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 $5x^2-2kx+1&lt;0$ has exactly one integral solution.


My attempt is as follows:

$$\left(x-\dfrac{2k-\sqrt{4k^2-20}}{10}\right)\left(x-\dfrac{2k+\sqrt{4k^2-20}}{10}\right)&lt;0$$
$$\left(x-\dfrac{k-\sqrt{k^2-5}}{5}\right)\left(x-\dfrac{k+\sqrt{k^2-5}}{5}\right)&lt;0$$
$$x\in\left(\dfrac{k-\sqrt{k^2-5}}{5},\dfrac{k+\sqrt{k^2-5}}{5}\right)$$

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.
Your idea is good; you want to find all positive integers $k$ for which there is precisely on integer between the roots of
$$5x^2-2kx+1=0.$$
Then the distance between the roots can be at most $2$, where the distance between the roots is precisely
$$\frac{1}{5}\sqrt{(-2k)^2-4\cdot1\cdot5}=\frac25\sqrt{k^2-5},$$
as you already found. This is at most $2$ if and only if $\sqrt{k^2-5}\leq5$, or equivalently $k\leq5$. This leaves only $5$ values of $k$ to check.
hint

Observe that

$$x_2=\frac{k+\sqrt{k^2-5}}{5}=\frac{1}{k-\sqrt{k^2-5}}$$

$ x_2 $ is integral if  $ (k -\sqrt{k^2-5})=\pm 1$ , 

By the same, $ x_1 $ is integral if
$ (k+\sqrt{k^2-5}) =\pm 1$.

This gives the possible values for $ k$ :
$-4, -3, 3, 4$.
the sum of positives values is $7$.