If $f(x) =5x^2 - 2kx + 1 < 0$ has exactly one integral solution, find the sum of all positive integral values of $k$. Question :  If $f(x) =5x^2 - 2kx + 1 < 0$ has exactly one integral solution, find the sum of all positive integral values of $k$.
My Attempt
Corresponding Equation of Inequality, $f(x) =5x^2 - 2kx + 1 = 0$. Let $\alpha,\beta$ be the roots. $f(x)<0$ for exactly one integral value of $x$.

Possible Graphs

Conditions Required

*

*$D>0 \\ 4k^2-20>0 \\k^2-5>0\\k  \in(-\infty,-\sqrt5) \;\cup \;(\sqrt5,\infty) \qquad\text{(1)}$

*$|\alpha-\beta|\leq2 \\ |\frac{\sqrt D}a|\leq 2 \\\\ 2\frac{\sqrt{k^2-5}}{5}\leq2 \;\;\&\;\; 2\frac{\sqrt{k^2-5}}{5}\geq-2 \\\sqrt{k^2-5}\leq{5}\;\;\&\;\;\sqrt{k^2-5}\geq{-5}\\$
Solving we get $k\in[-\sqrt{30},\sqrt{30}]\qquad{(2)}\\$
and $k\in(-\infty,-\sqrt5] \; \cup \;[\sqrt5,\infty) \qquad\text{(3)}$
From $(1),(2),(3)$, we get $\boxed{k\in[-\sqrt{30},-\sqrt5) \; \cup \;(\sqrt5,\sqrt{30}]}$
Hence, sum of all positive integral values of $x$ is $3+4+5=12$
Issue: I think I have solved this problem correctly. But the answer at the back of my book is: $4+5=9$.
I mean, $\{3\}$ does not belong to the solution set, which I find strange. I might have made a mistake somewhere while solving it, but couldn't find it. I am just asking why $\{3\}$ does not belong to the solution set. Because it should, from what I see. Please help me with it. Thanks.
 A: We have $f(x) =5x^2 - 2kx + 1=5(x^2-2\frac{k}{5}x +\frac{k^2}{25})-\frac{k^2}{5}+1=5(x-\frac{k}{5})^2 - \frac{k^2}{5}+1 \geq  1-\frac{k^2}{5}$
So it takes it's minimum at $\frac{k}{5}$ which is $1-\frac{k^2}{5}$
We want $1-\frac{k^2}{5}<0$ in order for the inequality to have solution, hence $k<-\sqrt{5}$ and $k>\sqrt{5}$ $(1)$
The solutions to $f(x)=0$ are 
$x_1=\frac{k-\sqrt{k^2-5}}{5}$ and $x_2=\frac{k+\sqrt{k^2-5}}{5}$
As you mentioned there must be the following requirement:
$x_2-x_1<2$ which by solving gives $-\sqrt{30}< k < \sqrt{30}$ $(2)$
From $(1)$ and $(2)$ we get the required solutions which are $k=4$ and $k=5$
The problem with $k=3$ is that $f(x)<0$ then $x \in (0.2, 1)$ which has no integral solution
A: Using your notations:
$$\alpha=\frac{k-\sqrt{k^2-5}}{5}; \beta=\frac{k+\sqrt{k^2-5}}{5}.$$
The integral solution of $5x^2 - 2kx + 1 < 0$ must be in $(\alpha,\beta)$.
Consider only $k>0$ by the requirement and note that:
$$\\
0<\alpha<1 \iff 0<k-\sqrt{k^2-5}<5 \iff k-5<\sqrt{k^2-5} \Rightarrow k>0
$$
Hence:
$$\beta>1 \Rightarrow k+\sqrt{k^2-5}>5 \Rightarrow \sqrt{k^2-5}>5-k \Rightarrow k>3.$$
This is an additional condition that is missed by your $|\alpha-\beta|<2$.
For $k=3$, the inequality $5x^2 - 6x + 1 < 0$ has a solution $(0.2,1)$, which does not have an integral solution.
A: Just plot the thing for $k = 4$ and $k = 5$.(Use Desmos (https://www.desmos.com/calculator/g3r9safvji), for instance); you'll see that $x = 1$ is the sole integer solution in those two cases; for smaller $k$ there are no integer solutions; for larger $k$ there are at least two. 
Those plots should let you see the error in your "solution".
