Polynomial Roots - 2

By Vieta's formula , we have $\begin{cases} p+q+r = 11 \\ pq+qr+rp = 13 \\ pqr = 4 \end{cases}$

Given that $k = \sqrt p + \sqrt q + \sqrt r$ , then

$\begin{aligned} k^2 & = \red{p+q+r} + 2(\sqrt{pq} + \sqrt{qr} + \sqrt{rp}) = \red{11} + 2(\sqrt{pq} + \sqrt{qr} + \sqrt{rp}) & \small \blue{\implies 2(\sqrt{pq} + \sqrt{qr} + \sqrt{rp}) = k-11} \end{aligned}$

$\begin{aligned} k^4 & = 121 + 44\left(\sqrt{pq} + \sqrt{qr} + \sqrt{rp}\right) + 4\left(\red{pq+qr+rp} + 2(\sqrt p \cdot \sqrt{\blue{pqr}} + \sqrt q \cdot \sqrt{\blue{pqr}} + \sqrt r \cdot \sqrt{\blue{pqr}})\right) & \small \blue{\text{Note }pqr = 4} \\ & = 121 + 44\left(\blue{\sqrt{pq} + \sqrt{qr} + \sqrt{rp}}\right) + 4\left(\red{13} + 4(\blue{\sqrt p + \sqrt q + \sqrt r}) \right) \\ & = 173 + 22\left(\blue{k^2 - 11} \right) + 16\blue k \\ & = 22k^2 + 16k - 69 \end{aligned}$

Therefore,

$\begin{aligned} \implies k^4 - 22k^2 - 16k + 69 & = 0 \\ k^4 - 22k^2 - 16k + 1001 & = \boxed{932} \end{aligned}$

From the given equation we can write $p+q+r=11, pq+qr+rp=13, pqr=4$ . Now, from the expression for $k$ , we get $k^2=p+q+r+2(\sqrt {pq}+\sqrt {qr}+\sqrt {rp})=11+2(\sqrt {pq}+\sqrt {qr}+\sqrt {rp}\implies (k^2-11)^2=4(\sqrt {pq}+\sqrt {qr}+\sqrt {rp})^2=4\left (pq+qr+rp+2(\sqrt {pq^2r}+\sqrt {pqr^2}+\sqrt {p^2qr})\right ) =4\left (13+4(\sqrt p+\sqrt q+\sqrt r) \right ) =52+16k\implies k^4-22k^2+121-16k=52\implies k^4-22k^2-16k+1001=52-121+1001=\boxed {932}$ .