The key is that it must always have rational roots for any $p, q$.
Claim: Let $f(x) = x^2 + bx + c$, where $b$ and $c$ are rational numbers. If $f(x)$ satisfies the condition that for all rational numbers $r$, $f(r)$ is the square of a rational number, then $f(x)$ must be the square of a polynomial.
Proof: By completing the square, we simply need to consider polynomials of the form $x^2 + \frac{n}{m}$, where $m,n$ are integers and $\gcd(n,m) = 1$.
Consider $x = 0$ and $\frac{1}{m}$, which gives us $\frac{n}{m}$ and $\frac{1+mn}{m^2}$ are both squares of rational numbers. Multiplying both terms by $m^2$, we get that $mn$ and $mn+1$ are both integers that are squares of a rational number, hence they are perfect squares, so $mn=0$.
Since $m\neq 0$, hence $n=0$ and thus $f(x) = x^2$ and we are done. $_\square$
Corollary: $b=0$ or $1$.
Proof: Consider the polynomial $f(x) = \left( \frac{p}{q} \right)^2 + 2(1-2b) \frac{p}{q} + 1$. From the conditions, for rational numbers, this always evaluates to a rational square. Hence, we must have $ f(x) = (ax+b)^2$.
Comparing coefficients, we see that $ a ^2 = 1, b^2 = 1$, which implies that $f(x) = (x+1)^2 $ or $(x-1)^2$, which hence correspond to $b=0$ and $b=1$ respectively. $_\square$
Note: Franklin shows that the claim is true for all polynomials, though it uses more machinery.
Hint:
the discriminant of original polynomial can be written as $(2k+1)^2 = 4(k^2+k)+1$
EDIT: If one still can not understand one can expand the discriminant and get $4m+1$ and then
$4m+1 = 4(k^2+k)+1$ which means $m = k^2+k\leq 50 \implies 1 \leq k \leq 6$.
There are 6 rational roots.
Best Answer
Note that the solutions of $6x^2 - 11x + a = 0$ are solutions of $(6x)^2 -11(6x) + 6a = 0$ and replacing $6x$ by $y$ these roots are rational if and only if roots to $y^2 -11y + 6a$ are rational. But this is monic integral polynomial hence rational roots must be integers. Now let's say $m, n$ are the roots then $m + n = 11$ and $mn = 6a$ so all possible values of $a$ can be given as $mn/6$ when $m + n = 11$ and $6$ divides $mn$.
Now assume $6$ divides $mn$ and by symmetry let's assume $3\vert m$ or in other words $m = 3k$ for some integer $k$. Thus $n = (11 - 3k)$ and $mn = 3k(11-3k)$. This would be divisible by $6$ if and only if $k(11-3k)$ is divisible by $2$ which is true for any value of $k$ so all possible values of $a$ are $\dfrac{3k(11-3k)}{6} = \dfrac{k(11-3k)}{2}$ for any integer $k$.