"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?
16x^2+16y^2-16x+24y-3=0
How do I solve this?
5 answers
yes
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.
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.
thank you very much
The coordinate of all relative maxima 2x^4+x^3-33x^2-16x+16
1 answer