Finding the minimum for this formula I think if we want to calculate $\min_x \sum_i (b_i - x)^2$, the answer should be the mean of $b_i$, right? Now if we add a weight to each term and make it $\min_x \sum_i w_i(b_i - x)^2$, what's the closed form solution for $x$ then?
 A: It's just the weighted average
$$
\sum_i w_i b_i
$$
(I assume $\sum_i w_i=1$).
To see this, do some algebra, as follows.  (For convenience we'll let $n$ be the number of terms, thereby sparing ourselves the need to repeatedly write $\sum\limits_i\ 1$.)
$$
\begin{align}
& {} \qquad \sum_{i=1}^n w_i(x - b_i)^2 = \sum_{i=1}^n w_i(x^2 - 2xb_i + b_i^2) = x^2 - \left(2x\sum_{i=1}^n (w_i b_i)\right) + \left(\sum_{i=1}^n w_i b_i^2\right) \\  \\  \\
& = \underbrace{\left(x^2 -2x\sum_{i=1}^n w_i b_i + \left(\sum_{i=1}^n w_i b_i \right)^2 \right)}_\text{a square} + \sum_{i=1}^n w_i b_i^2 - \left(\sum_{i=1}^n w_i b_i \right)^2\tag{completing the square} \\  \\  \\
& = \underbrace{\left( x - \sum_{i=1}^n w_i b_i \right)^2}_\text{a square} + \sum_{i=1}^n w_i b_i^2 - \left(\sum_{i=1}^n w_i b_i \right)^2.
\end{align}
$$
The term labeled "a square" is $0$ if $x =\text{the aformentioned weighted average}$, and is otherwise positive.  In the succeeding terms, "${}\;x\;{}$" does not appear at all.  Therefore the value of $x$ that minimizes the whole expression is the value of $x$ that minimizes the term labaled "a square", and that is the weighted average.  the actual value of the minimum is then given by those last two terms.
