Answer
See the detailed answer below.
Work Step by Step
First of all, we need to draw the force diagram exerted on $+q$-charge. Recalling that the like charges repel and the unlike charge attract.
$$\color{blue}{\bf [a]}$$
As we see in the first figure below, at any position that $|x|\lt a$, the net force exerted on $q$ is toward the left.
Hence,
$$\sum F_{ x,{\rm on} \;q}=-F_{-Q\;{\rm on}\;q}-F_{Q\;{\rm on}\;q}$$
We will neglect the signs of the charges since we are focusing on the forces' directions.
$$\sum F_{ x,{\rm on} \;q}=-\dfrac{kQq}{(a+x)^2}-\dfrac{kQq}{(a-x)^2}$$
$$\sum F_{ x,{\rm on} \;q}= \dfrac{- kQq(a-x)^2-kQq(a+x)^2}{(a+x)^2(a-x)^2} $$
Recalling that $(a-x)(a+b)=a^2-b^2$:
$$\sum F_{ x,{\rm on} \;q}= \dfrac{-kQq[(a-x)^2+(a+x)^2]}{ (a^2-x^2)^2} $$
$$\sum F_{ x,{\rm on} \;q}= \dfrac{-kQq[a^2-2ax+x^2+a^2+2ax+x^2]}{ (a^2-x^2)^2} $$
$$\sum F_{ x,{\rm on} \;q}= \dfrac{-kQq[2a^2+2x^2]}{ (a^2-x^2)^2} $$
$$\boxed{\sum F_{ x,{\rm on} \;q}= \dfrac{-2kQq(a^2+x^2)}{ (a^2-x^2)^2} }$$
$$\color{blue}{\bf [b]}$$
When $|x|\gt a$, we have two cases. when $q$ is on the right side and when it is on the left side.
Let's start with the right side: As we see in the second figure below, we cans ee that the force from $-Q$ is toward the left and the force from $Q$ is toward the right.
Hence,
$$\sum F_{ x,{\rm on} \;q}=-F_{-Q\;{\rm on}\;q}+F_{Q\;{\rm on}\;q}$$
$$\sum F_{ x,{\rm on} \;q}=-\dfrac{kQq}{(a+x)^2}+\dfrac{kQq}{(x-a)^2}$$
As we mentioned above we are neglecting the charge sings since we know the direction of the forces.
Rearranging:
$$\sum F_{ x,{\rm on} \;q}=\dfrac{kQq}{(x-a)^2}-\dfrac{kQq}{(x+a)^2}$$
$$\sum F_{ x,{\rm on} \;q}=\dfrac{kQq (x+a)^2 - kQq(x-a)^2}{(x-a)^2(x+a)^2} $$
$$\sum F_{ x,{\rm on} \;q}=\dfrac{kQq [ x^2+2ax+a^2 - (x^2-2ax+a^2)]}{(x-a)^2(x+a)^2} $$
$$\sum F_{ x,{\rm on} \;q}=\dfrac{kQq [ x^2+2ax+a^2 - x^2+2ax-a^2]}{(x-a)^2(x+a)^2} $$
$$\boxed{\sum F_{ x,{\rm on} \;q}= \dfrac{4axkQq }{ (x^2-a^2 )^2} }$$
Now let's start with the left side: As we see in the third figure below, we cans ee that the force from $-Q$ is toward the right and the force from $Q$ is toward the left.
Hence,
$$\sum F_{ x,{\rm on} \;q}= F_{-Q\;{\rm on}\;q}-F_{Q\;{\rm on}\;q}$$
$$\sum F_{ x,{\rm on} \;q}= \dfrac{kQq}{(x-a)^2}-\dfrac{kQq}{(x+a)^2}$$
Rearranging:
$$\sum F_{ x,{\rm on} \;q}= \dfrac{kQq (x+a)^2 -kQq (x-a)^2 }{(x-a)^2(x+a)^2} $$
$$\sum F_{ x,{\rm on} \;q}= \dfrac{kQq [ x^2+2ax+a^2 -(x^2-2ax+a^2)] }{(x-a)^2(x+a)^2} $$
$$\sum F_{ x,{\rm on} \;q}= \dfrac{kQq [ x^2+2ax+a^2 -x^2+2ax-a^2] }{(x^2-a^2)^2} $$
$$\boxed{\sum F_{ x,{\rm on} \;q}= \dfrac{4axkQq }{(x^2-a^2)^2} }$$
We can see from the last two boxed formula above, that in both cases, when $|x|\gt a$, the net force is always to the right.
