z^2+\sqrt{2}z=0$
First we factor a 3 (the lead coefficient of the x squared term) out of only the x terms this gives
$\displaystyle 3\left(x^2+\frac{\sqrt{2}}{3}x \quad \right)+2y^2+z^2+\sqrt{2}z=0$
Now we take half of the coefficient of the x term and square it.
$\displaystyle \left( \frac{\sqrt{2}}{6}\right)^2=\frac{1}{18}$
Now we add this inside the parentheses on the left to get
$\displaystyle 3\left(x^2+\frac{\sqrt{2}}{3}x +\frac{1}{18} \right)+2y^2+z^2+\sqrt{2}z=0$
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
$\displaystyle 3\left(x^2+\frac{\sqrt{2}}{3}x +\frac{1}{18} \right)+2y^2+z^2+\sqrt{2}z=3\cdot \frac{1}{18}=\frac{1}{6}$
Now factor the x's to get
$\displaystyle 3\left(x+\frac{\sqrt{2}}{6}\right)^2+2y^2+z^2+\sqr t{2}z=3\cdot \frac{1}{18}=\frac{1}{6}$
Now you just need to do the z's the y is already a perfect square.
Thanks
$\displaystyle {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'+$\displaystyle \frac{\sqrt{2} }{6 } $, y" = y' and z" = z+$\displaystyle \frac{\sqrt{2} }{2 } $
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.
$\displaystyle {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