1. **State the problem:** We need to find the value(s) of $x$ such that $x^2 = 8$. 2. **Formula and rules:** To solve for $x$ when given $x^2 = a$, we use the square root property: $$x = \pm \sqrt{a}$$ This means $x$ can be either the positive or negative square root of $a$. 3. **Apply the formula:** Here, $a = 8$, so $$x = \pm \sqrt{8}$$ 4. **Simplify the square root:** Since $8 = 4 \times 2$, we can write $$x = \pm \sqrt{4 \times 2} = \pm \sqrt{4} \times \sqrt{2} = \pm 2\sqrt{2}$$ 5. **Final answer:** The solutions to $x^2 = 8$ are $$x = 2\sqrt{2} \quad \text{or} \quad x = -2\sqrt{2}$$ This means $x$ can be either positive or negative $2\sqrt{2}$, which are the two real roots of the equation.