Solve Quadric Equation: 4x^2+y^2+4z^2-4y-24z+36=0

[Winzer]

Homework Statement

Identify the quadric:

Homework Equations

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

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?

 

[CaptainZappo]

Check your algebra in step 3.

More specifically, check the terms associated with Z.

 

[Winzer]

Check your algebra in step 3.

More specifically, check the terms associated with Z.

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

 

[VietDao29]

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

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. :)

 

[Winzer]

Shouldn't it be:

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

 

[Winzer]

Shouldn't it be:

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

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

 

[VietDao29]

Shouldn't it be:

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

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:

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

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.

 

[Winzer]

lol. I feel like a moron.