Question

Reduce the equation to one of the standard forms. This is Calculus 3.
4x^(2) + y^(2) +4z^(2) - 4y - 24z + 36 = 0
Please show steps and don't skip any.

Answer 1

So let's look at each variable independently.
If we first look at the `y`'s, we need to complete the square on them.\[\large y^2-4y\]To complete the square, we'll take the constant attached the the y term, \(-4\),
Divide it by 2 and then square it. \(\large \left(\dfrac{-4}{2}\right)^2 \quad = \quad 4\).

So we'll add a 4 for a complete square, \(\large y^2-4y+4\).

Which can be written as a square binomial \(\large (y-2)^2\).
Since we added a 4 into the equation, we need to also add a 4 to the other side,

\[\large 4x^2+\color{royalblue}{y^2}+4z^2\color{royalblue}{-4y}-24z+36=0\]\[\large 4x^2+4z^2+\color{royalblue}{y^2}\color{royalblue}{-4y}+\color{royalblue}{4}-24z+36=0+\color{royalblue}{4}\]\[\large 4x^2+4z^2+\color{royalblue}{(y-2)^2}-24z+36=\color{royalblue}{4}\]
That takes care of the y's. Any confusion there? :O

Answer 2

No confusion there, it's the z part that is bothering me.

Answer 3

\[\large 4z^2-24z\]Factoring out a 4 gives us,\[\large 4(z^2-6z)\]Now we'll complete the square within the brackets. Half of \(-6\) squared gives us \(9\).
Instead of adding 9 to each side, we have to be a little bit more careful since we're working within a set of brackets.
So to be extra careful, let's just `add` 9, and `subtract` 9 at the same time.\[\large 4(\color{orangered}{z^2-6z+9}-9)\]The orange is the part that will give us a perfect square.\[\large 4(\color{orangered}{(z-3)^2}-9)\]Now we can go ahead and distribute the 4 back in.\[\large 4(z-3)^2-36\]

Answer 4

Oh there was already a 36 in the problem.. .that was stupid of me.. woops.

Answer 5

We should have just included the 36 in the z group, that would have completed the square much easier.

Answer 6

\[\large 4z^2-24z+36\qquad \rightarrow \qquad 4(z^2-6z+9) \qquad \rightarrow \qquad 4(z-3)^2\]

Answer 7

I got it, thank you!

Answer 8

cool c: