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.
The product of the roots $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$ is $1$. That part was easy! The sum will be more work.
The sum $\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}$ of the roots simplifies to $\dfrac{\alpha^2+\beta^2}{\alpha\beta}$.
But $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$. Thus the sum of the roots is $\dfrac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$.
Substituting the known values of $\alpha+\beta$ and $\alpha\beta$ we find that the sum of the roots is $\dfrac{(-8)^2-2(-5)}{-5}$.
This simplifies to $-\dfrac{74}{5}$.
Thus the equation is
$$x^2+\frac{74}{5}x+1=0.$$
We can multiply through by $5$ if we wish.
Best Answer
To prove that no perfect square $k^2=7n+6$, we work with $k=7m\pm l; l\in0,1,2,3$
$$(7m+0)^2=49m^2=7(7m^2)+0$$ $$(7m\pm1)^2=49m^2\pm14m+1=7(7m^2\pm2m)+1$$ $$(7m\pm2)^2=49m^2\pm28m+4=7(7m^2\pm4m)+4$$ $$(7m\pm3)^2=49m^2\pm42m+9=7(7m^2\pm6m+1)+2$$
None of these are of the form $7n+6$, thusly it is not possible