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Mathematics 23 Online
OpenStudy (anonymous):

Write 3x^2+2y^2+24x-4y+26=0 in standard form.

OpenStudy (anonymous):

Do you know what standard form means?

OpenStudy (anonymous):

Yes. I tried the problem. But, I'm stuck mid-through and don't know how to complete it.

OpenStudy (anonymous):

Okay, can you tell me what standard form means then?

OpenStudy (anonymous):

For a circle, it's (x-h)^2 + (y-k)^2 = r^2.

OpenStudy (anonymous):

Yep, so let's rearrange your polynomial a bit: \[3x^2+24x+2y^2-4y=26\]Then we can begin to pull factors out: \[3(x^2+8x)+2(y^2-4y)=26\]It's starting to take shape, but to pull the power out of the parenthesis we need to complete the square: \[3((x+4)^2-16)+2((y-2)^2-4)=26 \]Now we can expand the parenthesis again: \[3(x+4)^2-48+2(y-2)^2 -8 = 26 \]Move the numbers across to the right hand side: \[3(x+4)^2-2(y-2)^2=82\]And there's your answer :)

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