Let p(x) be a quadratic polynomial such that for distinct α and β , p(α) = α and p(β) =β prove that α and β are roots of p[p(x)]−x=0 Find the remaining roots .


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Question Number 61269 by alphaprime last updated on 31/May/19

$L e t p (x) b e a q u a d r a t i c p o l y n o m i a l s u c h$ $t h a t f o r d i s t i n c t α a n d β,$ $p (α) = α a n d p (β) = β$ $p r o v e t h a t α a n d β a r e r o o t s o f p [p (x)] - x = 0$ $F i n d t h e r e m a i n i n g r o o t s .$
	Answered by ajfour last updated on 31/May/19
	$p(x)=a(x−h)^2 +k α, β are roots of a(r−h)^2 −r+k=0 ...(i) p[p(x)]−x=a{a(x−h)^2 +k−h}^2 +k−x And since from (i) a(r−h)^2 +k=r ....(ii) p[p(r)]−r=a{a(x−r)^2 +k−h}^2 −r+k using (ii) we get p[p(r)]−r=a{r−h}^2 −r+k =0 using (i) hence α,β are roots of p[p(x)]−x=0 . Again p[p(x)]−x=a{a(x−h)^2 +k−h}^2 +k−x or = a^3 (x−h)^4 +2a^2 (k−h)(x−h)^2 +a(k−h)^2 +k−x =a^3 x^4 −4a^3 hx^3 +[6a^3 h^2 +2a^2 (k−h)]x^2 −[4a^3 h^3 −+4a^2 h(k−h)+1]x +a^3 h^4 +2a^2 (k−h)h^2 +a(k−h)^2 +k Also lets assume p[p(x)]−x=a^3 [x^2 −(2ah+1)x+h^2 +(k/a)](x−γ)(x−δ) =a^3 {x^4 −(γ+δ+2ah+1)x^3 +[γδ+(2ah+1)(γ+δ)+h^2 +(k/a)]x^2 +(h^2 +(k/a))γδ} comparing the two expressions for p[p(x)]−x we conclude γ+δ = 4h−2ah−1 γδ(h^2 +(k/a))=a^3 h^4 +2a^2 (k−h)h^2 +a(k−h)^2 +k but from (i) 2h+(1/a)=α+β h^2 +(k/a)=αβ .....$
	$p (x) = a {(x - h)}^{2} + k$ $α, β a r e r o o t s o f$ $a {(r - h)}^{2} - r + k = 0 . . . (i)$ $p [p (x)] - x = a {a {(x - h)}^{2} + k - h}^{2} + k - x$ $A n d s i n c e f r o m (i)$ $a {(r - h)}^{2} + k = r . . . . (i i)$ $p [p (r)] - r = a {a {(x - r)}^{2} + k - h}^{2} - r + k$ $u s i n g (i i) w e g e t$ $p [p (r)] - r = a {r - h}^{2} - r + k$ $= 0 u s i n g (i)$ $h e n c e α, β a r e r o o t s o f p [p (x)] - x = 0 .$ $A g a i n$ $p [p (x)] - x = a {a {(x - h)}^{2} + k - h}^{2} + k - x$ $o r = a^{3} {(x - h)}^{4} + 2 a^{2} (k - h) {(x - h)}^{2}$ $+ a {(k - h)}^{2} + k - x$ $= a^{3} x^{4} - 4 a^{3} {h x}^{3} + [6 a^{3} h^{2} + 2 a^{2} (k - h)] x^{2}$ $- [4 a^{3} h^{3} - + 4 a^{2} h (k - h) + 1] x$ $+ a^{3} h^{4} + 2 a^{2} (k - h) h^{2} + a {(k - h)}^{2} + k$ $A l s o l e t s a s s u m e$ $p [p (x)] - x = a^{3} [x^{2} - (2 a h + 1) x + h^{2} + \frac{k}{a}] (x - γ) (x - δ)$ $= a^{3} {x^{4} - (γ + δ + 2 a h + 1) x^{3}$ $+ [γ δ + (2 a h + 1) (γ + δ) + h^{2} + \frac{k}{a}] x^{2}$ $+ (h^{2} + \frac{k}{a}) γ δ}$ $c o m p a r i n g t h e t w o e x p r e s s i o n s$ $f o r p [p (x)] - x w e c o n c l u d e$ $γ + δ = 4 h - 2 a h - 1$ $γ δ (h^{2} + \frac{k}{a}) = a^{3} h^{4} + 2 a^{2} (k - h) h^{2} + a {(k - h)}^{2} + k$ $b u t f r o m (i)$ $2 h + \frac{1}{a} = α + β$ $h^{2} + \frac{k}{a} = α β$ $. . . . .$
	Answered by perlman last updated on 31/May/19
	$p(x)=(x−a)(x−b)+x p[p(x)]=(p(x)−a)(p(x)−b)+p(x) pp(x) −x=0 ==> (p(x)−a)(p(x)−b)+p(x)−x=0 =((x−a)(x−b)+x−a)((x−a)(x−b)+x−b)+(x−a)(x−b)=0 (x−a)(x−b+1)(x−b)(x−a+1)+(x−a)(x−b)=0 (x−a)(x−b)[(x−b+1)(x−a+1)+1]=0 let solve (x−b+1)(x−a+1)+1=0 x^2 −(b+a−2)x+(a−1)(b−1)+1=0 Δ=(b+a−2)^2 −4(ab−a−b+2)=(b^2 +a^2 +4+2ba−4a−4b)−4ab+4a+4b−8 =b^2 +a^2 −2ab−4=(b−a−2)(b−a+2) let b>a we can do this cause a#b ===>Δ≥0<==>b≥a+2 if b≥a+2 X_(1.2) =(((b+a−2)+_− (√((b−a−2)(b−a+2))))/2) if a<b<=a+2 X_(1.2) =(((b+a−2)+_− i(√((a+2−b)(b−a+2))))/2) solution are if b∈]a+2;+∞[ solutions are {b,a,(((b+a−2)+_− (√((b−a−2)(b−a+2))))/2)} if b∈]a;a+2[ solutions are {b,a,(((b+a−2)+_− i(√((b−a−2)(b−a+2))))/2)}$
	$p (x) = (x - a) (x - b) + x$ $p [p (x)] = (p (x) - a) (p (x) - b) + p (x)$ $p p (x) - x = 0$ $==> (p (x) - a) (p (x) - b) + p (x) - x = 0$ $= ((x - a) (x - b) + x - a) ((x - a) (x - b) + x - b) + (x - a) (x - b) = 0$ $(x - a) (x - b + 1) (x - b) (x - a + 1) + (x - a) (x - b) = 0$ $(x - a) (x - b) [(x - b + 1) (x - a + 1) + 1] = 0$ $l e t s o l v e (x - b + 1) (x - a + 1) + 1 = 0$ $x^{2} - (b + a - 2) x + (a - 1) (b - 1) + 1 = 0$ $Δ = {(b + a - 2)}^{2} - 4 (a b - a - b + 2) = (b^{2} + a^{2} + 4 + 2 b a - 4 a - 4 b) - 4 a b + 4 a + 4 b - 8$ $= b^{2} + a^{2} - 2 a b - 4 = (b - a - 2) (b - a + 2)$ $You can't use 'macro parameter character #' in math mode$ $===> Δ ⩾ 0 <==> b ⩾ a + 2$ $i f b ⩾ a + 2$ $X_{1.2} = \frac{(b + a - 2) \underline{+} \sqrt{(b - a - 2) (b - a + 2)}}{2}$ $i f a < b <= a + 2$ $X_{1.2} = \frac{(b + a - 2) \underline{+} i \sqrt{(a + 2 - b) (b - a + 2)}}{2}$ $s o l u t i o n a r e i f b \in] a + 2; + \infty [s o l u t i o n s a r e$ ${b, a, \frac{(b + a - 2) \underline{+} \sqrt{(b - a - 2) (b - a + 2)}}{2}}$ $i f b \in] a; a + 2 [s o l u t i o n s a r e$ ${b, a, \frac{(b + a - 2) \underline{+} i \sqrt{(b - a - 2) (b - a + 2)}}{2}}$
	Commented by alphaprime last updated on 31/May/19
	Tanmay sir generalized and solution is pretty nice , but yours is valid in entire domain , But comment something on both the solutions , any remarks you wanna add , please ...
	Commented by perlman last updated on 31/May/19

	$i t a c k p (x) = (x - a) (x - b) + x$ $b u t g e n e r a l y i m u s t t a c k p = s (x - a) (x - b) + x$ $w i t h e s c o n s t a n t$
	Commented by alphaprime last updated on 31/May/19
	You mentioned it finally , In Tanmay sir's solution It's dependent on the parameter A you've left , which you haven't taken as arbitrary constant, so does it affect his or your solution or neither ones ?! it's amazing
	Answered by tanmay last updated on 31/May/19
	$p(x)=A(x−α)(x−β)+x p[p(x)] =A[A(x−α)(x−β)+x−α](A(x−α)(x−β)+x−β]+ A(x−α)(x−β) +x−x so p[p(α)]=0 p[p(β)]=0 nxt part... p[p(x)] =A[(x−α){A(x−β)+1}][(x−β){A(x−α)+1}]+if A(x−α)(x−β) for other two root p[p(x)]=0 x−α=m x−β=n A[m(An+1)][n(Am+1]+Amn=0 (Amn+m)(Amn+n)+mn=0 A^2 (mn)^2 +Amn(m+n)+mn+mn=0 mn[A^2 (mn)+A(m+n)+2]=0 when A^2 (mn)+A(m+n)+2=0 A^2 (x−α)(x−β)+A(x−α+x−β)+2=0 A^2 (x^2 −xα−xβ+αβ)+2Ax−A(α+β)+2=0 x^2 (A^2 )+x(2A−A^2 α−A^2 β)+A^2 αβ−A(α+β)+2=0 (Ax+1)^2 +x(−A^2 α−A^2 β)+(A^2 αβ−Aα−Aβ+1)=0 (Ax+1)^2 −A^2 x(α+β)+Aα(Aβ−1)−(Aβ−1)=0 (Ax+1)^2 −A^2 x(α+β)+(Aα−1)(Aβ−1)=0 A^2 x^2 +2Ax+1−A^2 x(α+β)+(Aα−1)(Aβ−1)=0 A^2 x^2 −Ax(α+β−2)+(Aα−1)(Aβ−1)+1=0 x=((A(α+β−2)±(√(A^2 (α+β−2)^2 −4A^2 [(Aα−1)(Aβ−1)+1])))/(2A^2 )) x=((α+β−2±(√((α+β−2)^2 −4[(Aα−1)(Aβ−1)+1)))/(2A))$
	$p (x) = A (x - α) (x - β) + x$ $p [p (x)]$ $= A [A (x - α) (x - β) + x - α] (A (x - α) (x - β) + x - β] +$ $A (x - α) (x - β) + x - x$ $s o$ $p [p (α)] = 0$ $p [p (β)] = 0$ $n x t p a r t . . .$ $p [p (x)]$ $= A [(x - α) {A (x - β) + 1}] [(x - β) {A (x - α) + 1}] + i f$ $A (x - α) (x - β)$ $f o r o t h e r t w o r o o t p [p (x)] = 0$ $x - α = m x - β = n$ $A [m (A n + 1)] [n (A m + 1] + A m n = 0$ $(A m n + m) (A m n + n) + m n = 0$ $A^{2} {(m n)}^{2} + A m n (m + n) + m n + m n = 0$ $m n [A^{2} (m n) + A (m + n) + 2] = 0$ $w h e n A^{2} (m n) + A (m + n) + 2 = 0$ $A^{2} (x - α) (x - β) + A (x - α + x - β) + 2 = 0$ $A^{2} (x^{2} - x α - x β + α β) + 2 A x - A (α + β) + 2 = 0$ $x^{2} (A^{2}) + x (2 A - A^{2} α - A^{2} β) + A^{2} α β - A (α + β) + 2 = 0$ ${(A x + 1)}^{2} + x (- A^{2} α - A^{2} β) + (A^{2} α β - A α - A β + 1) = 0$ ${(A x + 1)}^{2} - A^{2} x (α + β) + A α (A β - 1) - (A β - 1) = 0$ ${(A x + 1)}^{2} - A^{2} x (α + β) + (A α - 1) (A β - 1) = 0$ $A^{2} x^{2} + 2 A x + 1 - A^{2} x (α + β) + (A α - 1) (A β - 1) = 0$ $A^{2} x^{2} - A x (α + β - 2) + (A α - 1) (A β - 1) + 1 = 0$ $x = \frac{A (α + β - 2) \pm \sqrt{A^{2} {(α + β - 2)}^{2} - 4 A^{2} [(A α - 1) (A β - 1) + 1]}}{2 A^{2}}$ $x = \frac{α + β - 2 \pm \sqrt{{(α + β - 2)}^{2} - 4 [(A α - 1) (A β - 1) + 1}}{2 A}$
	Commented by alphaprime last updated on 31/May/19
	Very Nice cognizant
	Commented by tanmay last updated on 31/May/19

	$t h a n k y o u s i r . . .$
	Commented by alphaprime last updated on 31/May/19
	Question no.61211 , I seriously need your help sir , I can't even come on telegram due to issues, I'm seeking intense help , I don't expect much but there's a hope , Mjs sir is with me in the workspace and has promised to provide solution till Monday, I need you and other intellectuals to join him and resolve the system Indeed help me out please sir ��
	Commented by ajfour last updated on 31/May/19

	$W e t o o a r e g l a d t o h a v e y o u h e r e$ $o n t h e f o r u m S i r . Y o u r q u e s t i o n s$ $a r e p r e c i o u s .$
	Commented by alphaprime last updated on 31/May/19
	Sir I'm welcoming you in my workspace and please don't disappoint me , I'll love to notice presence of higher intellectual faculty , Just provide me your email ��
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