Find the three positive values of p for which the equation $px^2-4x+1=0$ will have rational roots 

Question: Find the three positive values of p for which the equation $$px^2-4x+1=0$$
will have rational roots.



My attempt (Algebraically):
Usually if it has to have rational roots then the discriminant must equal zero so
$$b^2 - 4ac = 0$$
$$ (-4)^4-4(p)(1)=0$$
$$ 16-4p = 0$$
$$ p = 4 $$
But the answers given are $3 , \frac{7}{4} , \frac{15}{4}$ how do they get that algebraically?

Given the original and solving for x, we get
$$px^2-4x+1=0$$
$$x=\frac{4\pm\sqrt{16-4(p)(1)}}{2p}=\frac{2\pm\sqrt{4-p}}{p}$$
From this is appears that there are many answers where $4-p$ is a perfect square but one is bogus.
$$p=0\implies x=\frac{2+2}{0}\not\in \mathbb{N}$$
$$p=3\implies x=\frac{2\pm1}{3}=\big\{1, \frac{1}{3}\big\}\quad \land\quad p=4\implies x=\frac{2+0}
{4}=\frac{1}{2}\qquad$$
$$p=-5\implies x=\frac{2+\pm3}{-5}=\big\{-1,\frac{1}{5}\big\}\quad \land\quad p=-12\implies x=\frac{2\pm4}{-12}=\big\{\frac{-1}{2},\frac{1}{6}\big\}$$
$$p=-21\implies x=\frac{2\pm5}{-21}=\big\{\frac{-1}{3},\frac{1}{7}\big\}\quad \land\quad p=-32\implies x=\frac{2\pm6}{-32}=\big\{\frac{-1}{4},\frac{1}{8}\big\}$$
Solving for p: we get $$p=\frac{4x-1}{x^2}$$
$$x=1\implies p=\frac{3}{1}\qquad x=2\implies p=\frac{7}{4}\qquad x=3\implies p=\frac{11}{9}\qquad x=4\implies p=\frac{15}{16}$$
It looks like there are infinite answers with $x$ as the independent variable but I don't know if they will fit back into the original solution for $x$.
