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:tongue:

For example, 3x^2 + 7x + 3y^2 -6y +3z^2 + 2z - 12..

On the quiz..

I first did it by..

(3x^2 + 7x + __) + (3y^2 - 6y + __) + (3z^2 + 2z + __) = 12

..

3(x^2 + 7x/3 + __) + 3(y^2 - 2y + __) + 3(z^2 + 2z/3 + __) = 12

..

3(x^2 + 7x/3 + 49/12) + 3(y^2 - 2y + 3) + 3(z^2 + 2z/3 + 1/3) = 12 + 49/12 + 3 + 1/3 = 19.42

(x+7/6)^2 + (y-1)^2 + (z-1/3)^2=19.42=233/12

Yeah.. I don't really know what to do with the 3s so I kind of ignored them.. Obviously I just divide by 3?..

==========

I redid the problem like..

3[(x^2+7x/3+__) + (y^2-2y+__)..etc

You get the deal..

Then I divided 12/3=4 and then yeah got answer to be 233/36.[real answer]

K. NOW to the problem:tongue: What if it were say.. 3x^2 +5x +7y^2 blah blah.. Anyone know how to complete the square of a equation with multiple variables/coefficients?

## Homework Statement

For example, 3x^2 + 7x + 3y^2 -6y +3z^2 + 2z - 12..

## Homework Equations

## The Attempt at a Solution

On the quiz..

I first did it by..

(3x^2 + 7x + __) + (3y^2 - 6y + __) + (3z^2 + 2z + __) = 12

..

3(x^2 + 7x/3 + __) + 3(y^2 - 2y + __) + 3(z^2 + 2z/3 + __) = 12

..

3(x^2 + 7x/3 + 49/12) + 3(y^2 - 2y + 3) + 3(z^2 + 2z/3 + 1/3) = 12 + 49/12 + 3 + 1/3 = 19.42

(x+7/6)^2 + (y-1)^2 + (z-1/3)^2=19.42=233/12

Yeah.. I don't really know what to do with the 3s so I kind of ignored them.. Obviously I just divide by 3?..

==========

I redid the problem like..

3[(x^2+7x/3+__) + (y^2-2y+__)..etc

You get the deal..

Then I divided 12/3=4 and then yeah got answer to be 233/36.[real answer]

K. NOW to the problem:tongue: What if it were say.. 3x^2 +5x +7y^2 blah blah.. Anyone know how to complete the square of a equation with multiple variables/coefficients?

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