An angle bisector of length $4$ units, creates a $6$ unit line segment on the base of the triangle. What are possible integer values of its side? 

What is the sum of the possible integer values of $x$

My Solution
I know that $\angle DAC \lt 90^o$ because it is an angle bisector, therefore:
$$4^2+x^2 \lt 6^2 \quad \quad  6-4 \lt x \lt 6+4$$
Combining these two I get:
$$2 \lt x \lt 5 \quad x\in Z$$
But this is wrong according to the answer key. How can I solve this problem?
Edit: Inequality should have been $4^2+x^2 \gt 6^2$, though still I'm getting a wrong answer (The sum should be $24$):
$$4 \lt x \lt 10 \quad (\sum x =35)\quad  x \in Z$$
 A: Clearly $x < 10.$ It is not hard to construct a valid figure in which $x = 9$ and all of the given conditions hold.
So the question is what the minimum value of $x$ is.
Note that if $\angle ADC$ is less than a right angle, as it certainly would be if 
$x < \sqrt{6^2 + 4^2} = \sqrt{52},$
then $\angle ADB$ is greater than a right angle and
$\lvert AB\rvert > \lvert BD\rvert.$
But since $AD$ bisects $\angle BAC,$
we have $\frac x6 = \frac{\lvert AB\rvert}{\lvert BD\rvert} > 1.$
Therefore $x > 6.$
Another way to find the same constraint is that in order for the sides $AB$ and $BC$ to intersect as shown in the figure, we require that $\angle ADC > \angle CAD,$
which implies that $x > 6.$
For good measure you can confirm that the figure can be constructed if $x=7,$
$x=8,$ or $x=9.$  But you will not find any stricter constraints than $6 < x < 10.$
So the sum of possible values of $x$ is $7+8+9=24.$
A: We are missing the forest for the trees.  The true bound has nothing to do with the Pythagorean Theorem or the Law of Cosines.
In $\triangle ABD$, the exterior angle ADC must have a greater measure than the opposing interior angle DAB.  By the bisection requirement, $\angle DAB$ is in turn congruent to $\angle CAD$.
So, in $\triangle ADC$, $\angle ADC$ must be larger than $\angle CAD$, forcing the opposite sides to have the same ordering:  $CA=x$ must be greater than $DC=6$.
