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Question Number 31804 by rahul 19 last updated on 15/Mar/18
A quadratic equation p(x)=0 having coefficient of x^2 unity is such that p(x)=0 and p(p(p(x)))=0 have a common root then, prove that : p(0)×p(1)=0.
Aquadraticequationp(x)=0havingcoefficientofx2unityissuchthatp(x)=0andp(p(p(x)))=0haveacommonrootthen,provethat:p(0)×p(1)=0.
Answered by MJS last updated on 15/Mar/18
p(x)=x^2 +bx+c p(0)×p(1)=0 ⇒ p(0)=0∨p(1)=0 case 1 p(0)=0 ⇒ c=0 p(x)=(x+b)x p(p(x))=q(x) p(p(p(x)))=p(q(x))=r(x) q(x)=(p(x)+b)p(x) ⇒ ⇒ q(0)=0 r(x)=(q(x)+b)q(x) ⇒ ⇒ r(0)=0 ⇒ ⇒ 0= common root of p(x) and r(x) case 2 p(1)=0 p(x)=(x−1)(x−c) [with b=−c−1; p(x)=x^2 −(c+1)x+c] q(x)=(p(x)−1)(p(x)−c)= =p^2 (x)−(c+1)p(x)+c r(x)=[p^2 (x)−(c+1)p(x)+c−1]× ×[p^2 (x)−(c+1)p(x)]= =[...]×[p(x)−(c+1)]×p(x) ⇒ ⇒ r(1)=0 ⇒ ⇒ 1= common root of p(x) and r(x)
p(x)=x2+bx+cp(0)×p(1)=0⇒p(0)=0∨p(1)=0case1p(0)=0⇒c=0p(x)=(x+b)xp(p(x))=q(x)p(p(p(x)))=p(q(x))=r(x)q(x)=(p(x)+b)p(x)⇒⇒q(0)=0r(x)=(q(x)+b)q(x)⇒⇒r(0)=0⇒⇒0=commonrootofp(x)andr(x)case2p(1)=0p(x)=(x−1)(x−c)[withb=−c−1;p(x)=x2−(c+1)x+c]q(x)=(p(x)−1)(p(x)−c)==p2(x)−(c+1)p(x)+cr(x)=[p2(x)−(c+1)p(x)+c−1]××[p2(x)−(c+1)p(x)]==[…]×[p(x)−(c+1)]×p(x)⇒⇒r(1)=0⇒⇒1=commonrootofp(x)andr(x)

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