Question

Как выделить квадрат в данном уравнении:
2х^2+6х+2у^2-8у+9/2=0

Answer 1

2x^2 + 6x + 2y^2 - 8y + 9/2 = 0
x^2 + 3x + y^2 - 4y + 9/4 = 0
(x^2 + 3x + 9/4) + (y^2 - 4y + 4) = 4
(x + 3/2)^2 + (y - 2)^2 = 2^2

Answer 2

$(x+a)^2=x^2+2ax+a^2\(y+a)^2=y^2+2ay+a^2\2x^2+6x+2y^2-8y+frac{9}{2}=0\(sqrt{2}x)^2+2*sqrt2x*frac{3}{sqrt2}+(frac{3}{sqrt2})^2+(sqrt2y)^2-2*sqrt2y*2sqrt2+8-frac{9}{2}+frac{9}{2}-8=0\(sqrt2x+frac{3}{sqrt{2}})^2+(sqrt2y-2sqrt{2})^2=8\(x+frac{3}{2})^2+(y-2)^2=4$