If the roots of $x^2-7x+1=0$ are $p$ and $q$, find a quadratic equation with roots $p^2+q^2$ and $\sqrt{p}+\sqrt{q}$ 
Let $p$ and $q$ be the roots of $x^2-7x+1=0$. Find a quadratic equation with roots of $p^2+q^2$ and $\sqrt{p} + \sqrt{q}$.

Attempt to solve:
What I got from the previous quadratic equation
$$p+q=-\frac{b}{a}=7 \text{ and } p\cdot q=\frac{c}{a}=1$$
I've got no problem for solving the first root, all I need is doing a little expansion
$$p^2+q^2=(p+q)^2-2pq=7^2-2(1)=47$$
The problem starts when I'm trying to find the second root. I tried to do this
$$p-q=\left(\sqrt{p} + \sqrt{q}\right)\left(\sqrt{p} - \sqrt{q}\right)$$
Well, how do I get the value of both $p-q$ and $\left(\sqrt{p} - \sqrt{q}\right)$ or maybe there are other method(s) to get the second root.

Let $x^2=y$ and $x$ be a root of $$ax^2+bx+c=0$$
$$(ax^2+c)^2=(-bx)^2$$
$$\implies(ay+c)^2=b^2y\iff a^2y^2+y(2ca-b^2)y+c^2=0$$
So, in the first case, $$a=c=1,b=-7$$
For the second, $$\dfrac{a^2}1=\dfrac{b^2-2ca}7=\dfrac{c^2}1$$
$\implies a^2=c^2\iff c=\pm a$
But $\dfrac ca=\sqrt p\sqrt q\ge0$ for real $p,q$
So, $c=a,$ $$\dfrac{b^2-2ca}7=c^2\implies\dfrac{b^2-2c(a)}7=(-a)^2\iff b^2=9a^2\iff b=? $$
But $-\dfrac ba=\sqrt p+\sqrt q\ge0$ for real $p,q$
Hope you can take it from here?
