Perimeter equation - 2, 3, 4 solutions.

[man111]

Find value of [tex]p[/tex] for which the equation [tex](x^2-2x)^2-(p+3)(x^2-2x)+(p-2)=0[/tex] has

(i) 4 real solutions

(ii) 3 real solutions only

(iii) 2 real solutions only

(iv) 1 real solution only

(v) No real solution.

[mkmarinov]

[tex]x^2-2x=u \ge -1[/tex]
4 real solutions => 2 real u, larger than -1.
3 real solutions => [tex]u_1=-1; u_2>-1[/tex].
2 real solutions => [tex]u_1<-1; u_2>-1[/tex].
1 real solution => [tex]u_1<-1; u_2=-1[/tex].
No real solutions => [tex]u_1, u_2 < -1[/tex].

[man111]

can anyone explain it.............

[kerry]

(i) 4 real solutions p≥-1

(ii) 3 real solutions only p=-1

(iii) 2 real solutions only p≤-1

(iv) 1 real solution only - never

(v) No real solution - never

[man111]

Thanks for all.

[garion]

1 real solution => [tex]u_1<-1; u_2=-1[/tex].

for 1 real solution also [tex]u_1=u_2=-1[/tex].