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

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