Question

Identify the standard form of the equation by completing the square.
3x^2 − 2y^2 − 12x − 16y + 4 = 0

Answer 1

I know how to begin by factoring the 3 and -2 so you're left with:
3(x^2-4x+4) -2(y+8y+16)= -4

Answer 2

And the next step is to make the constants equal so you get: 3(x^2-4) - 2(y+8)= -24

Answer 3

I would start like this:

group the x and y terms separately: \((3x^2-12x) +(-2y^2-16y)+4=0\)

Then i'd send 4 to the other side so i have all my variables on one side and my constants on the other: \((3x^2-12x)+(-2y^2+16y)=-4\)

Factor out the LCM for both sets of ( ) on the LHS : \(3(x^2-4x)-2(y^2+8y)=-4\)

The stuff inside the () is in the form \((ax^2+bx)\) So in order to complete the quadratic, we need to find new c values for both sets of ( ). We're goign to find \(c_1\) which correlates to the first set of ( ) and \(c_2\) which correlates to the second ( ) :
\(c_1 = \left(\dfrac{-4}{2}\right)^2 = (-2)^2 = 4\)
\(c_2 = \left( \dfrac{8}{2}\right)^2 = (4)^2 = 16\)

Now we plug these in to the ( ) respectively : \(3(x^2-4x\color{red}{+4})-2(y^2+8y\color{blue}{+16})=-4\)

But now here's the problem child, you have to remember whatever you do on the LHS you also have to do on the RHS. because we pulled out the LCMs \(3\) and \(-2\), we're going to multiply both \(c_1\) and \(c_2\) with these LCMs and add them on the RHS:

\(3(x^2-4x\color{red}{+4})-2(y^2+8y\color{blue}{+16}=-4\color{red}{+12} \color{blue}{-32} \)

Answer 4

Now we just simplify both the LHS and RHS : \(3(x-2)^2 -2(y+4)^2 = -24\)

this completes the standard form of the equation: \((x-h)^2+(y-k)^2=r^2\)

Answer 5

\(3(x^2-4x\color{red}{+4})-2(y^2+8y\color{blue}{+16})=-4\color{red}{+12} \color{blue}{-32}\)*

fixed a typo.

Answer 6

Okay, I just realized that I typed in a different answer than what I originally had, and I had the same answer as you and that's why I was confused because these are my choices for the correct answer.

Answer 7

Okay, let me see, one minute.

Answer 8

Ah, I got it! (y+4)^2/12 - (x-2)^2/8 =1 because you divide -24 to both sides!

Answer 9

Yeah :D

Answer 10

It's asking for the elliptical form

Answer 11

You're amazing! Thanks for your help!!!!

Answer 12

No problem. Did you understand my explanation, or were there any questions about it?

Answer 13

Nope, I understood it perfectly!

Answer 14

Nice :D