Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution. 
Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.

My attempt is as follows:
$$\left(x-\dfrac{2k-\sqrt{4k^2-20}}{10}\right)\left(x-\dfrac{2k+\sqrt{4k^2-20}}{10}\right)<0$$
$$\left(x-\dfrac{k-\sqrt{k^2-5}}{5}\right)\left(x-\dfrac{k+\sqrt{k^2-5}}{5}\right)<0$$
$$x\in\left(\dfrac{k-\sqrt{k^2-5}}{5},\dfrac{k+\sqrt{k^2-5}}{5}\right)$$
As it is given that it has got only one integral solution, so there must be exactly one integer between $\dfrac{k-\sqrt{k^2-5}}{5}$ and $\dfrac{k+\sqrt{k^2-5}}{5}$
Let $x_1=\dfrac{k-\sqrt{k^2-5}}{5}$ and $x_2=\dfrac{k+\sqrt{k^2-5}}{5}$ , then $[x_2]-[x_1]=1$ where [] is a greater integer function.
But from here, how to proceed? Please help me in this.

Well one way of looking at the solution is for $n, k \in Z$ it will have only one integral solution if 
$$n-1 \le \dfrac{k-\sqrt{k^2-5}}{5}<n< \dfrac{k+\sqrt{k^2-5}}{5} \le n+1$$
Now, $D \ge 0 \Rightarrow |k| \ge \sqrt5$ and $|\alpha -\beta| \le 2 \Rightarrow k\le \sqrt{30}$
Combining both the conditions we get $k \in ${3,4,5}.
If k=3, then we get
$$n-1 \le \dfrac{1}{5}<n< 1 \le n+1 \Rightarrow n \notin Z$$
Wolfram alpha provides the following integral solutions of the problem

