Find the sum of all values of $k$ for which $x^2 + kx - 9x + 25 - 9$ is the square of a binomial.
As in Solution 1, we factor $x^2 + kx - 9x + 25 - 9 = (x + a)^2$ for some number $a.$ Then
\[x^2 + (k - 9) x + 16 = x^2 + 2ax + a^2 = 0.\]We see that $a^2 = 16,$ so $a = \pm 4.$ Then $k = 9 \pm 8.$ The sum of all possible values of $k$ is $9 + 8 = \boxed{17}.$
To find the sum of all values of $k$ for which $x^2 + kx - 9x + 25 - 9$ is the square of a binomial, we need to determine the values of $k$ that make the given expression a perfect square trinomial.
Let's start by looking at the given expression:
$x^2 + (k-9)x + 16$
For this expression to be the square of a binomial, the second term of the trinomial ($k-9$) must be twice the product of the square roots of the first and last terms ($x^2$ and $16$). In other words, we need to satisfy the equation:
$k - 9 = 2\sqrt{x^2 \cdot 16}$.
Since $x^2$ can be positive or negative, we take the absolute value of $x$:
$k - 9 = 2|x|\sqrt{16}$.
Simplifying, we have:
$k - 9 = 2|x| \cdot 4$.
$k - 9 = 8|x|$.
Now, let's consider the two cases separately:
Case 1: $x \geq 0$
When $x \geq 0$, $|x| = x$. Substituting this into the equation, we get:
$k - 9 = 8x$.
Case 2: $x < 0$
When $x < 0$, $|x| = -x$. Substituting this into the equation, we get:
$k - 9 = -8x$.
In both cases, we can solve for $k$:
Case 1: $k - 9 = 8x$
$k = 8x + 9$.
Case 2: $k - 9 = -8x$
$k = -8x + 9$.
To find the sum of all values of $k$, we consider both cases together:
Sum of $k = (8x + 9) + (-8x + 9)$
Sum of $k = 16$.
Therefore, the sum of all values of $k$ for which $x^2 + kx - 9x + 25 - 9$ is the square of a binomial is $16$.