Posted by **Soto** on Thursday, April 8, 2010 at 2:44pm.

16x^2+16y^2-16x+24y-3=0

How do I solve this?

- Algebra 2 -
**Reiny**, Thursday, April 8, 2010 at 3:14pm
"Solving" means finding ordered pairs that satisfy the equation.

Your equation is a circle, and there are an infinite number of ordered pairs that will work.

Did you mean to put it in the standard form of a circle equation?

- Algebra 2 -
**Soto**, Thursday, April 8, 2010 at 4:04pm
yes

- Algebra 2 -
**Reiny**, Thursday, April 8, 2010 at 4:36pm
16x^2+16y^2-16x+24y-3=0

16(x^2 - x + .....) + 16(y^2 + (3/2)y + ... = 3

16(x^2 - x + 1/4) + 16(y^2 + (3/2)y + 9/16) = 3 + 4 + 9

16(x-1/2)^2 + 16(y+3/4) = 16

(x-1/2)^2 + (y+3/4)^2 = 1

all vital info about the circle is now obvious.

- Algebra 2 -
**Soto**, Thursday, April 8, 2010 at 4:41pm
thank you very much

- Algebra 2 -
**Anonymous**, Monday, September 19, 2011 at 12:13pm
The coordinate of all relative maxima 2x^4+x^3-33x^2-16x+16

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