Given, (p2+q2)x2−2q(p+r)x+q2+r2=0
On simplifying we get, (px−q)2+(qx−r)2=0
\Rightarrow px-q=0&qx-r=0
⇒x=pq=qr
⇒x=pq=qr=4 [because roots of equation x2−2x−8=0 are 4 or −2]
As p,q,r are positive, so x must be 4.
Now, q=4p and r=4q=16p
So, p2q2+r2=p2(4p)2+(4×4p)2=16+256=272.