Let the mean and the variance of 20 observations $$x_1, x_2, \ldots, x_{20}$$ be 15 and 9, respectively. For $$\alpha \in \mathbb{R}$$, if the mean of $$(x_1 + \alpha)^2, (x_2 + \alpha)^2, \ldots, (x_{20} + \alpha)^2$$ is 178, then the square of the maximum value of $$\alpha$$ is equal to _______

We are given 20 observations $$x_1, x_2, \ldots, x_{20}$$ with mean $$\bar{x} = 15$$ and variance $$\sigma^2 = 9$$. We know that $$\frac{1}{20}\sum x_i = 15$$ and $$\frac{1}{20}\sum x_i^2 - 15^2 = 9$$, so $$\frac{1}{20}\sum x_i^2 = 234$$.

We need the mean of $$(x_1+\alpha)^2, (x_2+\alpha)^2, \ldots, (x_{20}+\alpha)^2$$ to equal 178. This mean is:

$$\frac{1}{20}\sum(x_i+\alpha)^2 = \frac{1}{20}\sum(x_i^2 + 2\alpha x_i + \alpha^2) = \frac{1}{20}\sum x_i^2 + 2\alpha\cdot\frac{1}{20}\sum x_i + \alpha^2$$

$$= 234 + 2\alpha(15) + \alpha^2 = \alpha^2 + 30\alpha + 234$$

Setting this equal to 178: $$\alpha^2 + 30\alpha + 234 = 178$$, so $$\alpha^2 + 30\alpha + 56 = 0$$.

By the quadratic formula: $$\alpha = \frac{-30 \pm \sqrt{900 - 224}}{2} = \frac{-30 \pm \sqrt{676}}{2} = \frac{-30 \pm 26}{2}$$.

So $$\alpha = \frac{-30+26}{2} = -2$$ or $$\alpha = \frac{-30-26}{2} = -28$$. The maximum value of $$\alpha$$ is $$-2$$, and its square is $$(-2)^2 = 4$$.

Hence, the correct answer is 4.