coulombs-lawelectromagnetism
Coulombs law $F=K_e\frac{|q_1||q_2|}{r^2}$
This gives no indication of the direction of the force should you not have the absolute value in this equation
$(-)*(+)=-$
$(-)*(-)=+$
$(+)*(+)=+$
Thus ignoring the absolute value signs will automatically indicate that if the result is negative the force is attractive and if positive is repulsive
Adding absolute values to the equation leaves the user to figure out the direction manually
Pay close attention to how you defined $x$: it's not the coordinate of the desired point on the $x$ axis, it's how far left of the positive charge that point is. So when your equation tells you $x = 4.83$, that means the desired point is $4.83$ units left of the positive charge. But, presumably the question you've been asked wants the coordinate of the point, not how far left of the positive charge it is. So the last step in solving (which you skipped) is to figure out, if you have a point that is $4.83$ units left of the positive charge, what is the $x$-coordinate of that point?
In the example you mentioned at the end, they defined $x$ differently, such that is is the $x$-coordinate of the desired point. That's why that equation gets you directly to the coordinate.
It's important to be clear on what these variables mean: specifically, this equation is for calculating the force exerted on one charge by the other. The unit vector $\hat{r}$ points from the charge exerting the force to the charge experiencing the force. E.g. if you're calculating the force experienced by $q_1$, you should have this image in mind:
Now, the force on $q_1$ might be toward or away from $q_2$, depending on the signs of the charges. Clearly, to represent a force away from $q_2$, $\hat{r}$ should be multiplied by something positive, and to represent a force toward $q_2$, $\hat{r}$ should be multiplied by something negative. You know that like charges repel and opposite charges attract, so if $q_1$ and $q_2$ have the same sign, $\hat{r}$ needs to have a positive coefficient, and if they have opposite signs, $\hat{r}$ should have a negative coefficient.
Your formula (1) does exactly this. The coefficient $\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}$ is positive when $q_1$ and $q_2$ have the same sign, and negative when they don't. If you take the absolute values, as in formula (2), the coefficient is positive in all cases, so that will give the wrong result for opposite-signed charges.
I'm not quite sure how you found that equation (1) leads to contradictions; there's no need to alter the unit vector $\hat{r}$.
You could take the opposite convention for the direction of $\hat{r}$, i.e. take it to point from the charge experiencing the force to the charge exerting the force. This is not conventional, and it will confuse a lot of people, but it is physically valid. In that case, you would need to put an overall negative sign in the formula, like this: $$\vec{F} = -\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}\hat{r}\tag{opposite convention}$$ But still, the same argument as above will show you that the results you get are incorrect if you take the absolute values of the charges.
What is often done is to use Coulomb's law to calculate the magnitude of the force only, and then figure out the direction by physical reasoning. If you calculate the magnitude, $\lVert \vec{F}\rVert$, then you do take the absolute value of everything. $$\lVert\vec{F}\rVert = \left\lVert\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}\hat{r}\right\rVert = \left\lvert\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}\right\rvert \underbrace{\left\lVert\hat{r}\right\rVert}_{1} = \frac{1}{4\pi\epsilon_0}\frac{\lvert q_1\rvert\,\lvert q_2\rvert}{r^2}$$
Best Answer
Forces are vector quantities and therefore have magnitude (or modulus or length) and direction (with sign).
The expression $$ F=k_e{|q_1q_2|\over r^2}$$ is only the magnitude of the force.
The expression written explicitly in vector notation is $$ \mathbf F_1=k_e\frac{q_1q_2}{{|\mathbf r_{12}|}^2} \mathbf{\hat{r}}_{21}$$ where $ \mathbf{{r}}_{21} $ is the vectorial distance between the charges and $\mathbf{\hat{r}}_{21} =\mathbf{{r}}_{21}/|\mathbf{{r}}_{21}| $.
If the system is only one dimensional, then you can trivially consider the vectorial nature of the force formally by just removing the absolute values to the charges as you suggest.