Posted by Soto on Thursday, April 8, 2010 at 2:44pm.
16x^2+16y^216x+24y3=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^216x+24y3=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(x1/2)^2 + 16(y+3/4) = 16
(x1/2)^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^333x^216x+16
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