The problem asks us to find the geometric mean (or the equal ratio middle term) of $\sqrt{2}$ and $\sqrt{8}$. We need to find a number $x$ such that $\sqrt{2}$, $x$, and $\sqrt{8}$ form a geometric progression.

1. Problem Description

The problem asks us to find the geometric mean (or the equal ratio middle term) of $\sqrt{2}$ and $\sqrt{8}$. We need to find a number $x$ such that $\sqrt{2}$, $x$, and $\sqrt{8}$ form a geometric progression.

2. Solution Steps

In a geometric progression $a$, $b$, $c$, the middle term $b$ is the geometric mean of $a$ and $c$, and $b^2 = ac$. In our case, we have $a = \sqrt{2}$ and $c = \sqrt{8}$. We are looking for $x$ such that $x^2 = \sqrt{2} \cdot \sqrt{8}$.

$x^2 = \sqrt{2} \cdot \sqrt{8}$

$x^2 = \sqrt{2 \cdot 8}$

$x^2 = \sqrt{16}$

$x^2 = 4$

$x = \pm \sqrt{4}$

$x = \pm 2$

3. Final Answer

The geometric mean of $\sqrt{2}$ and $\sqrt{8}$ is $\pm 2$. Therefore, the answer is [A].