A common factor is a factor of the least degree in all of the terms. We have one here. $(x+9)$ is a factor in both terms. The powers are $\frac {1}{2}$ and $-\frac {1}{2}$. The least of which is $-\frac {1}{2}$.

We will use the distributive property to factor out $(x+9)^{\answer {-\frac {1}{2}}}$.

$(x+9)^{-\tfrac {1}{2}} \left ((x+9) + x \cdot \frac {1}{2}\right ) = 0$

We could, in turn, write this as a fraction

$$\frac {\left ((x+9) + x \cdot \frac {1}{2}\right )}{\sqrt {x+9}} = 0$$

$$\frac { \frac {3}{2} x + 9}{\sqrt {x+9}} = 0$$

This is a fraction, so it equals $0$ when the numerator equals $0$ and the denominator does not equal $0$.

$\answer {\frac {3}{2} x + 9} = 0$

$x = 6$, which is in the domain of $\sqrt {x+9} + x \cdot \frac {1}{2} (x+9)^{-\tfrac {1}{2}}$.

The numerator is a difference of two terms and $9 x^2$ is a common factor. We can use the Distributive Property to factor it out.

$$\frac {9x^2 (2x^2 - 3(x^2-3))}{x^6} = 0$$

$$\frac {9 \left ( \answer {9 - x^2} \right )}{x^4} = 0$$

This is a fraction, so it equals $0$ when the numerator equals $0$ and the denominator does not equal $0$.

$9 - x^2 = 0$

This has two solutions: $\{ -3, 3 \}$, both of which are in the domain.

The Distributive Property (factoring) gives us the product $12 x^2 (x-1)$.

We have two solutions: $\{ 0, 1 \}$, both of which are in the domain.

Converting to exponents gives us $2 - \frac {2}{x^{\tfrac {1}{3}}} = 0$

$2 = \frac {2}{x^{\tfrac {1}{3}}}$

The only way this can happen is if $x^{\tfrac {1}{3}} = \answer {1}$.

This has one solution: $\{ 1 \}$