1. Jun 25, 2007

### Winzer

1. The problem statement, all variables and given/known data

2. Relevant equations
$$4x^2+y^2+4z^2-4y-24z+36=0$$

3. The attempt at a solution

$$4x^2+(y-2)^2+4(z-3)^2=-36+9+4$$
Right?
Then
$$4x^2+(y-2)^2+4(z-3)^2=-23$$
But this is wrong. How?

2. Jun 25, 2007

### CaptainZappo

Check your algebra in step 3.

More specifically, check the terms associated with Z.

3. Jun 25, 2007

### Winzer

Mmm. I am not quite sure what you mean. If your talking about the 4 I just factored that out.

4. Jun 25, 2007

### VietDao29

Yes, that's what he means. Since you factor the 4 out, and you complete the square, like this:
$$4z ^ 2 - 24z = 4 (z ^ 2 - 6z) = 4 (z ^ 2 - 6z) + 36 - 36 = 4 (z ^ 2 - 6z + 9) - 36 = 4 (z - 3) ^ 2 \textcolor{red}{- 36}$$

In fact, you should add and subtract 36, instead of 4, as you did. :)

5. Jun 25, 2007

### Winzer

Shouldn't it be:

$$4x^2+y^2+4z^2-4y-24z+36=0$$
$$4x^2+(y^2-4y+4)+(4z^2-24z+144)=-36+144+4$$?

6. Jun 25, 2007

### Winzer

Actually I think I see it:
I have to complete the square first, then I can factor, then subtract.

7. Jun 25, 2007

### VietDao29

Nope, note that 4z2 = (2z)2

So, it should be:

$$4x^2+(y^2-4y+4)+((2z)^2- 2 \times (2z) \times 6 + 6 ^ 2)=-36 \textcolor{red}{+ 36} + 4 \Rightarrow ...$$

Can you take it from here? :)

-----------------

Edit:
No, you can factor first, like what I've done in the previous post. That's okay, but you should be extremely careful when doing this.

Last edited: Jun 25, 2007
8. Jun 25, 2007

### Winzer

lol. I feel like a moron.