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(C) The given equation can be rewritten as $x^2 + 2(\sqrt{2}y + 2)x + (2y^2 + 4\sqrt{2}y + 1) = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt

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$2$

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(C) The given equation can be rewritten as $x^2 + 2(\sqrt{2}y + 2)x + (2y^2 + 4\sqrt{2}y + 1) = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:$x = \frac{-2(\sqrt{2}y + 2) \pm \sqrt{4(\sqrt{2}y + 2)^2 - 4(2y^2 + 4\sqrt{2}y + 1)}}{2}$$x = -(\sqrt{2}y + 2) \pm \sqrt{(\sqrt{2}y + 2)^2 - (2y^2 + 4\sqrt{2}y + 1)}$$x = -\sqrt{2}y - 2 \pm \sqrt{2y^2 + 4\sqrt{2}y + 4 - 2y^2 - 4\sqrt{2}y - 1}$$x = -\sqrt{2}y - 2 \pm \sqrt{3}$.Thus,the lines are $x + \sqrt{2}y + 2 - \sqrt{3} = 0$ and $x + \sqrt{2}y + 2 + \sqrt{3} = 0$.These are parallel lines of the form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$.The distance between them is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.$d = \frac{|(2 + \sqrt{3}) - (2 - \sqrt{3})|}{\sqrt{1^2 + (\sqrt{2})^2}} = \frac{|2\sqrt{3}|}{\sqrt{3}} = 2$.

What is the distance between the lines represented by the equation $x^2 + 2\sqrt{2}xy + 2y^2 + 4x + 4\sqrt{2}y + 1 = 0$?

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