One possibility is for each quadratic to share a root.
Multiplying (xp)(xq) and (xq)(xr) give equations for c in terms of (p,q,r) with solutions c=3 and c=1.
The only other possibility is for one quadratic to have a repeated root.
Say the first quadratic has the dual root p. Since the discriminant is postive, p is real. But it's necessary too for p^2=4(c^2+1) which is impossible.
If it's the second quadratic with the repeated root then 2p=4 and p^2=2c(c^2+1). So p=2 and substituting gives a cubic equation with a real root c=1 by inspection and the remaining roots imaginary.
When c=3 (p,q,r)=(10,4,6).
When c=1 (p,q,r)=(2,1+sqrt(17),1sqrt(17)).

Posted by xdog
on 20160911 09:45:48 