(р-3)х^2-4рх+8р=0,
D=(-4p)^2-4(p-3)8p=16p^2-32p^2+96p=96p-16p^2,
D>0,
96p-16p^2>0,
96p-16p^2=0,
16p(6-p)=0,
p=0 или p=6,
-16p(p-6)>0,
p(p-6)<0,</p>
0
x1=(4p-4√(6p-p^2))/(2(p-3))>0,
x2=(4p+4√(6p-p^2))/(2(p-3))>0,
p-3≠0, p≠3;
(2p-2√(6p-p^2))(p-3)>0,
(2p+2√(6p-p^2))(p-3)>0,
2p-2√(6p-p^2)>0,
2p+2√(6p-p^2)>0,
p-3>0,
√(6p-p^2)
√(6p-p^2)>-p,
p>3,
6p-p^2
2p^2-6p>0,
2p^2-6p=0,
2p(p-3)=0,
p=0 или р=3,
p(p-3)>0,
p<0, p>3, p∈(-∞;0)U(3;+∞);
p∈(3,6);
2p-2√(6p-p^2)<0,</p>
2p+2√(6p-p^2)<0,</p>
p-3<0,</p>
√(6p-p^2)>p,
√(6p-p^2)<-p,</p>
p<3,</p>
2p^2-6p<0,</p>
p<0,</p>
p<3,</p>
0
p<0,</p>
p<3,</p>
p∈Ф.
Ответ: p∈(3,6).