The given equation can be re-written as p+q−xr+1+q+r−xp+1+r+p−xq+1 = 4−4xp+q+r or (p+q+r−x) (1p+1q+1r) = 4(p+q+r−x)p+q+r or (p+q+r−x) (1p+1q+1r) - 4p+q+r =0 Hence p+q+r−x=0 or x=p+q+r.
If P(Q−r)x2+Q(r−P)x+r(P−Q)=0 has equal roots then 2Q=(where P,Q,r ϵ R)