Find the Domain of y=(x+√2)²: Analyzing Positive and Negative Inputs

To find the positive and negative domains of the function $y = (x + \sqrt{2})^2$, we begin by noting that this function is a quadratic expression in vertex form. Here, the vertex is located at $x = -\sqrt{2}$, and since the coefficient of the squared term is positive, the parabola opens upwards.

The expression $(x + \sqrt{2})^2$ represents the square of the term $x + \sqrt{2}$. A squared term is always greater than or equal to zero for any real $x$. Therefore, $y$ is never negative for any real $x$.

Since we want to determine where the function is positive, we need $(x + \sqrt{2})^2 > 0$. This inequality holds true for all $x$ except at the point where the expression equals zero, which is $x = -\sqrt{2}$.

Therefore, the positive domain is where $x \neq -\sqrt{2}$, corresponding to $x > 0$ and $x \neq -\sqrt{2}$. However, the condition $x > 0$ is sufficient for this particular problem, indicating the positive domain requirement.

The negative domain does not exist, as the function cannot be negative.

Based on the analysis, the correct answer is:

$x < 0 :$ none

$x > 0 : x \neq -\sqrt{2}$